JEE Alternating Currents Questions

Using AC voltage divider relation with complex impedance $$Z_C = \frac{1}{j\omega C}$$:

$$V_A = V_{in} \left( \frac{Z_C}{R + Z_C} \right) = V_{in} \left( \frac{1}{1 + j\omega RC} \right)$$

$$V_B = V_{in} \left( \frac{R}{R + Z_C} \right) = V_{in} \left( \frac{j\omega RC}{1 + j\omega RC} \right)$$

Evaluating differential potential difference phasor ($$V_B - V_A$$): $$V_B - V_A = V_{in} \left( \frac{j\omega RC - 1}{1 + j\omega RC} \right)$$

$$\angle(V_B - V_A) - \angle V_{in} = \angle(j\omega RC - 1) - \angle(1 + j\omega RC)$$

$$\Delta \phi = (\pi - \tan^{-1}(\omega RC)) - \tan^{-1}(\omega RC) = \pi - 2\tan^{-1}(\omega RC)$$

$$\frac{\pi}{2} = \pi - 2\tan^{-1}(\omega RC) \implies 2\tan^{-1}(\omega RC) = \frac{\pi}{2} \implies \tan^{-1}(\omega RC) = \frac{\pi}{4}$$

$$\omega RC = 1 \implies \omega = \frac{1}{RC}$$

$$\omega = \frac{1}{10^5 \times 10^{-10}} = \frac{1}{10^{-5}} = 10^5\text{ rad/sec}$$

$$10^5 = 10^x \implies x = 5$$

$$V(t) = 220\sin(100\pi t)$$, $$R = 50\Omega$$.

Peak current: $$I_0 = V_0/R = 220/50 = 4.4$$ A.

Current: $$I = I_0\sin(100\pi t)$$.

Half peak: $$I_0/2$$ when $$\sin(100\pi t_1) = 1/2 \Rightarrow 100\pi t_1 = \pi/6 \Rightarrow t_1 = \frac{1}{600}$$ s.

Peak: when $$\sin(100\pi t_2) = 1 \Rightarrow 100\pi t_2 = \pi/2 \Rightarrow t_2 = \frac{1}{200}$$ s.

$$\Delta t = t_2 - t_1 = \frac{1}{200} - \frac{1}{600} = \frac{3-1}{600} = \frac{2}{600} = \frac{1}{300}$$ s $$= 3.33$$ ms.

The answer is Option (2): $$\boxed{3.3 \text{ ms}}$$.

The average power dissipated in an AC circuit is given by: $$P_{avg} = \frac{V_0 I_0}{2} \cos\phi$$, where $$V_0$$ is the peak voltage, $$I_0$$ is the peak current, and $$\phi$$ is the phase difference between voltage and current.

From the given equations:

$$V = 100\sin(100t)$$ V, so $$V_0 = 100$$ V

$$I = 100\sin\left(100t + \frac{\pi}{3}\right)$$ mA $$= 0.1\sin\left(100t + \frac{\pi}{3}\right)$$ A, so $$I_0 = 0.1$$ A

The phase difference is $$\phi = \frac{\pi}{3} = 60°$$, so $$\cos\phi = \cos 60° = \frac{1}{2}$$.

Substituting into the formula:

$$P_{avg} = \frac{100 \times 0.1}{2} \times \frac{1}{2} = \frac{10}{2} \times \frac{1}{2} = \frac{10}{4} = 2.5$$ W

Therefore, the average power dissipated is $$2.5$$ W, which is Option (3).

Primary voltage: $$V_p = 230$$ V. Turn ratio $$N_p : N_s = 10 : 1$$.

Secondary voltage: $$V_s = V_p \times \frac{N_s}{N_p} = 230 \times \frac{1}{10} = 23$$ V.

Load resistance: $$R_L = 46 \; \Omega$$.

Power consumed: $$P = \frac{V_s^2}{R_L} = \frac{23^2}{46} = \frac{529}{46} = 11.5$$ W.

The answer is $$11.5$$ W, which corresponds to Option (3).

Find the secondary current of a transformer with 80% efficiency, 10 V primary, 4 kW input power, and 240 V secondary voltage.

Recall that efficiency $$\eta = \frac{P_{\text{output}}}{P_{\text{input}}}$$.

With an input power of 4000 W and an efficiency of 0.80, the output power is $$P_{\text{output}} = \eta \times P_{\text{input}} = 0.80 \times 4000 = 3200\text{ W}$$.

Applying $$P = VI$$ to the secondary winding gives $$I_2 = \frac{P_{\text{output}}}{V_2} = \frac{3200}{240} = 13.33\text{ A}$$.

The correct answer is Option B: 13.33 A.

Given:
L = 1 H
C = 20 μF = 20 × 10⁻⁶ F
R = 300 Ω
v(t) = 50√2 sin(100t)

So:
$$V_{rms}=50V$$
ω = 100 rad/s

step 1: calculate reactances

Inductive reactance:
$$X_L=ωL=100\times1=100Ω$$

Capacitive reactance:
$$X_C=1/(ωC)$$
= 1 / (100 × 20×10⁻⁶)
= 1 / (2×10⁻³)
= 500 Ω

step 2: impedance of circuit

$$Z=\sqrt{\ \left(R^2+(X_L−X_C)^2\right)}$$
$$=\sqrt{\ \left(300^2+(100−500)^2\right)}$$
$$=\sqrt{\ 250000}$$
$$=500Ω$$

step 3: current

$$I_{rms}=V_{rms}/Z=50/500=0.1A$$

step 4: voltage across capacitor

$$V_C=I\times X_C=0.1\times500=50V$$

From DC supply: $$V = 20$$ V, $$I = 5$$ A. Resistance of coil: $$R = V/I = 4\Omega$$.

From AC supply: $$V = 20$$ V, $$I = 4$$ A. Impedance: $$Z = V/I = 5\Omega$$.

For a coil (R-L circuit): $$Z = \sqrt{R^2 + (\omega L)^2}$$

$$25 = 16 + \omega^2 L^2$$

$$\omega^2 L^2 = 9$$

$$\omega L = 3$$

$$\omega = 2\pi f = 2\pi \times 50 = 2 \times 3 \times 50 = 300$$ rad/s (using $$\pi = 3$$).

$$L = \frac{3}{300} = 0.01 \text{ H} = 10 \text{ mH}$$

The answer is $$\boxed{10}$$ mH.

An inductor is connected first to DC (100 V, 5 A) and then to AC (peak 200 V, $$X_L = 20\sqrt{3} \Omega$$). Under DC the inductor acts as a pure resistance at zero frequency, so the resistance is $$ R = \frac{V_{DC}}{I_{DC}} = \frac{100}{5} = 20 \,\Omega $$.

Under AC the impedance is $$ Z = \sqrt{R^2 + X_L^2} = \sqrt{20^2 + (20\sqrt{3})^2} = \sqrt{400 + 1200} = \sqrt{1600} = 40 \,\Omega $$.

The RMS voltage is $$V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{200}{\sqrt{2}}$$ V, so the RMS current is $$ I_{rms} = \frac{V_{rms}}{Z} = \frac{200}{\sqrt{2}\times 40} = \frac{200}{40\sqrt{2}} = \frac{5}{\sqrt{2}} \text{ A} $$.

Since power is dissipated only in the resistance, $$ P = I_{rms}^2 \times R = \frac{25}{2} \times 20 = 250 \text{ W} $$. Therefore the answer is $$250$$ W.

We need to find the power dissipation in a series RC circuit connected to an AC source.

Capacitive reactance: $$X_C = 4\sqrt{3} \text{ } \Omega$$

Resistance: $$R = 4 \text{ } \Omega$$

Peak voltage: $$V_0 = 8\sqrt{2} \text{ V}$$

For a series RC circuit, the impedance is given by:

$$Z = \sqrt{R^2 + X_C^2}$$

Substituting the values:

$$Z = \sqrt{(4)^2 + (4\sqrt{3})^2} = \sqrt{16 + 48} = \sqrt{64} = 8 \text{ } \Omega$$

The relationship between peak and RMS voltage is:

$$V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{8\sqrt{2}}{\sqrt{2}} = 8 \text{ V}$$

Using Ohm's law for AC circuits:

$$I_{rms} = \frac{V_{rms}}{Z} = \frac{8}{8} = 1 \text{ A}$$

In an AC circuit, power is dissipated only in the resistive component (capacitors and inductors do not dissipate power on average). The power dissipation formula is:

$$P = I_{rms}^2 \times R$$

This can also be written as $$P = V_{rms} \cdot I_{rms} \cdot \cos\phi$$, where $$\cos\phi = R/Z$$ is the power factor. Both forms give the same result.

$$P = (1)^2 \times 4 = 4 \text{ W}$$

Verification using power factor:

$$\cos\phi = \frac{R}{Z} = \frac{4}{8} = 0.5$$

$$P = V_{rms} \times I_{rms} \times \cos\phi = 8 \times 1 \times 0.5 = 4$$ W. This confirms our answer.

The answer is 4 W.

Step-down transformer: $$V_p = 2300$$ V, $$N_p = 3000$$, $$V_s = 230$$ V, $$I_p = 5$$ A, efficiency = 90%.

$$\frac{N_s}{N_p} = \frac{V_s}{V_p} = \frac{230}{2300} = \frac{1}{10}$$, so $$N_s = 300$$.

Input power = $$V_p I_p = 2300 \times 5 = 11500$$ W.

Output power = $$0.9 \times 11500 = 10350$$ W.

$$I_s = \frac{P_s}{V_s} = \frac{10350}{230} = 45$$ A.

The output current is 45 A.

We have resistance $$R = 100$$ $$\Omega$$, inductance $$L = 1$$ H, and capacitance $$C = 6.25$$ $$\mu$$F = $$6.25 \times 10^{-6}$$ F.

The quality factor of an RLC series circuit is given by:

$$Q = \frac{1}{R}\sqrt{\frac{L}{C}}$$

Substituting the values:

$$Q = \frac{1}{100}\sqrt{\frac{1}{6.25 \times 10^{-6}}}$$

$$Q = \frac{1}{100}\sqrt{\frac{10^6}{6.25}}$$

$$Q = \frac{1}{100}\sqrt{1.6 \times 10^5}$$

$$Q = \frac{1}{100} \times 400 = 4$$

So, the answer is $$4$$.

R=10Ω,L=3H,C=27μF

Step 1: Bandwidth

$$\Delta \omega = \frac{R}{L}$$​

Step 2: Quality factor

Q = $$\frac{\omega_0}{\Delta \omega} = \frac{\omega_0 L}{R}$$​

Step 3: Required ratio

$$\frac{Q}{\Delta \omega} = \frac{\omega_0 L}{R} \cdot \frac{L}{R}$$

= $$\omega_0 \frac{L^2}{R^2}$$

$$\omega_0 = \frac{1}{\sqrt{LC}}$$1​:

$$\frac{Q}{\Delta \omega} = \frac{1}{\sqrt{LC}} \cdot \frac{L^2}{R^2}$$

= $$\frac{L^2}{R^2 \sqrt{LC}}$$

Step 4: Substitute values

$$\sqrt{LC} = \sqrt{3 \times 27 \times 10^{-6}} = \sqrt{81 \times 10^{-6}} = 9 \times 10^{-3}$$

$$\frac{Q}{\Delta \omega} = \frac{3^2}{10^2 \times 9 \times 10^{-3}}$$

=$$ \frac{9}{100 \times 9 \times 10^{-3}}$$

= $$\frac{1}{10^{-1}} = 10\,\text{s}$$

Final Answer:

10 

The applied voltage is $$v = 45 \sin \left( \omega t \right)\;{\rm V}$$, so its peak (amplitude) value is $$V_0 = 45\;{\rm V}$$.

For a series LCR circuit the current amplitude at any angular frequency $$\omega$$ is
$$I(\omega) = \frac{V_0}{Z},\qquad Z = \sqrt{\,R^{2} + \left( \omega L - \frac{1}{\omega C} \right)^{2}}.$$

At resonance $$\omega_0 = \frac{1}{\sqrt{LC}} = 10^{5}\;{\rm rad\,s^{-1}}$$, the impedance is purely resistive, so
$$I_0 = I(\omega_0) = \frac{V_0}{R} \qquad -(1)$$

Using $$\omega_0^2 = \dfrac{1}{LC}\;$$ we get the capacitance
$$C = \frac{1}{L\omega_0^{2}}.$$

At $$\omega = 8\times10^{4}\;{\rm rad\,s^{-1}}$$ the current amplitude becomes $$I = 0.05\,I_0$$. Hence
$$\frac{I}{I_0} = \frac{R}{\sqrt{\,R^{2} + \left(\omega L - \frac{1}{\omega C}\right)^{2}}} = 0.05.$$ Squaring and rearranging, $$R^{2} = \frac{0.0025}{0.9975}\left(\omega L - \frac{1}{\omega C}\right)^{2}$$ $$\Longrightarrow\; R = 0.05006\left|\omega L - \frac{1}{\omega C}\right|.$$(2)

Now $$\omega L - \frac{1}{\omega C} = \,L\left(\omega - \frac{\omega_0^{2}}{\omega}\right) = L\frac{\omega^{2}-\omega_0^{2}}{\omega}.$$ For $$\omega = 8\times10^{4}\;{\rm rad\,s^{-1}}$$, $$\omega_0 = 10^{5}\;{\rm rad\,s^{-1}}$$ and $$L = 50\;{\rm mH}=0.05\;{\rm H}$$, $$\left|\omega^{2}-\omega_0^{2}\right| = \left| (8\times10^{4})^{2}-(10^{5})^{2}\right| = \left|6.4\times10^{9}-1.0\times10^{10}\right| = 3.6\times10^{9},$$ $$\left|\omega L - \frac{1}{\omega C}\right| = 0.05 \times \frac{3.6\times10^{9}}{8\times10^{4}} = 0.05 \times 4.5\times10^{4} = 2.25\times10^{3}\;\Omega.$$ Thus from (2) $$R = 0.05006 \times 2250 \approx 1.13\times10^{2}\;\Omega \approx 113\;\Omega.$$

Using $$I_0 = \dfrac{V_0}{R}$$ from (1): $$I_0 = \frac{45}{113}\,{\rm A} \approx 0.40\;{\rm A} = 400\;{\rm mA}.$$ So, P $$\to$$ 400 mA corresponding to List-II (3).

The quality factor is $$Q = \frac{\omega_0 L}{R} = \frac{10^{5}\times0.05}{113} \approx 44.4.$$ Thus, Q $$\to$$ 44.4 corresponding to List-II (1).

The bandwidth is $$\Delta\omega = \frac{\omega_0}{Q} = \frac{10^{5}}{44.4} \approx 2.25\times10^{3}\;{\rm rad\,s^{-1}}.$$ Hence, R $$\to$$ 2250 rad s$$^{-1}$$ corresponding to List-II (4).

At resonance the voltage and current are in phase, so the instantaneous power is $$p = v\,i.$$ Its maximum (peak) value is $$P_{\text{peak}} = V_0\,I_0 = 45 \times 0.40 = 18\;{\rm W}.$$ Therefore, S $$\to$$ 18 W corresponding to List-II (2).

Collecting the matches:
P $$\to$$ 3, Q $$\to$$ 1, R $$\to$$ 4, S $$\to$$ 2.

Hence the correct option is Option B: P $$\to$$ 3, Q $$\to$$ 1, R $$\to$$ 4, S $$\to$$ 2.

Given: $$V(t) = 210\sin 3000t$$ volt, so the angular frequency is $$\omega = 3000$$ rad/s.

Given: $$L = 10$$ mH $$= 10 \times 10^{-3}$$ H, $$C = 25$$ $$\mu$$F $$= 25 \times 10^{-6}$$ F, $$R = 100$$ $$\Omega$$.

The inductive reactance is:

$$X_L = \omega L = 3000 \times 10 \times 10^{-3} = 30 \text{ }\Omega$$

The capacitive reactance is:

$$X_C = \frac{1}{\omega C} = \frac{1}{3000 \times 25 \times 10^{-6}} = \frac{1}{0.075} = 13.33 \text{ }\Omega$$

The net reactance is:

$$X_L - X_C = 30 - 13.33 = 16.67 \text{ }\Omega$$

The phase difference is:

$$\tan\phi = \frac{X_L - X_C}{R} = \frac{16.67}{100} = 0.1667 \approx 0.17$$

$$\phi = \tan^{-1}(0.17)$$

Hence, the correct answer is Option A.

We have an alternating emf $$E = 440\sin(100\pi t)$$ applied to a pure inductor of inductance $$L = \frac{\sqrt{2}}{\pi}$$ H.

From the emf expression, the peak voltage is $$E_0 = 440$$ V and the angular frequency is $$\omega = 100\pi$$ rad/s.

The inductive reactance is $$X_L = \omega L = 100\pi \times \frac{\sqrt{2}}{\pi} = 100\sqrt{2}$$ $$\Omega$$.

The peak current through the inductor is $$I_0 = \frac{E_0}{X_L} = \frac{440}{100\sqrt{2}} = \frac{4.4}{\sqrt{2}}$$ A.

An a.c. ammeter reads the rms value of current, which is $$I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{4.4}{\sqrt{2} \times \sqrt{2}} = \frac{4.4}{2} = 2.2$$ A.

Hence, the correct answer is Option C.

A DC of 4 A flows through $$3 \Omega$$ and an AC of peak value 4 A flows through $$2 \Omega$$. We need the ratio of heat produced.

Since the DC current is 4 A, the heat produced in the resistor of $$R_1 = 3 \Omega$$ over time t is given by $$H_1 = I^2 R_1 t = (4)^2 \times 3 \times t = 48t$$.

For the AC circuit, heat is produced by the RMS current. Since the peak current is 4 A, we have $$I_{\text{rms}} = \frac{I_0}{\sqrt{2}} = \frac{4}{\sqrt{2}}$$. Substituting this into the power formula for the resistor $$R_2 = 2 \Omega$$ gives $$H_2 = I_{\text{rms}}^2 \times R_2 \times t = \frac{16}{2} \times 2 \times t = 16t$$.

From the above expressions, the ratio of the heat produced is $$\frac{H_1}{H_2} = \frac{48t}{16t} = \frac{3}{1}$$.

Therefore, the correct answer is Option B: $$3:1$$.

A series LCR circuit has $$L = 0.01$$ H, $$R = 10 \ \Omega$$, $$C = 1 \ \mu F$$, and amplitude $$V_m = 50$$ V. We are asked to find the current amplitude at a frequency 60% lower than the resonant frequency.

Since the resonant angular frequency is given by $$\omega_0 = \frac{1}{\sqrt{LC}},$$ substituting $$L = 0.01$$ and $$C = 10^{-6}$$ yields $$\omega_0 = \frac{1}{\sqrt{0.01 \times 10^{-6}}} = \frac{1}{\sqrt{10^{-8}}} = \frac{1}{10^{-4}} = 10^4 \text{ rad/s}.$$

Because the frequency of interest is 60% lower than resonance, we have $$\omega = \omega_0 - 0.6\omega_0 = 0.4\omega_0 = 0.4 \times 10^4 = 4000 \text{ rad/s}.$$

Next, the inductive reactance is $$X_L = \omega L = 4000 \times 0.01 = 40 \ \Omega,$$ and the capacitive reactance is $$X_C = \frac{1}{\omega C} = \frac{1}{4000 \times 10^{-6}} = \frac{1}{4 \times 10^{-3}} = 250 \ \Omega.$$

From the above reactances, the impedance of the circuit follows as $$Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{10^2 + (40 - 250)^2} = \sqrt{100 + (-210)^2} = \sqrt{100 + 44100} = \sqrt{44200} \approx 210.24 \ \Omega.$$

Therefore, the amplitude of the current is $$I_m = \frac{V_m}{Z} = \frac{50}{210.24} \approx 0.2378 \text{ A} \approx 238 \text{ mA}.$$

Answer: Option C: 238 mA

Given, primary voltage $$V_p = 8\,\text{kV} = 8000\,\text{V}$$, secondary voltage $$V_s = 160\,\text{V}$$ and load power $$P = 80\,\text{kW} = 80000\,\text{W}$$.

Since the load is purely resistive, the power factor is 1.

For an ideal transformer, input power is equal to output power.

The secondary current is

$$I_s = \frac{P}{V_s} = \frac{80000}{160} = 500\,\text{A}$$

Therefore, the load resistance in the secondary circuit is

$$R_s = \frac{V_s}{I_s} = \frac{160}{500} = 0.32\,\Omega$$

The primary current is

$$I_p = \frac{P}{V_p} = \frac{80000}{8000} = 10\,\text{A}$$

Therefore, the equivalent load resistance as seen from the primary side is

$$R_p = \frac{V_p}{I_p} = \frac{8000}{10} = 800\,\Omega$$

We can also verify the result using the resistance transformation relation.

The turns ratio is

$$\frac{N_p}{N_s} = \frac{V_p}{V_s} = \frac{8000}{160} = 50$$

Hence,

$$R_p = R_s\left(\frac{N_p}{N_s}\right)^2$$

$$R_p = 0.32 \times 50^2 = 0.32 \times 2500 = 800\,\Omega$$

Thus, the load resistance in the primary circuit is $$800\,\Omega$$ and in the secondary circuit is $$0.32\,\Omega$$.

Answer: Option C

We need to find the voltage equation across an inductor given the current equation $$i = 5\sin(49\pi t - 30°)$$ and inductance $$L = 30 \text{ mH}$$.

First, we identify the parameters: the peak current $$I_0 = 5$$ A, the angular frequency $$\omega = 49\pi$$ rad/s, and the inductance $$L = 30 \text{ mH} = 30 \times 10^{-3}$$ H.

