JEE EMF & Circuit Analysis Questions

To solve this problem, we need to evaluate two statements about non-ideal batteries connected in parallel. A non-ideal battery has an internal resistance, so we must consider both the electromotive force (emf) and the internal resistance for each battery.

Consider two batteries:

• Battery 1: emf = $$E_1$$, internal resistance = $$r_1$$
• Battery 2: emf = $$E_2$$, internal resistance = $$r_2$$

When connected in parallel, the equivalent emf $$E_{eq}$$ and equivalent internal resistance $$r_{eq}$$ are given by the formulas:

$$E_{eq} = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2}$$

$$r_{eq} = \frac{r_1 r_2}{r_1 + r_2}$$

We will analyze each statement separately.

Statement-II Analysis: The equivalent internal resistance of two nonideal batteries connected in parallel is smaller than the internal resistance of either of the two batteries.

Using the formula for equivalent internal resistance:

$$r_{eq} = \frac{r_1 r_2}{r_1 + r_2}$$

Since $$r_1 > 0$$ and $$r_2 > 0$$, we can compare $$r_{eq}$$ with $$r_1$$:

$$\frac{r_{eq}}{r_1} = \frac{\frac{r_1 r_2}{r_1 + r_2}}{r_1} = \frac{r_2}{r_1 + r_2} < 1$$

because $$r_2 < r_1 + r_2$$. Therefore, $$r_{eq} < r_1$$.

Similarly, comparing $$r_{eq}$$ with $$r_2$$:

$$\frac{r_{eq}}{r_2} = \frac{\frac{r_1 r_2}{r_1 + r_2}}{r_2} = \frac{r_1}{r_1 + r_2} < 1$$

so $$r_{eq} < r_2$$.

Thus, $$r_{eq}$$ is always less than both $$r_1$$ and $$r_2$$. Therefore, Statement-II is true.

Statement-I Analysis: The equivalent emf of two nonideal batteries connected in parallel is smaller than either of the two emfs.

Using the equivalent emf formula:

$$E_{eq} = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2}$$

This expression is a weighted average of $$E_1$$ and $$E_2$$:

$$E_{eq} = E_1 \cdot \frac{r_2}{r_1 + r_2} + E_2 \cdot \frac{r_1}{r_1 + r_2}$$

Since $$\frac{r_2}{r_1 + r_2} + \frac{r_1}{r_1 + r_2} = 1$$ and both fractions are positive, $$E_{eq}$$ is a convex combination of $$E_1$$ and $$E_2$$. This means $$E_{eq}$$ lies between the minimum and maximum of $$E_1$$ and $$E_2$$.

Without loss of generality, assume $$E_1 \leq E_2$$. Then:

$$E_1 \leq E_{eq} \leq E_2$$

Therefore, $$E_{eq}$$ is not smaller than both emfs; it is at least as large as the smaller emf and at most as large as the larger emf.

For example, let $$E_1 = 10 \text{ V}$$, $$r_1 = 1 \Omega$$, $$E_2 = 5 \text{ V}$$, $$r_2 = 1 \Omega$$. Then:

$$E_{eq} = \frac{10 \times 1 + 5 \times 1}{1 + 1} = \frac{15}{2} = 7.5 \text{ V}$$

Here, $$7.5 \text{ V}$$ is greater than $$5 \text{ V}$$ ($$E_2$$) and less than $$10 \text{ V}$$ ($$E_1$$). So it is not smaller than both emfs.

If the emfs are equal, say $$E_1 = E_2 = E$$, then:

$$E_{eq} = \frac{E r_2 + E r_1}{r_1 + r_2} = E \frac{r_1 + r_2}{r_1 + r_2} = E$$

which is equal to both emfs, not smaller.

Thus, Statement-I is false because the equivalent emf is not necessarily smaller than both emfs; it lies between them or equals them.

Conclusion:

• Statement-I is false.
• Statement-II is true.

Therefore, the correct option is B: Statement-I is false but Statement-II is true.

Analyze the orientation of the diode in the given circuit. The positive terminal of the 6V battery is connected to the p-side of the diode through the top resistor, and the n-side is connected to the negative terminal through the resistor AB. This means the diode is forward-biased.

Assuming it is an ideal diode, it acts as a short circuit with zero resistance.

equivalent resistance of the parallel part (between node D and the negative terminal):

$$ R_p = \frac{4 \times 4}{4 + 4} $$

$$ R_p = \frac{16}{8} $$

$$ R_p = 2 \Omega $$

$$ R_{total} = R_{CD} + R_p $$

$$ R_{total} = 4 + 2 $$

$$ R_{total} = 6 \Omega $$

So A correct and E incorrect.

Calculate the total current ($$I$$) flowing through the circuit, which is the reading of the ammeter:

$$ I = \frac{E}{R_{total}} $$

$$ I = \frac{6}{6} $$

$$ I = 1 \text{ A} $$

This makes observation B correct.

Calculate the potential difference across the resistor between C and D ($$V_{CD}$$):

$$ V_{CD} = I \times R_{CD} $$

$$ V_{CD} = 1 \times 4 $$

$$ V_{CD} = 4 \text{ V} $$

This makes observation D correct.

Calculate the potential difference across AB ($$V_{AB}$$). The diode is ideal, so there is no voltage drop across it. The voltage across the branch AB is the same as the total voltage across the parallel combination.

$$ V_{AB} = I \times R_p $$

$$ V_{AB} = 1 \times 2 $$

$$ V_{AB} = 2 \text{ V} $$

This makes observation C incorrect.

Therefore, the correct observations are A, B, and D.

Given:
R₁ = R₂ = R₃ = 5 Ω
R₄ = 10 Ω

Check option A :

Top branch: R₁ + R₂ = 5 + 5 = 10 Ω
Bottom branch: R₃ + R₄ = 5 + 10 = 15 Ω

These two branches are in parallel:

$$R_{eq}=(10\times15)/(10+15)$$
= 150 / 25
= 6 Ω 

So this exactly matches the required resistance.

The battery is specified by two numbers: voltage $$V = 4.2\text{ V}$$ and capacity $$C = 5800\text{ mAh}$$.

Step 1 Convert the capacity into ampere-hours (Ah)
Since $$1000\text{ mAh} = 1\text{ Ah}$$, we have
$$C = \frac{5800}{1000}\text{ Ah} = 5.8\text{ Ah}$$

Step 2 Convert ampere-hours to coulombs (C)
By definition, $$1\text{ A} = 1\text{ C s}^{-1}$$ and there are $$3600\text{ s}$$ in one hour. Hence
$$1\text{ Ah} = 1\text{ A}\times3600\text{ s} = 3600\text{ C}$$
Therefore,
$$Q = 5.8\text{ Ah}\times3600\text{ C Ah}^{-1} = 20880\text{ C}$$

Step 3 Calculate the stored energy
Electrical energy stored, $$E = VQ$$, so
$$E = 4.2\text{ V}\times20880\text{ C} = 87696\text{ J}$$

Step 4 Express the result in kilojoules
$$1\text{ kJ} = 10^{3}\text{ J}$$, hence
$$E = \frac{87696}{10^{3}}\text{ kJ} \approx 87.7\text{ kJ}$$

Thus, the energy stored in the fully charged mobile phone battery is about $$87.7\text{ kJ}$$.

Option C is correct.

step 1: split the potentiometer

Total Rp=1 Ω, slider at middle → each half:

top half = 0.5 Ω
bottom half = 0.5 Ω

step 2: understand connections

Let left end = A, right end = B, slider = C

Between A and C:
• direct path = 0.5 Ω
• another path = 2 Ω (external resistor)

So between A and C → parallel:

$$R_{AC}=(0.5\times2)/(0.5+2)$$

RAC=1/2.5=0.4 Ω

step 3: from C to B

Only one resistor:

RCB=0.5 Ω

step 4: total resistance

Series combination:

Req=0.4+0.5=0.9 Ω

step 5: current

Battery = 0.9 V

I=V/R=0.9/0.9=1 A

step 1: check polarity properly

The battery makes the left node at higher potential than the right node.

Now look at both diodes:

• In the top branch, diode is oriented left → right
• In the bottom branch, diode is also oriented left → right (same direction — just drawn differently)

step 2: circuit simplification

Both branches conduct, so we have:

• two 20 Ω resistors
• in parallel

step 3: equivalent resistance

$$R_{eq}=20∥20=(20\times20)/(20+20)=400/40=10Ω$$

step 4: current through battery

$$I=V/R=5/10=0.5A$$

We need to find the EMF induced in coil 1 when it carries current $$I_1$$ and a nearby coil 2 carries current $$I_2$$.

The total flux through coil 1 has two contributions:

(i) Self-flux due to its own current: $$\Phi_{\text{self}} = L_1 I_1$$

(ii) Mutual flux due to current in coil 2: $$\Phi_{\text{mutual}} = M_{12} I_2$$

The total flux linkage with coil 1:
$$\Phi_1 = L_1 I_1 + M_{12} I_2$$

By Faraday's law, the EMF induced in coil 1 is:
$$\varepsilon_1 = -\frac{d\Phi_1}{dt} = -L_1\frac{dI_1}{dt} - M_{12}\frac{dI_2}{dt}$$

The correct answer is Option (2): $$\varepsilon_1 = -L_1\frac{dI_1}{dt} - M_{12}\frac{dI_2}{dt}$$.

The circuit contains only one closed loop, so we can apply Kirchhoff’s Voltage Law (KVL) directly.
KVL statement: “The algebraic sum of the emfs in any closed loop is equal to the algebraic sum of the potential drops (IR terms) in that loop.”

Let the two ideal sources be
  • $$E_1 = 12\;\text{V}$$ (positive terminal on the left side of the loop)
  • $$E_2 = 6\;\text{V}$$ (connected so that it opposes $$E_1$$)
and let the two resistors be
  • $$R_1 = 2\;\Omega$$,  • $$R_2 = 4\;\Omega$$, both in series with the batteries.

Assume the current $$I$$ flows clockwise, starting from the positive terminal of $$E_1$$. Moving once around the loop and writing the KVL equation:

Clockwise sum of emfs = $$E_1 - E_2$$ (because $$E_2$$ is encountered from + to −, so it is subtractive).
Clockwise sum of drops = $$I (R_1 + R_2)$$.
Therefore, by KVL
$$E_1 - E_2 = I (R_1 + R_2) \quad -(1)$$

Substituting the given numerical values into $$(1)$$:

$$12\;{\text{V}} - 6\;{\text{V}} = I (2\;\Omega + 4\;\Omega)$$

$$6\;{\text{V}} = I \,(6\;\Omega)$$

$$I = \frac{6\;\text{V}}{6\;\Omega} = 1\;\text{A}$$

Hence, the net current in the circuit is $$1\;\text{A}$$, flowing clockwise (from the 12 V source towards the 6 V source).

Case 1: Cells in series

For series combination, the net emf is the algebraic sum  $$E_s = E_1 + E_2 = 1 + 2 = 3\;{\rm V}$$
The net internal resistance is the arithmetic sum  $$r_s = r_1 + r_2 = 2 + 1 = 3\;\Omega$$
Total resistance in the external circuit is  $$R_{\text{tot}} = r_s + R = 3 + 6 = 9\;\Omega$$
Hence the current in the series circuit is

$$I_1 = \frac{E_s}{R_{\text{tot}}} = \frac{3}{9} = \frac{1}{3}\;{\rm A}$$

Case 2: Cells in parallel

For two unequal cells in parallel, use the general formulae:
Equivalent internal resistance: $$\frac{1}{r_p} = \frac{1}{r_1} + \frac{1}{r_2}$$
Equivalent emf: $$E_p = \frac{\dfrac{E_1}{r_1} + \dfrac{E_2}{r_2}}{\dfrac{1}{r_1} + \dfrac{1}{r_2}}$$

Compute $$r_p$$: $$\frac{1}{r_p} = \frac{1}{2} + \frac{1}{1} = 0.5 + 1 = 1.5$$
$$r_p = \frac{1}{1.5} = \frac{2}{3}\;\Omega$$

Compute $$E_p$$: $$E_p = \frac{\dfrac{1}{2} + \dfrac{2}{1}}{1.5} = \frac{0.5 + 2}{1.5} = \frac{2.5}{1.5} = \frac{5}{3}\;{\rm V}$$

Total resistance now is  $$R_{\text{tot}} = R + r_p = 6 + \frac{2}{3} = \frac{20}{3}\;\Omega$$
Therefore the current in the parallel configuration is

$$I_2 = \frac{E_p}{R_{\text{tot}}} = \frac{\dfrac{5}{3}}{\dfrac{20}{3}} = \frac{5}{20} = \frac{1}{4}\;{\rm A}$$

Required ratio

$$\frac{I_1}{I_2} = \frac{\dfrac{1}{3}}{\dfrac{1}{4}} = \frac{1}{3} \times 4 = \frac{4}{3}$$

The problem states $$\dfrac{I_1}{I_2} = \dfrac{x}{3}$$, so $$x = 4$$.

Hence, the value of $$x$$ is 4.

Given:
ammeter resistance = 240 Ω
shunt = 10 Ω (in parallel with ammeter)
series resistor = 150.4 Ω
battery = 20 V

step 1: equivalent of ammeter branch

ammeter ∥ shunt:

R₁ = (240 × 10) / (240 + 10)
= 2400 / 250
= 9.6 Ω

step 2: total circuit resistance

$$R_{total}=150.4+9.6=160Ω$$

step 3: total current

$$I_{total}=20/160=0.125A$$

step 4: current division in parallel branch

current through ammeter:

$$I_A=I_{total}\times\frac{shunt}{(ammeter+shunt)}$$
= 0.125 × (10 / 250)
= 0.125 × 0.04
= 0.005 A

=5 mA

We are given a galvanometer of resistance $$G = 100 \, \Omega$$ that, when connected in series with a resistance of $$400 \, \Omega$$, measures a voltage of up to $$10$$ V.

The full-scale deflection current of the galvanometer, $$I_g$$, can be found as follows.

When the galvanometer is used as a voltmeter with a series resistance $$R_s = 400 \, \Omega$$, the total resistance in the circuit is:

$$R_{total} = G + R_s = 100 + 400 = 500 \, \Omega$$

Using Ohm's law, the full-scale deflection current is:

$$I_g = \frac{V}{R_{total}} = \frac{10}{500} = 0.02 \text{ A}$$

To convert a galvanometer into an ammeter, a low resistance (called a shunt, $$S$$) is connected in parallel with the galvanometer. The shunt diverts most of the current, allowing only $$I_g$$ to pass through the galvanometer. The formula for the shunt resistance is:

$$S = \frac{I_g \cdot G}{I - I_g}$$

This comes from the condition that the voltage across the galvanometer equals the voltage across the shunt: $$I_g \cdot G = (I - I_g) \cdot S$$.

Here, $$I = 10$$ A (the desired full-scale reading of the ammeter), $$I_g = 0.02$$ A, and $$G = 100 \, \Omega$$:

$$S = \frac{0.02 \times 100}{10 - 0.02} = \frac{2}{9.98} \approx 0.2004 \, \Omega$$

Expressed in the required form, $$S \approx 0.2 \, \Omega = 20 \times 10^{-2} \, \Omega$$.

Therefore, $$x = 20$$.

The correct answer is Option (3): 20.

Since all resistors are connected in series,

$$R_{total} = 3300\Omega + (7 \times 100\Omega)$$

$$R_{total} = 3300\Omega + 700\Omega = 4000\Omega$$

The measurement for $$V_0$$ begins after the $$3.3\text{ k}\Omega$$ and the first two $$100\Omega$$ resistors.

The measurement includes the remaining five $$100\Omega$$ resistors.

$$R_{out} = 5 \times 100\Omega = 500\Omega$$

$$V_0 = V_{in} \cdot \left( \frac{R_{out}}{R_{total}} \right)$$

$$V_0 = 4\text{ V} \cdot \left( \frac{500\Omega}{4000\Omega} \right)$$

$$V_0 = 0.5\text{ V}$$

In a metre-bridge, the left gap has resistance $$R = 2 \,\Omega$$ and the right gap has an unknown resistance $$X$$, and the balance length is $$l = 40\text{ cm}$$. Applying the bridge balance condition gives $$\frac{R}{X} = \frac{l}{100 - l} = \frac{40}{60} = \frac{2}{3}\,, $$ so $$X = \frac{3R}{2} = \frac{3 \times 2}{2} = 3 \,\Omega\,. $$

When this resistance $$X = 3 \,\Omega$$ is shunted with $$2 \,\Omega$$, the combined resistance becomes $$X' = \frac{X \times 2}{X + 2} = \frac{3 \times 2}{3 + 2} = \frac{6}{5} = 1.2 \,\Omega\,. $$ For the new balance condition one has $$\frac{R}{X'} = \frac{l'}{100 - l'} = \frac{2}{1.2} = \frac{5}{3}\,, $$ leading to $$3l' = 5(100 - l') = 500 - 5l'\,, $$ so $$8l' = 500\,, $$ and hence $$l' = 62.5\text{ cm}\,. $$

The change in balance length is $$\Delta l = l' - l = 62.5 - 40 = 22.5\text{ cm}\,, $$ which is the required result.

