Chapter 7: Alternating Current

Given: Power rating $$P = 100 \, \mathrm{W}$$, supply voltage (rms) $$V = 220 \, \mathrm{V}$$.

(a) Resistance of the bulb.

The bulb behaves as a pure resistor. The average power dissipated in a resistor on an ac supply is $$P = \dfrac{V^2}{R}$$, where $$V$$ is the rms voltage. Solving for $$R$$:

$$R = \dfrac{V^2}{P} = \dfrac{(220)^2}{100} = \dfrac{48400}{100} = 484 \, \Omega$$

(b) Peak voltage of the source.

The rms and peak values of a sinusoidal voltage are related by $$V = \dfrac{V_0}{\sqrt{2}}$$, so

$$V_0 = \sqrt{2}\,V = 1.414 \times 220 = 311 \, \mathrm{V}$$

(c) rms current through the bulb.

$$I = \dfrac{V}{R} = \dfrac{220}{484} = 0.455 \, \mathrm{A}$$

(Equivalently, $$I = \dfrac{P}{V} = \dfrac{100}{220} = 0.455 \, \mathrm{A}$$.)

Given: Inductance $$L = 25.0 \, \mathrm{mH} = 25.0 \times 10^{-3} \, \mathrm{H}$$, rms voltage $$V = 220 \, \mathrm{V}$$, frequency $$f = 50 \, \mathrm{Hz}$$.

Inductive reactance. The opposition offered by a pure inductor to ac is its inductive reactance:

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

$$X_L = 2 \times 3.14 \times 50 \times 25.0 \times 10^{-3}$$

$$X_L = 2 \times 3.14 \times 50 \times 0.025 = 7.85 \, \Omega$$

rms current. For a pure inductor the rms current is

$$I = \dfrac{V}{X_L} = \dfrac{220}{7.85} = 28.0 \, \mathrm{A}$$

DC connection. When the lamp–capacitor series combination is connected to a dc source, the capacitor charges up to the source voltage. Once it is fully charged, no steady current can flow through it — a capacitor blocks dc. Hence in the steady state the lamp does not glow.

AC connection. With an ac source, the voltage reverses every half cycle, so the capacitor is continually charged and discharged. An alternating current therefore flows in the circuit at all times, and the lamp glows.

Effect of reducing the capacitance.

DC case: The capacitor still blocks dc, so there is no change — the lamp continues not to glow.

AC case: The capacitive reactance is $$X_C = \dfrac{1}{\omega C}$$. Reducing $$C$$ increases $$X_C$$, so the current $$I = \dfrac{V}{X_C}$$ decreases. The lamp therefore glows less brightly (dimmer).

Given: $$C = 15.0 \, \mu\mathrm{F} = 15.0 \times 10^{-6} \, \mathrm{F}$$, $$V = 220 \, \mathrm{V}$$, $$f = 50 \, \mathrm{Hz}$$.

Capacitive reactance.

$$X_C = \dfrac{1}{\omega C} = \dfrac{1}{2\pi f C}$$

$$X_C = \dfrac{1}{2 \times 3.14 \times 50 \times 15.0 \times 10^{-6}} = \dfrac{1}{4.712 \times 10^{-3}} = 212.3 \, \Omega$$

rms current.

$$I = \dfrac{V}{X_C} = \dfrac{220}{212.3} = 1.04 \, \mathrm{A}$$

Peak current.

$$I_0 = \sqrt{2}\,I = 1.414 \times 1.04 = 1.47 \, \mathrm{A}$$

If the frequency is doubled. Since $$X_C = \dfrac{1}{2\pi f C}$$ is inversely proportional to $$f$$, doubling the frequency halves the capacitive reactance, $$X_C \to 106.2 \, \Omega$$. Because $$I = V/X_C$$, the current then doubles (rms current $$\approx 2.08 \, \mathrm{A}$$, peak current $$\approx 2.93 \, \mathrm{A}$$).

The light bulb (which behaves as a resistor $$R$$) and the open coil inductor are in series across the ac source. The current in the circuit is

$$I = \dfrac{V}{Z}, \qquad Z = \sqrt{R^2 + X_L^2}, \qquad X_L = \omega L$$

An open coil (air-core) inductor has a relatively small inductance $$L$$. When an iron rod is inserted into its interior, the iron — being ferromagnetic — greatly increases the magnetic flux linked with the coil, so the inductance $$L$$ increases substantially.