Since the circuit is purely inductive, the peak voltage is given by $$V_0 = I_0 \times \omega L$$. Substituting the known values yields:

$$V_0 = 5 \times 49\pi \times 30 \times 10^{-3}$$

$$= 5 \times 49 \times 3.1416 \times 0.03$$

$$= 5 \times 4.618$$

$$= 23.09 \approx 23.1 \text{ V}$$

In a purely inductive circuit, the voltage leads the current by 90°. Therefore, the phase of the voltage is shifted by +90° relative to the current, giving:

$$V = V_0 \sin(\omega t - 30° + 90°) = V_0 \sin(49\pi t + 60°)$$

Substituting the calculated peak voltage into this expression provides the voltage equation:

$$V = 23.1\sin(49\pi t + 60°)$$

Hence, the correct answer is Option D: $$23.1\sin(49\pi t + 60°)$$.

We need to determine how to increase the resonant frequency in a series LCR circuit.

Recall that the resonant frequency of a series LCR circuit is given by: $$f_0 = \frac{1}{2\pi\sqrt{LC}}$$

To increase $$f_0$$, we need to decrease either $$L$$ or $$C$$ (or both).

Evaluating the options, Option A: Increasing source frequency does not change the resonant frequency; it only changes the operating frequency.

Option B: Adding resistance does not affect the resonant frequency, since $$f_0$$ is independent of $$R$$.

Option C: Adding another capacitor in series with the first capacitor. For capacitors in series:

$$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}$$

This gives $$C_{eq} < C_1$$, so the equivalent capacitance decreases. Since $$f_0 \propto \frac{1}{\sqrt{C}}$$, decreasing $$C$$ increases $$f_0$$. This is correct.

Option D: Decreasing source frequency does not change the resonant frequency.

Hence, the correct answer is Option C.

Given: $$I = 5\sin(120\pi t)$$ A.

Comparing with the standard form $$I = I_0 \sin(\omega t)$$:

$$\omega = 120\pi \text{ rad/s}$$

The current starts from zero (at $$t = 0$$, $$I = 0$$) and reaches its peak value when $$\sin(120\pi t) = 1$$.

$$120\pi t = \frac{\pi}{2}$$

$$t = \frac{\pi}{2 \times 120\pi} = \frac{1}{240} \text{ s}$$

The correct answer is Option D.

The RMS value of conduction current in a parallel plate capacitor is $$I_{rms} = 6.9 \text{ } \mu A$$. The supply voltage is $$V_{rms} = 230 \text{ V}$$ and the angular frequency is $$\omega = 600 \text{ rad/s}$$.

For a purely capacitive circuit, the rms current and voltage are related by $$I_{rms} = V_{rms} \times \omega C$$.

Rearranging this relation to solve for capacitance gives $$C = \frac{I_{rms}}{V_{rms} \times \omega}$$.

Substituting the given values, we get $$C = \frac{6.9 \times 10^{-6}}{230 \times 600}$$, which simplifies to $$C = \frac{6.9 \times 10^{-6}}{138000}$$ and can also be expressed as $$C = \frac{6.9 \times 10^{-6}}{1.38 \times 10^5}$$.

Thus, the capacitance is $$C = 5 \times 10^{-11} \text{ F}$$, or $$C = 50 \text{ pF}$$.

Hence, the correct answer is Option B.

We have element X which passes peak current 5 A in phase with the peak voltage 100 V, so X is a resistor with resistance $$R = \frac{100}{5} = 20\;\Omega$$.

Element Y passes the same peak current of 5 A but lagging the voltage by $$\frac{\pi}{2}$$, so Y is an inductor with inductive reactance $$X_L = \frac{100}{5} = 20\;\Omega$$.

When X and Y are connected in series, the total impedance is $$Z = \sqrt{R^2 + X_L^2} = \sqrt{20^2 + 20^2} = 20\sqrt{2}\;\Omega$$.

The peak current in the series circuit is $$I_0 = \frac{V_0}{Z} = \frac{100}{20\sqrt{2}} = \frac{5}{\sqrt{2}}$$ A.

The rms current is $$I_{\text{rms}} = \frac{I_0}{\sqrt{2}} = \frac{5}{\sqrt{2} \cdot \sqrt{2}} = \frac{5}{2}$$ A.

Hence, the correct answer is Option 4.

We are given: resistance $$R = 40\ \Omega$$, AC source rated $$220$$ V, $$50$$ Hz. We need to find the time taken by the current to change from its maximum value to its rms value.

The current is given by $$i = I_0 \sin(\omega t)$$, where $$\omega = 2\pi f = 2\pi \times 50 = 100\pi$$ rad/s.

The maximum current occurs when $$\sin(\omega t_1) = 1$$, i.e., $$\omega t_1 = \frac{\pi}{2}$$. The rms value is $$I_{rms} = \frac{I_0}{\sqrt{2}}$$, so setting $$\frac{I_0}{\sqrt{2}} = I_0 \sin(\omega t_2)$$ gives $$\sin(\omega t_2) = \frac{1}{\sqrt{2}}$$. After the maximum (at $$\omega t = \frac{\pi}{2}$$), the next time this occurs is at $$\omega t_2 = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$$.

The difference in phase is $$\Delta(\omega t) = \frac{3\pi}{4} - \frac{\pi}{2} = \frac{\pi}{4}$$, so the time difference is $$\Delta t = \frac{\pi}{4\omega} = \frac{\pi}{4 \times 100\pi} = \frac{1}{400} \text{ s} = 2.5 \times 10^{-3} \text{ s} = 2.5 \text{ ms}$$.

Therefore, the correct answer is Option A.

We need to analyze both statements about AC circuits.

Statement I: The reactance of an AC circuit is zero. It is possible that the circuit contains a capacitor and an inductor. The net reactance of an AC circuit containing both an inductor and a capacitor is given by:

$$X_{net} = X_L - X_C = \omega L - \frac{1}{\omega C}$$

At resonance, when $$\omega = \frac{1}{\sqrt{LC}}$$, we get:

$$X_L = X_C \implies X_{net} = 0$$

Thus, it is indeed possible for the net reactance to be zero even when the circuit contains both a capacitor and an inductor. Therefore, Statement I is TRUE.

Statement II: In an AC circuit, the average power delivered by the source never becomes zero. The average power delivered in an AC circuit is given by:

$$P_{avg} = V_{rms} \cdot I_{rms} \cdot \cos\phi$$

where $$\cos\phi$$ is the power factor. For a purely inductive or purely capacitive circuit, the phase difference $$\phi = 90°$$, so:

$$\cos\phi = \cos 90° = 0$$

$$P_{avg} = V_{rms} \cdot I_{rms} \cdot 0 = 0$$

Hence, the average power can become zero. Thus, Statement II is FALSE.

From the above, since Statement I is true and Statement II is false, the correct option is Option C.

Given:
Source: 220 V,50 Hz
Lamp: 25 V,5 W

Step 1: Current through lamp
For rated operation:

$$P=VI⇒5=25⋅I⇒I=\ \frac{\ 5}{25}=\ \frac{\ 1}{5}A\ \ $$

Step 2: Voltage across resistance R

$$V_R=220−25=195V$$

Step 3: Calculate R

$$V_R=IR\Rightarrow195=\frac{1}{5}R\ ⇒R=195\times5=975Ω$$

Final Answer:

975 Ω

We need to find the frequency at which the current is in phase with the emf in an RLC series circuit (resonance condition). Current is in phase with emf when the circuit is at resonance. At resonance: $$\omega_0 = \frac{1}{\sqrt{LC}}$$ and $$f_0 = \frac{1}{2\pi\sqrt{LC}}$$.

Substituting the given values: $$L = 0.5$$ mH $$= 0.5 \times 10^{-3}$$ H, $$C = 200\ \mu$$F $$= 200 \times 10^{-6}$$ F gives $$LC = 0.5 \times 10^{-3} \times 200 \times 10^{-6} = 100 \times 10^{-9} = 10^{-7}$$ and $$\sqrt{LC} = \sqrt{10^{-7}} = 10^{-3.5} = \frac{10^{-3}}{\sqrt{10}}$$. Then $$f_0 = \frac{1}{2\pi \times \frac{10^{-3}}{\sqrt{10}}} = \frac{\sqrt{10}}{2\pi \times 10^{-3}}$$, which equals $$= \frac{3.162}{6.2832 \times 10^{-3}} = \frac{3.162}{6.2832} \times 10^3$$, yielding $$= 0.5033 \times 10^3 \approx 503.3 \text{ Hz}$$ or approximately $$\approx 5 \times 10^2 \text{ Hz}$$. The answer is 5 $$\times 10^2$$ Hz.

We need to find the value of $$a$$ for a purely inductive AC circuit where an inductor is connected across an alternating electromotive force (emf) source.

1. Identify the Given Data

• Inductance ($$L$$) = $$200\text{ mH} = 200 \times 10^{-3}\text{ H} = 0.2\text{ H}$$
• RMS Voltage ($$V_{rms}$$) = $$220\text{ V}$$
• Frequency ($$f$$) = $$50\text{ Hz}$$
• Peak Current ($$I_0$$) = $$\frac{\sqrt{a}}{\pi}\text{ A}$$

Note: In a purely inductive circuit, the current lags the voltage by $$90^\circ$$ ($$\frac{\pi}{2}$$ radians). Consequently, when the instantaneous voltage of the source drops to $$0\text{ V}$$, the current reaches its maximum magnitude (peak value $$I_0$$).

2. Calculate the Inductive Reactance ($$X_L$$)

The opposition offered by the inductor to the AC flow is given by:

$$X_L = \omega L = 2\pi f L$$

Substituting the given values:

$$X_L = 2 \times \pi \times 50 \times 0.2 = 20\pi\ \Omega$$

3. Calculate the Peak Voltage ($$V_0$$)

The relationship between the alternating source's peak voltage and its RMS value is:

$$V_0 = V_{rms} \times \sqrt{2} = 220\sqrt{2}\text{ V}$$

4. Calculate the Peak Current ($$I_0$$)

Using Ohm's law for an alternating inductive circuit network:

$$I_0 = \frac{V_0}{X_L}$$

Substituting our expressions for $$V_0$$ and $$X_L$$ yields:

$$I_0 = \frac{220\sqrt{2}}{20\pi} = \frac{11\sqrt{2}}{\pi}\text{ A}$$

5. Solve for $$a$$

We equate our derived value of peak current to the given mathematical form:

$$\frac{\sqrt{a}}{\pi} = \frac{11\sqrt{2}}{\pi}$$

Canceling out the common denominator $$\pi$$ from both sides gives:

$$\sqrt{a} = 11\sqrt{2}$$

Squaring both sides of the equation to isolate $$a$$:

$$a = (11\sqrt{2})^2 = 121 \times 2 = 242$$

Therefore, the value of $$a$$ is 242.

We have a capacitor of capacitance $$C = 500 \, \mu\text{F} = 500 \times 10^{-6}$$ F charged to $$V = 100$$ V, now connected to an inductor of inductance $$L = 50$$ mH $$= 50 \times 10^{-3}$$ H to form an LC circuit.

In an LC circuit, energy oscillates between the capacitor and the inductor. By conservation of energy, the maximum energy stored in the inductor equals the initial energy stored in the capacitor: $$\frac{1}{2}LI_{\max}^2 = \frac{1}{2}CV^2$$

Solving for $$I_{\max}$$: $$I_{\max} = V\sqrt{\frac{C}{L}} = 100 \times \sqrt{\frac{500 \times 10^{-6}}{50 \times 10^{-3}}} = 100 \times \sqrt{\frac{500}{50000}} = 100 \times \sqrt{0.01} = 100 \times 0.1 = 10$$ A.

Hence, the correct answer is 10.

Given that $$|V_L| = |V_C| = 2|V_R|$$, it implies:

$$I \cdot X_L = I \cdot X_C = 2(I \cdot R)$$

$$X_L = X_C = 2R$$

$$X_L = 2 \times 5 = 10\ \Omega$$

$$X_L = 2\pi f L$$

$$10 = 2 \times \pi \times 50 \times L$$

$$L = \frac{1}{10\pi} \times 1000\ \text{mH} = \frac{100}{\pi}\ \text{mH}$$

$$\frac{100}{\pi} = \frac{1}{K\pi}$$

$$100 = \frac{1}{K} \implies \mathbf{K = 0.01}$$

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In an LCR series circuit, the current amplitude becomes $$\frac{1}{\sqrt{2}}$$ times its maximum value at the half-power frequencies $$\omega_1$$ and $$\omega_2$$. Here $$\omega_1 = 212$$ rad/s, $$\omega_2 = 232$$ rad/s, and the resistance is $$R = 5$$ $$\Omega$$.

The bandwidth of the circuit is given by the difference between these frequencies: $$\Delta\omega = \omega_2 - \omega_1 = 232 - 212 = 20 \text{ rad/s}.$$

In a series LCR circuit, the bandwidth is also related to $$R$$ and $$L$$ by $$\Delta\omega = \frac{R}{L}.$$ Solving for $$L$$ gives $$L = \frac{R}{\Delta\omega} = \frac{5}{20} = 0.25 \text{ H}.$$

Converting to millihenry yields $$L = 0.25 \text{ H} = 250 \text{ mH}.$$

Therefore, the self inductance in the circuit is 250 mH.

We have a $$50 \text{ W}$$, $$100 \text{ V}$$ lamp connected in series with a capacitor of capacitance $$C = \dfrac{50}{\pi\sqrt{x}} \text{ }\mu\text{F}$$ to a $$200 \text{ V}$$, $$50 \text{ Hz}$$ AC source. We need to find $$x$$.

The lamp is rated $$50 \text{ W}$$ at $$100 \text{ V}$$:

$$R = \frac{V^2}{P} = \frac{100^2}{50} = 200 \text{ }\Omega$$

For the lamp to glow at full brightness, the voltage across it must be $$100 \text{ V}$$:

$$I = \frac{P}{V} = \frac{50}{100} = 0.5 \text{ A}$$

The total voltage is $$200 \text{ V}$$, so the impedance is:

$$Z = \frac{V_s}{I} = \frac{200}{0.5} = 400 \text{ }\Omega$$

For a series R-C circuit:

$$Z = \sqrt{R^2 + X_C^2}$$

$$400 = \sqrt{200^2 + X_C^2}$$

$$160000 = 40000 + X_C^2$$

$$X_C^2 = 120000$$

$$X_C = 200\sqrt{3} \text{ }\Omega$$

$$X_C = \frac{1}{2\pi f C}$$

$$200\sqrt{3} = \frac{1}{2\pi \times 50 \times C}$$

$$C = \frac{1}{200\sqrt{3} \times 100\pi} = \frac{1}{20000\sqrt{3}\pi}$$

Converting to $$\mu\text{F}$$:

$$C = \frac{10^6}{20000\sqrt{3}\pi} = \frac{50}{\sqrt{3}\pi} \text{ }\mu\text{F}$$

Comparing with $$C = \dfrac{50}{\pi\sqrt{x}} \text{ }\mu\text{F}$$:

$$\frac{50}{\pi\sqrt{x}} = \frac{50}{\sqrt{3}\pi}$$

$$\sqrt{x} = \sqrt{3}$$

$$x = 3$$

The value of $$x$$ is $$3$$.

The solution to find the time required to heat the resistance in an LCR circuit involves calculating the circuit impedance, current, and the heat produced.

1. Calculation of Net Reactance and Impedance (Z)

First, we find the difference between inductive reactance ($$X_L$$) and capacitive reactance ($$X_C$$):

$$|X_L - X_C| = |10 - 100| = 90\ \Omega$$

The total impedance Z of the circuit is given by:

$$Z = \sqrt{(X_L - X_C)^2 + R^2}$$

Substituting the values ($$R = 120\ \Omega$$):

$$Z = \sqrt{(90)^2 + (120)^2} = \sqrt{8100 + 14400} = \sqrt{22500}$$

$$Z = 150\ \Omega$$

2. Calculation of RMS Current ($$I_{rms}$$)

Using Ohm's law for AC circuits ($$V_{rms} = 20\text{ V}$$ based on common context for this problem):

$$I_{rms} = \frac{V_{rms}}{Z} = \frac{20}{150} = \frac{2}{15}\text{ A}$$

3. Energy Balance and Time ($\Delta t$)

The electrical energy dissipated as heat in the resistor ($$I_{rms}^2 R \Delta t$$) is equal to the heat energy required to raise the temperature ($$ms\Delta T$$):

$$I_{rms}^2 R \Delta t = C_{thermal} (\Delta T)$$

Given:

• Thermal capacity (ms or $$C_{thermal}$$) $$= 2\text{ J °C}^{-1}$$
• Temperature change ($$\Delta T$$) = $$16\text{ °C}$$

Substituting the values:

$$\left( \frac{2}{15} \right)^2 \times 120 \times \Delta t = 2 \times 16$$

$$\frac{4}{225} \times 120 \times \Delta t = 32$$

$$\frac{480}{225} \Delta t = 32$$

$$\Delta t = \frac{32 \times 225}{480}$$

$$\Delta t = \frac{7200}{480}$$

$$\boxed{\Delta t = 15\text{ sec}}$$

$$\omega = 2\pi f = 2 \times 3.14 \times 50 = 314\ \text{rad/s}$$

$$X_L = \omega L = 314 \times 100 \times 10^{-3} = 31.4\ \Omega$$

$$X_C = \frac{1}{\omega C} = \frac{1}{314 \times 100 \times 10^{-6}} = \frac{10^4}{314} \approx 31.85\ \Omega$$

$$Z = \sqrt{R^2 + (X_L - X_C)^2}$$

$$\implies Z = \sqrt{10^2 + (31.4 - 31.85)^2} = \sqrt{100 + (-0.45)^2} = \sqrt{100 + 0.2025} \approx 10\ \Omega$$

$$I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{220}{10} = 22\ \text{A}$$

We need to find the inductance required for minimum impedance (resonance condition) in a telegraph line. The line length is $$100$$ km and the capacitance per km is $$0.01$$ $$\mu$$F/km, so the total capacitance is $$C = 100 \times 0.01 = 1$$ $$\mu$$F $$= 10^{-6}$$ F. The frequency is $$f = 0.5$$ kHz $$= 500$$ Hz and $$\pi = \sqrt{10}$$.

The resonance condition for minimum impedance in a series LC circuit states that:

$$ \omega L = \frac{1}{\omega C} $$

This gives:

$$ L = \frac{1}{\omega^2 C} $$

Since $$\omega = 2\pi f = 2\sqrt{10} \times 500 = 1000\sqrt{10}$$, we have $$\omega^2 = (1000\sqrt{10})^2 = 1000000 \times 10 = 10^7$$. Substituting into the expression for $$L$$ with $$C = 10^{-6}$$ F yields:

$$ L = \frac{1}{10^7 \times 10^{-6}} = \frac{1}{10} = 0.1 \text{ H} = 100 \text{ mH} $$

Hence, the value of inductance required is 100 mH.

We begin with the expression for the impedance of a series $$LCR$$ circuit

$$Z=\sqrt{R^{2}+\left(X_L-X_C\right)^{2}}.$$

The rms current is $$I=\dfrac{V}{Z},$$ so the average (true) power delivered to the circuit is

$$P_{\text{avg}} = V_{\text{rms}}\,I_{\text{rms}}\cos\phi = V_{\text{rms}} \left(\dfrac{V_{\text{rms}}}{Z}\right)\left(\dfrac{R}{Z}\right) = \dfrac{V_{\text{rms}}^{2}\,R}{Z^{2}}.$$

Because $$R$$ is fixed, $$P_{\text{avg}}$$ becomes maximum when the impedance $$Z$$ is minimum. Looking at the formula for $$Z,$$ the minimum value occurs when the bracketed term vanishes, that is, when

$$X_L = X_C.$$

This condition is called resonance, and at resonance the circuit behaves as a pure resistance: the phase angle $$\phi$$ becomes zero and $$\cos\phi = 1,$$ giving the greatest possible average power.

We are given the inductive reactance

$$X_L = 250\pi\; \Omega.$$

To satisfy $$X_L = X_C,$$ we must have

$$X_C = 250\pi\; \Omega.$$

The capacitive reactance is related to the capacitance by the formula

$$X_C = \frac{1}{\omega C},$$

where the angular frequency $$\omega$$ equals $$2\pi f.$$ The supply frequency is $$f = 50\;\text{Hz},$$ so

$$\omega = 2\pi f = 2\pi \times 50 = 100\pi\;\text{rad s}^{-1}.$$

Substituting $$X_C = 250\pi$$ and $$\omega = 100\pi$$ into $$X_C = \dfrac{1}{\omega C},$$ we get

$$250\pi = \frac{1}{100\pi\,C}.$$

Rearranging for $$C$$ gives

$$C = \frac{1}{100\pi \times 250\pi} = \frac{1}{25000\pi^{2}}\;\text{farad}.$$

The problem states that $$\pi^{2} = 10,$$ so

$$C = \frac{1}{25000 \times 10} = \frac{1}{250000}\;\text{F}.$$

To convert to microfarads, recall that $$1\;\text{F} = 10^{6}\;\mu\text{F},$$ therefore

$$C = \frac{10^{6}}{250000}\;\mu\text{F} = 4\;\mu\text{F}.$$

Hence, the correct answer is Option B.

We are given an LCR circuit with $$R = 110 \; \Omega$$, supply voltage $$V = 220$$ V, and angular frequency $$\omega = 300$$ rad/s.

When only the capacitance is removed (leaving R and L in series), the current lags the voltage by 45°. This means $$\tan 45° = \frac{\omega L}{R}$$, so $$\omega L = R = 110 \; \Omega$$.

When only the inductor is removed (leaving R and C in series), the current leads the voltage by 45°. This means $$\tan 45° = \frac{1/(\omega C)}{R}$$, so $$\frac{1}{\omega C} = R = 110 \; \Omega$$.

Since $$\omega L = \frac{1}{\omega C} = 110 \; \Omega$$, the inductive reactance equals the capacitive reactance. This is the condition for resonance in an LCR circuit.

At resonance, the impedance of the circuit equals the resistance alone: $$Z = R = 110 \; \Omega$$.

The RMS current is $$I = \frac{V}{Z} = \frac{220}{110} = 2$$ A.

We need to determine the type of transformer used in the given circuit based on its input and output parameters.