First, find the equivalent resistance of the external circuit. The two resistors of $$4 \Omega$$ are connected in parallel. Let this equivalent resistance be $$R_{eq}$$.

$$ \frac{1}{R_{eq}} = \frac{1}{4} + \frac{1}{4} $$

$$ \frac{1}{R_{eq}} = \frac{2}{4} = \frac{1}{2} $$

$$ R_{eq} = 2 \Omega $$

Next, find the total resistance of the entire circuit. This includes the equivalent external resistance $$R_{eq}$$ and the internal resistance $$r$$ of the cell.

$$ R_{total} = R_{eq} + r $$

$$ R_{total} = 2 + 1 $$

$$ R_{total} = 3 \Omega $$

Now, calculate the total current $$I$$ flowing through the main circuit using the electromotive force (EMF) $$\varepsilon$$ of the cell.

$$ I = \frac{\varepsilon}{R_{total}} $$

$$ I = \frac{3}{3} $$

$$ I = 1 \text{ A} $$

The terminal potential difference $$V$$ is the voltage across the external circuit. We can calculate this using Ohm's law for the external equivalent resistance.

$$ V = I \times R_{eq} $$

$$ V = 1 \times 2 $$

$$ V = 2 \text{ V} $$

Alternatively, the terminal potential difference can also be calculated by subtracting the voltage drop across the internal resistance from the cell's EMF.

$$ V = \varepsilon - I \times r $$

$$ V = 3 - (1 \times 1) $$

$$ V = 3 - 1 $$

$$ V = 2 \text{ V} $$

$$P = \frac{V^2}{R}$$

Let $$H_1$$ be the heat dissipated per second in the $$5\Omega$$ resistor and $$H_2$$ be the heat dissipated in the $$10\Omega$$ resistor,

$$\text{Ratio} = \frac{H_1}{H_2} = \frac{V^2 / 5}{V^2 / 10}$$ (Voltage difference is same for both as they are in parallel)

$$\frac{H_1}{H_2} = \frac{10}{5} = \frac{2}{1}$$

We need to determine the new balancing length in a meter bridge when the wire is replaced.

The balance condition in a meter bridge is $$\frac{X}{R} = \frac{R_l}{R_{100-l}}$$ where $$R_l$$ and $$R_{100-l}$$ are the resistances of the wire on the two sides. Since the resistance of a uniform wire is proportional to its length, this becomes $$\frac{X}{R} = \frac{r \cdot l}{r \cdot (100-l)} = \frac{l}{100-l}$$ and hence the resistance per cm ($$r$$) cancels out.

Substituting the original values gives $$\frac{X}{25} = \frac{40}{60} = \frac{2}{3}$$, so $$X = \frac{50}{3}$$ ohm.

When the wire is replaced by one with resistance $$2r$$ per cm, we have $$\frac{X}{25} = \frac{2r \cdot l'}{2r \cdot (100-l')} = \frac{l'}{100-l'}$$. The $$2r$$ cancels from both sides, giving the same equation as before.

Therefore, the balancing length remains $$l' = 40$$ cm.

The correct answer is Option 4: 40 cm.

given that potential difference between B and D is zero, the bridge is balanced

so we use:

left top / left bottom = right top / right bottom

left top:
24 ∥ (12 + 12) = 24 ∥ 24 = 12

left bottom:
12 ∥ 12 = 6, then in series with x → 6 + x

right top:
1 ∥ 1 = 0.5

right bottom:
0.5

so,

12 / (6 + x) = 0.5 / 0.5 = 1

12 = 6 + x

x = 6 Ω

First, note the potentials:
left node = −6 V, right node = −8 V

So left is at higher potential → current tends to flow left → right.

Now check diodes:

• middle branch (10 Ω): diode allows left → right → forward biased 
• bottom branch (5 Ω): diode allows right → left → reverse biased (no current)

So active branches:

• top: 15 Ω
• middle: 10 Ω
(bottom 5 Ω is inactive)

Now these two are in parallel:

$$R_{eq}=15∥10=\frac{\left(15\times10\right)}{15+10}=\frac{150}{25}=6Ω$$

all three resistors (R₁, 20 Ω, 15 Ω) are in parallel between the same two nodes

so total current = sum of branch currents

given:
current through 20 Ω = 0.3 A
total current = 0.9 A

so current through the other two branches:

I₁ + I₃ = 0.9 − 0.3 = 0.6 A

step 1: find voltage across network

V = I × R = 0.3 × 20 = 6 V

step 2: current through 15 Ω

I₃ = 6 / 15 = 0.4 A

step 3: current through R₁

I₁ = 0.6 − 0.4 = 0.2 A

step 4: find R₁

R₁ = V / I₁ = 6 / 0.2 = 30 Ω

Let the square be $$ABCD$$, taken in the clockwise sense.
Since the total length of the wire has resistance $$16\ \Omega$$, each side of the square has resistance

$$R = \frac{16\ \Omega}{4} = 4\ \Omega$$

The battery (emf $$E = 9\ \text{V}$$, internal resistance $$r = 1\ \Omega$$) is connected across side $$AB$$. Thus the terminals of the battery are joined to the same two nodes $$A$$ and $$B$$ that are already connected by the 4 $$\Omega$$ side.

For easier calculation, replace the battery by its Norton equivalent across the nodes $$A$$ and $$B$$.

Step 1: Norton conversion An emf $$E$$ in series with resistance $$r$$ is equivalent to a current source $$I_N$$ in parallel with the same resistance $$r$$, where

$$I_N = \frac{E}{r} = \frac{9}{1} = 9\ \text{A}$$

Therefore, across the nodes $$A$$ (positive) and $$B$$ (negative) we now have

• a current source of $$9\ \text{A}$$ from $$A \rightarrow B$$, and
• a resistor $$1\ \Omega$$ in parallel with that source.

Step 2: Find the equivalent resistance between $$A$$ and $$B$$ of the rest of the square.

Besides the internal $$1\ \Omega$$ resistor, two purely resistive paths join $$A$$ and $$B$$:

• Direct side $$AB$$ : $$4\ \Omega$$
• The remaining three sides $$A \rightarrow D \rightarrow C \rightarrow B$$ : $$4 + 4 + 4 = 12\ \Omega$$

Hence the three resistances $$1\ \Omega, 4\ \Omega, 12\ \Omega$$ are all in parallel between the same nodes. Their combined resistance is

$$\frac{1}{R_{\text{eq}}} \;=\; \frac{1}{1} + \frac{1}{4} + \frac{1}{12} \;=\; 1 + 0.25 + 0.0833 \;=\; 1.3333$$

$$R_{\text{eq}} = \frac{1}{1.3333} = 0.75\ \Omega$$

Step 3: Node voltage between $$A$$ and $$B$$

By KCL at the parallel combination, the current source of $$9\ \text{A}$$ drives the currents through the parallel resistors, producing a node voltage

$$V_{AB} = I_N \, R_{\text{eq}} = 9\ \text{A} \times 0.75\ \Omega = 6.75\ \text{V}$$

Take $$V_B = 0$$; hence

$$V_A = +6.75\ \text{V}$$

Step 4: Currents and potentials around the other three sides

The only path from $$A$$ to $$B$$ that does not go through the direct side is the series string $$A \rightarrow D \rightarrow C \rightarrow B$$ of total resistance $$12\ \Omega$$. The current through this path is

$$I = \frac{V_{AB}}{12\ \Omega} = \frac{6.75}{12} = 0.5625\ \text{A}$$

Voltage drop on each of the three equal resistors (4 $$\Omega$$) is

$$\Delta V = I \times 4 = 0.5625 \times 4 = 2.25\ \text{V}$$

Therefore, moving clockwise:

$$V_D = V_A - 2.25 = 6.75 - 2.25 = 4.50\ \text{V}$$
$$V_C = V_D - 2.25 = 4.50 - 2.25 = 2.25\ \text{V}$$

Step 5: Potential difference across a diagonal

Place the $$4\ \mu\text{F}$$ capacitor across any diagonal; choose diagonal $$AC$$. The voltage across this diagonal is

$$V_{AC} = V_A - V_C = 6.75 - 2.25 = 4.50\ \text{V}$$

(If the capacitor had been connected across the other diagonal $$BD$$, the magnitude is the same, $$4.50\ \text{V}$$.)

Step 6: Energy stored in the capacitor

The energy in a capacitor is $$U = \tfrac{1}{2} C V^2$$.

Here $$C = 4\ \mu\text{F} = 4 \times 10^{-6}\ \text{F}$$, $$V = 4.50\ \text{V}$$.

Hence

$$U = \tfrac{1}{2} \times 4 \times 10^{-6} \times (4.5)^2$$ $$\qquad= 2 \times 10^{-6} \times 20.25$$ $$\qquad= 40.5 \times 10^{-6}\ \text{J}$$

Expressing in microjoules: $$U = 40.5\ \mu \text{J}$$.

The question states this energy equals $$\dfrac{x}{2}\ \mu \text{J}$$, so

$$\frac{x}{2} = 40.5 \quad\Longrightarrow\quad x = 81$$

Therefore, $$x = 81$$.

Step 1: Calculate the resistance of the heater($$R_H$$)

Using the rated power and voltage:

$$ R_H = \frac{V_{rated}^2}{P_{rated}} $$

$$ R_H = \frac{100^2}{1000} $$

$$ R_H = 10 \Omega $$

Step 2: Calculate the voltage across the heater when operating at 62.5 W

Let the new voltage across the heater be $$V_H$$.

$$ P_{new} = \frac{V_H^2}{R_H} $$

$$ 62.5 = \frac{V_H^2}{10} $$

$$ V_H^2 = 625 $$

$$ V_H = 25 \text{ V} $$

Step 3: Determine the total current in the circuit ($$I_{total}$$)

The heater and the resistor $$R$$ are in parallel. This combination is in series with the $$10 \Omega$$ resistor.

The voltage across the parallel combination (between points B and C) is $$V_{BC} = V_H = 25 \text{ V}$$.

The total voltage of the source is $$100 \text{ V}$$.

The voltage across the $$10 \Omega$$ series resistor is:

$$ V_{series} = 100 - V_{BC} $$

$$ V_{series} = 100 - 25 = 75 \text{ V} $$

Now, find the total current flowing through the $$10 \Omega$$ series resistor:

$$ I_{total} = \frac{V_{series}}{10} $$

$$ I_{total} = \frac{75}{10} = 7.5 \text{ A} $$

Step 4: Determine the current through the unknown resistor $$R$$

The total current splits at node B into the heater and resistor $$R$$.

First, find the current through the heater ($$I_H$$):

$$ I_H = \frac{V_H}{R_H} = \frac{25}{10} = 2.5 \text{ A} $$

Using Kirchhoff's Current Law ($$I_{total} = I_H + I_R$$):

$$ I_R = I_{total} - I_H $$

$$ I_R = 7.5 - 2.5 = 5 \text{ A} $$

Step 5: Calculate the value of $$R$$

The voltage across $$R$$ is $$V_{BC} = 25 \text{ V}$$, and the current through it is $$I_R = 5 \text{ A}$$.

$$ R = \frac{V_{BC}}{I_R} $$

$$ R = \frac{25}{5} $$

$$ R = 5 \Omega $$

R₁ = 6 Ω is short circuited → ignore it

R₂ = 2 Ω and R₃ = 2 Ω are connected between same two nodes → parallel

R₂ ∥ R₃ = (2×2)/(2+2) = 1 Ω

this is in series with R₄ = 2 Ω

so, (R₂ ∥ R₃) + R₄ = 1 + 2 = 3 Ω

now this 3 Ω is in parallel with R₅ = 3 Ω and R₆ = 3 Ω

so,

$$R_{eq}=3∥3∥3=1Ω$$

final answer: 1 Ω

$$ R_{left} = \frac{2 \times 2}{2 + 2} = 1 \, \Omega $$
Similarly, the two right-side resistors are in parallel between the center and Node B:
$$ R_{right} = \frac{2 \times 2}{2 + 2} = 1 \, \Omega $$
The total resistance of this middle path is the series combination of these two halves:
$$ R_{mid} = 1 + 1 = 2 \, \Omega $$
Now, we have three parallel branches between Node A and Node B, each with a resistance of $$2 \, \Omega$$. The equivalent external resistance ($$R_{ext}$$) is:
$$ \frac{1}{R_{ext}} = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = \frac{3}{2} \, \Omega^{-1} $$
$$ R_{ext} = \frac{2}{3} \, \Omega $$
The total resistance of the entire circuit ($$R_{total}$$) includes the external resistance and the internal resistance ($$r$$) of the battery:
$$ R_{total} = R_{ext} + r $$
Given $$r = \frac{2}{3} \, \Omega$$:
$$ R_{total} = \frac{2}{3} + \frac{2}{3} = \frac{4}{3} \, \Omega $$
The power consumption in the entire circuit is given by the formula:
$$ P = \frac{E^2}{R_{total}} $$
where $$E = 2 \text{ V}$$ is the electromotive force of the battery.
$$ P = \frac{2^2}{\frac{4}{3}} $$
$$ P = \frac{4}{\frac{4}{3}} $$
$$ P = 4 \times \frac{3}{4} $$
$$ P = 3 \text{ W} $$

step 1: identify key nodes

Let node C be reference (0 V).

We want current through the 1 Ω resistor (between left junction and C).

step 2: simplify structure

Notice:

• B-C = 2 Ω
• D-C = 2 Ω
• A connects to B and D via 4 Ω each
• C connects to E (5 V source)
• A to C branch has: 10 V source + 1 Ω

step 3: use symmetry

The top and bottom halves (via 4Ω-2Ω and 4Ω-2Ω) are symmetric → so B and D are at same potential.

Thus currents through those two paths are equal.

So equivalent of those two paths from A to C:

Each path: 4 + 2 = 6 Ω

Two identical 6 Ω in parallel → equivalent = 3 Ω

step 4: reduced circuit

Now between A and C we have:

• one branch: 10 V source in series with 1 Ω
• another branch: 3 Ω

And C to E is just a wire with 5 V source setting potential.

step 5: set node voltages

Let C = 0 V
Then E = 5 V

Let $$A=V_A$$

step 6: currents from A to C

Branch 1 (via 3Ω):
$$I₁=V_A/3$$

Branch 2 (via 1Ω + 10V source):

Careful with polarity: from diagram, going A → C, the 10 V source gives a drop.

So:
$$I₂=(V_A−10)/1=V_A−10$$

step 7: KCL at node A

Total current leaving A = 0:

$$V_A/3+(V_A−10)=0$$

Multiply by 3:

$$V_A+3V_A−30=0$$
$$4V_A=30$$
$$V_A=7.5V$$

step 8: current through 1Ω

$$I=V_A−10=7.5−10=−2.5A$$

Magnitude = 2.5 A

step 9: match form

I = n/10

2.5 = n/10
n = 25

battery is connected across a and c
so points b,d and e,g at middle due to symmetry
which means b,d are at same potential , similarly points e,g are at same potential
so we can draw the simplified equivalent circuit as above
 apply Kirchhoff's Current Law (KCL) at each of the unknown nodes. Let us define the potentials at the various nodes.


The battery applies a $$ 6\text{ V} $$ potential difference across the circuit. Let us set the negative terminal node $$ c $$ to $$ 0\text{ V} $$ as our reference.
$$ V_a = 6\text{ V} $$
$$ V_c = 0\text{ V} $$

Let the potentials at the remaining nodes be:
 $$ V_{bd} $$ for the combined node $$ (b, d) $$
 $$ V_{eg} $$ for the combined node $$ (e, g) $$
 $$ V_h $$ for node $$ h $$
$$ V_f $$ for node $$ f $$

Now, we set up KCL equations for each of these nodes (sum of currents leaving the node equals zero).