A larger $$L$$ gives a larger inductive reactance $$X_L = \omega L$$, and hence a larger impedance $$Z$$. Since $$I = V/Z$$, the current in the circuit decreases.

The power delivered to the bulb is $$P = I^2 R$$; with a smaller current the bulb receives less power and its glow falls.

Therefore the glow of the bulb decreases — option (b).

Given data

• Resistance: $$R = 200\,\Omega$$
• Capacitance: $$C = 15.0\,\mu\mathrm{F}=15.0\times10^{-6}\,\mathrm{F}$$
• Supply voltage (rms): $$V = 220\,\mathrm{V}$$
• Frequency: $$f = 50\,\mathrm{Hz}$$

Angular frequency

$$\omega = 2\pi f = 2\pi(50) = 100\pi \;\mathrm{rad\,s^{-1}} \approx 3.14\times10^{2}\,\mathrm{rad\,s^{-1}}$$

Capacitive reactance

$$X_{C}=\frac{1}{\omega C}=\frac{1}{(3.14\times10^{2})(15.0\times10^{-6})} =2.12\times10^{2}\,\Omega \;(\text{about } 212\,\Omega)$$

Impedance of the series RC circuit

Magnitude: $$Z = \sqrt{R^{2}+X_{C}^{2}} =\sqrt{(200)^{2}+(212)^{2}} =\sqrt{40000+44944} =\sqrt{84944} \approx 2.92\times10^{2}\,\Omega\;(291.5\,\Omega)$$

(a) rms current

$$I = \frac{V}{Z}=\frac{220\,\mathrm{V}}{291.5\,\Omega}=0.755\,\mathrm{A}$$

(b) rms voltages across the elements

• Across the resistor: $$V_{R}=IR=(0.755\,\mathrm{A})(200\,\Omega)=1.51\times10^{2}\,\mathrm{V} \;(151\,\mathrm{V})$$
• Across the capacitor: $$V_{C}=IX_{C}=(0.755\,\mathrm{A})(212\,\Omega)=1.60\times10^{2}\,\mathrm{V} \;(160\,\mathrm{V})$$

The simple algebraic sum of the magnitudes is

$$V_{R}+V_{C}=151\,\mathrm{V}+160\,\mathrm{V}=311\,\mathrm{V}>V_{\text{source}}$$

Why is it larger than 220 V?

The two component voltages are not in phase. In a series RC circuit the resistor voltage $$V_{R}$$ is in phase with the current, while the capacitor voltage $$V_{C}$$ lags the current by $$90^{\circ}$$. Therefore they must be added vectorially (phasor addition), not algebraically. The right-angled triangle formed by $$V_{R}$$ (along the current) and $$V_{C}$$ (lagging by $$90^{\circ}$$) has a hypotenuse

$$V=\sqrt{V_{R}^{2}+V_{C}^{2}}\approx\sqrt{(151)^{2}+(160)^{2}}\;\mathrm{V}\approx220\,\mathrm{V},$$

which exactly equals the applied rms voltage. Thus there is no contradiction.

Background formulae for an a.c. line

• Instantaneous voltage and current: $$v = V_0 \sin \omega t, \;\; i = I_0 \sin (\omega t - \phi)$$
• r.m.s. values: $$V = V_0/\sqrt 2, \; I = I_0/\sqrt 2$$
• Average (real) power delivered: $$P = VI \cos \phi$$. Here $$\cos\phi$$ is called the power factor.

When electricity is sent from a generating station (source) to a consumer (load), the two relevant quantities are

• the load current $$I$$ which the station must supply so that the consumer receives the desired real power $$P$$,
• the line (ohmic) loss in the transmission resistance $$R$$: $$P_{\text{loss}} = I^{2} R$$.

(a) Why a low power factor implies large power loss

Suppose the consumer needs a fixed real power $$P$$ and the r.m.s. transmission voltage is fixed at $$V$$ (this is normally the case for a given line rating).