From the given circuit diagram, we have the following data for the primary and secondary sides:

Primary voltage, $$V_1 = 220\text{ V}$$

Secondary power delivered to the load, $$P_2 = 60\text{ W}$$

Secondary load current, $$i_L = 0.11\text{ A}$$

The power on the secondary side of an ideal transformer can be expressed as: $$P_2 = V_2 \times i_L$$, where $$V_2$$ is the secondary output voltage.

Rearranging the formula to solve for the secondary voltage yields: $$V_2 = \frac{P_2}{i_L}$$.

Substituting the given values into the expression gives: $$V_2 = \frac{60}{0.11} \approx 545.45\text{ V}$$.

Comparing the secondary voltage with the primary voltage, we find that: $$V_2 > V_1$$ since $$545.45\text{ V} > 220\text{ V}$$.

Since the output voltage on the secondary side is greater than the input voltage on the primary side, the transformer increases the voltage level.

Therefore, the correct answer is Option C: Step-up transformer.

In a purely capacitive circuit with $$E_g = E_{g0}\sin\omega t$$, the current leads the voltage by $$90°$$. This is because in a capacitor, charge builds up as voltage rises, and maximum current flows when the voltage is zero (rate of change is maximum), while zero current flows when the voltage is at its peak.

Specifically, the current is: $$I_C = I_0\sin\left(\omega t + \frac{\pi}{2}\right) = I_0\cos(\omega t)$$

while the voltage across the capacitor is $$V_C = V_0\sin(\omega t)$$ (in phase with the applied emf).

In the phasor diagram, $$V_C$$ (or the emf $$E_g$$) is taken as the reference phasor, and $$I_C$$ leads $$V_C$$ by $$90°$$ (i.e., $$I_C$$ is drawn $$90°$$ ahead of $$V_C$$ in the direction of rotation). The correct phasor diagram shows $$I_C$$ perpendicular to and leading $$V_C$$, which corresponds to phasor diagram (3).

The given current is $$I = I_1 \sin\omega t + I_2 \cos\omega t$$. A hot wire ammeter reads the RMS (root mean square) value of the current.

The RMS value is $$I_{\text{rms}} = \sqrt{\langle I^2 \rangle}$$, where $$\langle \cdot \rangle$$ denotes the time average. Expanding $$I^2 = I_1^2 \sin^2\omega t + I_2^2 \cos^2\omega t + 2I_1 I_2 \sin\omega t \cos\omega t$$.

Taking the time average: $$\langle \sin^2\omega t \rangle = \frac{1}{2}$$, $$\langle \cos^2\omega t \rangle = \frac{1}{2}$$, and $$\langle \sin\omega t \cos\omega t \rangle = 0$$. Therefore $$\langle I^2 \rangle = \frac{I_1^2}{2} + \frac{I_2^2}{2} = \frac{I_1^2 + I_2^2}{2}$$.

Thus $$I_{\text{rms}} = \sqrt{\frac{I_1^2 + I_2^2}{2}}$$, and the correct answer is option 2.

In a series LCR circuit, the impedance is $$Z = \sqrt{R^2 + (X_L - X_C)^2}$$.

Substituting the given values: $$Z = \sqrt{6^2 + (10 - 4)^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2}$$ $$\Omega$$.

The power factor is $$\cos\phi = \frac{R}{Z} = \frac{6}{6\sqrt{2}} = \frac{1}{\sqrt{2}}$$.

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We need to find the output wave pattern monitored by a CRO for a series $$RC$$ circuit driven by an AC source generating a square wave, where the output is taken across the capacitor ($$C$$).

When a square wave voltage is applied to an $$RC$$ circuit, the input voltage alternates between a constant high value and a constant low value. The behavior of the voltage across the capacitor ($$V_C$$) is governed by the processes of charging and discharging through the resistor ($$R$$):

1. During the Positive Half-Cycle of the Square Wave (Charging Phase):

• When the input voltage jumps to its maximum constant value, the capacitor does not instantly reach this voltage. Instead, it begins to charge up exponentially.
• The voltage across the capacitor rises according to the exponential charging formula: $$V_C(t) = V_0 \left(1 - e^{-\frac{t}{RC}}\right)$$.
• This results in a smooth, upward-curving (convex) profile that gradually flattens out as it approaches the peak input voltage.

2. During the Negative/Zero Half-Cycle of the Square Wave (Discharging Phase):

• When the input voltage drops instantly to zero or its minimum value, the capacitor begins to release its stored charge through the resistor.
• The voltage across the capacitor drops exponentially according to the discharging formula: $$V_C(t) = V_0 e^{-\frac{t}{RC}}$$.
• This results in a smooth, downward-sloping curve that decays rapidly at first and then tapers off.

Conclusion

Combining these alternating cycles, the waveform across the capacitor consists of repeating segments of exponential growth followed by exponential decay. This distinctive rounded, sawtooth-like appearance is characteristic of a low-pass filter integrating a square wave input.

Therefore, the output wave pattern monitored by the CRO closely matches Option A, which depicts these smooth exponential charging and discharging transitions.

We need to determine the total rms current $$I$$ flowing through the given parallel alternating current (AC) circuit arrangement.

1. Identify the Given Parameters

From the problem, the parallel circuit consists of two parallel branches connected across an AC voltage source of $$V_{\text{rms}} = 200\text{ V}$$ with an angular frequency of $$\omega = 100\text{ rad s}^{-1}$$:

• Branch 1 (Top RC series path): Resistance $$R_1 = 100\ \Omega$$ and Capacitance $$C = 100\ \mu\text{F}$$.
• Branch 2 (Bottom RL series path): Resistance $$R_2 = 50\ \Omega$$ and Inductance $$L = 0.50\text{ H}$$.

2. Calculate Reactances and Branch Impedances

We first calculate the capacitive reactance ($$X_C$$) and inductive reactance ($$X_L$$), followed by the total impedance of each path:

• Capacitive Branch (Branch 1):
  The reactance of the ideal capacitor is:

  $$X_C = \frac{1}{\omega C} = \frac{1}{100 \times 100 \times 10^{-6}} = 100\ \Omega$$

  The impedance ($$Z_1$$) of the series RC branch is:

  $$Z_1 = \sqrt{R_1^2 + X_C^2} = \sqrt{100^2 + 100^2} = 100\sqrt{2}\ \Omega \approx 141.42\ \Omega$$

• Inductive Branch (Branch 2):
  The reactance of the ideal inductor is:

  $$X_L = \omega L = 100 \times 0.50 = 50\ \Omega$$

  The impedance ($$Z_2$$) of the series RL branch is:

  $$Z_2 = \sqrt{R_2^2 + X_L^2} = \sqrt{50^2 + 50^2} = 50\sqrt{2}\ \Omega \approx 70.71\ \Omega$$

3. Determine Individual Branch Currents and Phase Angles

The voltage drop across both parallel branches is equal to the source voltage ($$V_{\text{rms}} = 200\text{ V}$$):

• Current in Branch 1 ($$I_1$$):
  

  $$I_1 = \frac{V_{\text{rms}}}{Z_1} = \frac{200}{100\sqrt{2}} = \sqrt{2}\text{ A} \approx 1.414\text{ A}$$

  Since it is an RC circuit, the current leads the voltage by a phase angle of $$\phi_1 = \tan^{-1}\left(\frac{X_C}{R_1}\right) = \tan^{-1}(1) = +45^\circ$$.
• Current in Branch 2 ($$I_2$$):
  

  $$I_2 = \frac{V_{\text{rms}}}{Z_2} = \frac{200}{50\sqrt{2}} = 2\sqrt{2}\text{ A} \approx 2.828\text{ A}$$

  Since it is an RL circuit, the current lags behind the voltage by a phase angle of $$\phi_2 = \tan^{-1}\left(\frac{X_L}{R_2}\right) = \tan^{-1}(1) = -45^\circ$$.

4. Calculate total Phasor Current ($$I$$)

The net phase difference between the leading branch current ($$I_1$$) and lagging branch current ($$I_2$$) is $$\Delta\phi = 45^\circ - (-45^\circ) = 90^\circ$$. Because the two current vectors are perpendicular (in quadrature), their phasor sum can be evaluated using the Pythagorean theorem:

$$I = \sqrt{I_1^2 + I_2^2}$$

$$I = \sqrt{(\sqrt{2})^2 + (2\sqrt{2})^2} = \sqrt{2 + 8} = \sqrt{10}\text{ A}$$

$$I \approx 3.16\text{ A}$$

Conclusion

The total current flowing through the AC circuit configuration is 3.16 A.

For a series LCR circuit, the inductive reactance is $$X_L = 2\pi f L = 2\pi \times 50 \times 0.5 \times 10^{-3} \approx 0.157 \,\Omega$$.

The capacitive reactance is $$X_C = \frac{1}{2\pi f C} = \frac{1}{2\pi \times 50 \times 0.1 \times 10^{-12}} \approx 3.18 \times 10^{10} \,\Omega$$.

Since $$X_C \gg X_L$$, the net reactance is dominantly capacitive. The phase angle between the voltage and the current is given by $$\tan\phi = \frac{X_L - X_C}{R}$$. Here $$X_L - X_C \approx -3.18 \times 10^{10} \,\Omega$$ and $$R = 100 \,\Omega$$, so $$\tan\phi \approx -3.18 \times 10^8$$, giving $$\phi \approx -90^\circ$$. The negative sign means the current leads the voltage by $$\approx 90^\circ$$, which is characteristic of a predominantly capacitive circuit.

Therefore the phase angle between current and supplied voltage is $$\approx 90^\circ$$ and the circuit is predominantly capacitive.

We begin by identifying the two given simple harmonic motions:

$$y_1 = 10\sin\!\left(3\pi t + \frac{\pi}{3}\right),$$

$$y_2 = 5\left(\sin 3\pi t + \sqrt{3}\cos 3\pi t\right).$$

For any expression of the form $$y = a\sin\omega t + b\cos\omega t,$$ the amplitude is found from the well-known relation

$$\text{Amplitude} = \sqrt{a^{2}+b^{2}}.$$

First we look at $$y_1.$$ Here, the coefficient of the sine term is already a single number, namely $$10,$$ and there is no separate cosine term. Thus the amplitude of $$y_1$$ is directly

$$A_1 = 10.$$

Now we turn to $$y_2.$$ We rewrite it in the separated form to make the two coefficients clear:

$$y_2 = 5\sin 3\pi t + 5\sqrt{3}\cos 3\pi t.$$

So, in the standard notation, we have

$$a = 5, \quad b = 5\sqrt{3}.$$

Applying the amplitude formula, we obtain

$$A_2 = \sqrt{a^{2}+b^{2}} = \sqrt{\left(5\right)^{2} + \left(5\sqrt{3}\right)^{2}} = \sqrt{25 + 25\cdot 3} = \sqrt{25 + 75} = \sqrt{100} = 10.$$

Thus the amplitudes come out equal:

$$A_1 = 10, \qquad A_2 = 10.$$

The required ratio of the amplitudes is therefore

$$\frac{A_1}{A_2} = \frac{10}{10} = 1.$$

This means

$$x : 1 = 1 : 1,$$

so

$$x = 1.$$

So, the answer is $$1$$.

We have a transformer that converts 220 V A.C. to 12 V D.C. The secondary coil has 24 turns.

For an ideal transformer, the voltage ratio equals the turns ratio: $$\frac{V_1}{V_2} = \frac{N_1}{N_2}$$.

Here, $$V_1 = 220$$ V, $$V_2 = 12$$ V, and $$N_2 = 24$$.

Substituting, $$\frac{220}{12} = \frac{N_1}{24}$$.

$$N_1 = \frac{220 \times 24}{12} = \frac{5280}{12} = 440$$

Hence, the number of turns in the primary coil is 440.

So, the answer is $$440$$.

We need to find the angular frequency ($$\omega$$) at which the average power dissipated per cycle is exactly the same for both alternating current (AC) circuits shown in the diagram.

1. Identify the Circuit Components

From the problem :

• Circuit (a): A pure resistive circuit containing only a resistor ($$R = 5\ \Omega$$) connected across an AC voltage source of $$220\text{ V}$$.
• Circuit (b): A series LCR circuit containing a capacitor ($$C = 40\ \mu\text{F} = 40 \times 10^{-6}\text{ F}$$), a resistor ($$R = 5\ \Omega$$), and an inductor ($$L = 0.1\text{ H}$$) connected across an identical AC voltage source of $$220\text{ V}$$.

2. Analyze Average Power Dissipated

The average power dissipated ($$P_{\text{avg}}$$) in an AC circuit depends on the root-mean-square voltage ($$V_{\text{rms}}$$), total impedance ($$Z$$), and the power factor ($$\cos\phi$$):

$$P_{\text{avg}} = V_{\text{rms}} I_{\text{rms}} \cos\phi = V_{\text{rms}} \left(\frac{V_{\text{rms}}}{Z}\right) \left(\frac{R}{Z}\right) = \frac{V_{\text{rms}}^2 R}{Z^2}$$

• For Circuit (a):
  Since it contains only a resistor, its impedance is simply $$Z_a = R$$. The power equation simplifies to:

  $$P_a = \frac{V_{\text{rms}}^2}{R}$$

• For Circuit (b):
  The impedance of a series LCR circuit is given by $$Z_b = \sqrt{R^2 + (X_L - X_C)^2}$$, making its power:

  $$P_b = \frac{V_{\text{rms}}^2 R}{R^2 + (X_L - X_C)^2}$$

3. Apply the Condition for Equal Power

Setting the average power values equal to each other ($$P_a = P_b$$):

$$\frac{V_{\text{rms}}^2}{R} = \frac{V_{\text{rms}}^2 R}{R^2 + (X_L - X_C)^2}$$

Cancel out $$V_{\text{rms}}^2$$ and cross-multiply to solve:

$$R^2 + (X_L - X_C)^2 = R^2$$

$$(X_L - X_C)^2 = 0 \implies X_L = X_C$$

This shows that for the power to be identical, circuit (b) must be operating at its resonance frequency, where the inductive reactance matches the capacitive reactance.

4. Calculate the Angular Frequency ($$\omega$$)

Express the reactances in terms of angular frequency ($$\omega$$):

$$\omega L = \frac{1}{\omega C} \implies \omega^2 = \frac{1}{LC}$$

$$\omega = \frac{1}{\sqrt{LC}}$$

Substitute the given values of $$L$$ and $$C$$ into the formula:

$$\omega = \frac{1}{\sqrt{0.1 \times 40 \times 10^{-6}}} = \frac{1}{\sqrt{4 \times 10^{-6}}} = \frac{1}{2 \times 10^{-3}}$$

$$\omega = \frac{1000}{2} = 500\text{ rad s}^{-1}$$

Conclusion

The average power dissipated in one cycle will be the same in both circuits at an angular frequency of 500 $$\text{rad s}^{-1}$$.

We have a resonance circuit with inductance $$L = 2 \times 10^{-4}$$ H, resistance $$R = 6.28$$ $$\Omega$$, and resonant frequency $$f = 10$$ MHz $$= 10 \times 10^6$$ Hz.

The quality factor of a resonant circuit is given by $$Q = \frac{\omega L}{R}$$, where $$\omega = 2\pi f$$ is the angular frequency.

Calculating $$\omega$$: $$\omega = 2\pi \times 10 \times 10^6 = 2 \times 3.14 \times 10^7 = 6.28 \times 10^7$$ rad/s.

Now substituting into the quality factor formula: $$Q = \frac{6.28 \times 10^7 \times 2 \times 10^{-4}}{6.28}$$.

$$Q = \frac{6.28 \times 2 \times 10^3}{6.28} = 2 \times 10^3 = 2000$$

So, the answer is $$2000$$.

At resonance in a series LCR circuit, the impedance is purely resistive: $$Z = R$$. Therefore, the current at resonance is maximum and equals:

$$I_{\text{rms}} = \frac{V_{\text{rms}}}{R} = \frac{250}{5} = 50 \text{ A}$$

The power dissipated at resonance is:

$$P = I_{\text{rms}}^2 \times R = (50)^2 \times 5 = 2500 \times 5 = 12500 \text{ W}$$

Expressing in units of $$10^2$$ W: $$P = 125 \times 10^2$$ W.

Therefore the answer is $$125$$.

At resonance in a series LCR circuit, the inductive reactance equals the capacitive reactance, i.e., $$X_L = X_C$$. This means the impedance of the circuit is purely resistive, so $$Z = R = 8 \; \Omega$$.

The peak value of the voltage is given as $$V_0 = 250$$ V. The RMS voltage is $$V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{250}{\sqrt{2}}$$ V.

The power dissipated at resonance is given by $$P = \frac{V_{rms}^2}{R}$$. Substituting the values, we get $$P = \frac{\left(\frac{250}{\sqrt{2}}\right)^2}{8} = \frac{\frac{250^2}{2}}{8} = \frac{62500}{16} = 3906.25$$ W.

Converting to kilowatts, $$P = 3906.25 \div 1000 \approx 3.9$$ kW. Therefore, to the nearest integer, $$x = 4$$.

Based on the provided images, here is the formatted, step-by-step solution to the problem.

Problem Statement

At very high frequencies, find the effective impedance of the given AC circuit.

Step-by-Step Solution

To find the effective impedance at very high frequencies ($$f \to \infty$$), we need to evaluate how the reactive components (inductors and capacitors) behave.

1. Behavior of Capacitors at High Frequency

The capacitive reactance ($$X_C$$) is given by the formula:

$$X_C = \frac{1}{2\pi f C}$$

• As frequency ($$f$$) becomes very large, $$X_C$$ approaches zero ($$X_C \to 0$$).
• Therefore, all capacitors act as a short circuit (ideal wires).
• As frequency ($$f$$) becomes very large, $$X_L$$ approaches infinity ($$X_L \to \infty$$).
• Therefore, the inductor acts as an open circuit (broken wire/infinite resistance).
• Replace all capacitors ($$0.5\text{ F}$$) with simple connecting wires.
• Replace the branch containing the inductor ($$20\text{ H}$$) with an open break. Because this bottom-left branch is broken, no current flows through the $$1\ \Omega$$ resistor connected in series with the inductor.
• A single $$1\ \Omega$$ resistor in the top-left branch.
• Followed by a parallel combination of two $$2\ \Omega$$ resistors on the right.
• Parallel Combination:
• Total Impedance:

2. Behavior of Inductors at High Frequency

The inductive reactance ($$X_L$$) is given by the formula:

$$X_L = 2\pi f L$$

Simplified Circuit Analysis

Applying these conditions to the original network:

This leaves us with the following simplified equivalent circuit:

Calculating Equivalent Impedance ($$Z_{\text{eq}}$$)

The two $$2\ \Omega$$ resistors are in parallel. Their equivalent resistance ($$R_p$$) is:

$$R_p = \frac{2 \times 2}{2 + 2} = 1\ \Omega$$

This combination is in series with the remaining $$1\ \Omega$$ resistor:

$$Z_{\text{eq}} = 1\ \Omega + R_p = 1 + 1 = 2\ \Omega$$

Final Answer

The effective impedance of the given circuit at very high frequencies is $$2$$ $$\Omega$$.

We begin with the original series circuit that contains only a resistor of resistance $$R$$ and an inductor whose inductive reactance is $$X_L = 3R$$. In any series $$R$$-$$L$$ circuit, the impedance is given by the formula

$$Z = \sqrt{R^{2} + X_L^{2}}.$$

Substituting $$X_L = 3R$$, we obtain

$$Z_{\text{old}} = \sqrt{R^{2} + (3R)^{2}} = \sqrt{R^{2} + 9R^{2}} = \sqrt{10R^{2}} = R\sqrt{10}.$$

The power factor of an AC circuit is the cosine of the phase angle between the voltage and the current, and in a series circuit it is given by

$$\cos\phi = \frac{R}{Z}.$$

So, for the original circuit,

$$\cos\phi_{\text{old}} = \frac{R}{R\sqrt{10}} = \frac{1}{\sqrt{10}}.$$

Now a capacitor is connected in series, having a capacitive reactance $$X_C = 2R$$. In a series $$R$$-$$L$$-$$C$$ circuit, the net reactance is the difference between the inductive and capacitive reactances:

$$X = X_L - X_C.$$

Substituting the given values, we get

$$X = 3R - 2R = R.$$

Using the impedance formula again, this time with the new net reactance $$X = R$$, we have

$$Z_{\text{new}} = \sqrt{R^{2} + X^{2}} = \sqrt{R^{2} + R^{2}} = \sqrt{2R^{2}} = R\sqrt{2}.$$

The new power factor is therefore

$$\cos\phi_{\text{new}} = \frac{R}{R\sqrt{2}} = \frac{1}{\sqrt{2}}.$$

We are asked for the ratio of the new power factor to the old power factor. Writing this ratio explicitly, we have

$$\frac{\cos\phi_{\text{new}}}{\cos\phi_{\text{old}}} = \frac{\dfrac{1}{\sqrt{2}}}{\dfrac{1}{\sqrt{10}}} = \frac{1}{\sqrt{2}}\times\frac{\sqrt{10}}{1} = \sqrt{\frac{10}{2}} = \sqrt{5}.$$

Hence,

$$\cos\phi_{\text{new}} : \cos\phi_{\text{old}} = \sqrt{5} : 1.$$

Comparing with the statement in the question, $$\sqrt{5} : x$$, we immediately identify $$x = 1.$$

Hence, the correct answer is Option 1.

In a series LCR circuit at resonance, the impedance of the circuit equals the resistance $$R$$ alone, because the inductive and capacitive reactances cancel each other out ($$X_L = X_C$$). Therefore, the circuit behaves as a purely resistive circuit at resonance.

The power drawn from the source at resonance is given by $$P = \frac{V^2}{R}$$, where $$V = 120$$ V is the source voltage and $$P = 16$$ W is the power consumed.

Solving for $$R$$: $$R = \frac{V^2}{P} = \frac{(120)^2}{16} = \frac{14400}{16} = 900 \; \Omega$$.