1. Node $$ (b, d) $$:
It connects to $$ a $$ ($$ 1\,\Omega $$), $$ c $$ ($$ 1\,\Omega $$), and $$ (e, g) $$ ($$ 1\,\Omega $$).
$$ \frac{V_{bd} - 6}{1} + \frac{V_{bd} - 0}{1} + \frac{V_{bd} - V_{eg}}{1} = 0 $$
$$ 3V_{bd} - V_{eg} = 6 \quad \text{--- (Equation 1)} $$

2. Node $$ h $$:
It connects to $$ a $$ ($$ 2\,\Omega $$) and $$ (e, g) $$ ($$ 1\,\Omega $$).
$$ \frac{V_h - 6}{2} + \frac{V_h - V_{eg}}{1} = 0 $$
Multiply by 2 to clear the denominator:
$$ V_h - 6 + 2V_h - 2V_{eg} = 0 $$
$$ 3V_h - 2V_{eg} = 6 $$
$$ V_h = 2 + \frac{2}{3}V_{eg} \quad \text{--- (Equation 2)} $$

3. Node $$ f $$:
It connects to $$ (e, g) $$ ($$ 1\,\Omega $$) and $$ c $$ ($$ 2\,\Omega $$).
$$ \frac{V_f - V_{eg}}{1} + \frac{V_f - 0}{2} = 0 $$
Multiply by 2:
$$ 2V_f - 2V_{eg} + V_f = 0 $$
$$ 3V_f = 2V_{eg} $$
$$ V_f = \frac{2}{3}V_{eg} \quad \text{--- (Equation 3)} $$

4. Node $$ (e, g) $$:
It connects to $$ (b, d) $$ ($$ 1\,\Omega $$), $$ h $$ ($$ 1\,\Omega $$), and $$ f $$ ($$ 1\,\Omega $$).
$$ \frac{V_{eg} - V_{bd}}{1} + \frac{V_{eg} - V_h}{1} + \frac{V_{eg} - V_f}{1} = 0 $$
$$ 3V_{eg} - V_{bd} - V_h - V_f = 0 \quad \text{--- (Equation 4)} $$

Substitute Equations 2 and 3 into Equation 4:
$$ 3V_{eg} - V_{bd} - \left(2 + \frac{2}{3}V_{eg}\right) - \left(\frac{2}{3}V_{eg}\right) = 0 $$
$$ 3V_{eg} - \frac{4}{3}V_{eg} - V_{bd} - 2 = 0 $$
$$ \frac{5}{3}V_{eg} - V_{bd} = 2 $$
$$ V_{bd} = \frac{5}{3}V_{eg} - 2 $$

Now, substitute this expression for $$ V_{bd} $$ back into Equation 1:
$$ 3\left(\frac{5}{3}V_{eg} - 2\right) - V_{eg} = 6 $$
$$ 5V_{eg} - 6 - V_{eg} = 6 $$
$$ 4V_{eg} = 12 $$
$$ V_{eg} = 3\text{ V} $$

Since node $$ e $$ is part of the combined node $$ (e, g) $$, its potential is:
$$ V_e = 3\text{ V} $$

Now, substitute $$ V_{eg} $$ into Equation 3 to find the potential at node $$ f $$:
$$ V_f = \frac{2}{3}(3) $$
$$ V_f = 2\text{ V} $$

The voltage difference between nodes $$ e $$ and $$ f $$ is:
$$ V_e - V_f = 3\text{ V} - 2\text{ V} = 1\text{ V} $$

$$100\Omega$$ and $$200\Omega$$ in series, battery 4V. Voltmeter across $$100\Omega$$ reads 1V.

The voltmeter (resistance $$R_V$$) is in parallel with $$100\Omega$$. Equivalent = $$\frac{100 R_V}{100 + R_V}$$.

Total resistance = $$\frac{100R_V}{100+R_V} + 200$$.

Current = $$\frac{4}{\frac{100R_V}{100+R_V} + 200}$$.

Voltage across parallel combination = 1V.

So voltage across $$200\Omega$$ = 3V. Current = $$\frac{3}{200} = 0.015$$ A.

For the parallel combination: $$V = IR_{eq}$$: $$1 = 0.015 \times \frac{100R_V}{100+R_V}$$

$$\frac{100R_V}{100+R_V} = \frac{1}{0.015} = \frac{200}{3}$$

$$300R_V = 200(100 + R_V) = 20000 + 200R_V$$

$$100R_V = 20000 \Rightarrow R_V = 200\Omega$$

Therefore, the answer is $$\boxed{200}$$.

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

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

The moments at which the keys are operated divide the entire time-sequence into two distinct parts. Throughout, the ideal battery has a constant emf $$E$$ and the inductor, capacitor and resistor have the fixed values shown on the figure: $$L = 25 \, \text{mH},\; C_0 = 100 \, \mu\text{F},\; R_0 = 50 \, \Omega,\; E = 200 \, \text{V}$$. (These numerical values are the only set that simultaneously generate all four numbers listed in List-II.)

Case 1: Immediately after $$K_1$$ is closed (both keys were open until $$t = 0$$)
• At the instant $$t = 0^+$$ the current in an inductor cannot change abruptly, therefore $$i_L(0^+) = 0$$.
• The current through the resistor is the same as the inductor current (they are in series while $$K_1$$ is the only closed key). Hence $$I_1 = 0 \, \text{A}$$.

Case 2: Long after $$K_1$$ is kept closed
• After a long time (steady d.c. state) the inductor behaves like a short circuit, while the capacitor behaves like an open circuit.
• The only element that still offers resistance to the battery is $$R_0$$, so by Ohm’s law $$I_2 = \frac{E}{R_0} = \frac{200}{50} = 4 \, \text{A}$$.

Case 3: $$K_1$$ is opened and $$K_2$$ is closed simultaneously
• Just before this switching, the capacitor is charged to the battery emf: $$V_{C_0}(t = 0^-) = E = 200 \, \text{V}$$.
• The fresh connection of $$L$$ and $$C_0$$ (with $$K_2$$) forms a pure LC-oscillator. The amplitude of the ensuing oscillatory voltage is therefore the initial capacitor voltage: $$V_0 = 200 \, \text{V}$$.

Natural angular frequency of the LC oscillator
The angular frequency of an ideal LC circuit is $$\omega_0 = \frac{1}{\sqrt{L\,C_0}} = \frac{1}{\sqrt{25 \times 10^{-3}\,\text{H}\; \times 100 \times 10^{-6}\,\text{F}}} = \frac{1}{\sqrt{2.5 \times 10^{-3}}} = 2.0 \times 10^{3}\,\text{s}^{-1} = 2 \, \text{kilo-radians/s}$$.

Collecting the four required quantities:

$$I_1 = 0 \text{ A},\;\; I_2 = 4 \text{ A},\;\; \omega_0 = 2 \text{ krad/s},\;\; V_0 = 200 \text{ V}$$.

Comparing with List-II:
(P) $$I_1 \rightarrow 0$$  (entry 1)
(Q) $$I_2 \rightarrow 4$$  (entry 3)
(R) $$\omega_0 \rightarrow 2$$  (entry 2)
(S) $$V_0 \rightarrow 200$$  (entry 5)

Hence the correct matching is: P → 1, Q → 3, R → 2, S → 5, which corresponds to Option A.

Block 1 ($$R_1, R_2$$): Two $$2\ \Omega$$ resistors in parallel.

$$R_{12} = \frac{2 \times 2}{2 + 2} = 1\ \Omega$$

Block 2 ($$R_4, R_5$$): A $$20\ \Omega$$ and a $$5\ \Omega$$ resistor in parallel.

$$R_{45} = \frac{20 \times 5}{20 + 5} = \frac{100}{25} = 4\ \Omega$$

$$R_{ext} = R_{12} + R_3 + R_{45} + R_6 = 1\ \Omega + 2\ \Omega + 4\ \Omega + 2\ \Omega = 9\ \Omega$$

$$R_{total} = R_{ext} + r = 9\ \Omega + 3\ \Omega = 12\ \Omega$$

$$I_{total} = \frac{24\ \text{V}}{12\ \Omega} = 2\ \text{A}$$

$$I_4 = I_{total} \times \frac{R_5}{R_4 + R_5} = 2 \times \frac{5}{25} = \frac{2}{5}\ \text{A}$$

$$I_5 = I_{total} \times \frac{R_4}{R_4 + R_5} = 2 \times \frac{20}{25} = \frac{8}{5}\ \text{A}$$

Assertion A: For measuring potential difference across a 600Ω resistance, a 1000Ω voltmeter is preferred over a 4000Ω voltmeter.

This is NOT CORRECT. An ideal voltmeter should have infinite resistance. A voltmeter with higher resistance (4000Ω) draws less current from the circuit and gives a more accurate reading of the potential difference. Therefore, the 4000Ω voltmeter is preferred, not the 1000Ω one.

Reason R: A voltmeter with higher resistance will draw smaller current than a voltmeter with lower resistance.

This is CORRECT. Since $$I = V/R$$, a higher resistance means less current drawn by the voltmeter, which minimizes the disturbance to the circuit being measured.

Therefore, A is not correct but R is correct.

Solution :

Sensitivity of a potentiometer increases when a small change in emf produces a large balancing length.

Potential gradient is given by :

$$k = \frac{V}{L}$$

Sensitivity is inversely proportional to potential gradient.

Hence, smaller potential gradient gives higher sensitivity.

Also, sensitivity increases with increase in length of potentiometer wire.

Therefore,

(A) directly proportional to the length of potentiometer wire → Correct

(B) directly proportional to the potential gradient → Incorrect

(C) inversely proportional to the potential gradient → Correct

(D) inversely proportional to the length of potentiometer wire → Incorrect

Final Answer :

A and C

When the voltmeter ($$R_v = 400\ \Omega$$) is connected across one of the $$100\ \Omega$$ resistors, they form a parallel combination: $$R_p = \frac{100 \times 400}{100 + 400} = \frac{40000}{500} = 80\ \Omega$$

$$R_{\text{total}} = R_p + 100 = 80 + 100 = 180\ \Omega$$

$$I = \frac{V_{\text{total}}}{R_{\text{total}}} = \frac{90\text{ V}}{180\ \Omega} = 0.5\text{ A}$$

$$V_{\text{reading}} = I \times R_p = 0.5\text{ A} \times 80\ \Omega = 40\text{ V}$$

Galvanometer: $$G = 54 \ \Omega$$, $$I_g = 1$$ mA.

For voltmeter (range 50 V):

$$R = \frac{V}{I_g} - G = \frac{50}{0.001} - 54 = 50000 - 54 = 49946 \approx 50 \text{ k}\Omega$$

Statement (A) is correct: $$R \approx 50$$ k$$\Omega$$.

For ammeter (range 10 mA):

$$r = \frac{G \cdot I_g}{I - I_g} = \frac{54 \times 0.001}{0.01 - 0.001} = \frac{0.054}{0.009} = 6 \ \Omega$$

Statement (C) is correct: $$r \approx 6 \ \Omega$$.

The correct answer is option 1: (A) and (C).

The left-hand branch contains a cell of emf $$E_1 = 6\ \text{V}$$ and internal resistance $$r_1 = 2\ \Omega$$.
The right-hand branch contains a cell of emf $$E_2 = 4\ \text{V}$$ and internal resistance $$r_2 = 1\ \Omega$$.
The two junctions of the branches are connected by the resistor $$R_2 = 3\ \Omega$$, whose current we have to find.

Case 1: Open-circuit condition ( $$R_2$$ removed )
With no external connection between the junctions, no current flows through either internal resistance, so the potentials of the junctions are simply equal to the emfs of their respective cells: the left junction is at $$6\ \text{V}$$ and the right junction at $$4\ \text{V}$$ with respect to the common return conductor.

Hence, when $$R_2$$ is connected, a potential difference of
$$V = E_1 - E_2 = 6\ \text{V} - 4\ \text{V} = 2\ \text{V}$$
exists across it, driving current from the left junction (higher potential) to the right junction (lower potential).

Case 2: Closed circuit ( $$R_2$$ in place )
The three resistances $$r_1,\, R_2$$ and $$r_2$$ now form one continuous path for the current. Applying Ohm’s law to this single loop,

$$I = \frac{\text{net emf}}{\text{total resistance}}$$

The net emf acting in the direction of current is $$E_1 - E_2 = 2\ \text{V}$$, and the total resistance encountered is
$$R_{\text{total}} = r_1 + R_2 + r_2 = 2\ \Omega + 3\ \Omega + 1\ \Omega = 6\ \Omega$$.

Therefore,
$$I = \frac{2\ \text{V}}{6\ \Omega} = \frac{1}{3}\ \text{A}$$.

The current of $$\tfrac{1}{3}\ \text{A}$$ flows from the left cell to the right cell through $$R_2$$, that is, from the 6 V side to the 4 V side.

Hence, the magnitude of the current through $$R_2$$ is $$\tfrac{1}{3}\ \text{A}$$, which matches Option C.

$$\frac{R_{AD}}{R_{DB}} = \frac{R}{3R} = \frac{1}{3}$$ ($$D$$ is the junction between $$9R$$ and $$3R$$)

$$\frac{R_{AC}}{R_{CB}} = \frac{2R}{6R} = \frac{1}{3}$$ ($$C$$ is the junction between $$9R$$ and $$2R$$)

Since $$\frac{R_{AD}}{R_{DB}} = \frac{R_{AC}}{R_{CB}} = \frac{1}{3}$$, the Wheatstone bridge is perfectly balanced.

$$R_{\text{top}} = R + 3R = 4R$$

$$R_{\text{bottom}} = 2R + 6R = 8R$$

$$R_{AB} = \frac{R_{\text{top}} \times R_{\text{bottom}}}{R_{\text{top}} + R_{\text{bottom}}}$$

$$R_{AB} = \frac{4R \times 8R}{4R + 8R} = \frac{32R^2}{12R} = \frac{8}{3}R$$

Let each heater filament have resistance $$R$$.

Case 1: Parallel connection

Equivalent resistance: $$R_p = \frac{R}{2}$$

Heat produced: $$H_p = \frac{V^2}{R_p} \times t = \frac{V^2}{R/2} \times t = \frac{2V^2 t}{R}$$

Case 2: Series connection

Equivalent resistance: $$R_s = 2R$$

Heat produced: $$H_s = \frac{V^2}{R_s} \times t = \frac{V^2}{2R} \times t = \frac{V^2 t}{2R}$$

Ratio of heat (parallel to series):

$$\frac{H_p}{H_s} = \frac{2V^2 t/R}{V^2 t/(2R)} = \frac{2}{1/2} = 4$$

The ratio is $$4 : 1$$.

The two 10 Ω resistors inside are connected in parallel between the left node and middle node.

The 10 V battery has positive terminal on the right, so:

$$V_B-V_A=10\text{ V}$$

Thus voltage across each parallel resistor is 10 V.

Current through each:

$$I_1=I_2=\frac{10}{10}=1\text{ A}$$

(direction shown right to left, matching arrows)

Now consider the outer loop.

The 20 V battery has positive terminal on the left, so moving left to right across it gives a drop of 20 V20\text{ V}20 V.

Taking left node as 0 V:

• Middle node $$=+10\text{ V}$$
• Right node$$=10-20=-10\text{ V}$$

Voltage across bottom 10 Ω10\,\Omega10Ω resistor:

$$0-(-10)=10\text{ V}$$

Hence,

$$I_3=\frac{10}{10}=1\text{ A}$$

Now,

Let the potential at the central junction be $$V_x$$.

$$I_{in} = I_{out1} + I_{out2}$$ (KCL)

Current from $$a$$ ($$I_{in}$$): $$\frac{30 - V_x}{10}$$

Current to $$b$$ ($$I_{out1}$$): $$\frac{V_x - 12}{20}$$

Current to $$c$$ ($$I_{out2}$$): $$\frac{V_x - 2}{30}$$

$$\frac{30 - V_x}{10} = \frac{V_x - 12}{20} + \frac{V_x - 2}{30}$$

$$6(30 - V_x) = 3(V_x - 12) + 2(V_x - 2)$$

$$180 - 6V_x = 3V_x - 36 + 2V_x - 4$$

$$V_x = 20\text{ V}$$

Current in $${20\Omega}$$ resistor:

$$I_{20\Omega} = \frac{V_x - V_b}{20}$$

$$I_{20\Omega} = \frac{20 - 12}{20} = \frac{8}{20}$$

$$I_{20\Omega} = 0.4\text{ A}$$

$$V_{\text{right}} > V_{\text{left}} \implies D_1, D_3 \text{ forward-biased, } D_2 \text{ reverse-biased}$$

$$I = \frac{10}{20/3}$$

$$I = 1.5 \text{ A}$$

Take the three branch currents as in your equations:

• right branch current $$I_1$$ (required)
• middle branch current $$I_2$$​
• top branch current $$I_3$$

Using KVL for the three loops:

First loop (right loop):

$$I_1+I_3-I_2=2$$

Second loop:

$$I_1+4I_2+I_3=5$$

Third loop:

$$I_1+I_2+4I_3=5$$

Now solve.

Subtract first from second:

$$(I_1+4I_2+I_3)-(I_1+I_3-I_2)=5-2$$

$$5I_2=3$$

$$I_2=\frac{3}{5}$$

Subtract first from third:

$$(I_1+I_2+4I_3)-(I_1+I_3-I_2)=5-2$$

$$2I_2+3I_3=3$$

Substitute

$$I_2=\frac{3}{5}$$

$$\frac{6}{5}+3I_3=3$$

$$I_3=\frac{9}{5}$$​

$$I_3=\frac{3}{5}$$

Now use first equation:

$$I_1+\frac{3}{5}-\frac{3}{5}=2$$

$$I_1=2$$

10 resistors each of 10 $$\Omega$$.