From the power relation $$P = VI \cos \phi$$, the required line current is

$$I = \dfrac{P}{V \cos \phi}.$$

This current has to pass through the (unavoidable) resistance $$R$$ of the transmission line. The ohmic loss is therefore

$$P_{\text{loss}} = I^{2} R = \left( \dfrac{P}{V \cos \phi} \right)^{\!2} R = \dfrac{P^{2} R}{V^{2} \cos^{2} \phi}.$$

Hence, for fixed $$P,\,V,\,R$$,

$$P_{\text{loss}} \propto \dfrac{1}{\cos^{2} \phi}.$$

If the power factor is low (i.e. $$\cos \phi \ll 1$$), the denominator $$\cos^{2}\phi$$ becomes small, the required current $$I$$ becomes large, and the resistive loss $$P_{\text{loss}}$$ increases quadratically. Therefore, a low power factor leads directly to a large power loss during transmission.

(b) How a capacitor improves the power factor

Most industrial and commercial loads (induction motors, transformers, fluorescent lamps, etc.) are dominantly inductive. An inductive reactance $$X_L = \omega L$$ makes the current lag behind the voltage by an angle $$\phi_L$$, so the power factor $$\cos \phi_L$$ is less than unity.

Attach a capacitor of reactance $$X_C = \dfrac{1}{\omega C}$$ either in series with the load (for simple single-device circuits) or, more usually, in parallel (shunt) across the load (power-factor correction banks used in distribution networks):

• The capacitor current leads the voltage by $$90^{\circ}$$, exactly opposite to the lag introduced by the inductor.
• Because the two reactive currents are out of phase, they partially cancel each other.

Quantitatively, if the inductive and capacitive reactances appear in series with a resistance $$R$$, the net reactance becomes

$$X = X_L - X_C = \omega L - \dfrac{1}{\omega C}.$$

By choosing $$C$$ so that $$X_C \approx X_L$$, we can make $$X \to 0$$. The impedance $$Z$$ then approaches the pure resistance $$R$$:

$$Z = \sqrt{R^{2}+X^{2}} \longrightarrow R, \qquad \phi = \tan^{-1}\!\left(\dfrac{X}{R}\right) \longrightarrow 0.$$

Therefore $$\cos \phi \longrightarrow 1$$ and the power factor is said to be improved (brought as close to unity as economically feasible). With a higher power factor:

• The load draws a smaller current for the same real power.
• Transmission losses $$P_{\text{loss}} = I^{2}R$$ fall accordingly.

Capacitor banks are inexpensive, have negligible energy dissipation, and supply only the necessary leading reactive power. Hence they are the standard devices for power-factor correction in a.c. power systems.

Given data
Peak (maximum) source voltage: $$V_{m}=283\;\text{V}$$   Frequency: $$f=50\;\text{Hz}$$
Resistance: $$R=3\;\Omega$$   Inductance: $$L = 25.48\,\text{mH}=0.02548\,\text{H}$$
Capacitance: $$C = 796\,\mu\text{F}=7.96\times10^{-4}\,\text{F}$$

The root-mean-square (r.m.s.) source voltage is
$$V_{\text{rms}}=\frac{V_{m}}{\sqrt 2}=\frac{283}{\sqrt 2}\;\text{V}\approx 200\;\text{V}$$

The angular frequency is
$$\omega = 2\pi f = 2\pi(50) = 100\pi\;\text{rad s}^{-1}\approx 314.16\;\text{rad s}^{-1}$$

Reactances
Inductive reactance:
$$X_{L}=\omega L = 314.16\times0.02548\;\Omega \approx 8.00\;\Omega$$
Capacitive reactance:
$$X_{C}=\frac{1}{\omega C}=\frac{1}{314.16\times7.96\times10^{-4}}\;\Omega\approx 4.00\;\Omega$$

(a) Impedance
The net reactance: $$X = X_{L}-X_{C} = 8.00-4.00 = 4.00\;\Omega$$
$$Z = \sqrt{R^{2}+X^{2}} = \sqrt{3^{2}+4^{2}}\;\Omega \approx 5.0\;\Omega$$

(b) Phase difference
$$\tan\phi = \frac{X}{R}=\frac{4.00}{3}=1.33$$
$$\phi \approx \arctan(1.33) \approx 53^{\circ}$$
(because $$X_{L}>X_{C}$$, the current lags the source voltage by this angle).