Therefore, the value of resistance $$R$$ in the circuit is $$900 \; \Omega$$.

In an LCR series circuit with $$L = 30$$ mH $$= 30 \times 10^{-3}$$ H, $$R = 1\ \Omega$$, and angular frequency $$\omega = 300$$ rad s$$^{-1}$$, we need to find the capacitance for which the current leads the voltage by $$45^\circ$$.

The inductive reactance is $$X_L = \omega L = 300 \times 30 \times 10^{-3} = 9\ \Omega$$.

For the current to lead the voltage, the circuit must be capacitively dominant, meaning $$X_C > X_L$$. The phase angle $$\phi$$ satisfies:

$$\tan\phi = \frac{X_C - X_L}{R}$$

With $$\phi = 45^\circ$$, $$\tan 45^\circ = 1$$, so:

$$X_C - X_L = R \implies X_C - 9 = 1 \implies X_C = 10\ \Omega$$

Since $$X_C = \frac{1}{\omega C}$$:

$$\frac{1}{\omega C} = 10 \implies C = \frac{1}{300 \times 10} = \frac{1}{3000} = \frac{1}{3} \times 10^{-3}\ \text{F}$$

Comparing with $$C = \frac{1}{x} \times 10^{-3}$$ F, we get $$x = 3$$.

Therefore, the value of $$x$$ is $$\boxed{3}$$.

We have the instantaneous current as

$$i(t)=\sqrt{42}\,\sin\!\left(\frac{2\pi}{T}\,t\right)+10.$$

To find its r.m.s. (root-mean-square) value, we begin with the basic definition: for a periodic current of period $$T$$, the r.m.s. value is

$$I_{\text{rms}}=\sqrt{\frac1T\int_0^T i^2(t)\,dt}.$$

So we must first square the given current:

$$i^2(t)=\Bigl(\sqrt{42}\,\sin\!\left(\tfrac{2\pi}{T}t\right)+10\Bigr)^2.$$

Expanding the square term-by-term, we obtain

$$i^2(t)=\left(\sqrt{42}\,\sin\!\left(\tfrac{2\pi}{T}t\right)\right)^2 +2\left(\sqrt{42}\,\sin\!\left(\tfrac{2\pi}{T}t\right)\right)(10) +10^2.$$

Simplifying each part separately gives

$$i^2(t)=42\,\sin^2\!\left(\tfrac{2\pi}{T}t\right) +20\sqrt{42}\,\sin\!\left(\tfrac{2\pi}{T}t\right) +100.$$

We now take the average of this expression over one complete cycle. To do so, we recall two standard time-averages for a pure sine wave over its period:

$$\frac1T\int_0^T\sin\!\left(\tfrac{2\pi}{T}t\right)\,dt=0,$$

$$\frac1T\int_0^T\sin^2\!\left(\tfrac{2\pi}{T}t\right)\,dt=\frac12.$$

Using these results, the mean (average) of each term is

$$\frac1T\int_0^T 42\,\sin^2\!\left(\tfrac{2\pi}{T}t\right)\,dt =42\left(\frac12\right)=21,$$

$$\frac1T\int_0^T 20\sqrt{42}\,\sin\!\left(\tfrac{2\pi}{T}t\right)\,dt =20\sqrt{42}\,(0)=0,$$

$$\frac1T\int_0^T 100\,dt=100.$$

Adding these average values we have the mean square of the current:

$$\langle i^2(t)\rangle =21+0+100=121.$$

Finally, the r.m.s. value is the square root of this mean square:

$$I_{\text{rms}}=\sqrt{121}=11\ \text{A}.$$

So, the answer is $$11\text{ A}.$$

The transmitting station releases waves of wavelength $$\lambda = 960\,\text{m}$$. The frequency of the wave is $$f = \frac{c}{\lambda} = \frac{3 \times 10^8}{960}\,\text{Hz}$$.

For a resonant circuit, the resonant frequency is $$f = \frac{1}{2\pi\sqrt{LC}}$$. At resonance, this must equal the frequency of the transmitted wave.

Therefore: $$\frac{c}{\lambda} = \frac{1}{2\pi\sqrt{LC}}$$, which gives $$\sqrt{LC} = \frac{\lambda}{2\pi c}$$.

Squaring both sides: $$LC = \frac{\lambda^2}{4\pi^2 c^2}$$.

Solving for $$L$$: $$L = \frac{\lambda^2}{4\pi^2 c^2 C}$$.

Substituting $$\lambda = 960\,\text{m}$$, $$c = 3 \times 10^8\,\text{m/s}$$, and $$C = 2.56 \times 10^{-6}\,\text{F}$$:

$$L = \frac{(960)^2}{4\pi^2 \times (3 \times 10^8)^2 \times 2.56 \times 10^{-6}}$$

Computing the numerator: $$960^2 = 921600$$. Computing the denominator: $$4\pi^2 \approx 39.478$$, $$(3 \times 10^8)^2 = 9 \times 10^{16}$$. So the denominator is $$39.478 \times 9 \times 10^{16} \times 2.56 \times 10^{-6} = 39.478 \times 23.04 \times 10^{10} \approx 9.096 \times 10^{12}$$.

Therefore: $$L = \frac{921600}{9.096 \times 10^{12}} \approx 1.013 \times 10^{-7}\,\text{H} \approx 10 \times 10^{-8}\,\text{H}$$.

Comparing with $$x \times 10^{-8}$$ H, we get $$x = 10$$.

The source has frequency $$f = 750 \text{ Hz}$$ and rms voltage $$V_{\text{rms}} = 20 \text{ V}$$.

First we convert frequency to angular frequency using the relation $$\omega = 2\pi f$$. Hence

$$\omega = 2\pi \times 750 \; \text{rad s}^{-1} = 1500\pi \; \text{rad s}^{-1} \approx 4712 \; \text{rad s}^{-1}.$$

For the series combination we need the reactances. The inductive reactance is defined by the formula $$X_L = \omega L$$. Substituting $$L = 0.1803\ \text{H}$$,

$$X_L = 4712 \times 0.1803 \; \Omega \approx 849.6 \; \Omega.$$

The capacitive reactance is given by $$X_C = \dfrac{1}{\omega C}$$. With $$C = 10\ \mu\text{F} = 10 \times 10^{-6}\ \text{F},$$

$$X_C = \dfrac{1}{4712 \times 10 \times 10^{-6}} \; \Omega = \dfrac{1}{0.0471239} \; \Omega \approx 21.23 \; \Omega.$$

The net reactance of the series circuit is therefore

$$X = X_L - X_C = 849.6 \; \Omega - 21.23 \; \Omega \approx 828.4 \; \Omega.$$

The magnitude of the total impedance of a series $$RLC$$ circuit is obtained from the relation

$$Z = \sqrt{R^2 + X^2}.$$

Putting $$R = 100\ \Omega$$ and $$X = 828.4\ \Omega$$,

$$Z = \sqrt{(100)^2 + (828.4)^2} = \sqrt{10000 + 686263} = \sqrt{696263} \approx 834.5\ \Omega.$$

The rms current in the circuit is given by Ohm’s law $$I_{\text{rms}} = \dfrac{V_{\text{rms}}}{Z}$$. Using $$V_{\text{rms}} = 20\ \text{V},$$

$$I_{\text{rms}} = \dfrac{20}{834.5}\ \text{A} \approx 2.396 \times 10^{-2}\ \text{A}$$

or $$I_{\text{rms}} \approx 0.02396\ \text{A}.$$

The power actually dissipated (converted to heat) is only that in the resistor. The average power in the resistor is

$$P = I_{\text{rms}}^{2} R.$$

Substituting,

$$P = (0.02396)^2 \times 100\ \text{W} = 0.000574 \times 100\ \text{W} \approx 0.0574\ \text{W}.$$

The resistor has a heat capacity of $$2\ \text{J}/^{\circ}\text{C}$$. The energy needed to raise its temperature by $$\Delta T = 10^{\circ}\text{C}$$ is

$$Q = (\text{heat capacity}) \times \Delta T = 2 \times 10\ \text{J} = 20\ \text{J}.$$

If the whole of the electrical power goes into heating (no losses to the surroundings are assumed), the time required is

$$t = \dfrac{Q}{P} = \dfrac{20\ \text{J}}{0.0574\ \text{W}} \approx 3.47 \times 10^{2}\ \text{s} \approx 347\ \text{s}.$$

Rounded to the nearest whole second this is about $$348\ \text{s}$$.

Hence, the correct answer is Option D.

We first recall the loop equation for a series $$L\,C\,R$$ circuit. Applying Kirchhoff’s voltage law around the closed loop, the algebraic sum of the potential drops across the inductor, resistor and capacitor is zero, i.e.

$$V_L+V_R+V_C = 0.$$

For each element we write its constitutive relation: the inductor obeys $$V_L = L\,\dfrac{dI}{dt},$$ the resistor obeys Ohm’s law $$V_R = R\,I,$$ and the capacitor obeys $$V_C = \dfrac{q}{C},$$ where $$q$$ is the charge on the capacitor and $$I=\dfrac{dq}{dt}$$ is the circuit current. Substituting these three expressions into the loop equation gives

$$L\,\dfrac{dI}{dt}+R\,I+\dfrac{q}{C}=0.$$

Now we eliminate the current in favour of charge. Because $$I=\dfrac{dq}{dt},$$ we have $$\dfrac{dI}{dt}=\dfrac{d^2q}{dt^2}.$$ Writing everything in terms of $$q(t)$$ yields the second-order differential equation

$$L\,\dfrac{d^2q}{dt^2}+R\,\dfrac{dq}{dt}+\dfrac{1}{C}\,q=0.\qquad(1)$$

This equation is the electrical analogue of the mechanical damped spring-mass system. For the mechanical oscillator of mass $$m,$$ damping constant $$b$$ and spring constant $$k,$$ Newton’s second law gives

$$m\,\dfrac{d^2x}{dt^2}+b\,\dfrac{dx}{dt}+k\,x=0.\qquad(2)$$

Comparing equation (1) term by term with equation (2), we match the coefficients of the second derivative, the first derivative and the displacement/charge terms. We obtain

$$\begin{aligned} \text{Coefficient of }\dfrac{d^2}{dt^2}:&\quad L \longleftrightarrow m,\\[6pt] \text{Coefficient of }\dfrac{d}{dt}:&\quad R \longleftrightarrow b,\\[6pt] \text{Coefficient of the variable itself}:&\quad \dfrac{1}{C} \longleftrightarrow k. \end{aligned}$$

The last line tells us that $$\dfrac{1}{C}=k,$$ or equivalently $$C=\dfrac{1}{k}.$$ Therefore the complete correspondence is

$$L \leftrightarrow m,\qquad C \leftrightarrow \dfrac{1}{k},\qquad R \leftrightarrow b.$$

This mapping is precisely the one stated in Option D.

Hence, the correct answer is Option D.

We have a series $$RLC$$ circuit whose resistance, capacitance and inductance are given as $$R = 100\,\Omega,\; C = 2\,\mu\text{F} = 2 \times 10^{-6}\,\text{F},\; L = 80\,\text{mH} = 0.08\,\text{H}.$$

The quality factor $$Q$$ of a series resonant circuit is defined by the standard relation

$$Q \;=\; \dfrac{\omega_0 L}{R},$$

where $$\omega_0$$ is the angular resonant frequency. First, we find $$\omega_0$$. For a series $$RLC$$ circuit, the resonance condition gives

$$\omega_0 = \dfrac{1}{\sqrt{LC}}.$$

Substituting the given inductance and capacitance, we get

$$\omega_0 = \dfrac{1}{\sqrt{\,L\,C\,}} = \dfrac{1}{\sqrt{\,0.08 \times 2 \times 10^{-6}\,}} = \dfrac{1}{\sqrt{1.6 \times 10^{-7}}}.$$

Now, $$1.6 \times 10^{-7} = 16 \times 10^{-8} = (4^2) \times 10^{-8},$$ so

$$\sqrt{1.6 \times 10^{-7}} = 4.0 \times 10^{-4}\; \text{(approximately)}.$$

Hence,

$$\omega_0 = \dfrac{1}{4.0 \times 10^{-4}} = 2.5 \times 10^{3}\,\text{rad s}^{-1}.$$

Now we return to the quality-factor formula:

$$Q = \dfrac{\omega_0 L}{R} = \dfrac{\left(2.5 \times 10^{3}\right)\left(0.08\right)}{100}.$$

Simplifying numerator first,

$$\left(2.5 \times 10^{3}\right)\left(0.08\right) = 2.5 \times 0.08 \times 10^{3} = 0.20 \times 10^{3} = 200.$$

Now dividing by the resistance:

$$Q = \dfrac{200}{100} = 2.$$

So, the quality factor of the given series $$RLC$$ circuit is $$2$$.

Hence, the correct answer is Option A.

We have an ideal series $$LC$$ circuit fed by the sinusoidal voltage

$$v(t)=10\sin(314t)\;{\rm volt}$$

Comparing with the standard form $$V_0\sin(\omega t)$$, we identify the angular frequency

$$\omega = 314\;{\rm rad\,s^{-1}}\; , \qquad V_0 = 10\;{\rm V}.$$

The given component values are

$$L = 40\;{\rm mH}=40\times10^{-3}\;{\rm H}=0.04\;{\rm H},$$

$$C = 100\;\mu{\rm F}=100\times10^{-6}\;{\rm F}=1.0\times10^{-4}\;{\rm F}.$$

For a series circuit containing only $$L$$ and $$C$$, the impedance is purely reactive. We must therefore find the inductive reactance $$X_L$$ and the capacitive reactance $$X_C$$ and then combine them.

Formulae stated first:

Inductive reactance: $$X_L = \omega L.$$

Capacitive reactance: $$X_C = \dfrac{1}{\omega C}.$$

Substituting the numerical values

$$X_L = 314 \times 0.04 = 12.56\;\Omega,$$

$$X_C = \dfrac{1}{314 \times 1.0\times10^{-4}} = \dfrac{1}{3.14\times10^{-2}} = 31.85\;\Omega.$$

Because the inductor and capacitor are in series, the combined reactance is

$$X = X_L - X_C.$$

So,

$$X = 12.56 - 31.85 = -19.29\;\Omega.$$

The negative sign shows that the net reactance is capacitive. Its magnitude (which is the impedance, since there is no resistance) is

$$Z = |X| = 19.29\;\Omega.$$

Current amplitude

Ohm’s law for AC gives $$I_0 = \dfrac{V_0}{Z}.$$ Substituting,

$$I_0 = \dfrac{10}{19.29} \approx 0.52\;{\rm A}.$$

Phase of the current

For a purely capacitive reactance, current leads voltage by $$\dfrac{\pi}{2}$$ (90°). Mathematically, if the voltage is $$\sin(\omega t)$$, the current will be $$\sin(\omega t + \dfrac{\pi}{2}) = \cos(\omega t).$$

Hence the complete time‐domain expression for the current is

$$i(t) = 0.52\cos(314t)\;{\rm A}.$$

Matching this with the options provided, we see that it corresponds to Option A.

Hence, the correct answer is Option A.

We are told that the coil offers an opposition of $$100\ \Omega$$ to the alternating current and that the current lags behind the applied voltage by $$45^{\circ}$$. In AC circuit language the total opposition is called the impedance and is denoted by $$Z$$, while the opposition due purely to the inductance is called the inductive reactance and is denoted by $$X_L$$. The phase angle $$\phi$$ between voltage and current satisfies the trigonometric relations

$$\tan\phi = \dfrac{X_L}{R},\qquad\; X_L = Z\sin\phi,\qquad\; R = Z\cos\phi.$$

Here the given quantity $$100\ \Omega$$ is the magnitude of the impedance, so we write

$$Z = 100\ \Omega.$$

The phase angle is given as $$\phi = 45^{\circ}$$, and we recall that

$$\sin 45^{\circ} = \dfrac{1}{\sqrt2}.$$

Using the relation $$X_L = Z\sin\phi$$ we substitute the known values:

$$X_L = 100 \times \dfrac{1}{\sqrt2} = \dfrac{100}{\sqrt2}\ \Omega = 70.7\ \Omega$$

(the numerical value $$100/\sqrt2$$ is $$70.7$$ to three significant figures).

Next we connect the inductive reactance to the self-inductance by the standard formula

$$X_L = \omega L,$$

where $$\omega$$ is the angular frequency of the applied signal. Angular frequency is related to ordinary frequency $$f$$ by

$$\omega = 2\pi f.$$

The problem states that $$f = 1000\ \text{Hz}$$, so

$$\omega = 2\pi \times 1000 = 2000\pi\ \text{rad s}^{-1}.$$

Now we solve for the self-inductance $$L$$:

$$L = \dfrac{X_L}{\omega} = \dfrac{70.7}{2000\pi}\ \text{H}.$$

Carrying out the division,

$$L = \dfrac{70.7}{6283}\ \text{H} \approx 1.125 \times 10^{-2}\ \text{H}.$$

Writing this with two significant figures gives

$$L \approx 1.1 \times 10^{-2}\ \text{H}.$$

Hence, the correct answer is Option A.

We have an a.c. source of rms voltage $$V = 250\; \text{V}$$ and frequency $$f = 50\; \text{Hz}$$ feeding a series combination of a resistor $$R$$ and an inductor $$L$$.

The average power consumed in any a.c. circuit is given by the formula

$$P = V I \cos\phi = I^{2} R = \dfrac{V^{2} R}{R^{2} + X_{L}^{2}},$$

where $$\phi$$ is the phase angle and $$X_{L} = \omega L = 2\pi f L$$ is the inductive reactance. For the data given we know directly:

$$P = 400\; \text{W}, \qquad V = 250\; \text{V}, \qquad \cos\phi = 0.8.$$

First we calculate the current. From the definition of power factor,

$$\cos\phi = \dfrac{R}{Z},$$

where $$Z$$ is the magnitude of the series impedance. Therefore

$$R = Z\cos\phi = 0.8\,Z.$$

The power in terms of impedance is obtained by substituting $$I = V/Z$$ in the expression $$P = I^{2}R$$:

$$P = \left(\dfrac{V}{Z}\right)^{2} R = \dfrac{V^{2}}{Z^{2}}\,(0.8\,Z) = 0.8\,\dfrac{V^{2}}{Z}.$$

Re-arranging gives

$$Z = 0.8\,\dfrac{V^{2}}{P}.$$

Substituting the numerical values,

$$Z = 0.8\,\dfrac{(250)^{2}}{400} = 0.8\,\dfrac{62500}{400} = 0.8 \times 156.25 = 125\;\Omega.$$

Using $$R = 0.8\,Z$$, we find

$$R = 0.8 \times 125 = 100\;\Omega.$$

The inductive reactance is obtained from

$$Z^{2} = R^{2} + X_{L}^{2} \;\;\Longrightarrow\;\; X_{L} = \sqrt{Z^{2} - R^{2}} = \sqrt{125^{2} - 100^{2}} = \sqrt{15625 - 10000} = \sqrt{5625} = 75\;\Omega.$$

Now a capacitor is added in series to make the overall power factor unity. Unity power factor means the net reactance is zero, so the capacitive reactance must exactly cancel the inductive reactance:

$$X_{C} = X_{L} = 75\;\Omega.$$

For a capacitor, reactance is given by the formula

$$X_{C} = \dfrac{1}{\omega C}, \qquad \omega = 2\pi f = 2\pi \times 50 = 100\pi\;\text{rad/s}.$$

Setting $$X_{C} = 75\;\Omega$$, we get

$$75 = \dfrac{1}{\omega C} \;\;\Longrightarrow\;\; C = \dfrac{1}{\omega \times 75} = \dfrac{1}{100\pi \times 75} = \dfrac{1}{7500\pi}\;\text{F}.$$

To express the capacitance in micro-farads, we multiply by $$10^{6}$$:

$$C = \dfrac{10^{6}}{7500\pi}\;\mu\text{F}.$$

The problem statement writes the capacitance in the form

$$C = \left(\dfrac{n}{3\pi}\right)\,\mu\text{F}.$$

To match this with our value, equate the two numerators:

$$\dfrac{10^{6}}{7500} = \dfrac{n}{3} \;\;\Longrightarrow\;\; n = 3 \times \dfrac{10^{6}}{7500} = \dfrac{3000000}{7500} = 400.$$

So, the answer is $$400$$.

We have a transformer with $$N_p = 300$$ turns in the primary winding and $$N_s = 150$$ turns in the secondary winding. For an ideal transformer, the voltage ratio is given by the basic transformer turns-ratio formula

$$\frac{V_s}{V_p} = \frac{N_s}{N_p}.$$

Here $$V_p$$ is the primary (input) voltage and $$V_s$$ is the secondary (output) voltage. Substituting the given numbers, we get

$$\frac{V_s}{V_p} = \frac{150}{300} = \frac12.$$

So the secondary voltage is precisely one-half of the primary voltage:

$$V_s = \frac12\,V_p \quad\Longrightarrow\quad V_p = 2\,V_s.$$

Now, the question supplies the output power of the transformer as $$P_{\text{out}} = 2.2\ \text{kW} = 2200\ \text{W}$$ and also tells us that the current in the secondary coil is $$I_s = 10\ \text{A}.$$ Using the definition of electrical power,

$$P_{\text{out}} = V_s I_s,$$

we can determine the secondary voltage:

$$V_s = \frac{P_{\text{out}}}{I_s} = \frac{2200\ \text{W}}{10\ \text{A}} = 220\ \text{V}.$$

With $$V_s = 220\ \text{V}$$ already in hand and the earlier relation $$V_p = 2\,V_s,$$ we immediately obtain the primary voltage:

$$V_p = 2 \times 220\ \text{V} = 440\ \text{V}.$$

For an ideal transformer (neglecting losses), the input power equals the output power:

$$P_{\text{in}} = P_{\text{out}}.$$

Therefore, $$P_{\text{in}} = V_p I_p,$$ where $$I_p$$ is the primary current. Substituting the known values,

$$V_p I_p = P_{\text{out}} \quad\Longrightarrow\quad 440\ \text{V}\, \times I_p = 2200\ \text{W}.$$

Solving for $$I_p$$ gives

$$I_p = \frac{2200\ \text{W}}{440\ \text{V}} = 5\ \text{A}.$$

So the transformer must be supplied with $$440\ \text{V}$$ and will draw $$5\ \text{A}$$ from the primary source.