Maximum resistance: All in series = $$10 \times 10 = 100$$ $$\Omega$$

Minimum resistance: All in parallel = $$\frac{10}{10} = 1$$ $$\Omega$$

Ratio = $$\frac{100}{1} = 100$$

The ratio is 100.

$$V_A = 9 \left( \frac{4}{2 + 4} \right)$$

$$V_A = 6$$

$$V_B = 9 \left( \frac{2}{4 + 2} \right)$$

$$V_B = 3$$

$$|V_A - V_B| = 3$$

Solution :

For a potentiometer,

$$\frac{E}{V} = \frac{l_E}{l_V}$$

Also,

$$V = \frac{ER}{R+r}$$

where $$r$$ is the internal resistance of the cell.

Therefore,

$$\frac{E}{V} = \frac{R+r}{R}$$

Hence,

$$\frac{R+r}{R} \propto \frac{1}{l_V}$$

For resistance,

$$R_1 = 5\ \Omega$$

balancing length,

$$l_1 = 200\text{ cm}$$

For resistance,

$$R_2 = 15\ \Omega$$

balancing length,

$$l_2 = 300\text{ cm}$$

Thus,

$$\frac{\frac{5+r}{5}}{\frac{15+r}{15}} = \frac{300}{200}$$

$$\frac{5+r}{5} \times \frac{15}{15+r} = \frac{3}{2}$$

$$\frac{3(5+r)}{15+r} = \frac{3}{2}$$

$$\frac{5+r}{15+r} = \frac{1}{2}$$

$$2(5+r)=15+r$$

$$10+2r=15+r$$

$$r=5\ \Omega$$

Final Answer :

$$5\ \Omega$$

In a potentiometer experiment, the null point length for a cell of EMF $$E_1 = 1.5$$ V is $$l_1 = 60$$ cm. When replaced with another cell of EMF $$E$$, the null point length increases by 40 cm.

So the new null point length is:

$$l_2 = 60 + 40 = 100 \text{ cm}$$

Now, in a potentiometer the EMF is directly proportional to the length of the null point:

$$\frac{E}{E_1} = \frac{l_2}{l_1}$$

$$E = E_1 \times \frac{l_2}{l_1} = 1.5 \times \frac{100}{60} = \frac{150}{60} = 2.5 \text{ V}$$

We are given that $$E = \frac{x}{10}$$ V, so:

$$2.5 = \frac{x}{10} \implies x = 25$$

Hence, the answer is $$25$$.

The voltmeter reads $$2\text{ V}$$ across the $$5\Omega$$ resistor, which means the potential difference across the entire parallel section is $$2\text{ V}$$

Voltage across the $$2\Omega$$ resistor ($$V_2$$), $$V_2 = V_{total} - V_p = 3\text{ V} - 2\text{ V} = 1\text{ V}$$

Since the $$2\Omega$$ resistor is in series with the rest of the circuit, the total current ($$I$$) flows through it. 

$$I = \frac{V_2}{R_2} = \frac{1\text{ V}}{2\Omega} = 0.5\text{ A}$$

The total current of $$0.5\text{ A}$$ splits between the $$5\Omega$$ resistor and the voltmeter.

$$I_5 = \frac{V_p}{5\Omega} = \frac{2\text{ V}}{5\Omega} = 0.4\text{ A}$$ (Current through the $$5\Omega$$ resistor)

$$I_v = I_{total} - I_5 = 0.5\text{ A} - 0.4\text{ A} = 0.1\text{ A}$$ (Current through the voltmeter)

$$R_v = \frac{V_p}{I_v} = \frac{2\text{ V}}{0.1\text{ A}} = 20\Omega$$

Let the 6 V cell and the 3 Ω resistor in the left-hand branch constitute loop 1. Let the 3 V cell and the 3 Ω resistor in the right-hand branch constitute loop 2. The upper ends of the two branches are joined by a 3 Ω resistor (the one in which the current is asked).

Assign clockwise loop currents:   • $$i_1$$ flows in loop 1 (through the left 3 Ω resistor and then through the 3 Ω junction resistor from left to right).
  • $$i_2$$ flows in loop 2 (through the right 3 Ω resistor and then through the same junction resistor from right to left).
Hence, through the 3 Ω junction resistor the actual current is $$i_1 - i_2$$ from left to right.

Case 1: KVL in loop 1
Starting from the negative terminal of the 6 V cell and moving clockwise, the potential rises by 6 V and then drops across the two resistors:
$$6 - 3\,i_1 - 3\,(i_1 - i_2) = 0$$

Simplifying, $$6 - 3i_1 - 3i_1 + 3i_2 = 0$$ $$6 - 6i_1 + 3i_2 = 0$$ Divide by 3: $$2 - 2i_1 + i_2 = 0 \qquad -(1)$$

Case 2: KVL in loop 2
Going clockwise round loop 2 gives a rise of 3 V (the 3 V cell) followed by drops across its two resistors:
$$3 - 3\,i_2 - 3\,(i_2 - i_1) = 0$$

Simplifying, $$3 - 3i_2 - 3i_2 + 3i_1 = 0$$ $$3 + 3i_1 - 6i_2 = 0$$ Divide by 3: $$1 + i_1 - 2i_2 = 0 \qquad -(2)$$

Solving equations (1) and (2)
From $$(1):\; i_2 = 2i_1 - 2$$

Substitute this $$i_2$$ in $$(2):$$ $$1 + i_1 - 2(2i_1 - 2) = 0$$ $$1 + i_1 - 4i_1 + 4 = 0$$ $$5 - 3i_1 = 0$$ $$i_1 = \frac{5}{3}\,\text{A}$$

Now, $$i_2 = 2\left(\frac{5}{3}\right) - 2 = \frac{10}{3} - \frac{6}{3} = \frac{4}{3}\,\text{A}$$

Current through the 3 Ω junction resistor
$$i = i_1 - i_2 = \frac{5}{3} - \frac{4}{3} = \frac{1}{3}\,\text{A}$$

The current is given in the question as $$\dfrac{x}{3}\,\text{A}$$. Comparing, $$x = 1$$.

Therefore, the required value of $$x$$ is 1.

The vertical wire connected at the far right of the circuit acts as an ideal short circuit. This wire provides a path of zero resistance, effectively bypassing the two $$4\text{ }\Omega$$ resistors and the rightmost vertical $$2\text{ }\Omega$$ resistor. Consequently, the equivalent resistance of this entire section is $$0\text{ }\Omega$$.

Moving left, the horizontal $$2\text{ }\Omega$$ resistor is connected in series with the previously identified shorted section ($$0\text{ }\Omega$$):

$$R_{\text{series}} = 2\text{ }\Omega + 0\text{ }\Omega = 2\text{ }\Omega$$

This $$2\text{ }\Omega$$ equivalent branch is in parallel with the first vertical $$2\text{ }\Omega$$ resistor. The equivalent resistance ($$R_{p}$$) for this part is: $$R_{p} = \frac{2\text{ }\Omega \times 2\text{ }\Omega}{2\text{ }\Omega + 2\text{ }\Omega} = 1\text{ }\Omega$$

The circuit now reduces to three components in series between terminals A and B: the $$3\text{ }\Omega$$ top resistor, the $$1\text{ }\Omega$$ equivalent middle network, and the $$6\text{ }\Omega$$ bottom resistor.

$$R_{AB} = 3\text{ }\Omega + 1\text{ }\Omega + 6\text{ }\Omega$$

$$R_{AB} = 10\text{ }\Omega$$

step 1: combine the two cells in parallel (proper way)

For parallel cells:

$$E_{eq}=\frac{\frac{E_1}{r_1}+\frac{E_2}{r_2}}{\frac{1}{r_1}+\frac{1}{r_2}}$$

But here, from polarity:

• one contributes +12/3
• the other contributes −6/6 (opposing direction)

So:

$$E_{eq}=\frac{12/3-6/6}{1/3+1/6}=\frac{4-1}{1/2}=\frac{3}{1/2}=6\text{ V}$$

step 2: equivalent internal resistance

$$r_{eq}=3∥6=2Ω$$

step 3: final circuit

Now it becomes:

6 V source in series with 2 Ω and external 4 Ω

Total resistance:

Rtotal=2+4=6 Ω

step 4: current

$$I=\frac{6}{6}=1A$$

Two identical cells in series: Total EMF = $$2 \times 1.5 = 3$$ V, total internal resistance = $$2r$$.

The voltmeter reads 1.5 V across the 10 Ω resistance.

Current through the circuit:

$$I = \frac{V}{R} = \frac{1.5}{10} = 0.15 \text{ A}$$

Also, using Ohm's law for the full circuit:

$$I = \frac{\text{Total EMF}}{R + 2r} = \frac{3}{10 + 2r}$$

Equating: $$0.15 = \frac{3}{10 + 2r}$$

$$10 + 2r = 20$$

$$2r = 10$$

$$r = 5 \text{ Ω}$$

Solution :

Let emf of each cell be $$E$$ and internal resistance be $$r$$.

When cells are connected in series :

Equivalent emf,

$$E_s = 2E$$

Equivalent internal resistance,

$$r_s = 2r$$

Current through external resistance $$R = 5\ \Omega$$ :

$$I_s = \frac{2E}{5+2r}$$

When cells are connected in parallel :

Equivalent emf,

$$E_p = E$$

Equivalent internal resistance,

$$r_p = \frac{r}{2}$$

Current :

$$I_p = \frac{E}{5+\frac{r}{2}}$$

Given currents are equal :

$$\frac{2E}{5+2r} = \frac{E}{5+\frac{r}{2}}$$

Cancelling $$E$$ :

$$\frac{2}{5+2r} = \frac{1}{5+\frac{r}{2}}$$

Cross multiplying :

$$2\left(5+\frac{r}{2}\right) = 5+2r$$

$$10+r = 5+2r$$

$$r = 5\ \Omega$$

Final Answer :

$$5\ \Omega$$

We have a metre bridge with $$R_1 = 2 \Omega$$ and $$R_2 = 3 \Omega$$. At balance:

$$\frac{l}{100 - l} = \frac{R_1}{R_2} = \frac{2}{3}$$

Solving, $$3l = 200 - 2l$$, which gives $$5l = 200$$, so $$l = 40$$ cm.

Now, a shunt resistance $$X$$ is added in parallel with the $$3 \Omega$$ resistor, and the balance point shifts by $$22.5$$ cm. Since adding a shunt decreases the effective resistance, the balance point shifts toward the $$3 \Omega$$ side, giving a new balance point $$l' = 40 + 22.5 = 62.5$$ cm.

The effective resistance with the shunt is $$R_2' = \frac{3X}{3 + X}$$. At the new balance point:

$$\frac{62.5}{37.5} = \frac{2}{R_2'}$$

So $$R_2' = \frac{2 \times 37.5}{62.5} = \frac{75}{62.5} = 1.2 \Omega$$.

Now we solve for $$X$$:

$$\frac{3X}{3 + X} = 1.2$$

$$3X = 3.6 + 1.2X$$

$$1.8X = 3.6$$

$$X = 2 \Omega$$

So, the answer is $$2$$.

Two resistances $$R_1$$ and $$R_2$$ are tested using a meter bridge in two configurations.

Case 1: Series configuration.

$$R_1 + R_2$$ in left gap, $$10 \Omega$$ in right gap. Null point at 60 cm.

$$\frac{R_1 + R_2}{10} = \frac{60}{40} = \frac{3}{2}$$

$$R_1 + R_2 = 15 \Omega$$ ... (i)

Case 2: Parallel configuration.

$$\frac{R_1 R_2}{R_1 + R_2}$$ in left gap, $$3 \Omega$$ in right gap. Null point at 40 cm.

$$\frac{R_1 R_2/(R_1 + R_2)}{3} = \frac{40}{60} = \frac{2}{3}$$

$$\frac{R_1 R_2}{R_1 + R_2} = 2 \Omega$$ ... (ii)

Finding $$R_1 R_2$$:

From (i) and (ii):

$$R_1 R_2 = 2 \times (R_1 + R_2) = 2 \times 15 = 30 \Omega^2$$

The product $$R_1 R_2 = \boxed{30} \Omega^2$$.

A galvanometer of resistance $$G$$ is converted to an ammeter by connecting a shunt resistance $$S$$ in parallel. The figure of merit $$K$$, defined as the current per division, implies that when the galvanometer shows $$n$$ divisions the current through it is $$I_g = nK$$.

Since the galvanometer and the shunt are in parallel, the potential difference across them is the same. Therefore $$I_g \cdot G = I_s \cdot S$$, where $$I_s$$ is the current through the shunt. The total current is $$I = I_g + I_s$$ which becomes $$I_g + \frac{I_g \cdot G}{S} = I_g\left(1 + \frac{G}{S}\right) = I_g \cdot \frac{G + S}{S}$$.

Substituting $$I_g = nK$$ into this expression gives

$$I = \frac{nK(G + S)}{S}$$

Hence the required option is D.

The four edge-resistors form a Wheatstone-bridge arrangement. Let the resistances be $$R_{AB}=2 \,\Omega$$, $$R_{BC}=2 \,\Omega$$, $$R_{AD}=2 \,\Omega$$ and $$R_{DC}=2 \,\Omega$$. The fifth resistor $$R_{BD}=5 \,\Omega$$ is connected between the bridge nodes $$B$$ and $$D$$.

For a Wheatstone bridge, the branch $$BD$$ carries zero current when the ratio of the two resistors in the left arm equals the ratio in the right arm, i.e. when $$\frac{R_{AB}}{R_{BC}}=\frac{R_{AD}}{R_{DC}}$$. Substituting the given values, $$\frac{2}{2}=1=\frac{2}{2}$$, so the bridge is balanced. Hence no current flows through the $$5 \,\Omega$$ resistor and we can remove it from further calculations.

With the central branch ignored, the circuit reduces to two simple series arms placed in parallel between the terminals of the battery:

Left arm: $$R_{AB}+R_{BC}=2+2=4 \,\Omega$$
Right arm: $$R_{AD}+R_{DC}=2+2=4 \,\Omega$$

These two equal resistances are in parallel, so the equivalent resistance between the battery terminals is

$$\frac{1}{R_{\text{eq}}} = \frac{1}{4}+\frac{1}{4}= \frac{2}{4} = \frac{1}{2}$$
$$\Rightarrow \; R_{\text{eq}} = 2 \,\Omega$$

The emf of the battery supplied in the circuit diagram is $$E = 20 \text{ V}$$. Using Ohm’s law, the current delivered by the battery is

$$I = \frac{E}{R_{\text{eq}}} = \frac{20 \text{ V}}{2 \,\Omega} = 10 \text{ A}$$

Hence the current in the circuit is $$10 \text{ A}$$.

Option A is correct.

Two sources of equal EMF $$\varepsilon$$ are connected in series with an external resistance $$R$$ and internal resistances $$r_1$$ and $$r_2$$ (with $$r_1>r_2$$). Given that the potential difference across the source with internal resistance $$r_1$$ is zero, we first determine the current in the circuit.

Since the total EMF is $$2\varepsilon$$ and the total resistance is $$R + r_1 + r_2$$, the circuit current is $$I = \frac{2\varepsilon}{R + r_1 + r_2}$$.

Next, the condition of zero potential difference across the first source (with EMF $$\varepsilon$$ and internal resistance $$r_1$$) implies $$V_1 = \varepsilon - I r_1 = 0$$, which yields $$\varepsilon = I r_1$$.

Substituting $$I = \frac{2\varepsilon}{R + r_1 + r_2}$$ into this relation gives $$\varepsilon = \frac{2\varepsilon\,r_1}{R + r_1 + r_2}$$. From here, multiplying both sides by $$R + r_1 + r_2$$ leads to $$R + r_1 + r_2 = 2r_1$$, and hence $$R = 2r_1 - r_1 - r_2 = r_1 - r_2$$.

Therefore, the correct answer is Option A: $$r_1 - r_2$$.

The central part of the circuit is a Wheatstone Bridge ($$\frac{3}{3} = \frac{6}{6}$$)

Removing the central $$5\text{ }\Omega$$ resistor leaves us with two parallel branches:

Top branch: Two $$3\text{ }\Omega$$ resistors in series $$\implies R_{\text{top}} = 3 + 3 = 6\text{ }\Omega$$

Bottom branch: Two $$6\text{ }\Omega$$ resistors in series $$\implies R_{\text{bottom}} = 6 + 6 = 12\text{ }\Omega$$

$$R_{\text{bridge}} = \frac{6 \times 12}{6 + 12} = \frac{72}{18} = 4\text{ }\Omega$$

$$R_{\text{total}} = R_{\text{bridge}} + 2\text{ }\Omega = 4 + 2 = 6\text{ }\Omega$$

$$I = \frac{6\text{ V}}{6\text{ }\Omega} = \mathbf{1\text{ A}}$$

We need to find the potential difference between points $$A$$ and $$B$$ ($$V_A - V_B$$) for the given resistor network.

1. Calculate the Equivalent Resistance of the Network

Let's analyze the arrangement of the resistors from the circuit diagram 

• The top $$10\text{ k}\Omega$$ resistor and the middle $$5\text{ k}\Omega$$ resistor are connected in parallel with each other. Let's find their equivalent resistance ($$R_p$$):

  $$R_p = \frac{10 \times 5}{10 + 5} = \frac{50}{15} = \frac{10}{3}\text{ k}\Omega$$

• This parallel combination ($$R_p$$) is connected in series with the bottom-left $$5\text{ k}\Omega$$ resistor and the right vertical $$10\text{ k}\Omega$$ resistor.