(c) Power dissipated
r.m.s. current:
$$I_{\text{rms}} = \frac{V_{\text{rms}}}{Z}=\frac{200}{5.0}\;\text{A}\approx 40\;\text{A}$$
Power in a series $$LCR$$ circuit is lost only in $$R$$:
$$P = I_{\text{rms}}^{2}R = (40\,\text{A})^{2}(3\,\Omega) \approx 4.8\times10^{3}\,\text{W} = 4.8\,\text{kW}$$

(d) Power factor
$$\cos\phi = \frac{R}{Z}=\frac{3}{5.0}\approx 0.60$$ (lagging).

Data taken from the previous example
Series $$RLC$$ circuit with the values

• Resistance  $$R = 400\,\Omega$$
• Inductance  $$L = 0.30\,\text{H}$$
• Capacitance  $$C = 40\,\mu\text{F}=40\times10^{-6}\,\text{F}$$
• Applied (r.m.s.) voltage  $$V_{\text{rms}} = 220\,\text{V}$$

(a) Resonant frequency
At resonance the inductive and capacitive reactances cancel: $$\omega_0L = \dfrac1{\omega_0C} \;\Rightarrow\; \omega_0 = \dfrac1{\sqrt{LC}}$$
Numerically, $$LC = 0.30\times40\times10^{-6}=1.2\times10^{-5}\;\text{H F}$$
$$\sqrt{LC}=\sqrt{1.2\times10^{-5}} = 3.464\times10^{-3}$$
$$\omega_0 = \dfrac{1}{3.464\times10^{-3}} = 2.887\times10^{2}\,\text{rad\,s}^{-1}$$
$$f_0 = \dfrac{\omega_0}{2\pi}=\dfrac{2.887\times10^{2}}{2\pi}=45.9\,\text{Hz}\;(\text{≈ }46\,\text{Hz})$$

(b) Impedance, current and power at resonance
Because the net reactance is zero, the impedance is purely resistive:
$$Z_0 = R = 400\,\Omega$$
The r.m.s. current is therefore
$$I_0 = \dfrac{V_{\text{rms}}}{Z_0}=\dfrac{220}{400}=0.55\,\text{A}$$
The phase angle is zero, so the power factor is unity and the average power dissipated is
$$P_0 = V_{\text{rms}} I_{\text{rms}}\cos\phi = 220\times0.55\times1 = 1.21\times10^{2}\,\text{W}\;(=121\,\text{W})$$

Result
(a) $$f_0\approx46\,\text{Hz}$$
(b) At resonance  $$Z = 400\,\Omega,\; I = 0.55\,\text{A},\; P = 121\,\text{W}.$$

Goal. Explain the physical principle behind the walk-through metal detector used at airports.

1  Construction of the detector

• The doorway houses an induction coil that forms the inductive element $$L$$ of a tuned RLC oscillator.
• A small capacitor $$C$$ is connected so that the coil and the capacitor constitute a tank circuit having natural (angular) resonant frequency
  $$\omega_0 = \tfrac{1}{\sqrt{LC}}.$$
• The circuit is kept in steady oscillation by a low-power transistor amplifier; as long as the electrical parameters remain unchanged, the amplitude of the alternating magnetic field is constant.

2  Basic physics involved

1. Faraday’s law of electromagnetic induction:
   Whenever the magnetic flux $$\Phi_B$$ linked with a conducting loop changes, an emf
   $$\mathcal{E} = -\dfrac{d\Phi_B}{dt}$$ gets induced.
2. Eddy currents: In a bulk metallic piece (not a single wire) these induced emf’s circulate as closed loops of current inside the body; they are called eddy currents.
3. Lenz’s law: The eddy currents set up their own magnetic field that opposes the change responsible for them. Hence they always try to weaken the original alternating field produced by the coil.

3  What happens when a person carrying metal walks in

• The metal article is suddenly exposed to the coil’s alternating magnetic field; the flux through it becomes time-varying.
• By Faraday’s law, appreciable eddy currents are induced in the object. These currents behave, from the coil’s point of view, as an additional load.
• Electrically this load can be represented by an extra resistance $$R_e$$ and a small change in effective inductance $$\Delta L$$ coupled to the original circuit.

4  Effect on the tuned circuit

Before the metal arrives, the impedance of the series RLC is

$$Z_0 = \sqrt{R^2 + (\omega L - \tfrac{1}{\omega C})^2}.$$

At resonance $$\omega = \omega_0$$, so $$Z_0 = R$$ and the oscillating current reaches a large, constant amplitude.