Hence, the correct answer is Option C.

We first note that the source voltage is written as $$V = 200\sqrt{2}\,\sin(100t)$$. Comparing this with the standard form $$V = V_{0}\sin(\omega t)$$ we can at once read the angular frequency as $$\omega = 100\;{\rm rad\,s^{-1}}.$$

There are two independent series branches connected across this same source.

Branch 1 (R-L branch)

The impedance of a resistor $$R_1$$ in series with an inductor $$L$$ is, by definition,

$$Z_1 = R_1 + j\omega L,$$

where the symbol $$j$$ stands for $$\sqrt{-1}$$ (conventional electrical engineering notation).

Substituting the numerical data

$$R_1 = 10\;\Omega,\qquad L = \frac{\sqrt{3}}{10}\;{\rm H},\qquad \omega = 100\;{\rm rad\,s^{-1}},$$

we obtain

$$\omega L \;=\; 100 \times \frac{\sqrt{3}}{10}\;=\;10\sqrt{3}\;\Omega.$$

Hence

$$Z_1 = 10 + j\,10\sqrt{3}\;\Omega.$$

The phase angle by which the branch current lags its voltage for an R-L circuit is given by the well-known relation

$$\phi_1 = \tan^{-1}\!\Bigl(\frac{\omega L}{R_1}\Bigr).$$

Substituting,

$$\phi_1 \;=\; \tan^{-1}\!\Bigl(\frac{10\sqrt{3}}{10}\Bigr) \;=\; \tan^{-1}(\sqrt{3}) \;=\; 60^{\circ}.$$

Because the term $$+j\omega L$$ is inductive, this angle represents lag. Therefore

$$I_1 \;\text{lags}\; V \;\text{by}\; 60^{\circ},$$

or equivalently, the phase of $$I_1$$ is

$$\angle I_1 = -60^{\circ}.$$

Branch 2 (R-C branch)

The impedance of a resistor $$R_2$$ in series with a capacitor $$C$$ is

$$Z_2 = R_2 - \dfrac{j}{\omega C}.$$

Using the given values

$$R_2 = 20\;\Omega,\qquad C = \frac{\sqrt{3}}{2}\,\mu{\rm F} = \frac{\sqrt{3}}{2}\times 10^{-6}\;{\rm F},$$

we first compute the capacitive reactance $$X_C$$:

$$\omega C = 100 \times \frac{\sqrt{3}}{2}\times 10^{-6} \;=\; 50\sqrt{3}\times 10^{-6}.$$

Taking the reciprocal,

$$X_C \;=\; \frac{1}{\omega C} \;=\; \frac{1}{50\sqrt{3}\times 10^{-6}} \;=\; \frac{10^{6}}{50\sqrt{3}} \;=\; \frac{20\,000}{\sqrt{3}}\;\Omega \;=\; \frac{20\,000\sqrt{3}}{3}\;\Omega.$$

For later numerical clarity we also write

$$X_C \;\approx\; 11\,546.7\;\Omega.$$

Thus

$$Z_2 = 20 \;-\; j\,X_C = 20 - j\,11\,546.7\;\Omega.$$

The phase angle of the current with respect to the voltage for an R-C series circuit is given by

$$\phi_2 = \tan^{-1}\!\Bigl(\frac{X_C}{R_2}\Bigr),$$

and because the reactance term carries a minus sign, this angle represents a lead (capacitive currents lead the voltage).

Substituting,

$$\phi_2 \;=\; \tan^{-1}\!\Bigl(\frac{11\,546.7}{20}\Bigr) \;=\; \tan^{-1}(577.335) \;\approx\; 89.9^{\circ}.$$

To a very good approximation we may treat this as exactly

$$\angle I_2 = +90^{\circ}\;,$$

i.e. the current $$I_2$$ leads the source voltage by $$90^{\circ}.$$

Phase difference between the two branch currents

The current $$I_1$$ has phase $$-60^{\circ}$$, while $$I_2$$ has phase $$+90^{\circ}$$. The absolute phase difference $$\Delta\phi$$ is therefore

$$\Delta\phi = (+90^{\circ}) - (-60^{\circ}) = 90^{\circ} + 60^{\circ} = 150^{\circ}.$$

This is the required result.

Hence, the correct answer is Option D.

We begin by recalling that in an AC circuit the average (real) power dissipated as heat in the resistor is given by the well-known relation

$$P = I_{\text{rms}}^{2}\,R,$$

where $$I_{\text{rms}}$$ is the root-mean-square current in the circuit and $$R$$ is the resistance. The total electrical energy converted into heat in a time interval $$t$$ is therefore

$$E = P\,t = I_{\text{rms}}^{2}\,R\,t.$$

So our main task is to find $$I_{\text{rms}}$$ for the given series $$RLC$$ circuit.

For a series combination we may write the impedance $$Z$$ as

$$Z = \sqrt{R^{2} + \left(X_{L}-X_{C}\right)^{2}},$$

where

$$X_{L} = 2\pi f L \qquad\text{(inductive reactance)},$$

$$X_{C} = \frac{1}{2\pi f C} \qquad\text{(capacitive reactance)}.$$

Substituting the numerical data:

$$L = 20\ \text{mH} = 20 \times 10^{-3}\ \text{H},$$ $$C = 120\ \mu\text{F} = 120 \times 10^{-6}\ \text{F},$$ $$f = 50\ \text{Hz},$$ $$R = 60\ \Omega,\quad V_{\text{rms}} = 24\ \text{V}.$$

First we calculate $$X_{L}$$:

$$X_{L} = 2\pi f L = 2\pi \times 50 \times 20\times 10^{-3}$$

$$\;\;= 100\pi \times 0.02$$

$$\;\;= 6.283\ \Omega \;(\text{approximately}).$$

Next we calculate $$X_{C}$$:

$$X_{C} = \frac{1}{2\pi f C} = \frac{1}{2\pi \times 50 \times 120 \times 10^{-6}}$$

$$\;\;= \frac{1}{100\pi \times 120 \times 10^{-6}}$$

$$\;\;= \frac{1}{0.037699}$$

$$\;\;= 26.53\ \Omega \;(\text{approximately}).$$

The net reactance is therefore

$$X = X_{L} - X_{C} = 6.283 - 26.53 = -20.25\ \Omega.$$

(The negative sign merely tells us that the circuit is overall capacitive; we shall use the magnitude in the impedance expression.)

Thus the impedance magnitude is

$$Z = \sqrt{R^{2} + X^{2}} = \sqrt{60^{2} + (-20.25)^{2}}$$

$$\;\;= \sqrt{3600 + 410.06}$$

$$\;\;= \sqrt{4010.06}$$

$$\;\;= 63.36\ \Omega \;(\text{approximately}).$$

Now, Ohm’s law for AC says

$$I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{24}{63.36} = 0.3788\ \text{A}\;(\text{approximately}).$$

We can now find the average power dissipated in the resistor:

$$P = I_{\text{rms}}^{2}\,R = (0.3788)^{2} \times 60$$

$$\;\;= 0.1435 \times 60$$

$$\;\;= 8.61\ \text{W}\;(\text{approximately}).$$

The problem asks for the energy dissipated in $$t = 60\ \text{s}$$, so

$$E = P\,t = 8.61 \times 60$$

$$\;\;= 516.6\ \text{J}\;(\text{approximately}).$$

Expressed in scientific notation this is

$$E \approx 5.17 \times 10^{2}\ \text{J}.$$

Hence, the correct answer is Option A.

We are given the alternating (sinusoidal) voltage $$V(t)=220\sin(100\pi t)\ \text{volt}$$ which is applied across a purely resistive load of resistance $$R=50\ \Omega.$$

For a purely resistive circuit, current and voltage remain in the same phase, and we can use Ohm’s law in its instantaneous form

$$i(t)=\frac{v(t)}{R}.$$

Substituting the given voltage expression, we obtain

$$i(t)=\frac{220\sin(100\pi t)}{50}.$$

Simplifying the constant factor term-by-term,

$$i(t)=\left(\frac{220}{50}\right)\sin(100\pi t)=4.4\sin(100\pi t)\ \text{ampere}.$$

Here, the maximum (peak) value of the sine function is 1, so the peak current is

$$I_{\text{peak}}=4.4\ \text{A}.$$

The problem asks for the time required for the current to rise from one-half of this peak value to the full peak value, that is from

$$I_1=\frac{I_{\text{peak}}}{2}=\frac{4.4}{2}=2.2\ \text{A}$$

up to

$$I_2=I_{\text{peak}}=4.4\ \text{A}.$$

First, we locate the instant when the current equals the half-peak value. Setting $$i(t)=2.2\ \text{A}$$ in our current expression, we have

$$4.4\sin(100\pi t_1)=2.2.$$

Dividing both sides by 4.4 keeps every step explicit:

$$\sin(100\pi t_1)=\frac{2.2}{4.4}=\frac12.$$

A sine value of $$\frac12$$ corresponds to an argument of $$\frac{\pi}{6},\; 5\frac{\pi}{6}, \ldots$$ but for the first rising half-cycle we select the principal value

$$100\pi t_1=\frac{\pi}{6}.$$

Now solve for $$t_1$$ by dividing both sides by $$100\pi$$:

$$t_1=\frac{\pi/6}{100\pi}=\frac1{600}\ \text{s}.$$

Next, we find the instant of the peak current. A peak occurs when the sine function reaches 1, hence

$$\sin(100\pi t_2)=1.$$

The first positive solution for this equation is

$$100\pi t_2=\frac{\pi}{2}.$$

Again dividing by $$100\pi$$ gives

$$t_2=\frac{\pi/2}{100\pi}=\frac1{200}\ \text{s}.$$

The desired time interval is the difference $$t_2-t_1$$:

$$t_2-t_1=\frac1{200}-\frac1{600}.$$

Bringing the fractions to a common denominator of 600, we write

$$t_2-t_1=\frac{3}{600}-\frac{1}{600}=\frac{2}{600}=\frac1{300}\ \text{s}.$$

Converting seconds to milliseconds (recall that $$1\ \text{s}=1000\ \text{ms}$$), we have

$$t_2-t_1=\frac1{300}\times1000\ \text{ms}=3.33\ \text{ms}\ (\text{to two decimal places}).$$

Hence, the correct answer is Option D.

We have the primary (input) voltage $$V_p = 2300\ \text{V}$$ and the current in the primary coil $$I_p = 5\ \text{A}$$.

The electrical power delivered to the primary side of a transformer is given by the product of voltage and current, that is the formula

$$P_{\text{in}} = V_p \, I_p.$$

Substituting the given values,

$$P_{\text{in}} = 2300 \times 5 = 11500\ \text{W}.$$

The transformer is not ideal; its efficiency is stated to be $$\eta = 90\% = 0.90.$$ By definition of efficiency,

$$\eta = \frac{P_{\text{out}}}{P_{\text{in}}}.$$

Re-arranging to obtain the output power,

$$P_{\text{out}} = \eta \, P_{\text{in}}.$$

Substituting the known quantities,

$$P_{\text{out}} = 0.90 \times 11500 = 10350\ \text{W}.$$

The secondary (output) voltage is given as $$V_s = 230\ \text{V}.$$ The power leaving the secondary is also the product of this voltage and the secondary current $$I_s$$, so

$$P_{\text{out}} = V_s \, I_s.$$

Solving for the unknown current $$I_s$$, we write

$$I_s = \frac{P_{\text{out}}}{V_s}.$$

Substituting the calculated output power and the given secondary voltage,

$$I_s = \frac{10350}{230} = 45\ \text{A}.$$

Hence, the correct answer is Option A.

We begin with the standard series RLC circuit, in which the resistor $$R$$, inductor $$L$$ and capacitor $$C$$ are connected in series and are driven by a sinusoidal voltage source of angular frequency $$\omega$$ and peak (amplitude) value $$v_m$$.

For any series RLC circuit the total impedance is written as

$$Z = R + j\left(\omega L - \frac{1}{\omega C}\right),$$

where the symbol $$j$$ represents the imaginary unit $$\sqrt{-1}$$. The magnitude of this impedance is therefore

$$|Z| = \sqrt{R^{2} + \left(\omega L - \frac{1}{\omega C}\right)^{2}}.$$

Resonance in a series circuit occurs at that frequency $$\omega_0$$ at which the net reactance vanishes, i.e.

$$\omega_0 L - \frac{1}{\omega_0 C} = 0.$$

Solving this equation for the resonant angular frequency, we get

$$\omega_0 L = \frac{1}{\omega_0 C} \quad \Longrightarrow \quad \omega_0^2 = \frac{1}{LC} \quad \Longrightarrow \quad \omega_0 = \frac{1}{\sqrt{LC}}.$$

At resonance the impedance magnitude reduces to its purely resistive part, namely $$|Z| = R$$, and hence the current amplitude becomes maximum, signifying the phenomenon of current resonance.

Now we introduce the quality factor, denoted by $$Q$$. For a series resonant circuit, the definition used in most elementary courses is

$$Q = \frac{\text{Energy stored per cycle (maximum)}}{\text{Energy dissipated per cycle}}.$$

An alternative and very convenient expression—derived from that definition and widely quoted—is

$$Q = \frac{\omega_0 L}{R}.$$

We will show why this result holds. In an inductor the peak magnetic energy stored equals

$$U_L^{\text{(max)}} = \frac{1}{2} L I_m^{2},$$

while in a capacitor the peak electric energy stored is

$$U_C^{\text{(max)}} = \frac{1}{2} C V_m^{2}.$$

Because the inductor and capacitor exchange energy sinusoidally with each other, the total maximum energy stored in the reactive elements at resonance can be represented (for one half-cycle) by either of the above forms; customarily we use the inductive one, yielding

$$U_{\text{stored}}^{\text{(max)}} = \frac{1}{2} L I_m^{2}.$$

The average power dissipated in the resistor over a complete cycle is

$$P_{\text{avg}} = I_{\text{rms}}^{2} R,$$

and for a sinusoidal current the relation between peak and rms values is $$I_m = \sqrt{2}\,I_{\text{rms}}$$, so that

$$P_{\text{avg}} = \frac{I_m^{2}}{2}\,R.$$

The energy dissipated per cycle is therefore

$$W_{\text{diss}} = P_{\text{avg}} \times T,$$

where $$T = \tfrac{2\pi}{\omega_0}$$ is the period at resonance. Substituting, we obtain

$$W_{\text{diss}} = \frac{I_m^{2}}{2}\,R \times \frac{2\pi}{\omega_0} = \frac{\pi I_m^{2} R}{\omega_0}.$$

Using the definition $$Q = \dfrac{2\pi \times (\text{Maximum energy stored})}{\text{Energy dissipated per cycle}},$$ we write

$$Q = \frac{2\pi \times \frac{1}{2} L I_m^{2}}{\dfrac{\pi I_m^{2} R}{\omega_0}} = \frac{2\pi \times \frac{1}{2} L I_m^{2} \times \omega_0}{\pi I_m^{2} R} = \frac{\omega_0 L}{R}.$$

The current and its amplitude $$I_m$$ cancel out, leaving a remarkably simple relationship that ties together the three element parameters. This final algebraic form,

$$Q = \frac{\omega_0 L}{R},$$

is exactly the same as Option B in the list given.

None of the other options reproduces the factor $$\omega_0 L$$ in the numerator with $$R$$ in the denominator, so they are incorrect for a series RLC circuit at resonance.

Hence, the correct answer is Option B.

We have an alternating e.m.f. (voltage) and current described by

$$E(t)=100\sin 30t,\qquad I(t)=20\sin\!\left(30t-\frac{\pi}{4}\right).$$

The amplitude (maximum value) of the voltage is therefore

$$E_{0}=100,$$

and the amplitude of the current is

$$I_{0}=20.$$

First we convert these amplitudes to their r.m.s. (root-mean-square) values. The standard relation is

$$\text{r.m.s. value}=\frac{\text{amplitude}}{\sqrt{2}}.$$

So for the voltage

$$V_{\text{rms}}=\frac{E_{0}}{\sqrt{2}}=\frac{100}{\sqrt{2}},$$

and for the current

$$I_{\text{rms}}=\frac{I_{0}}{\sqrt{2}}=\frac{20}{\sqrt{2}}.$$

Next we identify the phase difference between the voltage and the current. From the given expressions, the voltage is $$\sin 30t$$ while the current is $$\sin(30t-\pi/4).$$ The current therefore lags the voltage by

$$\phi=\frac{\pi}{4}\;(\text{that is, }45^{\circ}).$$

The average (or real) power consumed by an a.c. circuit is found with the formula

$$P_{\text{avg}}=V_{\text{rms}}\,I_{\text{rms}}\cos\phi.$$

Substituting the numerical values we have just obtained:

$$P_{\text{avg}}=\left(\frac{100}{\sqrt{2}}\right)\!\left(\frac{20}{\sqrt{2}}\right)\cos\!\left(\frac{\pi}{4}\right).$$

Now evaluate step by step:

First multiply the numerators:

$$100\times20=2000.$$

Multiply the denominators:

$$\sqrt{2}\times\sqrt{2}=2.$$

Thus the product of the two r.m.s. values is

$$\frac{2000}{2}=1000.$$

Next, recall the trigonometric value

$$\cos\!\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}.$$

So the average power becomes

$$P_{\text{avg}}=1000\times\frac{1}{\sqrt{2}}=\frac{1000}{\sqrt{2}}\ \text{watt}.$$

This completes the calculation of the real (average) power.

Now we determine the watt-less (reactive) component of the current. The reactive current is the component that is out of phase by 90° with the voltage. Its magnitude is

$$I_{\text{watt-less}}=I_{\text{rms}}\sin\phi.$$

Again substituting the numbers:

$$I_{\text{watt-less}}=\left(\frac{20}{\sqrt{2}}\right)\sin\!\left(\frac{\pi}{4}\right).$$

We use the trigonometric value

$$\sin\!\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}.$$

Therefore

$$I_{\text{watt-less}}=\frac{20}{\sqrt{2}}\times\frac{1}{\sqrt{2}}=\frac{20}{2}=10\ \text{ampere}.$$

Thus, in one complete cycle of the given a.c. the circuit consumes an average power of $$\dfrac{1000}{\sqrt{2}}\,$$ watt and carries a watt-less current of $$10$$ ampere.

Hence, the correct answer is Option C.

First note what the arc-lamp itself needs. On direct current it draws $$I = 10 \text{ A}$$ when the potential difference across it is $$V = 80 \text{ V}$$. For a purely resistive device the resistance is defined by Ohm’s law

$$R = \dfrac{V}{I}.$$

Substituting the given values, we have

$$R = \dfrac{80}{10} = 8 \;\Omega.$$

Now the lamp must be operated from a sinusoidal mains supply of $$220 \text{ V(rms)},\; 50 \text{ Hz}$$, yet we still want the current through the lamp to be only $$10 \text{ A(rms)}$$. Because the mains voltage is higher than 80 V, an additional impedance must be placed in series to limit the current. The problem asks for an inductor, so that extra impedance will be the inductive reactance $$X_L$$ of the coil.

For a series circuit containing a resistance $$R$$ and an inductive reactance $$X_L$$, the magnitude of the total impedance $$Z$$ is given by

$$Z = \sqrt{R^{2} + X_L^{2}}.$$

But on alternating current we also have, from Ohm’s law extended to impedances,

$$Z = \dfrac{V_{\text{rms}}}{I_{\text{rms}}}.$$

Substituting the required rms values,

$$Z = \dfrac{220}{10} = 22 \;\Omega.$$

Equating the two expressions for $$Z$$ gives

$$\sqrt{R^{2} + X_L^{2}} = 22.$$

We already found $$R = 8 \;\Omega$$, so

$$\sqrt{8^{2} + X_L^{2}} = 22,$$

$$\sqrt{64 + X_L^{2}} = 22.$$

Now square both sides to remove the square root:

$$64 + X_L^{2} = 22^{2} = 484.$$

Solving for $$X_L^{2}$$, we obtain

$$X_L^{2} = 484 - 64 = 420,$$

so

$$X_L = \sqrt{420} \approx 20.49 \;\Omega.$$

The inductive reactance is related to the inductance $$L$$ by the formula

$$X_L = \omega L,$$

where $$\omega$$ is the angular frequency $$\omega = 2\pi f$$. For a supply frequency of $$f = 50 \text{ Hz}$$, we have

$$\omega = 2\pi \times 50 = 100\pi \;\text{rad s}^{-1} \approx 314.16 \;\text{rad s}^{-1}.$$

Substituting into the reactance formula gives

$$L = \dfrac{X_L}{\omega} = \dfrac{20.49}{314.16} \;\text{H}.$$

Carrying out the division,

$$L \approx 0.0652 \;\text{H}.$$

This value rounds to $$0.065 \text{ H}$$, matching option B.

Hence, the correct answer is Option B.

In an $$LR$$ circuit, the ratio $$\tau = \frac{L}{R}$$ is known as the inductive time constant.

The total current in such a circuit consists of two parts: the transient response (which decays exponentially) and the steady-state response.

The condition $$t_0 \gg \tau$$ means that several time constants have passed, causing the transient component ($$e^{-t/\tau}$$) to die out completely.

Once the transients have vanished, the circuit reaches a steady state where the current oscillates at the same frequency as the source voltage $$V(t) = V_0 \sin(\omega t)$$.

The impedance ($$Z$$) of the series $$LR$$ circuit is $$Z = \sqrt{R^2 + X_L^2} = \sqrt{R^2 + (\omega L)^2}$$

The steady-state current $$I(t)$$ is given by $$I(t) = I_0 \sin(\omega t - \phi)$$

Where Amplitude ($$I_0$$): $$I_0 = \frac{V_0}{Z}$$ (a constant value), and Phase Lag ($$\phi$$): $$\phi = \tan^{-1}\left(\frac{\omega L}{R}\right)$$.