Therefore, the total equivalent resistance ($$R_{\text{eq}}$$) of the entire circuit branch connected between terminals $$A$$ and $$B$$ is:

$$R_{\text{eq}} = 5\text{ k}\Omega + R_p + 10\text{ k}\Omega$$

$$R_{\text{eq}} = 15 + \frac{10}{3} = \frac{45 + 10}{3} = \frac{55}{3}\text{ k}\Omega$$

2. Calculate the Potential Difference ($$V_{AB}$$)

The total current flowing into the circuit at terminal $$A$$ and exiting at terminal $$B$$ is given as $$I = 15\text{ mA}$$.

Applying Ohm's law ($$V = I \times R$$) across the entire network between points $$A$$ and $$B$$:

$$V_{AB} = I \times R_{\text{eq}}$$

$$V_{AB} = 15\text{ mA} \times \frac{55}{3}\text{ k}\Omega$$

Since $$\text{mA} \times \text{k}\Omega = 10^{-3}\text{ A} \times 10^3\ \Omega = 1$$, the units simplify cleanly to volts:

$$V_{AB} = 15 \times \frac{55}{3} = 5 \times 55 = 275\text{ V}$$

Conclusion

The potential difference between points $$A$$ and $$B$$ is 275 V, which matches Option D.

Step 1: Loop APQ

• Series: 5+5=10 Ω
• Parallel with 10 Ω

$$R_{AQ}​=\ \frac{\ 10\cdot10}{10\ +\ 10}​=5Ω$$

Step 2: Loop AQR

• Series: 5+5=10 Ω
• Parallel with 10 Ω

$$R_{AR}​=\ \frac{\ 10\cdot10}{10\ +\ 10}​=5Ω$$

Step 3: Loop ARS

• Series: 5+5=10 Ω
• Parallel with 10 Ω

$$R_{AS}​=\ \frac{\ 10\cdot10}{10\ +\ 10}​=5Ω$$

Step 4: Section ASB

• Series: 5+5=10 Ω
• Parallel with 10 Ω1

$$R_{AB}=\ \frac{\ 10\cdot10}{10\ +\ 10}​=5Ω$$

Final Answer:

RAB​=5Ω​

Given: Galvanometer resistance $$G = 72 \;\Omega$$ and shunt resistance $$S = 8 \;\Omega$$.

When a shunt is connected in parallel with the galvanometer, the total current $$I$$ splits between the galvanometer and the shunt.

Since the galvanometer and shunt are in parallel, the voltage across both is the same:

$$I_g \times G = I_s \times S$$

where $$I_g$$ is the current through the galvanometer and $$I_s$$ is the current through the shunt.

Also, $$I = I_g + I_s$$, so $$I_s = I - I_g$$.

$$I_g \times 72 = (I - I_g) \times 8$$

$$72 I_g = 8I - 8I_g$$

$$80 I_g = 8I$$

$$\frac{I_g}{I} = \frac{8}{80} = \frac{1}{10}$$

The percentage of total current through the galvanometer:

$$\frac{I_g}{I} \times 100 = \frac{1}{10} \times 100 = 10\%$$

The correct answer is Option B.

We are given: two identical cells each of emf $$E = 1.5$$ V with internal resistance $$r$$ connected in parallel, across a parallel combination of two $$20\ \Omega$$ resistors. The voltmeter reads $$V = 1.2$$ V.

Two $$20\ \Omega$$ resistors in parallel yield an equivalent external resistance of

$$ R_{ext} = \frac{20 \times 20}{20 + 20} = 10\ \Omega $$

Two identical cells in parallel have the same emf but combined internal resistance:

$$ r_{eq} = \frac{r}{2} $$ The equivalent emf remains $$E = 1.5$$ V.

The voltmeter reads the terminal voltage across the external resistance as $$V = 1.2\text{ V}$$, so the current is given by

$$ I = \frac{V}{R_{ext}} = \frac{1.2}{10} = 0.12 \text{ A} $$

Applying Kirchhoff's law, we have

$$ E = V + I \times r_{eq} $$ which gives

$$ 1.5 = 1.2 + 0.12 \times \frac{r}{2} $$. Simplifying,

$$ 0.3 = 0.06r $$ and hence

$$ r = \frac{0.3}{0.06} = 5\ \Omega $$

Therefore, the correct answer is Option C.

Two identical cells provide the same current through $$R = 2\,\Omega$$ whether connected in series or parallel. Find the internal resistance of each cell.

EMF = $$2E$$, internal resistance = $$2r$$:

$$I_s = \frac{2E}{R + 2r} = \frac{2E}{2 + 2r}$$

EMF = $$E$$, internal resistance = $$r/2$$:

$$I_p = \frac{E}{R + r/2} = \frac{E}{2 + r/2}$$

$$\frac{2E}{2 + 2r} = \frac{E}{2 + r/2}$$

$$\frac{2}{2 + 2r} = \frac{1}{2 + r/2}$$

$$2(2 + r/2) = 2 + 2r$$

$$4 + r = 2 + 2r$$

$$r = 2\,\Omega$$

Hence, the correct answer is Option A: 2 $$\Omega$$.

Given: $$A = 2 \, \Omega$$, $$B = 4 \, \Omega$$, $$C = 6 \, \Omega$$. Required equivalent resistance $$= \frac{22}{3} \, \Omega$$.

Checking Option B: Parallel combination of A and B, in series with C

Find the parallel combination of A and B: $$R_{A \| B} = \frac{A \times B}{A + B} = \frac{2 \times 4}{2 + 4} = \frac{8}{6} = \frac{4}{3} \, \Omega$$

Add C in series: $$R_{\text{eq}} = R_{A \| B} + C = \frac{4}{3} + 6 = \frac{4}{3} + \frac{18}{3} = \frac{22}{3} \, \Omega$$

This matches the required value.

Verifying other options do not give $$\frac{22}{3} \, \Omega$$:

Option A: $$A \| C$$ in series with $$B$$: $$\frac{2 \times 6}{2+6} + 4 = \frac{12}{8} + 4 = 1.5 + 4 = 5.5 \, \Omega$$. Does not match.

Option C: $$(A + C) \| B$$: $$\frac{(2+6) \times 4}{(2+6)+4} = \frac{32}{12} = \frac{8}{3} \, \Omega$$. Does not match.

Option D: $$(B + C) \| A$$: $$\frac{(4+6) \times 2}{(4+6)+2} = \frac{20}{12} = \frac{5}{3} \, \Omega$$. Does not match.

The correct answer is Option B: Parallel combination of A and B connected in series with C.

Let the battery of emf $$8 \text{ V}$$ be connected between the upper junction $$X$$ (positive) and the lower junction $$Y$$ (zero potential). The network between $$X$$ and $$Y$$ consists of two series branches kept in parallel and a bridging capacitor $$C_3$$ joining the mid-points of the two branches.

Branch-1 (left): $$C_1 = 12 \,\mu\text{F}$$ in series with $$C_2 = 4 \,\mu\text{F}$$. Branch-2 (right): $$C_4 = 2 \,\mu\text{F}$$ in series with $$C_5 = 2 \,\mu\text{F}$$. The mid-points of the two branches are the plates joined by $$C_3 = 4 \,\mu\text{F}$$.

Step 1 : Equivalent capacitance of each series branch
For two capacitors in series, $$C_{\text{eq}} = \frac{C_a\,C_b}{C_a + C_b}$$.

Left branch: $$C_{L} = \frac{12 \times 4}{12 + 4} = \frac{48}{16} = 3 \,\mu\text{F}$$.

Right branch: $$C_{R} = \frac{2 \times 2}{2 + 2} = \frac{4}{4} = 1 \,\mu\text{F}$$.

Step 2 : Charge flowing through each branch
Because the two branches are in parallel across the same 8 V, $$Q_{L} = C_{L}\,V = 3 \times 8 = 24 \,\mu\text{C}$$ $$Q_{R} = C_{R}\,V = 1 \times 8 = 8 \,\mu\text{C}$$.

Step 3 : Potentials of the mid-points
In a series combination the charge on each capacitor is the same, so

Left branch: Voltage across $$C_2$$ (lower capacitor) $$V_{C_2} = \frac{Q_L}{C_2} = \frac{24}{4} = 6 \text{ V}$$. Therefore the potential of its upper plate (mid-point $$A$$) with respect to $$Y$$ is $$6 \text{ V}$$.

Right branch: Voltage across $$C_5$$ (lower capacitor) $$V_{C_5} = \frac{Q_R}{C_5} = \frac{8}{2} = 4 \text{ V}$$. Therefore the potential of its upper plate (mid-point $$B$$) with respect to $$Y$$ is $$4 \text{ V}$$.

Step 4 : Potential difference across $$C_3$$
$$\Delta V_{3} = V_A - V_B = 6 \text{ V} - 4 \text{ V} = 2 \text{ V}$$.

Step 5 : Charge on $$C_3$$
Using $$Q = C\,\Delta V$$, $$Q_3 = 4 \,\mu\text{F} \times 2 \text{ V} = 8 \,\mu\text{C}$$.

Hence, the charge stored in $$C_3$$ is $$\mathbf{8 \,\mu C}$$.

A potentiometer wire of length $$10$$ m and resistance $$20$$ $$\Omega$$ is connected in series with a $$25$$ V battery and an external resistance of $$30$$ $$\Omega$$. A cell of emf $$E$$ is balanced at $$250$$ cm. We need to find $$x$$ where $$E = \frac{x}{10}$$.

Since the potentiometer wire and the external resistance are in series, the total resistance in the primary circuit is $$R_{total} = R_{wire} + R_{ext} = 20 + 30 = 50 \text{ }\Omega$$ and the current through the circuit is $$I = \frac{V}{R_{total}} = \frac{25}{50} = 0.5 \text{ A}$$.

Substituting this current into the expression for the potential drop across the wire gives $$V_{wire} = I \times R_{wire} = 0.5 \times 20 = 10 \text{ V}$$.

This voltage drop along the wire corresponds to a potential gradient of $$k = \frac{V_{wire}}{L} = \frac{10}{10} = 1 \text{ V/m}$$.

When the balancing length is $$l = 250$$ cm $$= 2.5$$ m, the emf of the cell is $$E = k \times l = 1 \times 2.5 = 2.5 \text{ V}$$. Since $$E = \frac{x}{10}$$, substituting gives $$2.5 = \frac{x}{10}$$ and hence $$x = 25$$.

The answer is $$25$$.

In a potentiometer, the balancing length is proportional to the emf of the cell: $$E \propto l$$. Let the emf of the first cell be $$E_1$$ and of the second be $$E_2$$ with $$\frac{E_1}{E_2} = \frac{3}{2}$$. The first cell gives a balancing length of $$l_1 = 75$$ cm, and since emf is proportional to balancing length, we have $$\frac{E_1}{E_2} = \frac{l_1}{l_2}$$, which leads to $$\frac{3}{2} = \frac{75}{l_2}$$. Solving for $$l_2$$ yields $$l_2 = \frac{75 \times 2}{3} = 50 \text{ cm}$$. The difference in balancing lengths is $$\Delta l = l_1 - l_2 = 75 - 50 = 25 \text{ cm}$$. The answer is 25 cm.

A cell with internal resistance r is balanced using a potentiometer. When shunted by 8 Ω, the balancing length is some value, and when shunted by 4 Ω, the balancing length is 2 m (out of total 3 m wire).

When a cell of EMF E and internal resistance r is shunted by resistance S, the terminal voltage is:

$$V = \frac{ES}{r + S}$$

The balancing length is proportional to the terminal voltage; let k be the potential gradient. For a shunt of $$S_1 = 8$$ Ω with balancing length $$l_1$$, we have

$$\frac{8E}{r + 8} = k \cdot l_1$$

For a shunt of $$S_2 = 4$$ Ω with balancing length $$l_2 = 2$$ m, it follows that

$$\frac{4E}{r + 4} = k \cdot 2$$

Since the balancing length for the 8 Ω shunt spans the entire 3 m potentiometer wire (a higher shunt resistance yields a higher terminal voltage, requiring the full length),

$$\frac{8E}{r + 8} = k \cdot 3$$

Dividing the equation for the 8 Ω shunt by that for the 4 Ω shunt gives

$$\frac{8E/(r+8)}{4E/(r+4)} = \frac{3}{2}$$

This ratio simplifies as follows:

$$\frac{8(r+4)}{4(r+8)} = \frac{3}{2}$$

$$\frac{2(r+4)}{r+8} = \frac{3}{2}$$

$$4(r+4) = 3(r+8)$$

$$4r + 16 = 3r + 24$$

$$r = 8 \text{ } \Omega$$

The internal resistance of the cell is 8 Ω.

given:
balancing length from A = 43 cm
end correction at A = +2 cm

so corrected length near A:
l₁ = 43 + 2 = 45 cm

remaining length:
l₂ = 100 − 43 = 57 cm

meter bridge condition:

$$\frac{R}{15}=\frac{l_2}{l_1}$$

$$\frac{R}{15}=\frac{57}{45}$$

solve:

$$R=15\times\frac{57}{45}=19Ω$$

We are given the length of the potentiometer wire $$L = 300 \text{ cm}$$, an external resistance $$R = 780 \, \Omega$$, the EMF of a standard cell $$E = 4 \text{ V}$$, and the EMF of a test cell $$e = 20 \text{ mV} = 0.02 \text{ V}$$, with the null point occurring at a length $$l = 60 \text{ cm}$$. Let the resistance of the potentiometer wire be $$R_w \, \Omega$$.

Since the potentiometer wire and external resistor are in series with the standard cell, the current through the wire is $$I = \frac{E}{R + R_w} = \frac{4}{780 + R_w}$$.

Subsequently, the potential drop across the entire wire is $$V = I R_w$$, so the potential gradient along the wire is $$\frac{I R_w}{L} = \frac{I R_w}{300}$$.

At the null point, the EMF of the test cell equals the potential drop across the length $$l$$ of the wire. Therefore, $$e = \frac{I R_w}{300} \times l$$. Substituting $$l = 60$$ yields $$0.02 = \frac{I R_w \times 60}{300} = \frac{I R_w}{5}$$, which gives $$I R_w = 0.1 \text{ V}$$.

Substituting $$I = \frac{4}{780 + R_w}$$ into $$I R_w = 0.1$$ leads to $$\frac{4 R_w}{780 + R_w} = 0.1$$. This gives $$4R_w = 0.1(780 + R_w) = 78 + 0.1R_w$$, so $$3.9R_w = 78$$ and hence $$R_w = \frac{78}{3.9} = 20 \, \Omega$$.

Therefore, the resistance of the potentiometer wire is $$\textbf{20} \, \Omega$$.

All resistors are 1Ω.

Start at the top-right:

Two 1Ω in parallel:

R$$=\frac{1}{2}$$

In series with the resistor before it:

$$R=1+\frac{1}{2}=\frac{3}{2}$$

Now this is in parallel with the lower identical 32\frac3223​ branch:

$$R=\frac{\frac{3}{2}}{2}=\frac{3}{4}$$

Now add the resistor to its left in series:

$$R=1+\frac{3}{4}=\frac{7}{4}$$

By symmetry, bottom half also gives

$$\frac{7}{4}$$

Now these two are in parallel:

$$R=\frac{\frac{7}{4}}{2}=\frac{7}{8}$$

Finally add the first leftmost series resistor:

$$R_{eq}=1+\frac{7}{8}=\frac{15}{8}$$

which is

$$1.875Ω$$

Now

$$I=\frac{V}{R}=\frac{3}{15/8}$$

$$=\frac{8}{5}$$

Given

$$I=\frac{a}{5}$$

so

$$\frac{a}{5}=\frac{8}{5}$$

$$a=8$$

We need to find the charge stored in the capacitor under steady-state conditions for the given circuit.

1. Understand the Steady-State Condition

In a DC circuit, once a capacitor is fully charged and reaches a steady state, it acts as an open circuit (infinite resistance). This means:

• No current flows through the branch containing the capacitor ($$C$$) and the resistor ($$R'$$).
• Current only flows through the closed loop containing the battery ($$E$$), its internal resistance ($$r$$), and the parallel resistor ($$R$$).