• The appearance of $$R_e$$ raises the total resistance to $$R' = R + R_e$$.
• The small change $$\Delta L$$ shifts the resonant frequency to $$\omega' \approx \omega_0\left(1 - \tfrac{\Delta L}{2L}\right).$$
• Consequently the circuit is instantly driven off resonance and its current amplitude falls sharply.

5  Detection and alarm

An envelope detector continuously monitors the oscillator’s amplitude. The moment the amplitude drops below a preset threshold, the electronic circuit actuates a buzzer or beeper. Thus the sound is heard only when a conducting (metallic) object, capable of supporting eddy currents, is carried through the doorway.

6  Principle stated concisely

The metal detector operates on the principle of electromagnetic induction (Faraday’s law) and the eddy currents it produces in nearby conductors. The eddy currents load and detune a resonant AC circuit; the resulting change in amplitude/frequency is sensed and converted into an audible signal.

Given data

• Inductance: $$L = 44\,\text{mH} = 44 \times 10^{-3}\,\text{H}$$
• Supply voltage (rms): $$V_{\text{rms}} = 220\,\text{V}$$
• Frequency: $$f = 50\,\text{Hz}$$

Step 1: Calculate the inductive reactance

The reactance of a pure inductor is $$X_L = 2\pi f L$$.

Substituting the given values,

$$\begin{aligned}X_L &= 2\pi \times 50\,\text{Hz} \times 44 \times 10^{-3}\,\text{H}\\[2pt] &= 2\pi \times 50 \times 0.044\\[2pt] &= 6.283 \times 50 \times 0.044\\[2pt] &= 13.4\,\Omega\;(\text{approximately}).\end{aligned}$$

Step 2: Find the rms current

For a purely inductive circuit the rms current is given by $$I_{\text{rms}} = \dfrac{V_{\text{rms}}}{X_L}$$.

Hence,

$$I_{\text{rms}} = \dfrac{220\,\text{V}}{13.4\,\Omega} \approx 16\,\text{A}.$$

Result

The rms value of the current in the circuit is about $$16\,\text{A}$$.

The circuit contains only a capacitor, therefore the opposition to the alternating current is the capacitive reactance $$X_C$$.

Given values:
$$C = 60\,\mu\mathrm{F} = 60 \times 10^{-6}\;\mathrm{F}$$
$$f = 60\;\mathrm{Hz}$$
$$V_{\mathrm{rms}} = 110\;\mathrm{V}$$

1. Calculate the capacitive reactance:

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

Substituting the numbers,
$$X_C = \frac{1}{2\pi \,(60)\,(60 \times 10^{-6})}$$

First evaluate the denominator step by step:
$$2\pi = 6.2832$$
$$6.2832 \times 60 = 376.992$$
$$376.992 \times 60 \times 10^{-6} = 376.992 \times 0.000060 = 0.02261952$$

Hence,
$$X_C = \frac{1}{0.02261952}\;\Omega \approx 44.2\;\Omega$$

2. The r.m.s. current in a purely capacitive circuit is given by

$$I_{\mathrm{rms}} = \frac{V_{\mathrm{rms}}}{X_C}$$

Therefore,
$$I_{\mathrm{rms}} = \frac{110\;\mathrm{V}}{44.2\;\Omega} \approx 2.49\;\mathrm{A}$$

Rounding to two significant figures,

$$I_{\mathrm{rms}} \approx 2.5\;\mathrm{A}$$

Background (Exercises 7.3 and 7.4)

• Ex. 7.3 : a pure inductor $$L = 25\,\mathrm{mH}$$ is connected to a $$V_{\mathrm{rms}} = 220\,\mathrm{V}$$, $$f = 50\,\mathrm{Hz}$$ source.
• Ex. 7.4 : a pure capacitor $$C = 50\,\mu\mathrm{F}$$ is connected to a $$V_{\mathrm{rms}} = 200\,\mathrm{V}$$, $$f = 50\,\mathrm{Hz}$$ source.

For an a.c. circuit the average (net) power over one complete cycle is

$$P_{\text{avg}} = V_{\mathrm{rms}} I_{\mathrm{rms}} \cos\varphi,$$

where $$\varphi$$ is the phase difference between the voltage and the current.