After a very large time ($$t_0 \gg L/R$$), the current becomes a stable sinusoidal oscillation. Therefore, Option D is the correct graphical representation.

For a series $$R$$-$$L$$-$$C$$ circuit driven by an a.c. source of angular frequency $$\omega$$, the impedance is $$Z=\sqrt{R^{2}+\left(X_L-X_C\right)^{2}}$$, where the reactances are
  • Inductive: $$X_L=\omega L$$
  • Capacitive: $$X_C=\dfrac{1}{\omega C}$$.

The power factor is defined as the cosine of the phase angle $$\phi$$ between voltage and current:
$$\cos\phi=\dfrac{R}{Z}\qquad -(1)$$

The tangent of the same angle is therefore
$$\tan\phi=\dfrac{X_L-X_C}{R}\qquad -(2)$$

The problem states that the power factor is $$0.5$$, so
$$\cos\phi=0.5\;\;\Longrightarrow\;\;\phi=60^{\circ}$$ (since $$\cos60^{\circ}=0.5$$).

Hence
$$\tan\phi=\tan60^{\circ}=\sqrt{3}$$.

Substituting $$\tan\phi$$ and the given resistance $$R=100\,\Omega$$ into equation $$(2)$$ gives
$$X_L-X_C = R\,\tan\phi = 100 \times \sqrt{3} = 100\sqrt{3}\;\Omega\qquad -(3)$$

The statement of the question further writes this difference as
$$X_L-X_C = \sqrt{3}\,\alpha\;\Omega\qquad -(4)$$

Comparing $$(3)$$ and $$(4)$$:
$$\sqrt{3}\,\alpha = 100\sqrt{3}\;\Longrightarrow\;\alpha = 100$$.

Therefore, the required value of $$\alpha$$ is 100.

The given components are connected in series, so the circuit is a series $$R\text{-}L\text{-}C$$ combination.

Resistance: $$R = 50 \,\Omega$$
Inductive reactance: $$X_L = 100 \,\Omega$$
Capacitive reactance: $$X_C = 50 \,\Omega$$
Applied rms voltage: $$V_{\text{rms}} = 10 \text{ V}$$

Step 1 Find the net reactance.
For a series circuit, the net reactance is the algebraic difference of the individual reactances:
$$X = X_L - X_C = 100 - 50 = 50 \,\Omega$$

Step 2 Calculate the impedance.
For a series $$R\text{-}L\text{-}C$$ circuit, the impedance is
$$Z = \sqrt{R^{2} + X^{2}}$$
$$\Rightarrow Z = \sqrt{50^{2} + 50^{2}}$$
$$\Rightarrow Z = \sqrt{2500 + 2500}$$
$$\Rightarrow Z = \sqrt{5000}$$
$$\Rightarrow Z = 70.71 \,\Omega$$

Step 3 Find the rms current.
Ohm’s law for ac gives $$I_{\text{rms}} = \dfrac{V_{\text{rms}}}{Z}$$.
$$I_{\text{rms}} = \dfrac{10}{70.71} = 0.1414 \text{ A}$$

Step 4 Determine the power factor.
Power factor in a series circuit is $$\cos\phi = \dfrac{R}{Z}$$.
$$\cos\phi = \dfrac{50}{70.71} = 0.707$$

Step 5 Calculate the average (true) power.
Average power dissipated is
$$P = V_{\text{rms}} I_{\text{rms}} \cos\phi$$
Substitute the values:
$$P = 10 \times 0.1414 \times 0.707$$
$$P = 1.0 \text{ W}$$

Hence, the average power dissipated by the circuit is 1 W.

For a series LCR circuit, the current in the circuit is maximum at the resonant frequency. This condition occurs when the inductive reactance is exactly equal to the capacitive reactance:

$$ X_L = X_C $$

$$ \omega L = \frac{1}{\omega C} $$

From this, the resonant angular frequency $$\omega$$ is given by the formula:

$$ \omega = \frac{1}{\sqrt{LC}} $$

$$ \omega = \frac{1}{\sqrt{10^{-1} \times 25 \times 10^{-9}}} $$

$$ \omega = \frac{1}{\sqrt{25 \times 10^{-10}}} $$

$$ \omega = \frac{1}{5 \times 10^{-5}} $$

$$ \omega = 2 \times 10^4 \text{ rad s}^{-1} $$

Given
Self-inductance $$L = 1 \text{ H}$$, Resistance $$R = 100\pi \, \Omega$$, Frequency $$f = 50 \text{ Hz}$$, RMS supply voltage $$V_{rms}=100\pi \text{ V}$$.

First find the angular frequency.
Formula: $$\omega = 2\pi f$$ $$-(1)$$
Using $$f = 50$$ Hz,
$$\omega = 2\pi \times 50 = 100\pi \text{ rad s}^{-1}$$.

Inductive reactance $$X_L$$ is
$$X_L = \omega L$$ $$-(2)$$
Substituting $$\omega = 100\pi$$ and $$L = 1$$ H,
$$X_L = 100\pi \, \Omega$$.

The impedance of an $$R-L$$ series circuit is
$$Z = \sqrt{R^2 + X_L^2}$$ $$-(3)$$.
Here $$R = 100\pi$$ and $$X_L = 100\pi$$, so
$$Z = \sqrt{(100\pi)^2 + (100\pi)^2} = 100\pi\sqrt{2}\, \Omega$$.

RMS current:
$$I_{rms} = \frac{V_{rms}}{Z}$$ $$-(4)$$.
Putting $$V_{rms}=100\pi$$ V and $$Z = 100\pi\sqrt{2}$$ Ω,
$$I_{rms} = \frac{100\pi}{100\pi\sqrt{2}} = \frac{1}{\sqrt{2}} \text{ A}$$.

Maximum (peak) current relates to RMS current by
$$I_0 = \sqrt{2}\, I_{rms}$$ $$-(5)$$.
Hence
$$I_0 = \sqrt{2} \times \frac{1}{\sqrt{2}} = 1 \text{ A}$$.

Therefore, the maximum current flowing in the circuit is 1 A.

We need to find the maximum current in an LC oscillation circuit where a capacitor of $$100 \, \mu F$$ charged to $$12$$ V is connected to a $$6.4$$ mH inductor.

In an LC circuit, energy oscillates between the electric field of the capacitor and the magnetic field of the inductor. The total energy is conserved:

$$\frac{1}{2}CV^2 = \frac{1}{2}LI_{max}^2$$

The left side is the maximum energy stored in the capacitor (when current is zero), and the right side is the maximum energy stored in the inductor (when the capacitor is fully discharged).

From the energy conservation equation:

$$I_{max} = V\sqrt{\frac{C}{L}}$$

$$V = 12$$ V, $$C = 100 \, \mu F = 100 \times 10^{-6}$$ F, $$L = 6.4$$ mH $$= 6.4 \times 10^{-3}$$ H:

$$I_{max} = 12\sqrt{\frac{100 \times 10^{-6}}{6.4 \times 10^{-3}}}$$

$$= 12\sqrt{\frac{10^{-4}}{6.4 \times 10^{-3}}} = 12\sqrt{\frac{1}{64}}$$

$$= 12 \times \frac{1}{8} = 1.5 \text{ A}$$

The correct answer is Option (2): 1.5 A.

The series circuit consists of a pure resistor $$R = 90\,\Omega$$ and a coil whose resistance is negligible (so it behaves as a pure inductor with inductive reactance $$X_L$$). The applied rms supply is $$V = 120\,\text{V}$$ at frequency $$f = 60\,\text{Hz}$$.

The voltmeter shows the rms voltage across the resistor to be $$V_R = 36\,\text{V}$$. For a pure resistor, voltage and current are in phase, so the rms current in the entire series circuit is

$$I = \frac{V_R}{R} = \frac{36}{90} = 0.40\,\text{A}$$

In an $$R$$-$$L$$ series combination, the resistor voltage $$V_R$$ is in phase with the current while the inductor voltage $$V_L$$ leads the current by $$90^{\circ}$$. Therefore, the supply voltage is the vector (phasor) sum of $$V_R$$ and $$V_L$$, giving the rms relation

$$V = \sqrt{V_R^{2} + V_L^{2}} \quad -(1)$$

Rearranging $$(1)$$ for $$V_L$$:

$$V_L = \sqrt{V^{2} - V_R^{2}} = \sqrt{(120)^{2} - (36)^{2}}$$

$$V_L = \sqrt{14400 - 1296} = \sqrt{13104} \approx 1.145 \times 10^{2}\,\text{V}$$

Thus, $$V_L \approx 114\,\text{V}$$.

The inductive reactance is then obtained from Ohm’s law for the inductor:

$$X_L = \frac{V_L}{I} = \frac{114}{0.40} \approx 285\,\Omega$$

For a pure inductor, $$X_L = 2\pi f L$$. Therefore, the inductance $$L$$ is

$$L = \frac{X_L}{2\pi f} = \frac{285}{2\pi \times 60}$$

$$L = \frac{285}{120\pi} \approx \frac{285}{376.99} \approx 0.756\,\text{H}$$

Rounding to two significant figures, $$L \approx 0.76\,\text{H}$$.

Answer: Option B (0.76 H)

We are given a series LCR circuit with an AC signal of $$200$$ V and $$50$$ Hz. The inductor has inductance $$L = 10$$ mH, and the voltage across the inductor is $$V_L = 31.4$$ V. We need to find the current in the circuit.

Recall that the voltage across an inductor in an AC circuit is related to the current by $$V_L = I \times X_L$$, where $$X_L$$ is the inductive reactance.

The inductive reactance is given by the formula $$X_L = \omega L = 2\pi f L$$. Substituting the given values ($$f = 50$$ Hz, $$L = 10 \text{ mH} = 10 \times 10^{-3}$$ H) yields $$X_L = 2\pi \times 50 \times 10 \times 10^{-3}$$, so $$X_L = 2\pi \times 0.5 = \pi \approx 3.14 \, \Omega$$.

Using the relationship $$V_L = I \times X_L$$ and solving for $$I$$ gives $$I = \frac{V_L}{X_L} = \frac{31.4}{3.14} = 10 \text{ A}$$.

The correct answer is Option (3): 10 A.

For a series LR circuit, the power factor is given by:

$$\cos\phi = \frac{R}{Z} = \frac{R}{\sqrt{R^2 + \omega^2 L^2}}$$

Case 1: $$\omega_1 = 1000 \text{ rad/s}$$, power factor $$= \frac{1}{\sqrt{2}}$$

$$\frac{1}{\sqrt{2}} = \frac{R}{\sqrt{R^2 + (1000)^2 L^2}}$$

Squaring both sides:

$$\frac{1}{2} = \frac{R^2}{R^2 + 10^6 L^2}$$

$$R^2 + 10^6 L^2 = 2R^2$$

$$10^6 L^2 = R^2$$

$$\omega_1 L = R \implies L = \frac{R}{1000}$$

Case 2: $$\omega_2 = 2000 \text{ rad/s}$$

$$\omega_2 L = 2000 \times \frac{R}{1000} = 2R$$

The new power factor is:

$$\cos\phi' = \frac{R}{\sqrt{R^2 + (2R)^2}} = \frac{R}{\sqrt{R^2 + 4R^2}} = \frac{R}{\sqrt{5R^2}} = \frac{1}{\sqrt{5}}$$

The correct answer is $$\frac{1}{\sqrt{5}}$$.

$$V_0=20$$V, $$I_0=10$$A. Phase difference $$\phi=\pi/3$$.

$$P_{avg}=\frac{V_0I_0}{2}\cos\phi=\frac{20\times10}{2}\cos60°=100\times0.5=50$$ W.

The answer is Option (4): 50 W.

In an AC circuit, the instantaneous current is zero when the instantaneous voltage is maximum. This implies a 90-degree phase difference between voltage and current. We need to identify which circuit elements produce this behavior.

Recall: If $$V = V_0\sin(\omega t)$$ and $$I = I_0\sin(\omega t \pm 90°)$$, then when $$V$$ is at maximum ($$\sin(\omega t) = 1$$), $$I = I_0\sin(\omega t \pm 90°) = \pm I_0\cos(\omega t) = 0$$ (at the same instant).

A. Pure inductor: In a pure inductor, current lags voltage by 90°: $$I = I_0\sin(\omega t - 90°)$$. When voltage is maximum, current is zero. YES.

B. Pure capacitor: In a pure capacitor, current leads voltage by 90°: $$I = I_0\sin(\omega t + 90°)$$. When voltage is maximum, current is zero. YES.

C. Pure resistor: In a pure resistor, current and voltage are in phase: $$I = I_0\sin(\omega t)$$. When voltage is maximum, current is also maximum (not zero). NO.

D. Combination of inductor and capacitor (LC circuit): In a pure LC circuit (no resistance), the impedance is purely reactive: $$Z = j(\omega L - \frac{1}{\omega C})$$. The current either leads or lags the voltage by exactly 90°. Therefore, when voltage is maximum, current is zero. YES.

The correct answer is Option (4): A, B and D only.

Transformer voltage equation: $$\frac{V_s}{V_p} = \frac{N_s}{N_p}$$

Given parameters: $$V_p = 200\text{ V},\quad N_p = 100,\quad N_s = 10$$

Calculating total secondary voltage ($$V_s$$):

$$V_s = V_p \times \frac{N_s}{N_p} = 200 \times \frac{10}{100} = 22\text{ V}$$

Applying voltage divider rule across the series resistors in the secondary circuit to find $$V_0$$:

$$V_0 = V_s \times \frac{7\text{ k}\Omega}{15\text{ k}\Omega + 7\text{ k}\Omega}$$

$$V_0 = 22 \times \frac{7}{22} = 7\text{ V}$$

In Bohr's model, the radius of the $$n$$th orbit is proportional to $$n^2$$:

$$r_n \propto n^2$$

For the third orbit: $$r_3 = R \propto 3^2 = 9$$

For the fourth orbit: $$r_4 \propto 4^2 = 16$$

$$\frac{r_4}{r_3} = \frac{16}{9}$$

$$r_4 = \frac{16}{9}R$$

The answer is $$\frac{16}{9}R$$, which corresponds to Option (2).

The current is $$i = 6 + \sqrt{56}\sin(100\pi t + \pi/3)$$

This is a DC component plus an AC component.

DC component: $$I_{DC} = 6$$ A

AC peak value: $$I_0 = \sqrt{56}$$ A

RMS of AC component: $$I_{AC,rms} = \frac{\sqrt{56}}{\sqrt{2}} = \sqrt{28}$$ A

Total RMS: $$I_{rms} = \sqrt{I_{DC}^2 + I_{AC,rms}^2} = \sqrt{36 + 28} = \sqrt{64} = 8$$ A

The answer is 8.

Given: $$L = \frac{100}{\pi}$$ mH $$= \frac{0.1}{\pi}$$ H, $$C = \frac{10^{-3}}{\pi}$$ F, $$R = 10 \; \Omega$$, $$f = 50$$ Hz.

$$\omega = 2\pi f = 100\pi$$ rad/s.

$$X_L = \omega L = 100\pi \times \frac{0.1}{\pi} = 10 \; \Omega$$

$$X_C = \frac{1}{\omega C} = \frac{1}{100\pi \times \frac{10^{-3}}{\pi}} = \frac{1}{100 \times 10^{-3}} = \frac{1}{0.1} = 10 \; \Omega$$

Since $$X_L = X_C$$, the circuit is at resonance.

Impedance: $$Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{100 + 0} = 10 \; \Omega$$

Power factor: $$\cos\phi = \frac{R}{Z} = \frac{10}{10} = 1$$

The answer is $$\boxed{1}$$.

A capacitor of $$C = 150$$ $$\mu$$F is connected to an alternating voltage source described by $$E = 36\sin(120\pi t)$$ V.

From this expression, the peak emf is $$E_0 = 36$$ V, the angular frequency is $$\omega = 120\pi$$ rad/s, and the capacitance is $$C = 150 \times 10^{-6}$$ F.

Next, the capacitive reactance is given by

$$X_C = \dfrac{1}{\omega C} = \dfrac{1}{120\pi \times 150 \times 10^{-6}} = \dfrac{1}{120 \times 150 \times \pi \times 10^{-6}}$$

$$= \dfrac{1}{18000\pi \times 10^{-6}} = \dfrac{1}{0.018\pi} = \dfrac{1000}{18\pi} = \dfrac{500}{9\pi}$$

Having found $$X_C,$$ the maximum current is

$$I_0 = \dfrac{E_0}{X_C} = \dfrac{36}{\dfrac{500}{9\pi}} = \dfrac{36 \times 9\pi}{500} = \dfrac{324\pi}{500}$$

Using $$\pi \approx 3.14$$, one obtains $$I_0 \approx \dfrac{324 \times 3.14}{500} = \dfrac{1017.36}{500} \approx 2.03$$ A.

Thus the current is approximately 2 A, giving the final result $$2$$ A.

The resonant frequency of an LC oscillator is given by:

$$\omega_0 = \frac{1}{\sqrt{LC}}$$

When the inductance becomes twice ($$L' = 2L$$) and capacitance becomes eight times ($$C' = 8C$$):

$$\omega' = \frac{1}{\sqrt{L'C'}} = \frac{1}{\sqrt{2L \cdot 8C}} = \frac{1}{\sqrt{16LC}} = \frac{1}{4\sqrt{LC}} = \frac{\omega_0}{4}$$

Since $$\omega' = x \cdot \omega_0$$:

$$x = \frac{1}{4}$$

Therefore, the correct answer is Option A: $$\mathbf{\frac{1}{4}}$$.

We need to find the ratio of power factors $$P_1$$ to $$P_2$$ for a series LR circuit before and after adding a capacitor.

Case 1: Series LR circuit with $$X_L = R$$.

Impedance: $$Z_1 = \sqrt{R^2 + X_L^2} = \sqrt{R^2 + R^2} = R\sqrt{2}$$

Power factor: $$P_1 = \cos\phi_1 = \frac{R}{Z_1} = \frac{R}{R\sqrt{2}} = \frac{1}{\sqrt{2}}$$

Case 2: Series LCR circuit with $$X_C = X_L = R$$.

Net reactance: $$X_L - X_C = R - R = 0$$

Impedance: $$Z_2 = \sqrt{R^2 + 0} = R$$

Power factor: $$P_2 = \cos\phi_2 = \frac{R}{R} = 1$$

Ratio:

$$\frac{P_1}{P_2} = \frac{1/\sqrt{2}}{1} = \frac{1}{\sqrt{2}}$$

Therefore, $$P_1 : P_2 = 1 : \sqrt{2}$$.

The correct answer is Option 2: $$1 : \sqrt{2}$$.

We need to find the value of $$x$$ where the capacitance for maximum current is $$\dfrac{1}{x}$$ F.

Maximum current in a series LCR circuit occurs at resonance, where $$\omega^2 L C = 1 \implies C = \dfrac{1}{\omega^2 L}$$.

Given $$f = 7$$ kHz $$= 7000$$ Hz, $$L = 2\,\mu$$H $$= 2 \times 10^{-6}$$ H, and $$\pi = \dfrac{22}{7}$$, we have $$\omega = 2\pi f = 2 \times \dfrac{22}{7} \times 7000 = 2 \times 22000 = 44000 \text{ rad/s}$$.

Then $$\omega^2 = (44000)^2 = 1936 \times 10^6$$ and $$C = \dfrac{1}{\omega^2 L} = \dfrac{1}{1936 \times 10^6 \times 2 \times 10^{-6}} = \dfrac{1}{1936 \times 2} = \dfrac{1}{3872} \text{ F}$$.

Since $$C = \dfrac{1}{x}$$ F, it follows that $$x = 3872$$.

The answer is $$\boxed{3872}$$.

We are given a series LCR circuit with $$R = 80$$ $$\Omega$$, $$X_L = 100$$ $$\Omega$$, $$X_C = 40$$ $$\Omega$$, and input voltage $$V = 2500\cos(100\pi t)$$ V.

$$ Z = \sqrt{R^2 + (X_L - X_C)^2} $$

$$ Z = \sqrt{80^2 + (100 - 40)^2} = \sqrt{6400 + 3600} = \sqrt{10000} = 100 \text{ } \Omega $$

From $$V = 2500\cos(100\pi t)$$, the amplitude (peak voltage) is $$V_0 = 2500$$ V.

$$ I_0 = \frac{V_0}{Z} = \frac{2500}{100} = 25 \text{ A} $$

The amplitude of current in the circuit is 25 A.

We have an AC source of 220 V at 50 Hz, with $$R = 100$$ $$\Omega$$ and $$X_L = 79.6$$ $$\Omega$$.

The average rate of energy supplied (power) is maximized at resonance, when the impedance equals $$R$$. At resonance, $$X_C = X_L$$, so

$$\frac{1}{2\pi f C} = X_L$$

Solving for $$C$$,

$$C = \frac{1}{2\pi f X_L}$$

Now substituting the values,

$$C = \frac{1}{2\pi \times 50 \times 79.6} = \frac{1}{2 \times 3.14159 \times 50 \times 79.6} = \frac{1}{25000.5}$$

$$C \approx 4.0 \times 10^{-5} \text{ F} = 40 \text{ }\mu\text{F}$$

Hence, the correct answer is Option 2.

Solution :

Given :

$$L = 0.5\text{ mH} = 0.5 \times 10^{-3}\text{ H}$$

$$C = 20\ \mu\text{F} = 20 \times 10^{-6}\text{ F}$$

$$R = 20\ \Omega$$

Source voltage :

$$V_{rms} = 220\text{ V}$$

Current is in phase with emf.

Hence, circuit is in resonance.

At resonance :

$$X_L = X_C$$

and impedance,

$$Z = R$$

Therefore,

$$I_{rms} = \frac{V_{rms}}{R}$$

$$= \frac{220}{20}$$

$$= 11\text{ A}$$

Amplitude of current :

$$I_0 = \sqrt2\ I_{rms}$$

$$= 11\sqrt2\text{ A}$$

Given amplitude is of the form :

$$x^{1/2}$$

Therefore,

$$x = 242$$

Final Answer :

$$242$$

For maximum current amplitude in an LCR series circuit, the circuit must be at resonance, where $$X_L = X_C$$.