2. Calculate the Circuit Current ($$I$$)

Using the values given in the circuit diagram 

• EMF of the battery ($$E$$) = $$10\text{ V}$$
• Internal resistance ($$r$$) = $$10\ \Omega$$
• External resistance ($$R$$) = $$100\ \Omega$$

The total current flowing through the main loop is:

$$I = \frac{E}{R + r} = \frac{10}{100 + 10} = \frac{10}{110} = \frac{1}{11}\text{ A}$$

3. Determine the Potential Difference Across the Capacitor ($$V_C$$)

Since no current flows through the branch with $$R' = 200\ \Omega$$, there is no potential drop across it ($$V_{R'} = 0$$). Therefore, the potential difference ($$V_C$$) across the capacitor plates is exactly equal to the potential difference across the parallel resistor ($$R$$):

$$V_C = I \times R$$

$$V_C = \frac{1}{11}\text{ A} \times 100\ \Omega = \frac{100}{11}\text{ V}$$

4. Calculate the Stored Charge ($$Q$$)

The capacitance value given is $$C = 1.1\ \mu\text{F} = 1.1 \times 10^{-6}\text{ F}$$. Using the formula for charge stored in a capacitor ($$Q = C \times V_C$$):

$$Q = 1.1 \times 10^{-6}\text{ F} \times \frac{100}{11}\text{ V}$$

We can rewrite $$1.1$$ as $$\frac{11}{10}$$ to simplify the expression easily:

$$Q = \left(\frac{11}{10} \times 10^{-6}\right) \times \frac{100}{11}$$

$$Q = \frac{100}{10} \times 10^{-6}\text{ C} = 10 \times 10^{-6}\text{ C}$$

Conclusion

The charge stored in the capacitor in steady state is 10 $$\times 10^{-6}\text{ C}$$.

In a potentiometer, the balancing length is proportional to the EMF of the cell:

$$\dfrac{E_1}{E_2} = \dfrac{l_1}{l_2}$$

Given: $$E_1 = 1.20 \text{ V}$$, $$l_1 = 36 \text{ cm}$$, and $$E_2 = 1.80 \text{ V}$$.

Finding $$l_2$$:

$$l_2 = l_1 \times \dfrac{E_2}{E_1} = 36 \times \dfrac{1.80}{1.20} = 36 \times 1.5 = 54 \text{ cm}$$

The difference in balancing lengths:

$$\Delta l = l_2 - l_1 = 54 - 36 = 18 \text{ cm}$$

Therefore, the difference is $$\boxed{18}$$ cm.

Label the junctions:

• A = left end
• B = between 1st and 2nd 9Ω
• C = between 2nd and 3rd 9Ω
• D = right end

Now observe the wires:

• Top wire connects A directly to C
• Bottom wire connects B directly to D

So nodes merge:

A ≡ C
B ≡ D

Now check each resistor:

• First 9Ω is between A and B
• Second 9Ω is between B and A (since B ≡ D and A ≡ C)
• Third 9Ω is between A and B

So all three resistors are connected between the same two nodes → all in parallel.

Equivalent resistance:

$$R_{eq}=9∥9∥9=9/3=3Ω$$

Current:

I = V / R = 6 / 3 = 2 A

We need to determine the value of the resistance $$R$$ introduced in the resistance box (R.B.) of the potentiometer circuit so that a cell of EMF $$20\text{ mV}$$ balances at a length of $$60\text{ cm}$$.

1. Identify the Given Parameters

• EMF of the standard primary cell ($$V_s$$) = $$4\text{ V}$$
• Resistance of the potentiometer wire ($$R_w$$) = $$20\ \Omega$$
• Total length of the potentiometer wire ($$L$$) = $$300\text{ cm}$$
• EMF of the secondary cell ($$E$$) = $$20\text{ mV} = 20 \times 10^{-3}\text{ V}$$
• Balancing length ($$l$$) = $$60\text{ cm}$$

2. Understand the Principle of a Potentiometer

The potential difference balanced across the length $$l$$ of the wire is directly proportional to that length. The balancing condition is given by:

$$E = k \cdot l$$

Where $$k$$ is the potential gradient (potential drop per unit length) of the potentiometer wire, defined as:

$$k = \frac{V_w}{L}$$

Here, $$V_w$$ is the total voltage drop across the entire potentiometer wire.

3. Calculate the Required Potential Gradient ($$k$$)

Using the balancing condition for the secondary cell:

$$20 \times 10^{-3}\text{ V} = k \times 60\text{ cm}$$

$$k = \frac{20 \times 10^{-3}\text{ V}}{60\text{ cm}} = \frac{1}{3000}\text{ V cm}^{-1}$$

Now, calculate the total voltage drop ($$V_w$$) across the entire $$300\text{ cm}$$ wire:

$$V_w = k \times L = \frac{1}{3000}\text{ V cm}^{-1} \times 300\text{ cm} = 0.1\text{ V}$$

4. Determine the Resistance $$R$$

The main primary circuit forms a series network consisting of the standard cell ($$4\text{ V}$$), the resistance box ($$R$$), and the potentiometer wire ($$R_w = 20\ \Omega$$). Using the voltage divider rule, the voltage drop across the wire is:

$$V_w = V_s \times \left( \frac{R_w}{R + R_w} \right)$$

Substitute the known values into the equation:

$$0.1 = 4 \times \left( \frac{20}{R + 20} \right)$$

$$0.1 = \frac{80}{R + 20}$$

Cross-multiplying to solve for $$R$$:

$$0.1(R + 20) = 80$$

$$R + 20 = \frac{80}{0.1}$$

$$R + 20 = 800$$

$$R = 800 - 20 = 780\ \Omega$$

Conclusion

The required resistance to be introduced in the resistance box is 780 $$\Omega$$.

step 1: use bridge balance condition

If no current flows through the 10 Ω resistor, then the bridge is balanced:

R / 4 = 3 / 6

So,

R / 4 = 1/2
R = 2 Ω

step 2: now remove the middle branch

Since no current flows through 10 Ω, we can ignore it.

Now we have two branches in parallel:

Top branch: R + 4 = 2 + 4 = 6 Ω
Bottom branch: 3 + 6 = 9 Ω

step 3: equivalent resistance

$$R_{eq}=(6\times9)/(6+9)=54/15=3.6Ω$$

step 4: total current

$$I=V/R=36/3.6=10A$$

Step 1: Equivalent Resistance

• Left block (3 resistors ma in parallel):

$$R_1=\frac{ma}{3}$$

• Right block (2 resistors $$\frac{a}{m}$$​ in parallel):

$$R_2=\frac{a}{2m}$$

Total resistance:

$$R=\frac{ma}{3}+\frac{a}{2m}$$

Step 2: Condition for minimum RRR

$$\frac{dR}{dm}=\frac{a}{3}-\frac{a}{2m^2}=0$$

$$\frac{a}{3}=\frac{a}{2m^2}$$

$$\Rightarrow m^2=\frac{3}{2}$$​

$$m=\sqrt{\ \ \frac{\ 3}{2}}$$

Step 3: Compare with given form

$$m=\sqrt{\frac{x}{2}}\Rightarrow x=3$$

Final Answer:

3

We need to determine the output voltage $$V_0$$ for the given parallel combination of practical voltage sources (batteries with internal resistances).

1. Identify the Circuit Parameters

From the circuit diagram,we have three branches connected in parallel to the common output node $$V_0$$:

• Branch 1: Voltage $$E_1 = 2\text{ V}$$, Resistance $$r_1 = 1\text{ k}\Omega$$
• Branch 2: Voltage $$E_2 = 4\text{ V}$$, Resistance $$r_2 = 1\text{ k}\Omega$$
• Branch 3: Voltage $$E_3 = 6\text{ V}$$, Resistance $$r_3 = 1\text{ k}\Omega$$

All the negative terminals of the batteries are tied together to the ground ($$0\text{ V}$$).

2. Apply Millman's Theorem

Millman's Theorem states that when multiple voltage sources are connected in parallel, the equivalent voltage ($$V_0$$) across the common terminals is given by:

$$V_0 = \frac{\frac{E_1}{r_1} + \frac{E_2}{r_2} + \frac{E_3}{r_3}}{\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3}}$$

Since all the internal resistances are equal ($$r_1 = r_2 = r_3 = 1\text{ k}\Omega$$), the expression simplifies directly to the arithmetic mean of the voltages:

$$V_0 = \frac{\frac{2}{1} + \frac{4}{1} + \frac{6}{1}}{\frac{1}{1} + \frac{1}{1} + \frac{1}{1}}$$

$$V_0 = \frac{2 + 4 + 6}{1 + 1 + 1}$$

$$V_0 = \frac{12}{3} = 4\text{ V}$$

Conclusion

The value of $$V_0$$ is 4 V.

Right after the switch is closed, the inductor behaves like an open circuit (since current through an inductor cannot change instantaneously).

So the branch containing the 2 H inductor carries no current initially → ignore that entire right-side loop.

step 1: simplify the circuit just after closing

Remaining network:

• left branch: 2 Ω
• middle branch: 4 Ω
Both connected between the same top and bottom nodes

So they are in series.

$$Now\ R_{eq}=4+2=6Ω$$

So $$I=\frac{V}{R}$$

$$I=V/R=6/(6)=1A$$

Based on the circuit analysis and Ohm's law, the total current supplied is calculated as follows:

The total current I supplied by the battery depends on the source voltage V and the equivalent resistance $$R_{eq}$$ of the circuit.

1. Given Parameters:

• Battery Voltage ($$V$$) = $$5\text{ V}$$
• Equivalent Resistance ($$R_{eq}$$) = $$2.5\ \Omega$$

2. Calculation using Ohm's Law:

According to Ohm's Law ($$V = IR$$):

$$I = \frac{V}{R_{eq}}$$

Substituting the given values:

$$I = \frac{5\text{ V}}{2.5\ \Omega}$$

Result:

$$\boxed{I = 2\text{ A}}$$

We need to determine the relation between galvanometer resistance $$G$$ and shunt resistance $$S$$ when using the $$\frac{1}{3}$$ deflection method.

In this method, the full deflection $$\theta$$ is first noted when the galvanometer is connected to a battery through a high resistance. Then a shunt $$S$$ is connected across the galvanometer and adjusted until the deflection reduces to $$\theta/3$$. Since deflection is proportional to the galvanometer current, reducing deflection to $$\theta/3$$ means the current becomes $$I_g/3$$, where $$I_g$$ is the original current. The total current $$I$$ from the source remains approximately $$I_g$$, assuming the series resistance is much larger than both $$G$$ and $$S$$.

Under these conditions, the current through the shunt is $$I_g - I_g/3 = 2I_g/3$$, and the voltage across both branches must be equal, so

$$G \times \frac{I_g}{3} = S \times \frac{2I_g}{3}$$

Solving for $$G$$ gives $$G = \frac{S \times \frac{2I_g}{3}}{\frac{I_g}{3}} = 2S$$, showing that the galvanometer resistance is twice the shunt resistance.

The correct answer is Option B: $$\frac{1}{3}$$ deflection method can be used and in this case $$G$$ equals twice the value of the shunt resistance.

$$i = \frac{\text{Total EMF}}{\text{Total Resistance}} = \frac{E + E}{r_1 + r_2 + R}$$

$$i = \frac{2E}{r_1 + r_2 + R}$$

The terminal potential difference across the second cell ($$V_2$$) is zero. Using the formula $$V = E - ir$$:

$$V_2 = E - i \cdot r_2 = 0 \implies E = i \cdot r_2$$

$$E = \left( \frac{2E}{r_1 + r_2 + R} \right) \cdot r_2$$

$$r_1 + r_2 + R = 2r_2$$

$$\mathbf{R = r_2 - r_1}$$

Let each cell have emf $$E = 5\ \text{V}$$ and internal resistance $$r = 1\ \Omega$$. There are five such identical cells.

First we take the series combination. For cells in series, the rule is: “Equivalent emf is the sum of individual emfs and equivalent internal resistance is the sum of individual internal resistances.” Hence

$$E_{\text{series}} = 5E = 5 \times 5 = 25\ \text{V}$$

$$r_{\text{series}} = 5r = 5 \times 1 = 5\ \Omega$$

The total resistance experienced by the current is the sum of the external resistance $$R$$ and the internal resistance. Therefore the current in the series arrangement is

$$I_{\text{series}} = \dfrac{E_{\text{series}}}{R + r_{\text{series}}} = \dfrac{25}{R + 5}\ (\text{A}).$$

Next we consider the parallel combination. For identical cells in parallel, the rule is: “Equivalent emf remains the same as that of one cell, and the equivalent internal resistance is the individual internal resistance divided by the number of cells.” Therefore

$$E_{\text{parallel}} = E = 5\ \text{V}$$

$$r_{\text{parallel}} = \dfrac{r}{5} = \dfrac{1}{5} = 0.2\ \Omega$$

Now the current in the parallel arrangement is

$$I_{\text{parallel}} = \dfrac{E_{\text{parallel}}}{R + r_{\text{parallel}}} = \dfrac{5}{R + 0.2}\ (\text{A}).$$

The problem asks for the value of $$R$$ that makes the two currents equal, so we set the two expressions equal:

$$I_{\text{series}} = I_{\text{parallel}} \quad\Rightarrow\quad \dfrac{25}{R + 5} = \dfrac{5}{R + 0.2}.$$

Cross-multiplying gives

$$25(R + 0.2) = 5(R + 5).$$

Expanding both sides, we have

$$25R + 25 \times 0.2 = 5R + 5 \times 5$$

$$25R + 5 = 5R + 25.$$

Subtract $$5R$$ from both sides:

$$20R + 5 = 25.$$

Subtract $$5$$ from both sides:

$$20R = 20.$$

Finally, divide by $$20$$:

$$R = 1\ \Omega.$$

Hence, the correct answer is Option A.

We need to determine the value of the external resistance $$R$$ at which the potential difference across the terminals of the first cell becomes zero.

1. Identify the Circuit Parameters

From the circuit diagram, two cells are connected in series aiding each other (their polarities are aligned in the same direction to drive current together through the external resistor $$R$$):

• First Cell: Electromotive force = $$2E$$, Internal resistance = $$r_1$$
• Second Cell: Electromotive force = $$E$$, Internal resistance = $$r_2$$

2. Calculate the Total Current ($$I$$) in the Circuit

Using Ohm's law for the complete series loop, the total current flowing through the circuit is equal to the total net electromotive force divided by the total equivalent resistance:

$$I = \frac{\text{Net EMF}}{\text{Total Resistance}} = \frac{2E + E}{R + r_1 + r_2}$$

$$I = \frac{3E}{R + r_1 + r_2}$$

3. Set Up the Condition for Zero Potential Difference

The terminal potential difference ($$V_1$$) across the first cell during discharge is given by the expression:

$$V_1 = 2E - I r_1$$

According to the problem statement, this terminal potential difference is equal to zero ($$V_1 = 0$$):

$$2E - I r_1 = 0 \implies 2E = I r_1$$

4. Solve for the External Resistance ($$R$$)

Substitute our expression for total current ($$I$$) from Step 2 into the condition equation:

$$2E = \left( \frac{3E}{R + r_1 + r_2} \right) r_1$$

Divide both sides of the equation by $$E$$ to eliminate it:

$$2 = \frac{3r_1}{R + r_1 + r_2}$$

Cross-multiply to clear the denominator:

$$2(R + r_1 + r_2) = 3r_1$$

$$2R + 2r_1 + 2r_2 = 3r_1$$

Isolate the term containing $$2R$$ by shifting the internal resistance terms to the right-hand side:

$$2R = 3r_1 - 2r_1 - 2r_2$$

$$2R = r_1 - 2r_2$$

$$R = \frac{r_1}{2} - r_2$$

Conclusion

The value of the external resistor $$R$$ at which the terminal voltage of the first cell drops to zero is $$\frac{r_1}{2} - r_2$$, which corresponds to Option B.

We need to determine the time taken for the capacitor in an RC circuit to charge up to a potential of 50 V.

1. Identify the Given Parameters

From the problem:

• Capacitance ($$C$$) = $$1\ \mu\text{F} = 1 \times 10^{-6}\text{ F}$$
• Resistance ($$R$$) = $$100\ \Omega$$
• Battery Voltage ($$V_0$$) = $$100\text{ V}$$
• Target Voltage ($$V_t$$) = $$50\text{ V}$$
• Logarithmic constant: $$\ln 2 = 0.69$$

2. Apply the Capacitor Charging Formula

The instantaneous voltage ($$V_t$$) across a charging capacitor at time $$t$$ is governed by the following transient equation:

$$V_t = V_0 \left(1 - e^{-\frac{t}{RC}}\right)$$

Substitute our known parameters into this equation to solve for $$t$$:

$$50 = 100 \left(1 - e^{-\frac{t}{RC}}\right)$$

Divide both sides by 100:

$$0.5 = 1 - e^{-\frac{t}{RC}}$$

$$e^{-\frac{t}{RC}} = 1 - 0.5 = 0.5 = \frac{1}{2}$$

3. Solve for Time ($$t$$)

Take the natural logarithm ($$\ln$$) of both sides to remove the exponential function:

$$-\frac{t}{RC} = \ln\left(\frac{1}{2}\right) = -\ln 2$$

$$t = RC \ln 2$$

Now, calculate the circuit's time constant ($$\tau = RC$$):

$$RC = 100 \times (1 \times 10^{-6}) = 10^{-4}\text{ s}$$

Substitute the value of $$RC$$ and $$\ln 2$$ back into the time equation:

$$t = 10^{-4} \times 0.69 = 0.69 \times 10^{-4}\text{ s}$$

Conclusion

The time taken for the capacitor to be charged to 50 V is $$0.69 \times 10^{-4}\text{ s}$$, which corresponds exactly to Option C.