1. Pure inductor (Exercise 7.3)

The current lags the voltage by $$90^\circ$$: $$\varphi = 90^\circ$$.

Hence $$\cos\varphi = \cos 90^\circ = 0 \;\Rightarrow\; P_{\text{avg}} = 0.$$

(The source alternately stores energy in the magnetic field of the inductor and then gets it back; the net energy supplied in one cycle is therefore zero.)

2. Pure capacitor (Exercise 7.4)

The current leads the voltage by $$90^\circ$$: $$\varphi = -90^\circ$$ (or $$+90^\circ$$ in magnitude).

Again $$\cos\varphi = \cos 90^\circ = 0 \;\Rightarrow\; P_{\text{avg}} = 0.$$

(Here energy is cyclically stored in and returned from the electric field of the capacitor, so no net energy is consumed.)

Conclusion

For both circuits the average power absorbed in a complete cycle is zero because the phase difference between current and voltage is $$90^\circ$$, making the power factor $$\cos\varphi$$ equal to zero.

The circuit forms an ideal, loss‑free L C oscillator.

Convert the given components to SI units:
$$C = 30\,\mu\mathrm{F} = 30 \times 10^{-6}\,\mathrm{F}$$
$$L = 27\,\mathrm{mH} = 27 \times 10^{-3}\,\mathrm{H}$$

For free oscillations the angular frequency is

$$\omega = \dfrac{1}{\sqrt{L\,C}}$$

First evaluate the product $$L C$$:

$$L C = \left(27 \times 10^{-3}\right)\left(30 \times 10^{-6}\right)\,\mathrm{H\,F}$$ $$= 810 \times 10^{-9}\,\mathrm{s^{2}}$$ $$= 8.10 \times 10^{-7}\,\mathrm{s^{2}}$$

Take the square root:

$$\sqrt{L C} = \sqrt{8.10 \times 10^{-7}}$$ $$= \sqrt{8.10}\,\sqrt{10^{-7}}$$ $$\approx 2.85 \times 3.16 \times 10^{-4}\,\mathrm{s}$$ $$\approx 9.0 \times 10^{-4}\,\mathrm{s}$$

Finally,

$$\omega = \dfrac{1}{9.0 \times 10^{-4}}\,\mathrm{rad\,s^{-1}} \approx 1.1 \times 10^{3}\,\mathrm{rad\,s^{-1}}$$

Given data

• Resistance: $$R = 20\,\Omega$$
• Inductance: $$L = 1.5\,\mathrm{H}$$
• Capacitance: $$C = 35\,\mu\mathrm{F} = 35 \times 10^{-6}\,\mathrm{F}$$
• Applied ac (r.m.s.) voltage: $$V = 200\,\mathrm{V}$$

The circuit is operated at its natural (resonant) frequency.

Step 1 : Impedance at resonance

For a series $$LCR$$ circuit the impedance is $$Z = \sqrt{R^{2} + \left(\omega L - \dfrac{1}{\omega C}\right)^2}$$.

At resonance $$\omega = \omega_{0} = \dfrac{1}{\sqrt{LC}}$$, so $$\omega_{0}L = \dfrac{1}{\omega_{0}C}$$ and therefore $$Z = R$$ (purely resistive).

Step 2 : r.m.s. current in the circuit

Because $$Z = R$$, the r.m.s. current amplitude is

$$I = \dfrac{V}{R} = \dfrac{200\,\mathrm{V}}{20\,\Omega} = 10\,\mathrm{A}$$.

Step 3 : Average power over one complete cycle

The average power absorbed by an a.c. circuit is

$$P_{\text{avg}} = V I \cos \phi,$$

where $$\phi$$ is the phase angle between current and voltage. At resonance $$\phi = 0^{\circ}$$, hence $$\cos \phi = 1$$.

Thus,

$$P_{\text{avg}} = V I = (200\,\mathrm{V})(10\,\mathrm{A}) = 2.0 \times 10^{3}\,\mathrm{W}.$$

Alternative form :

$$P_{\text{avg}} = \dfrac{V^{2}}{R} = \dfrac{(200\,\mathrm{V})^{2}}{20\,\Omega} = 2.0 \times 10^{3}\,\mathrm{W},$$ identical to the value obtained above.

Hence, the average power delivered to the circuit in one complete cycle is $$2.0\,\text{kW}$$.