Given: $$C = 62.5$$ nF $$= 62.5 \times 10^{-9}$$ F, $$R = 50 \; \Omega$$, $$f = 2.0$$ kHz $$= 2000$$ Hz, and $$\pi^2 = 10$$.

At resonance:

$$\omega L = \frac{1}{\omega C}$$

$$L = \frac{1}{\omega^2 C}$$

The angular frequency:

$$\omega = 2\pi f = 2\pi \times 2000 = 4000\pi$$

Computing $$\omega^2$$:

$$\omega^2 = (4000\pi)^2 = 16 \times 10^6 \times \pi^2 = 16 \times 10^6 \times 10 = 1.6 \times 10^8$$

Therefore:

$$L = \frac{1}{1.6 \times 10^8 \times 62.5 \times 10^{-9}} = \frac{1}{1.6 \times 10^8 \times 6.25 \times 10^{-8}} = \frac{1}{10} = 0.1 \text{ H}$$

Converting to millihenry:

$$L = 0.1 \text{ H} = 100 \text{ mH}$$

Therefore, the answer is $$\mathbf{100}$$.

We are given an oscillating LC circuit with inductance $$L = 75$$ mH $$= 75 \times 10^{-3}$$ H, capacitance $$C = 1.2$$ $$\mu$$F $$= 1.2 \times 10^{-6}$$ F, and maximum charge $$Q_0 = 2.7$$ $$\mu$$C $$= 2.7 \times 10^{-6}$$ C.

Apply energy conservation. At maximum charge, all energy is in the capacitor. At maximum current, all energy is in the inductor:

$$ \frac{Q_0^2}{2C} = \frac{LI_0^2}{2} $$

Solve for maximum current: $$ I_0 = \frac{Q_0}{\sqrt{LC}} $$

Calculate $$\sqrt{LC}$$: $$ LC = 75 \times 10^{-3} \times 1.2 \times 10^{-6} = 90 \times 10^{-9} = 9 \times 10^{-8} $$

$$ \sqrt{LC} = \sqrt{9 \times 10^{-8}} = 3 \times 10^{-4} \text{ s} $$

Find $$I_0$$: $$ I_0 = \frac{2.7 \times 10^{-6}}{3 \times 10^{-4}} = 0.9 \times 10^{-2} = 9 \times 10^{-3} \text{ A} = 9 \text{ mA} $$

The maximum current in the circuit is 9 mA.

For a series LCR circuit, power factor is:

$$\cos\phi=\frac{R}{Z}$$

where impedance,

$$Z=\sqrt{R^2+(X_L-X_C)^2}$$

Given:

• R=80 Ω
• XL=70 Ω
• XC=130 Ω

Net reactance:

$$X=X_L-X_C=70-130=-60$$

Impedance:

$$Z=\sqrt{80^2+60^2}$$

$$Z=\sqrt{6400+3600}$$

$$Z=\sqrt{10000}=100$$

Therefore,

$$\cos\phi=\frac{80}{100}=0.8$$

Given power factor is written as:

$$\frac{x}{10}$$

So,

$$\frac{x}{10}=0.8$$

x=8

We need to identify the type of AC circuit in which wattless current flows.

Wattless current refers to a current that does not consume any power. The average power in an AC circuit is given by:

$$P = V_{rms} \times I_{rms} \times \cos\phi$$

where $$\phi$$ is the phase difference between voltage and current. For $$P = 0$$, we need $$\cos\phi = 0$$, which means $$\phi = \frac{\pi}{2}$$ (i.e., $$90°$$). A phase difference of $$90°$$ between voltage and current occurs in both a purely inductive circuit (current lags voltage by $$90°$$) and a purely capacitive circuit (current leads voltage by $$90°$$).

In a purely resistive circuit, $$\phi = 0$$, so full power is dissipated. In LCR and RC circuits, the phase angle is generally not $$90°$$, so some power is dissipated.

Among the given options, a purely inductive circuit has wattless current. The correct answer is Option B: Purely Inductive circuit.

In a series $$LR$$ circuit, $$X_L = R$$, and the power factor is $$P_1$$. When a capacitor with $$X_C = X_L$$ is added in series, the power factor becomes $$P_2$$. We need to find $$\dfrac{P_1}{P_2}$$.

The power factor is defined as:

$$P_1 = \cos\phi_1 = \frac{R}{Z_1}$$

For a series $$LR$$ circuit, the impedance is:

$$Z_1 = \sqrt{R^2 + X_L^2}$$

Given $$X_L = R$$:

$$Z_1 = \sqrt{R^2 + R^2} = R\sqrt{2}$$

$$P_1 = \frac{R}{R\sqrt{2}} = \frac{1}{\sqrt{2}}$$

When a capacitor with $$X_C = X_L$$ is added in series, resonance occurs since $$X_L = X_C$$:

$$Z_2 = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{R^2 + 0} = R$$

$$P_2 = \frac{R}{R} = 1$$

$$\frac{P_1}{P_2} = \frac{1/\sqrt{2}}{1} = \frac{1}{\sqrt{2}}$$

The correct answer is Option B: $$\dfrac{1}{\sqrt{2}}$$.

We match each device in List I with its correct description in List II.

(a) A rectifier is a device that converts alternating current (A.C.) voltage into direct current (D.C.) voltage. This matches description (ii).

(b) A stabilizer (voltage regulator) is a device used to maintain a constant output voltage even when the input voltage or load current changes. This matches description (iv).

(c) A transformer is a device used either for stepping up or stepping down the A.C. voltage. This matches description (i).

(d) A filter is a circuit used to remove any ripple in the rectified output voltage, producing a smoother D.C. output. This matches description (iii).

Therefore, the correct matching is (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii).

We have an a.c. circuit in which a resistor of resistance $$R$$, an inductor offering inductive reactance $$X_L$$ and a capacitor offering capacitive reactance $$X_C$$ are all connected in series.

The impedance of any series $$R$$-$$L$$-$$C$$ circuit is defined by the formula

$$Z \;=\; \sqrt{\,R^{2} \;+\; \bigl(X_L - X_C\bigr)^{2}}.$$

This expression comes directly from phasor analysis: the resistor contributes a real component $$R$$, while the inductor and capacitor contribute imaginary components $$+jX_L$$ and $$-jX_C$$ respectively; their algebraic sum in the imaginary direction is $$j\,(X_L - X_C)$$. Taking the magnitude of the complex quantity $$R + j\,(X_L - X_C)$$ gives the above square-root relation.

Now the question states that the magnitudes of all three quantities are equal, that is

$$X_L \;=\; R \;=\; X_C.$$

Substituting $$X_L = X_C$$ straight into the impedance formula, we obtain

$$Z = \sqrt{\,R^{2} + \bigl(X_L - X_C\bigr)^{2}} = \sqrt{\,R^{2} + \bigl(X_L - X_L\bigr)^{2}} = \sqrt{\,R^{2} + 0^{2}} = \sqrt{\,R^{2}} = R.$$

Thus the magnitude of the impedance of the circuit reduces exactly to the resistance itself.

Hence, the correct answer is Option B.

For a series $$RLC$$ circuit the impedance is $$Z = \sqrt{R^{2}+\left(\,\omega L-\frac{1}{\omega C}\right)^{2}}$$ and the phase angle $$\phi$$ between the current $$I$$ and the applied emf $$E$$ is given by

$$\tan \phi = \frac{\omega L-\dfrac{1}{\omega C}}{R}$$

Sign of $$\left(\omega L-\dfrac{1}{\omega C}\right)$$ decides whether the circuit behaves inductively or capacitively.

Case a: $$\omega L \gt \dfrac{1}{\omega C}$$ ⇒ $$\tan\phi \gt 0$$, so $$\phi$$ is positive. A positive phase angle means the current lags behind the emf.
Hence (a) matches with (ii) “Current lags behind the applied emf”.

Case b: $$\omega L = \dfrac{1}{\omega C}$$ ⇒ $$\tan\phi = 0$$, so $$\phi = 0$$. The current is exactly in phase with the emf and the circuit is at resonance.
Hence (b) matches with (i) “Current is in phase with emf”.

Case c: $$\omega L \lt \dfrac{1}{\omega C}$$ ⇒ $$\tan\phi \lt 0$$, so $$\phi$$ is negative. A negative phase angle means the current leads the emf.
Hence (c) matches with (iv) “Current leads the emf”.

Case d: “Resonant frequency” refers to the condition $$\omega L = \dfrac{1}{\omega C}$$, where the impedance is minimum ($$Z = R$$) and the circuit admits the largest current.
Hence (d) matches with (iii) “Maximum current occurs”.

Collecting all matches:
(a) - (ii); (b) - (i); (c) - (iv); (d) - (iii)

Therefore the correct option is Option A.

$$X_L = \omega L = 100 \times 0.1 = 10\ \Omega$$

$$X_C = \frac{1}{\omega C} = \frac{1}{100 \times 10^{-4}} = 100\ \Omega$$

$$Z = \sqrt{120^2 + (10 - 100)^2} = \sqrt{120^2 + (-90)^2} = 150\ \Omega$$

$$I_0 = \frac{V_0}{Z} \implies I_0 = \frac{30}{150} = 0.2\text{ A}$$

$$f_r = \frac{1}{2\pi\sqrt{LC}} \implies f_r = \frac{1}{2\pi\sqrt{0.1 \times 10^{-4}}} = \frac{1}{2\pi\sqrt{10^{-5}}} = \frac{100\sqrt{10}}{2\pi} \approx 50.3\text{ Hz} \approx 50\text{ Hz}$$

In a purely resistive AC circuit, the current and voltage are in phase, so the phase difference is zero. This matches (a) with (ii).

In a purely inductive AC circuit, the current lags the voltage by $$\frac{\pi}{2}$$. This matches (b) with (iii).

In a purely capacitive AC circuit, the current leads the voltage by $$\frac{\pi}{2}$$. This matches (c) with (i).

In an LCR series circuit, the phase difference between current and voltage is $$\phi = \tan^{-1}\left(\frac{X_C - X_L}{R}\right)$$. This matches (d) with (iv).

Therefore the correct matching is (a)-(ii), (b)-(iii), (c)-(i), (d)-(iv).

For a purely resistive circuit the alternating current has the same waveform as the alternating voltage, only scaled by the resistance. Thus, if the instantaneous voltage is written as

$$v(t)=V_{\text m}\sin(\omega t),$$

then the instantaneous current is

$$i(t)=\dfrac{v(t)}{R}=I_{\text m}\sin(\omega t),\qquad\text{where }I_{\text m}= \dfrac{V_{\text m}}{R}.$$

First we express the maximum (peak) and rms values.

Formula: $$V_{\text m}=\sqrt{2}\,V_{\text{rms}},\qquad I_{\text{rms}}=\dfrac{I_{\text m}}{\sqrt{2}}.$$

The supply gives $$V_{\text{rms}}=220\ \text V,$$ so

$$V_{\text m}=\sqrt{2}\times220\ \text V=220\sqrt{2}\ \text V.$$

With the resistance $$R=10\ \Omega$$ we have

$$I_{\text m}=\dfrac{V_{\text m}}{R}=\dfrac{220\sqrt{2}}{10}=22\sqrt{2}\ \text A,$$

and correspondingly

$$I_{\text{rms}}=\dfrac{I_{\text m}}{\sqrt{2}}=22\ \text A.$$

Now we look at how the current changes with time. We write

$$i(t)=I_{\text m}\sin(\omega t).$$

The maximum value occurs when $$\sin(\omega t)=1.$$ The smallest positive angle giving this is $$\omega t=\dfrac{\pi}{2},$$ so the time at that instant is

$$t_1=\dfrac{\pi/2}{\omega}.$$

Next we want the time when the current has fallen to its rms value. We set

$$i(t_2)=I_{\text{rms}}=\dfrac{I_{\text m}}{\sqrt{2}}.$$

Substituting into the waveform,

$$I_{\text m}\sin(\omega t_2)=\dfrac{I_{\text m}}{\sqrt{2}} \;\Longrightarrow\; \sin(\omega t_2)=\dfrac{1}{\sqrt{2}}.$$

The principal angles whose sine equals $$1/\sqrt{2}$$ are $$\omega t_2=\dfrac{\pi}{4}\ \text{or}\ \dfrac{3\pi}{4}.$$

Since the current is decreasing after reaching its maximum at $$\omega t=\pi/2$$, we select the later angle

$$\omega t_2=\dfrac{3\pi}{4}.$$

The angular frequency is determined by the supply frequency

Formula: $$\omega = 2\pi f.$$

With $$f = 50\ \text{Hz},$$

$$\omega = 2\pi \times 50 = 100\pi\ \text{rad s}^{-1}.$$

Hence,

$$t_1=\dfrac{\pi/2}{100\pi}=\dfrac{1}{200}\ \text s=0.005\ \text s,$$

$$t_2=\dfrac{3\pi/4}{100\pi}=\dfrac{3}{400}\ \text s=0.0075\ \text s.$$

The required time interval for the current to drop from its maximum to its rms value is

$$\Delta t = t_2 - t_1 = \left(\dfrac{3}{400}-\dfrac{2}{400}\right)\text s=\dfrac{1}{400}\ \text s=0.0025\ \text s.$$

Converting to milliseconds,

$$\Delta t = 0.0025\ \text s = 2.5\ \text{ms}.$$

Hence, the correct answer is Option A.

At resonance in a series $$RLC$$ circuit the inductive reactance equals the capacitive reactance. We first state this resonance condition:

$$X_L = X_C \quad\Longrightarrow\quad \omega L = \dfrac{1}{\omega C}$$

Here $$\omega$$ is the angular resonant frequency, related to the ordinary frequency $$f$$ by the well-known formula $$\omega = 2\pi f$$. Solving the above equality for the inductance $$L$$ gives

$$L = \dfrac{1}{\omega^{2} C}.$$

We are given

$$f = 60\ \text{Hz}, \qquad C = 0.1\ \mu\text{F}.$$

First we convert the capacitance into farads, remembering that $$1\ \mu\text{F} = 10^{-6}\ \text{F}:$$

$$C = 0.1\ \mu\text{F} = 0.1 \times 10^{-6}\ \text{F} = 1 \times 10^{-7}\ \text{F}.$$

Now we calculate the angular frequency:

$$\omega = 2\pi f = 2\pi \times 60\ \text{rad s}^{-1} = 120\pi\ \text{rad s}^{-1}.$$

For numerical work we use $$\pi \approx 3.1416$$, so

$$\omega = 120 \times 3.1416\ \text{rad s}^{-1} \approx 376.99\ \text{rad s}^{-1}.$$

Next we square this angular frequency, showing each step:

$$\omega^{2} = (376.99)^{2}\ \text{rad}^{2}\text{s}^{-2}.$$

Working out the square:

$$376.99^{2} = 376.99 \times 376.99 = 142\,129\ (\text{to the nearest unit}).$$

So

$$\omega^{2} \approx 1.42129 \times 10^{5}\ \text{rad}^{2}\text{s}^{-2}.$$

We now substitute $$\omega^{2}$$ and $$C$$ into the inductance formula:

$$L = \dfrac{1}{\omega^{2} C} = \dfrac{1}{\left(1.42129 \times 10^{5}\right) \left(1 \times 10^{-7}\right)}.$$

First multiply the denominator:

$$(1.42129 \times 10^{5})(1 \times 10^{-7}) = 1.42129 \times 10^{5-7} = 1.42129 \times 10^{-2}.$$

So we have

$$L = \dfrac{1}{1.42129 \times 10^{-2}}\ \text{henry}.$$

The reciprocal is obtained as follows:

$$\dfrac{1}{1.42129 \times 10^{-2}} = \dfrac{1}{0.0142129} \approx 70.36\ \text{H}.$$

Rounding to three significant figures gives

$$L \approx 70.3\ \text{H}.$$

Hence, the correct answer is Option D.

The inductive reactance is given by $$X_L = 2\pi f L = \omega L$$. If the frequency is halved, the new reactance is $$X_L' = 2\pi \left(\frac{f}{2}\right) L = \frac{X_L}{2}$$. So the inductive reactance is halved.

In a purely inductive circuit, the current is $$I = \frac{V}{X_L}$$. When the reactance is halved, the current becomes $$I' = \frac{V}{X_L/2} = \frac{2V}{X_L} = 2I$$. So the current is doubled.

Therefore, when the frequency is halved, the inductive reactance is halved and the current is doubled.

We have a series circuit containing an inductor of inductance $$L = 0.07\;\text{H}$$ and a resistor of resistance $$R = 12\;\Omega$$ connected to an AC source of root-mean-square (r.m.s.) voltage $$V = 220\;\text{V}$$ and frequency $$f = 50\;\text{Hz}$$.

First we need the inductive reactance. The formula for inductive reactance is stated as

$$X_L = \omega L$$

where $$\omega$$ is the angular frequency given by $$\omega = 2\pi f$$. Substituting the given frequency and taking $$\pi = \dfrac{22}{7}$$, we get

$$\omega = 2\pi f = 2 \times \dfrac{22}{7} \times 50 = \dfrac{44}{7} \times 50 = \dfrac{44 \times 50}{7} = \dfrac{2200}{7}\;\text{rad s}^{-1}.$$

Now substituting this value of $$\omega$$ and the given value of $$L$$ in the formula $$X_L = \omega L$$, we have

$$X_L = \left(\dfrac{2200}{7}\right) \times 0.07 = \dfrac{2200 \times 0.07}{7}.$$ Recognising that $$0.07 = \dfrac{7}{100}$$, we can write

$$X_L = \dfrac{2200}{7} \times \dfrac{7}{100} = \dfrac{2200}{100} = 22\;\Omega.$$

So, the inductive reactance is $$X_L = 22\;\Omega$$.

Next, the total impedance $$Z$$ of a series $$R\!-\!L$$ circuit is given by the relation

$$Z = \sqrt{R^2 + X_L^2}.$$

Substituting $$R = 12\;\Omega$$ and $$X_L = 22\;\Omega$$, we get

$$Z = \sqrt{(12)^2 + (22)^2} = \sqrt{144 + 484} = \sqrt{628}.$$

Evaluating the square root,

$$\sqrt{628} \approx 25.06\;\Omega.$$ For practical purposes this may be rounded to $$25.1\;\Omega$$.

The r.m.s. current $$I$$ in the circuit is obtained from Ohm’s law for AC circuits,

$$I = \dfrac{V}{Z}.$$

Substituting the numerical values,

$$I = \dfrac{220\;\text{V}}{25.06\;\Omega} \approx 8.78\;\text{A}.$$

Rounded to two significant figures, the current is $$I \approx 8.8\;\text{A}.$$

Finally, to find the phase angle $$\phi$$ between the current and the source voltage, we use the relation for a series $$R\!-\!L$$ circuit,

$$\tan\phi = \dfrac{X_L}{R}.$$

Substituting $$X_L = 22\;\Omega$$ and $$R = 12\;\Omega$$, we have

$$\tan\phi = \dfrac{22}{12} = \dfrac{11}{6}.$$

Therefore,

$$\phi = \tan^{-1}\!\left(\dfrac{11}{6}\right).$$

Because the circuit contains an inductor, the current lags the voltage by this angle.

Combining the two results, we conclude that the approximate current in the circuit is $$8.8\;\text{A}$$ and the phase angle between the current and the source voltage is $$\tan^{-1}\!\left(\dfrac{11}{6}\right).$$

Hence, the correct answer is Option A.

We have a power transmission line that supplies an input (primary) voltage of $$V_p = 2300 \text{ V}$$ to a step-down transformer. The number of turns on the primary winding is given as $$N_p = 4000$$. The transformer delivers an output (secondary) voltage of $$V_s = 230 \text{ V}$$.

The current in the primary circuit is stated to be $$I_p = 5 \text{ A}$$. The transformer works with an efficiency of $$\eta = 90\% = 0.90$$. Our goal is to compute the secondary (output) current $$I_s$$.

First, we recall the definition of efficiency for a transformer:

$$\eta \;=\; \dfrac{P_{\text{out}}}{P_{\text{in}}}$$

Here, $$P_{\text{in}}$$ is the input power and $$P_{\text{out}}$$ is the output power.

We now calculate the input power using the elementary electric power formula $$P = V I$$:

$$P_{\text{in}} \;=\; V_p \, I_p$$

Substituting the given values,

$$P_{\text{in}} \;=\; 2300 \text{ V} \times 5 \text{ A} \;=\; 11500 \text{ W}$$

Next, we find the output power from the efficiency relation. Rearranging the efficiency formula, we get

$$P_{\text{out}} \;=\; \eta \, P_{\text{in}}$$

Putting the numbers in,

$$P_{\text{out}} \;=\; 0.90 \times 11500 \text{ W}$$

$$P_{\text{out}} \;=\; 10350 \text{ W}$$

The transformer’s secondary voltage is already given as $$V_s = 230 \text{ V}$$. Again using the power formula on the secondary side,

$$P_{\text{out}} \;=\; V_s \, I_s$$

We solve for the unknown secondary current $$I_s$$:

$$I_s \;=\; \dfrac{P_{\text{out}}}{V_s}$$

Substituting $$P_{\text{out}} = 10350 \text{ W}$$ and $$V_s = 230 \text{ V}$$, we obtain

$$I_s \;=\; \dfrac{10350}{230} \text{ A}$$

Performing the division,

$$I_s \;=\; 45 \text{ A}$$

So, the secondary winding of the transformer delivers a current of $$45 \text{ A}$$.

Hence, the correct answer is Option D.

We are told that the applied alternating emf is $$e = e_0 \sin(100t)$$, where the time $$t$$ is in seconds.