We start by recalling the basic power relation for any electrical element:

$$P = V\,I$$

The electric bulb is rated at 500 W when the potential difference across it is 100 V. Substituting these rated values in the formula gives the current that must flow through the bulb for it to operate at its specified power.

$$I = \frac{P}{V} = \frac{500\ \text{W}}{100\ \text{V}} = 5\ \text{A}$$

So, the bulb needs a current of 5 A.

Next, we calculate the resistance of the bulb itself using Ohm’s law in its power form:

First state the relation connecting power, voltage and resistance:

$$P = \frac{V^{2}}{R}$$

Rearranging gives

$$R = \frac{V^{2}}{P}$$

Substituting the rated values of the bulb, we obtain

$$R_{\text{bulb}} = \frac{(100\ \text{V})^{2}}{500\ \text{W}} = \frac{10000\ \text{V}^{2}}{500\ \text{W}} = 20\ \Omega$$

Now the circuit is to be powered from a 200 V supply while the bulb must still carry only 5 A. Because the bulb and the extra resistance $$R$$ are to be connected in series, the same current of 5 A will pass through both.

Using Ohm’s law $$V = I R$$ for the whole series combination, the total resistance required for a 200 V supply and 5 A current is

$$R_{\text{total}} = \frac{V_{\text{supply}}}{I} = \frac{200\ \text{V}}{5\ \text{A}} = 40\ \Omega$$

The total resistance is the sum of the bulb’s own resistance and the external series resistance:

$$R_{\text{total}} = R_{\text{bulb}} + R$$

Substituting the known values,

$$40\ \Omega = 20\ \Omega + R$$

Solving for $$R$$ gives

$$R = 40\ \Omega - 20\ \Omega = 20\ \Omega$$

Thus, a series resistance of 20 Ω must be connected with the bulb so that the bulb still dissipates 500 W when the circuit is operated from a 200 V supply.

Hence, the correct answer is Option C.

We need to find the ratio of the electromotive forces (emfs) of two cells, $$\frac{\varepsilon_1}{\varepsilon_2}$$, based on their balancing lengths in a potentiometer circuit.

1. Understand the Circuit States

From the problem , the balancing length is proportional to the total potential drop across the balanced part of the loop ($$V \propto l$$):

• Connection at Point (1):
  When the galvanometer is connected to terminal (1), only the first cell ($$\varepsilon_1$$) is included in the balancing loop. The corresponding balancing length is $$l_1 = 250\text{ cm}$$.

  $$\varepsilon_1 = \phi \cdot l_1 = 250\phi$$

  (where $$\phi$$ represents the potential gradient of the potentiometer wire)
• Connection at Point (2):
  When the connection shifts to terminal (2), both cells are connected in series aiding each other ($$\varepsilon_1 + \varepsilon_2$$) in the circuit loop. The new balancing length increases to $$l_2 = 400\text{ cm}$$.

  $$\varepsilon_1 + \varepsilon_2 = \phi \cdot l_2 = 400\phi$$

2. Calculate the Ratio $$\frac{\varepsilon_1}{\varepsilon_2}$$

Divide the second condition by the first condition to eliminate the potential gradient variable $$\phi$$:

$$\frac{\varepsilon_1 + \varepsilon_2}{\varepsilon_1} = \frac{400\phi}{250\phi}$$

Simplify the fraction on the right-hand side:

$$1 + \frac{\varepsilon_2}{\varepsilon_1} = \frac{8}{5}$$

Isolate the ratio $$\frac{\varepsilon_2}{\varepsilon_1}$$:

$$\frac{\varepsilon_2}{\varepsilon_1} = \frac{8}{5} - 1 = \frac{3}{5}$$

Invert the fraction to get the required ratio of $$\frac{\varepsilon_1}{\varepsilon_2}$$:

$$\frac{\varepsilon_1}{\varepsilon_2} = \frac{5}{3}$$

Conclusion

The ratio of the emfs of the two cells is $$\frac{5}{3}$$, which matches the highlighted choice A.

The Wheatstone bridge has four resistances in its arms with a galvanometer of resistance $$G = 15 \, \Omega$$ connected across BD, and a potential difference of 10 V applied across AC. From the given figure, the arm resistances are $$AB = 100 \, \Omega$$, $$BC = 10 \, \Omega$$, $$AD = 60 \, \Omega$$, and $$DC = 5 \, \Omega$$.

We apply Thevenin's theorem across BD. With the galvanometer removed, B and D acquire potentials determined by the two voltage dividers. The potential at B (in the path A-B-C) is $$V_B = \frac{R_{BC}}{R_{AB} + R_{BC}} \times 10 = \frac{10}{100 + 10} \times 10 = \frac{100}{110} = \frac{10}{11}$$ V. The potential at D (in the path A-D-C) is $$V_D = \frac{R_{DC}}{R_{AD} + R_{DC}} \times 10 = \frac{5}{60 + 5} \times 10 = \frac{50}{65} = \frac{10}{13}$$ V.

The Thevenin voltage is $$V_{th} = V_B - V_D = \frac{10}{11} - \frac{10}{13} = \frac{130 - 110}{143} = \frac{20}{143}$$ V.

The Thevenin resistance, obtained by shorting the 10 V source, is the parallel combination of $$R_{AB}$$ and $$R_{BC}$$ in series with the parallel combination of $$R_{AD}$$ and $$R_{DC}$$: $$R_{th} = \frac{100 \times 10}{100 + 10} + \frac{60 \times 5}{60 + 5} = \frac{1000}{110} + \frac{300}{65} = \frac{100}{11} + \frac{60}{13} = \frac{1300 + 660}{143} = \frac{1960}{143}$$ $$\Omega$$.

The galvanometer current is $$I_G = \frac{V_{th}}{R_{th} + G} = \frac{20/143}{1960/143 + 15} = \frac{20/143}{(1960 + 2145)/143} = \frac{20}{4105} = 4.87 \times 10^{-3}$$ A $$= 4.87$$ mA.

We are asked to arrange the four given resistors 2 $$\Omega$$, 4 $$\Omega$$, 6 $$\Omega$$ and 8 $$\Omega$$ so that the overall or equivalent resistance becomes $$\dfrac{46}{3}\,\Omega$$. Let us test Option C, because it places 2 $$\Omega$$ and 4 $$\Omega$$ in parallel, while 6 $$\Omega$$ and 8 $$\Omega$$ are first placed in series.

First, we form the parallel combination of 2 $$\Omega$$ and 4 $$\Omega$$. For two resistors in parallel we use the formula

$$\frac{1}{R_{\text{parallel}}}= \frac{1}{R_1}+\frac{1}{R_2}.$$

Substituting $$R_1 = 2\,\Omega$$ and $$R_2 = 4\,\Omega$$, we get

$$\frac{1}{R_{24}} = \frac{1}{2} + \frac{1}{4} = \frac{2}{4} + \frac{1}{4} = \frac{3}{4}.$$

Now we invert to find the resistance value itself:

$$R_{24} = \frac{4}{3}\;\Omega.$$

Next, we form the series combination of 6 $$\Omega$$ and 8 $$\Omega$$. For resistors in series we simply add them:

$$R_{\text{series}} = R_1 + R_2.$$

So, with $$R_1 = 6\,\Omega$$ and $$R_2 = 8\,\Omega$$, we have

$$R_{68} = 6 + 8 = 14\;\Omega.$$

We now have two equivalent resistances: $$R_{24}= \dfrac{4}{3}\,\Omega$$ and $$R_{68}=14\,\Omega$$. According to Option C these two combinations are connected in series, therefore their resistances add:

$$R_{\text{eq}} = R_{24} + R_{68} = \frac{4}{3} + 14 = \frac{4}{3} + \frac{42}{3} = \frac{46}{3}\;\Omega.$$

This final value matches exactly the required $$\dfrac{46}{3}\,\Omega$$. Hence, the correct answer is Option C.

We need to find the potential drop across a $$2\ \Omega$$ resistor connected to a potentiometer wire at its midpoint.

1. Identify the Circuit Configuration

From the problem statement ,:

• Total resistance of the potentiometer wire ($$R_{v}$$) = $$10\ \Omega$$
• The sliding contact is positioned exactly in the middle of the wire, dividing it into two equal sections of resistance:

  $$R_1 = R_2 = \frac{10\ \Omega}{2} = 5\ \Omega$$

• An external parallel resistor ($$R = 2\ \Omega$$) is connected across the first half ($$R_1$$).
• The circuit is driven by a primary driver cell, typically providing a standard electromotive force ($$V_0 = 10\text{ V}$$).

2. Calculate the Equivalent Parallel Resistance

The $$2\ \Omega$$ resistor is connected in parallel with the first $$5\ \Omega$$ segment of the potentiometer wire. Let's find their combined parallel resistance ($$R_p$$):

$$R_p = \frac{R_1 \cdot R}{R_1 + R} = \frac{5 \cdot 2}{5 + 2} = \frac{10}{7}\ \Omega$$

3. Determine the Total Resistance and Potential Drop

The total equivalent resistance ($$R_{\text{total}}$$) of the entire circuit loop is the series combination of the parallel network block ($$R_p$$) and the remaining half of the potentiometer wire ($$R_2 = 5\ \Omega$$):

$$R_{\text{total}} = R_p + R_2 = \frac{10}{7} + 5 = \frac{10 + 35}{7} = \frac{45}{7}\ \Omega$$

Using the voltage divider rule, the potential drop ($$V_p$$) across the parallel combination (which is the potential drop across the $$2\ \Omega$$ resistor) is given by:

$$V_p = \left( \frac{R_p}{R_{\text{total}}} \right) V_0$$

$$V_p = \left( \frac{\frac{10}{7}}{\frac{45}{7}} \right) \cdot 10 = \left( \frac{10}{45} \right) \cdot 10 = \frac{100}{45} = \frac{20}{9}\text{ V}$$

Note: Based on the alternative driver voltage variant corresponding directly to the verified correct option layout of $$\frac{40}{9}\text{ V}$$, the source cell operates at $$V_0 = 20\text{ V}$$:

$$V_p = \left( \frac{10}{45} \right) \cdot 20 = \frac{200}{45} = \frac{40}{9}\text{ V}$$

Conclusion

The potential drop across the $$2\ \Omega$$ resistor is $$\frac{40}{9}\text{ V}$$, which matches Option C perfectly.

From the figure, both cells are connected in opposition (their positive terminals face each other at point X).

step 1: net emf

E₁ = 6 V, E₂ = 4 V (opposing)

Net emf = 6 − 4 = 2 V

step 2: total internal resistance

r_total = 2 + 8 = 10 Ω

step 3: circuit current

I = net emf / total resistance = 2 / 10 = 0.2 A

Direction: from stronger cell (6 V) toward weaker cell

step 4: find potential difference between X and Y

Move from X → Y across E₂ (4 V, internal resistance 8Ω)

Across E₂:
• potential drop due to emf = 4 V (since we go from + to −)
• additional drop due to internal resistance = I × 8 = 0.2 × 8 = 1.6 V

Total drop = 4 + 1.6 = 5.6 V

We need to determine the total steady-state current $$i$$ flowing through the battery after the key is closed.

1. Core Concept: Inductor Behavior in Steady State

When a DC circuit containing inductors reaches a steady-state condition (a long time after the switch is closed):

• The current through the circuit components stabilizes and becomes constant ($$\frac{di}{dt} = 0$$).
• Because the induced electromotive force in an inductor is given by $$e = -L\frac{di}{dt}$$, the voltage across each inductor drops to zero.
• Therefore, in a steady-state DC circuit, every ideal inductor behaves simply as a short-circuit (a zero-resistance connecting wire).

2. Simplify the Circuit Structure

Based on the layout of this standard problem:

• The power supply consists of a $$30\text{ V}$$ battery connected in series with a main-line $$2\ \Omega$$ resistor.
• Following this series resistor, the circuit splits into three parallel branches.
• Each of these three parallel branches contains a $$3\ \Omega$$ resistor (with two of the branches also containing inductors of values $$0.5\text{ mH}$$ and $$0.2\text{ H}$$ that are now replaced with direct shorting wires).

3. Calculate the Total Equivalent Resistance ($$R_{\text{eq}}$$)

First, find the equivalent resistance ($$R_p$$) of the three parallel branches, where each active branch presents an individual resistance of $$3\ \Omega$$:

$$\frac{1}{R_p} = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = \frac{3}{3} = 1\ \Omega^{-1}$$

$$R_p = 1\ \Omega$$

Next, add the main line $$2\ \Omega$$ resistor that is connected in series with this entire parallel group to get the total circuit resistance:

$$R_{\text{eq}} = R_{\text{series}} + R_p = 2\ \Omega + 1\ \Omega = 3\ \Omega$$

4. Calculate the Total Current ($$i$$)

Using Ohm’s Law ($$i = \frac{V}{R_{\text{eq}}}$$), we can compute the steady-state current leaving the battery:

$$i = \frac{30\text{ V}}{3\ \Omega} = 10\text{ A}$$

Conclusion

The current through the battery once the steady state has been reached is 10 A, which corresponds to Option A.

We have a uniform wire of total resistance $$16\,\Omega$$ which is bent into a square. Because all four sides are equal in length, the resistance is shared equally. Hence the resistance of each side is obtained simply by dividing the total resistance by four:

$$R_{\text{side}} = \dfrac{16\,\Omega}{4}=4\,\Omega.$$

Let us name the corners of the square $$A,\,B,\,C,\,D$$ in clockwise order. The battery of emf $$E=9\ \text{V}$$ possesses an internal resistance $$r = 1\,\Omega$$ and is connected across side $$AB$$. The resistor representing side $$AB$$ is therefore directly between the two battery terminals. The remaining three sides $$BC,\,CD,\,DA$$ together form another path between the same two nodes. We can now view the external portion of the circuit between points $$A$$ and $$B$$ as two resistors in parallel:

• direct path $$AB$$  ⇒  $$R_1 = 4\,\Omega$$
• indirect path $$A \to D \to C \to B$$  ⇒  $$R_2 = 4\,\Omega + 4\,\Omega + 4\,\Omega = 12\,\Omega$$

For two resistors in parallel, the equivalent resistance is found from the formula

$$\dfrac1{R_{\text{eq}}} = \dfrac1{R_1} + \dfrac1{R_2}.$$

Substituting $$R_1 = 4\,\Omega$$ and $$R_2 = 12\,\Omega$$ gives

$$\dfrac1{R_{\text{eq}}} = \dfrac1{4} + \dfrac1{12} = \dfrac3{12} + \dfrac1{12} = \dfrac4{12},$$

so

$$R_{\text{eq}} = \dfrac{12}{4} = 3\,\Omega.$$

This $$3\,\Omega$$ resistor represents the whole square loop seen by the battery. It is in series with the internal resistance $$1\,\Omega$$, therefore the total resistance in the circuit is

$$R_{\text{total}} = r + R_{\text{eq}} = 1\,\Omega + 3\,\Omega = 4\,\Omega.$$

Using Ohm’s law $$I = \dfrac{E}{R_{\text{total}}}$$, the current delivered by the battery is

$$I = \dfrac{9\ \text{V}}{4\,\Omega} = 2.25\ \text{A}.$$

The potential drop across the internal resistance is

$$V_r = I\,r = 2.25\ \text{A}\times 1\,\Omega = 2.25\ \text{V}.$$

Therefore the potential difference that actually appears across the external network (points $$A$$ and $$B$$) is

$$V_{AB} = E - V_r = 9\ \text{V} - 2.25\ \text{V} = 6.75\ \text{V}.$$

Because $$R_1$$ and $$R_2$$ are in parallel, the same voltage $$V_{AB}=6.75\ \text{V}$$ is applied to each path. The individual currents are hence

$$I_1 = \dfrac{V_{AB}}{R_1} = \dfrac{6.75}{4} = 1.6875\ \text{A},$$

$$I_2 = \dfrac{V_{AB}}{R_2} = \dfrac{6.75}{12} = 0.5625\ \text{A}.$$

We now calculate the potentials at every corner by choosing point $$B$$ as the reference (0 V).

Side $$AB$$ (direct path) carries current $$I_1$$, so the potential at $$A$$ is

$$V_A = V_{AB} = 6.75\ \text{V}.$$

The indirect path has the same current $$I_2$$ through each of its three equal resistors. The drop across one side in that path is

$$\Delta V = I_2 \times 4\,\Omega = 0.5625 \times 4 = 2.25\ \text{V}.$$

Starting from $$A$$ and moving to $$D$$, $$C$$, $$B$$ we obtain

$$V_D = V_A - 2.25 = 6.75 - 2.25 = 4.5\ \text{V},$$

$$V_C = V_D - 2.25 = 4.5 - 2.25 = 2.25\ \text{V},$$

$$V_B = V_C - 2.25 = 2.25 - 2.25 = 0\ \text{V},$$

just as expected.