Comparing $$e = e_0 \sin(\omega t)$$ with the standard form $$e = e_0 \sin(\omega t)$$, we recognise that the angular frequency is

$$\omega = 100\ \text{rad s}^{-1}\,.$$

The circuit to be chosen must give a phase difference of

$$\phi = \frac{\pi}{4} = 45^\circ$$

between the applied emf $$e$$ and the current $$i$$. We therefore examine how the phase angle arises in series $$RC$$ and $$RL$$ circuits.

For a series RC circuit the impedance is

$$Z = R - jX_C = R - j\frac{1}{\omega C},$$

so the phase of the impedance (with respect to the real axis) is

$$\theta = -\tan^{-1}\!\!\left(\frac{X_C}{R}\right) = -\tan^{-1}\!\!\left(\frac{1}{\omega R C}\right).$$

Because the current leads the voltage by the magnitude of this angle, the current-voltage phase difference is

$$\phi_{RC} = \tan^{-1}\!\!\left(\frac{1}{\omega R C}\right).$$

For a series RL circuit the impedance is

$$Z = R + jX_L = R + j\omega L,$$

so the phase of the impedance is

$$\theta = \tan^{-1}\!\!\left(\frac{X_L}{R}\right)=\tan^{-1}\!\!\left(\frac{\omega L}{R}\right).$$

This time the current lags the voltage by the same magnitude, giving

$$\phi_{RL} = \tan^{-1}\!\!\left(\frac{\omega L}{R}\right).$$

In both cases we require the magnitude of the phase angle to be

$$\phi = \frac{\pi}{4}\,,$$

which implies

$$\tan\phi = 1.$$

We now check each option.

Option A: RC circuit, $$R = 1\ \text{k}\Omega = 10^3\ \Omega$$, $$C = 1\ \mu\text{F} = 10^{-6}\ \text{F}$$.

Using $$\tan\phi = \dfrac{1}{\omega R C}$$, we have

$$\tan\phi_A = \frac{1}{(100)(10^3)(10^{-6})} = \frac{1}{0.1} = 10.$$

Since $$\tan\phi_A = 10 \neq 1$$, the phase angle is not $$\pi/4$$. This option is rejected.

Option B: RL circuit, $$R = 1\ \text{k}\Omega = 10^3\ \Omega$$, $$L = 1\ \text{mH} = 10^{-3}\ \text{H}$$.

Using $$\tan\phi = \dfrac{\omega L}{R}$$, we have

$$\tan\phi_B = \frac{(100)(10^{-3})}{10^3} = \frac{0.1}{10^3} = 1\times10^{-4}.$$

Since $$\tan\phi_B \neq 1$$, the required phase difference is not obtained. This option is rejected.

Option C: RL circuit, $$R = 1\ \text{k}\Omega = 10^3\ \Omega$$, $$L = 10\ \text{mH} = 10^{-2}\ \text{H}$$.

Again using $$\tan\phi = \dfrac{\omega L}{R}$$,

$$\tan\phi_C = \frac{(100)(10^{-2})}{10^3} = \frac{1}{10^3} = 0.001.$$

This is still far from unity, so the phase angle is not $$\pi/4$$. This option is also rejected.

Option D: RC circuit, $$R = 1\ \text{k}\Omega = 10^3\ \Omega$$, $$C = 10\ \mu\text{F} = 10^{-5}\ \text{F}$$.

Using $$\tan\phi = \dfrac{1}{\omega R C}$$, we get

$$\tan\phi_D = \frac{1}{(100)(10^3)(10^{-5})} = \frac{1}{1} = 1.$$

Because $$\tan\phi_D = 1$$, we have

$$\phi_D = \tan^{-1}(1) = 45^\circ = \frac{\pi}{4}.$$

This is exactly the required phase difference. Therefore option D satisfies the given condition.

Hence, the correct answer is Option D.

We begin by recalling the basic relations for a series $$LCR$$ circuit driven by an alternating voltage of angular frequency $$\omega$$.

Inductive reactance is given by the formula $$X_L=\omega L$$.

Capacitive reactance is given by the formula $$X_C=\dfrac{1}{\omega C}$$.

The net (effective) reactance is the difference of the two because the inductor and capacitor contribute opposite phasors:

$$X = X_L - X_C\;.$$

The magnitude of the total impedance is then obtained from the resistance $$R$$ and the net reactance $$X$$ via the impedance formula for a right-angled phasor triangle, i.e.

$$Z=\sqrt{R^2+X^2}\;.$$

The phase angle $$\phi$$ between the source voltage and the current follows from

$$\tan\phi=\dfrac{X}{R}\;.$$

Now we substitute the numerical values one by one.

The data are

$$R = 5\;\Omega,\;\; L = 25\text{ mH} = 25\times10^{-3}\text{ H},\;\; C = 1000\;\mu\text{F} = 1000\times10^{-6}\text{ F}=0.001\text{ F},\;\; \omega = 320\;\text{s}^{-1}.$$

First we calculate the individual reactances.

Inductive reactance:

$$X_L = \omega L = 320 \times 0.025 = 8.0\;\Omega.$$

Capacitive reactance:

$$X_C = \dfrac{1}{\omega C} = \dfrac{1}{320 \times 0.001} = \dfrac{1}{0.32} = 3.125\;\Omega.$$

Hence the net reactance is

$$X = X_L - X_C = 8.0 - 3.125 = 4.875\;\Omega.$$

Substituting $$R=5\;\Omega$$ and $$X=4.875\;\Omega$$ in the impedance expression, we have

$$Z = \sqrt{R^2 + X^2} = \sqrt{5^2 + (4.875)^2} = \sqrt{25 + 23.7656} = \sqrt{48.7656}\;\Omega.$$

Evaluating the square root,

$$Z \approx 6.99\;\Omega \approx 7\;\Omega.$$

For the phase angle, we insert the same values into $$\tan\phi=\dfrac{X}{R}$$:

$$\tan\phi = \dfrac{4.875}{5} = 0.975.$$

The arctangent of 0.975 is very close to $$45^\circ$$ (more precisely about $$44.2^\circ$$), so to the nearest whole degree we may write

$$\phi \approx 45^\circ.$$

Thus, the total impedance of the given series $$LCR$$ circuit is approximately $$7\;\Omega$$ and the phase difference between source voltage and current is about $$45^\circ$$.

Hence, the correct answer is Option B.

The discharging of a capacitor in a series LCR circuit is analogous to a damped harmonic oscillator, such as a damped pendulum. The resistor $$R$$ acts as the damping element, dissipating energy and causing the amplitude of the charge to decay exponentially over time.

The maximum charge $$Q_{Max}$$ on the capacitor at any time $$t$$ represents the envelope of the damped oscillation:

$$Q_{Max} = Q_0 e^{-\frac{Rt}{2L}}$$

$$Q_{Max}^2 = \left( Q_0 e^{-\frac{Rt}{2L}} \right)^2 = Q_0^2 e^{-\frac{Rt}{L}}$$

A larger inductance results in a smaller decay constant ($$\frac{R}{L}$$), meaning the charge amplitude decreases more slowly (slower decay).

Given that $$L_1 > L_2$$, the decay for $$L_1$$ is slower than for $$L_2$$. Consequently, for any time $$t > 0$$, the curve for $$L_1$$ will remain at a higher value than the curve for $$L_2$$.

The problem asks which instrument can measure an alternating-current (AC) voltage that appears across a pure resistance. To decide, we examine the working principle of each listed device, keeping in mind the special nature of AC signals.

First, recall that an AC quantity such as the current through the resistance can be written as $$i(t)=I_0\sin\omega t$$. The instantaneous value of $$i(t)$$ swings symmetrically above and below zero, so over a complete cycle the algebraic average of $$i(t)$$ itself is

$$\langle i(t)\rangle=\frac{1}{T}\int_0^T I_0\sin\omega t \,dt=0$$

However, the heating (thermal) effect of the current is determined by the mean square value, because the power converted to heat is

$$P=i^2 R$$

and the cycle average of $$i^2(t)$$ is

$$\langle i^2(t)\rangle=\frac{1}{T}\int_0^T I_0^2\sin^2\omega t\,dt=\frac{I_0^2}{2}=I_{\text{rms}}^2$$

So an instrument that depends on the heating effect of current will respond to $$I_{\text{rms}}$$ (and, correspondingly, to the rms value of the AC voltage across the resistance), whereas an instrument that attempts to respond to the instantaneous direction of the current will average out to zero.

Now we inspect each option.

Option A: Moving magnet galvanometer. A moving magnet or tangent galvanometer relies on the torque produced by a steady magnetic field generated by a direct current. Because the torque changes sign with the alternating direction of AC, the average torque over a cycle is zero, giving no deflection. Therefore, it cannot measure AC.

Option B: Moving coil galvanometer. A moving coil galvanometer (d'Arsonval type) likewise needs a unidirectional current through its coil. For an AC input $$i(t)=I_0\sin\omega t$$, the torque $$\tau\propto i(t)$$ changes sign every half cycle, again producing an average torque of zero. Hence, the pointer stays at rest and AC is not indicated.

Option C: Hot wire voltmeter. A hot wire voltmeter works on the principle that the wire’s expansion depends on the temperature rise caused by the heating power $$P=i^2R$$. As shown above, $$\langle i^2\rangle$$ is non-zero for AC; in fact it equals $$I_{\text{rms}}^2$$. Therefore the length of the hot wire (and thus the pointer reading) depends on the rms value of the alternating voltage, giving a correct measurement regardless of current direction. It can therefore measure AC as well as DC.

Option D: Potentiometer. A potentiometer is a null-balance device that compares an unknown emf to a known steady DC source. The balance condition requires a constant potential difference; if an AC source is connected, the potential difference reverses polarity, so balancing is impossible. Hence, a conventional potentiometer cannot measure AC.

From the above examination, only the hot wire voltmeter (Option C) operates on a principle—heating effect proportional to $$i^2$$—that yields a meaningful, non-zero indication for an AC voltage.

Hence, the correct answer is Option C.

A unity power factor ($$\cos \phi = 1$$) occurs when the circuit is in resonance.

$$X_L = X_{C_{eq}}$$

$$\omega L = \frac{1}{\omega C_{eq}}$$

$$C_{eq} = \frac{1}{\omega^2 L}$$

It is given that the current leads the applied voltage. This indicates that the circuit is predominantly capacitive.

The capacitive reactance ($$X_C$$) is greater than the inductive reactance ($$X_L$$) 

$$X_C > X_L$$ 

$$\frac{1}{\omega C} > \omega L \implies 1 > \omega^2 LC$$

Since the initial capacitance $$C$$ satisfies $$C < \frac{1}{\omega^2 L}$$, we need to increase the total capacitance to reach the resonance value $$C_{eq}$$.

To increase total capacitance, the additional capacitor $$C'$$ must be connected in parallel with $$C$$.

$$C_{eq} = C + C'$$

$$C + C' = \frac{1}{\omega^2 L}$$

$$C' = \frac{1}{\omega^2 L} - C$$

$$C' = \frac{1 - \omega^2 LC}{\omega^2 L}$$

$$V = \sqrt{V_R^2 + (V_C - V_L)^2}$$

The given ratio of the RMS voltages is $$V_L : V_C : V_R = 1 : 2 : 3$$.

$$V_L = \frac{1}{3}V_R$$ and $$V_C = \frac{2}{3}V_R$$

$$V = \sqrt{V_R^2 + \left(\frac{2}{3}V_R - \frac{1}{3}V_R\right)^2}$$

$$V = \sqrt{V_R^2 + \frac{1}{9}V_R^2} = \sqrt{\frac{10}{9}V_R^2} = \frac{\sqrt{10}}{3}V_R$$

$$100 = \frac{\sqrt{10}}{3}V_R$$

$$V_R = 30 \times 3.162 \approx 94.86\text{ V}$$ (close to 95 V)

In a series LCR circuit, resonance occurs when the inductive reactance $$X_L$$ equals the capacitive reactance $$X_C$$. At this frequency, the circuit behaves purely resistively. Let's evaluate each option step by step.

Option A states that the current in the circuit is in phase with the applied voltage. At resonance, the phase angle $$\phi$$ is given by $$\phi = \tan^{-1}\left(\frac{X_L - X_C}{R}\right)$$. Since $$X_L = X_C$$, $$\phi = 0$$, meaning the current and voltage are in phase. Thus, option A is correct.

Option B states that inductive and capacitive reactances are equal. This is the defining condition for resonance in a series LCR circuit, so $$X_L = X_C$$. Therefore, option B is correct.

Option C states that if resistance $$R$$ is reduced, the voltage across the capacitor will increase. At resonance, the impedance $$Z = R$$ (minimum), so the current amplitude $$I_0 = \frac{V_0}{R}$$, where $$V_0$$ is the peak applied voltage. The voltage across the capacitor is $$V_C = I \cdot X_C$$. Substituting, $$V_C = \left(\frac{V_0}{R}\right) \cdot X_C$$. Since $$X_C = \frac{1}{\omega C}$$ remains constant at the same resonant frequency, $$V_C$$ is inversely proportional to $$R$$. If $$R$$ decreases, $$V_C$$ increases. Thus, option C is correct.

Option D states that the impedance of the circuit is maximum. The impedance in a series LCR circuit is $$Z = \sqrt{R^2 + (X_L - X_C)^2}$$. At resonance, $$X_L = X_C$$, so $$Z = \sqrt{R^2 + 0} = R$$. This is the minimum possible impedance, not the maximum. Therefore, option D is not correct.

Hence, the correct answer is Option D.

First, we have a series LR circuit connected to an AC source of frequency ω. The inductive reactance $$X_L$$ is given as 2R, where R is the resistance. We need to find the power factor for this original circuit.

The power factor is defined as $$\cos \phi$$, where $$\phi$$ is the phase angle between the voltage and current. For a series LR circuit, the impedance $$Z_1$$ is calculated as:

$$ Z_1 = \sqrt{R^2 + X_L^2} $$

Substituting $$X_L = 2R$$:

$$ Z_1 = \sqrt{R^2 + (2R)^2} = \sqrt{R^2 + 4R^2} = \sqrt{5R^2} = R\sqrt{5} $$

The power factor for the original circuit is:

$$ \cos \phi_1 = \frac{R}{Z_1} = \frac{R}{R\sqrt{5}} = \frac{1}{\sqrt{5}} $$

Now, a capacitor is added in series with capacitive reactance $$X_C = R$$. The new circuit is an LCR series circuit with resistance R, inductive reactance $$X_L = 2R$$, and capacitive reactance $$X_C = R$$.

The net reactance $$X$$ is:

$$ X = X_L - X_C = 2R - R = R $$

The impedance $$Z_2$$ for the new circuit is:

$$ Z_2 = \sqrt{R^2 + X^2} = \sqrt{R^2 + (R)^2} = \sqrt{R^2 + R^2} = \sqrt{2R^2} = R\sqrt{2} $$

The power factor for the new circuit is:

$$ \cos \phi_2 = \frac{R}{Z_2} = \frac{R}{R\sqrt{2}} = \frac{1}{\sqrt{2}} $$

We need the ratio of the new power factor to the old power factor:

$$ \text{Ratio} = \frac{\cos \phi_2}{\cos \phi_1} = \frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{5}}} = \frac{1}{\sqrt{2}} \times \sqrt{5} = \frac{\sqrt{5}}{\sqrt{2}} = \sqrt{\frac{5}{2}} $$

Comparing with the options, $$\sqrt{\frac{5}{2}}$$ corresponds to option D.

Hence, the correct answer is Option D.

$$f = \frac{1}{2\pi\sqrt{LC}}$$ (Resonant Frequency)

$$f = \frac{1}{2 \times \pi \times \sqrt{24 \times 2 \times 10^{-6}}}$$

$$f = \frac{1}{2\pi \times \sqrt{48} \times 10^{-3}}$$

$$f \approx \frac{1000}{2 \times 3.14159 \times 6.928}$$

$$f \approx \frac{1000}{43.53}$$

$$f \approx 22.97 \text{ Hz}$$

The lower cutoff frequency ($$f_1$$) is halfway between $$3.5$$ and $$4.0$$, which is $$3.75\text{ kHz}$$.

The upper cutoff frequency ($$f_2$$) is halfway between $$6.0$$ and $$6.5$$, which is $$6.25\text{ kHz}$$

$$\Delta f = f_2 - f_1 = 6.25\text{ kHz} - 3.75\text{ kHz} = 2.5\text{ kHz}$$

The maximum power occurs exactly at $$f_0 = 5\text{ kHz}$$

$$Q = \frac{f_0}{\Delta f}$$

$$Q = \frac{5\text{ kHz}}{2.5\text{ kHz}} = 2.0$$

In the given series L-C-R circuit, the capacitance $$ C = 10^{-11} $$ Farad, inductance $$ L = 10^{-5} $$ Henry, and resistance $$ R = 100 $$ Ohm. Initially, a constant DC voltage $$ E $$ is applied, and the capacitor acquires a charge of $$ 10^{-9} $$ C. For a capacitor, the charge $$ Q $$ is related to the voltage $$ E $$ by the formula $$ Q = C \cdot E $$. Substituting the known values:

$$ 10^{-9} = (10^{-11}) \cdot E $$

Solving for $$ E $$:

$$ E = \frac{10^{-9}}{10^{-11}} = 10^{-9} \cdot 10^{11} = 10^{2} = 100 \, \text{Volts} $$

Thus, the DC voltage $$ E $$ is 100 Volts.

The DC source is replaced by a sinusoidal voltage source with peak voltage $$ E_0 $$ equal to the DC voltage $$ E $$, so $$ E_0 = 100 $$ V. We need to find the peak charge on the capacitor at resonance.

At resonance in a series L-C-R circuit, the impedance is minimum and purely resistive, equal to $$ R $$. The peak current $$ I_0 $$ is given by:

$$ I_0 = \frac{E_0}{R} = \frac{100}{100} = 1 \, \text{A} \, (\text{peak}) $$

At resonance, the capacitive reactance $$ X_C $$ is calculated using the resonant frequency $$ \omega_0 = \frac{1}{\sqrt{LC}} $$. The capacitive reactance is:

$$ X_C = \frac{1}{\omega_0 C} = \frac{1}{\left( \frac{1}{\sqrt{LC}} \right) C} = \frac{\sqrt{LC}}{C} = \sqrt{\frac{L}{C}} $$

Substituting the given values:

$$ \frac{L}{C} = \frac{10^{-5}}{10^{-11}} = 10^{-5} \cdot 10^{11} = 10^{6} $$

$$ \sqrt{\frac{L}{C}} = \sqrt{10^{6}} = 10^{3} \, \Omega $$

So, $$ X_C = 1000 \, \Omega $$.

The peak voltage across the capacitor $$ V_{C0} $$ is:

$$ V_{C0} = I_0 \cdot X_C = (1) \cdot (1000) = 1000 \, \text{V} \, (\text{peak}) $$

The peak charge $$ Q_0 $$ on the capacitor is given by $$ Q_0 = C \cdot V_{C0} $$:

$$ Q_0 = (10^{-11}) \cdot (1000) = 10^{-11} \cdot 10^{3} = 10^{-8} \, \text{C} $$

Therefore, at resonance, the peak value of the charge acquired by the capacitor is $$ 10^{-8} $$ C.

Comparing with the options:

A. $$ 10^{-15} $$ C

B. $$ 10^{-6} $$ C

C. $$ 10^{-10} $$ C

D. $$ 10^{-8} $$ C

Hence, the correct answer is Option D.

The resultant source voltage $$E$$ is the magnitude of the phasor difference between $$V_L$$ and $$V_C$$: $$E = |V_L - V_C|$$

$$E = |300 - 400|$$

$$E = 100\text{ V}$$

Using AC voltage phasor relationship: $$V_{\text{rms}} = \sqrt{V_R^2 + (V_L - V_C)^2}$$

Identifying condition from $$V_L = V_C$$: $$V_L - V_C = 0 \implies \text{Circuit is at resonance}$$

$$220 = \sqrt{V_R^2 + (300 - 300)^2} \implies V_R = 220\text{ V}$$

$$I_{\text{rms}} = \frac{V_R}{R} = \frac{220}{100} = 2.2\text{ A}$$

Given:

$$R=200\,\Omega,\qquad V=220\,V,\qquad f=50\,Hz$$

Let the applied voltage be

$$e=E_0\sin\omega t$$

For an RL circuit,

$$i=\frac{E_0}{\sqrt{R^2+X_L^2}}\sin(\omega t-\phi)$$

and

$$\tan\phi=\frac{X_L}{R}$$

Since the current lags by $$30^\circ$$,

$$X_L=R\tan30^\circ$$

Similarly, for the RC circuit,

$$i=\frac{E_0}{\sqrt{R^2+X_C^2}}\sin(\omega t+\phi)$$

Since the current leads by $$30^\circ$$,

$$X_C=R\tan30^\circ$$

Therefore,

$$X_L=X_C$$

So, in the original LCR circuit, the two reactances cancel:

$$Z=\sqrt{R^2+(X_L-X_C)^2}=R$$

Hence,

$$I_0=\frac{E_0}{\sqrt{R^2+(X_L-X_C)^2}}=\frac{E_0}{R}$$

Since $$V=\frac{E_0}{\sqrt2}$$,

$$I=\frac{V}{R}=\frac{220}{200}=1.1\,A$$

The circuit is effectively purely resistive, so

$$P=I^2R$$

$$P=(1.1)^2(200)=242\,W$$

Hence, the answer is $$242\,W$$.

$$E = \frac{1}{2} L I_{\text{rms}}^2$$ $$\implies$$ $$16 = \frac{1}{2} L (2)^2$$

$$L = 8\text{ H}$$

$$X_L = 2\pi f L$$ $$\implies X_L = 2 \times 3.14 \times 50 \times 8$$ $$\implies X_L = 2512\ \Omega$$

$$P = I_{\text{rms}}^2 R$$

$$32 = (2)^2 R$$ $$\implies R = 8\ \Omega$$

$$\text{Ratio} = \frac{X_L}{R}$$ $$\implies \text{Ratio} = \frac{2512}{8}$$

$$\text{Ratio} = 314$$