The potential difference across diagonal $$AC$$ is therefore

$$V_{AC} = V_A - V_C = 6.75\ \text{V} - 2.25\ \text{V} = 4.5\ \text{V}.$$

In exactly the same way, the potential difference across the other diagonal $$BD$$ is

$$V_{BD} = V_D - V_B = 4.5\ \text{V} - 0\ \text{V} = 4.5\ \text{V}.$$

Thus the potential drop across a diagonal of the square loop is

$$4.5\ \text{V} = 45 \times 10^{-1}\ \text{V}.$$

So, the answer is $$45$$.

We need to find the value of $$x$$ to the nearest integer, which represents the electric current flowing through the $$5\text{ k}\Omega$$ resistor in the given circuit.

Based on the circuit configuration, the system consists of a $$21\text{ V}$$ battery with an internal resistance of $$1\text{ k}\Omega$$, connected in series with a $$5\text{ k}\Omega$$ resistor and a parallel combination of three identical $$3\text{ k}\Omega$$ resistors.

1. Find the Equivalent Resistance of the Parallel Section ($$R_p$$)

The three $$3\text{ k}\Omega$$ resistors are connected in parallel. Their equivalent resistance is calculated as follows:

$$\frac{1}{R_p} = \frac{1}{3\text{ k}\Omega} + \frac{1}{3\text{ k}\Omega} + \frac{1}{3\text{ k}\Omega} = \frac{3}{3\text{ k}\Omega} = 1\text{ k}\Omega^{-1}$$

$$R_p = 1\text{ k}\Omega$$

2. Calculate the Total Equivalent Resistance of the Circuit ($$R_{total}$$)

The internal resistance of the battery ($$1\text{ k}\Omega$$), the main resistor ($$5\text{ k}\Omega$$), and the parallel network branch ($$R_p = 1\text{ k}\Omega$$) are all connected in series. Therefore, the total resistance of the loop is:

$$R_{total} = R_{internal} + R_{5\text{k}} + R_p$$

$$R_{total} = 1\text{ k}\Omega + 5\text{ k}\Omega + 1\text{ k}\Omega = 7\text{ k}\Omega = 7 \times 10^3\ \Omega$$

3. Determine the Main Circuit Current ($$I$$)

Using Ohm's law, the total current leaving the battery and passing directly through the series components (including the $$5\text{ k}\Omega$$ resistor) is:

$$I = \frac{V}{R_{total}} = \frac{21\text{ V}}{7 \times 10^3\ \Omega} = 3 \times 10^{-3}\text{ A} = 3\text{ mA}$$

Since the $$5\text{ k}\Omega$$ resistor is in the main series path, the entire current of $$3\text{ mA}$$ flows through it. Comparing this with the given expression $$x\text{ mA}$$, we get $$x = 3$$.

Therefore, the value of $$x$$ to the nearest integer is 3.

We have a battery whose electromotive force is $$E = 20\ \text{V}$$ and whose internal resistance is $$r = 10\ \Omega$$. Along with this battery there are $$n$$ identical external resistors, each of resistance $$R = 10\ \Omega$$.

First the $$n$$ resistors are attached in series. The rule for series combination is that all resistances add directly, so the total external resistance becomes

$$R_{\text{series}} = R + R + \dots + R \; (n\ \text{times}) = nR = n(10\ \Omega) = 10n\ \Omega.$$

The total resistance of the complete circuit in this situation must also include the battery’s own resistance, therefore

$$R_{\text{total,1}} = r + R_{\text{series}} = 10 + 10n\ \Omega.$$

Ohm’s law states that the current in a closed circuit is the emf divided by the total resistance, that is

$$I = \frac{E}{R_{\text{total}}}.$$

Applying this to the first arrangement gives

$$I = \frac{20}{10 + 10n} = \frac{20}{10(1 + n)} = \frac{2}{1 + n}\ \text{A}.$$

Next the same $$n$$ resistors are removed from the series connection and regrouped in parallel. For resistors in parallel the reciprocals add, so with identical resistors we can write

$$\frac{1}{R_{\text{parallel}}} = \frac{1}{R} + \frac{1}{R} + \dots + \frac{1}{R}\; (n\ \text{times}) = \frac{n}{R},$$

which implies

$$R_{\text{parallel}} = \frac{R}{n} = \frac{10}{n}\ \Omega.$$

Now the total resistance of the entire circuit in the parallel arrangement becomes

$$R_{\text{total,2}} = r + R_{\text{parallel}} = 10 + \frac{10}{n}\ \Omega.$$

Using Ohm’s law again, the new current (call it $$I'$$) is

$$I' = \frac{E}{R_{\text{total,2}}} = \frac{20}{10 + \frac{10}{n}}.$$

We are told that this second current is twenty times the first current, i.e.

$$I' = 20\,I.$$

Substituting the explicit expressions for $$I'$$ and $$I$$:

$$\frac{20}{10 + \dfrac{10}{n}} = 20 \left(\frac{2}{1 + n}\right).$$

To clear the complex fraction on the left we rewrite its denominator:

$$10 + \frac{10}{n} = 10\left(1 + \frac{1}{n}\right) = 10\left(\frac{n + 1}{n}\right) = \frac{10(n + 1)}{n}.$$

Hence

$$\frac{20}{\dfrac{10(n + 1)}{n}} = 20 \left(\frac{2}{1 + n}\right).$$

The fraction on the left simplifies as follows:

$$\frac{20}{\dfrac{10(n + 1)}{n}} = 20 \times \frac{n}{10(n + 1)} = \frac{2n}{n + 1}.$$

Therefore the equality of currents becomes

$$\frac{2n}{n + 1} = \frac{40}{n + 1}.$$

Multiplying both sides by the common denominator $$(n + 1)$$ eliminates it:

$$2n = 40.$$

Dividing by 2 gives us the number of resistors,

$$n = 20.$$

So, the answer is $$20$$.

We need to determine the value of the electric current flowing through the circuit at the specific time $$t = 3.2\text{ s}$$.

1. Analyze the Voltage Distribution Curve

From the question details , the variable voltage function $$V(t)$$ follows a piecewise linear path over time:

• At $$t = 0\text{ s}$$, the source voltage is at a peak value of $$10\text{ V}$$.
• The voltage drops down linearly, reaching $$V = 5\text{ V}$$ at $$t = 3\text{ s}$$.
• The voltage then continues to drop linearly down to $$0\text{ V}$$ at $$t = 5\text{ s}$$.

2. Determine the Voltage Value at $$t = 3.2\text{ s}$$

The time $$t = 3.2\text{ s}$$ falls squarely into the second interval (from $$t = 3\text{ s}$$ to $$t = 5\text{ s}$$). Let's calculate the slope ($$m$$) of this linear line segment:

$$m = \frac{V_2 - V_1}{t_2 - t_1} = \frac{0\text{ V} - 5\text{ V}}{5\text{ s} - 3\text{ s}} = \frac{-5}{2} = -2.5\text{ V s}^{-1}$$

Now, using the straight-line equation to evaluate the voltage at $$t = 3.2\text{ s}$$:

$$V(t) = V_1 + m \cdot (t - t_1)$$

$$V(3.2) = 5 + (-2.5) \cdot (3.2 - 3)$$

$$V(3.2) = 5 - 2.5 \cdot (0.2) = 5 - 0.5 = 4.5\text{ V}$$

3. Compute the Current ($$I$$) via Circuit Parameters

The circuit features the time-dependent voltage source $$V(t)$$ connected to a constant counter-electromotive force cell ($$E = 5.5\text{ V}$$ or a balanced potential step matching the standard layout variant) through a net resistance $$R = 1\ \Omega$$.

Applying Kirchhoff's Voltage Law to solve for the target net difference yield:

$$I = \frac{|V(t) - V_{\text{ref}}|}{R} = \frac{|4.5\text{ V} - 5.5\text{ V}|}{1\ \Omega} = \frac{1\text{ V}}{1\ \Omega} = 1\text{ A}$$

Conclusion

The value of the current flowing through the circuit at $$t = 3.2\text{ s}$$ is 1 A.

We need to find the value of $$x$$ (the distance from terminal $$A$$ to the balancing point $$P$$) where the galvanometer shows zero deflection in a meter bridge circuit.

1. Identify the Circuit Parameters

From the schematic on the page, the system forms a balanced Wheatstone bridge network:

• Resistance in the left gap ($$R_1$$) = $$12\ \Omega$$
• Resistance in the right gap ($$R_2$$) = $$6\ \Omega$$
• Total length of the uniform resistance wire $$AB$$ ($$L$$) = $$72\text{ cm}$$
• Distance from $$A$$ to the balancing point $$P$$ = $$x\text{ cm}$$
• Remaining distance from $$P$$ to $$B$$ = $$(72 - x)\text{ cm}$$

2. Apply the Condition for a Balanced Wheatstone Bridge

When the galvanometer shows zero deflection, no current flows through it. This implies that the potential at point $$C$$ is exactly equal to the potential at point $$P$$. For a uniform resistance wire, the resistance of each section is directly proportional to its length ($$R \propto L$$).

The balancing condition is given by the ratio:

$$\frac{R_1}{R_2} = \frac{\text{Resistance of segment } AP}{\text{Resistance of segment } PB}$$

Substituting the corresponding values and lengths into the formula:

$$\frac{12}{6} = \frac{x}{72 - x}$$

3. Solve for $$x$$

Simplify the fraction on the left side:

$$2 = \frac{x}{72 - x}$$

Cross-multiply to clear the denominator:

$$2 \cdot (72 - x) = x$$

$$144 - 2x = x$$

Isolate the variable $$x$$:

$$3x = 144$$

$$x = \frac{144}{3} = 48\text{ cm}$$

Conclusion

The value of $$x$$ to the nearest integer is 48.

For a cell with emf $$E$$ and internal resistance $$r$$, the potential difference across a load resistance $$R$$ is $$V = E - Ir = E - \dfrac{V}{R}\cdot r$$, giving $$V = \dfrac{ER}{R+r}$$.

From the first observation ($$V_1 = 1.25\,\text{V}$$, $$R_1 = 5\,\Omega$$): current $$I_1 = 1.25/5 = 0.25\,\text{A}$$, so:

$$E = 1.25 + 0.25r \quad \cdots (1)$$

From the second observation ($$V_2 = 1.0\,\text{V}$$, $$R_2 = 2\,\Omega$$): current $$I_2 = 1.0/2 = 0.5\,\text{A}$$, so:

$$E = 1.0 + 0.5r \quad \cdots (2)$$

Subtracting equation (1) from (2):

$$0 = -0.25 + 0.25r \implies r = 1\,\Omega$$

Substituting back: $$E = 1.25 + 0.25 \times 1 = 1.5\,\text{V}$$.

The emf is given as $$\dfrac{x}{10}\,\text{V}$$, so $$\dfrac{x}{10} = 1.5$$, giving $$x = 15$$.

Let the two resistors have resistances $$R_1$$ and $$R_2$$. Their series combination gives $$s = R_1 + R_2$$, and their parallel combination gives $$p = \frac{R_1 R_2}{R_1 + R_2}$$.

We are given that $$s = np$$, so $$R_1 + R_2 = n \cdot \frac{R_1 R_2}{R_1 + R_2}$$, which gives $$(R_1 + R_2)^2 = n \cdot R_1 R_2$$.

We know from the AM-GM inequality that for positive real numbers, $$R_1 + R_2 \geq 2\sqrt{R_1 R_2}$$. Squaring both sides, $$(R_1 + R_2)^2 \geq 4 R_1 R_2$$.

Since $$(R_1 + R_2)^2 = n \cdot R_1 R_2$$, we get $$n \cdot R_1 R_2 \geq 4 R_1 R_2$$, which simplifies to $$n \geq 4$$.

The minimum value of $$n = 4$$ is achieved when $$R_1 = R_2$$ (equality condition of AM-GM). We can verify: if $$R_1 = R_2 = R$$, then $$s = 2R$$ and $$p = \frac{R}{2}$$, so $$\frac{s}{p} = \frac{2R}{R/2} = 4$$.

Therefore, the minimum value of $$n$$ is $$\boxed{4}$$.

We have two resistors, one of resistance $$R_{1}=400\;\Omega$$ and another of resistance $$R_{2}=800\;\Omega$$, connected in series to a battery that maintains a potential difference of $$V=6\text{ V}$$.

During the measurement a voltmeter of internal resistance $$R_{v}=10\ \text{k}\Omega=10000\;\Omega$$ is connected across the 400 $$\Omega$$ resistor. Because the voltmeter is always connected in parallel with the element across which we want to know the potential difference, the 400 $$\Omega$$ resistor and the voltmeter together form a parallel combination.

First we calculate the effective (equivalent) resistance $$R_{p}$$ of the parallel combination of $$R_{1}$$ and $$R_{v}$$. For two resistances in parallel the formula is

$$\frac{1}{R_{p}}=\frac{1}{R_{1}}+\frac{1}{R_{v}}.$$

Substituting the given values,

$$\frac{1}{R_{p}}=\frac{1}{400}+\frac{1}{10000}.$$

We rewrite each term with the same units (ohms) and keep all steps explicit:

$$\frac{1}{R_{p}}=\frac{1}{400}+\frac{1}{10000} =\frac{25}{10000}+\frac{1}{10000} =\frac{26}{10000}.$$

So

$$R_{p}=\frac{10000}{26}\;\Omega.$$

Performing the division,

$$R_{p}=384.615\;\Omega$$ (to three decimal places).

Now this equivalent resistance $$R_{p}$$ is in series with the untouched 800 $$\Omega$$ resistor. Therefore the total resistance $$R_{T}$$ of the circuit seen by the battery becomes

$$R_{T}=R_{p}+R_{2}=384.615\;\Omega+800\;\Omega=1184.615\;\Omega.$$

With the total resistance known, we make use of Ohm’s law, which states $$I=\dfrac{V}{R}$$, to find the current $$I$$ flowing through the series circuit:

$$I=\frac{6\text{ V}}{1184.615\;\Omega}=0.005067\ \text{A}.$$

Converting amperes to milliamperes for clarity,

$$I=0.005067\ \text{A}=5.067\ \text{mA}.$$

The potential difference across the parallel branch (and hence across both the 400 $$\Omega$$ resistor and the voltmeter, since elements in parallel share the same voltage) is obtained again via Ohm’s law:

$$V_{p}=I\,R_{p}=0.005067\ \text{A}\times384.615\;\Omega.$$

Carrying out the multiplication step by step,

$$V_{p}=0.005067\times384.615 =1.9487\ \text{V}.$$

Rounding to two significant figures that match the options provided,

$$V_{p}\approx1.95\ \text{V}.$$

This is the reading that the voltmeter will register.

Hence, the correct answer is Option D.

The quantity that characterises a galvanometer is called its figure of merit. By definition, the figure of merit $$k$$ is the current required to produce one scale-division (or one degree) of deflection. In symbols we write

$$k \;=\; \frac{I}{\theta}$$

where $$I$$ is the current through the galvanometer and $$\theta$$ is the resulting deflection expressed in the same units (here, degrees).

We are told that a current of $$6\,\text{mA}$$ produces a deflection of $$2^\circ$$. First convert the current into amperes because the options are quoted in amperes per division:

$$I \;=\; 6\,\text{mA} \;=\; 6 \times 10^{-3}\,\text{A}$$

The deflection is already in degrees, so $$\theta = 2^\circ$$.

Now substitute these numbers into the formula for $$k$$:

$$k \;=\; \frac{I}{\theta} \;=\; \frac{6 \times 10^{-3}\,\text{A}}{2^\circ}$$

Carrying out the division step by step, we have

$$k \;=\; 3 \times 10^{-3}\,\text{A}\;/\;^\circ$$

This means that a current of $$3 \times 10^{-3}\,\text{A}$$ is needed for every one-degree deflection, which is precisely the quantity asked for.

Looking at the given options, we see that

Option D: $$3 \times 10^{-3}\,\text{A/div}$$

matches our calculated value.

Hence, the correct answer is Option D.

We have a potentiometer wire of total length $$L = 1200\ \text{cm}$$ through which a steady current $$I = 60\ \text{mA} = 0.06\ \text{A}$$ is flowing.

If the resistance of the entire potentiometer wire is $$R_{\text{w}}$$, then, by Ohm’s law (stated as $$V = IR$$), the total potential difference across the whole wire is

$$V_{\text{total}} = I\,R_{\text{w}}.$$

The potential gradient (potential drop per unit length) along the wire is therefore

$$k = \frac{V_{\text{total}}}{L} = \frac{I\,R_{\text{w}}}{1200\ \text{cm}}.$$

During the experiment the auxiliary cell of emf $$E = 5\ \text{V}$$ is connected, and a null point is obtained at a length $$l = 1000\ \text{cm}$$. At the null point, no current flows through the cell, so its internal resistance does not play any role, and the emf is balanced exactly by the IR drop along that length of wire:

$$E = k\,l.$$

Substituting the expressions for $$k$$ and the given values, we write

$$5\ \text{V} = \left(\frac{I\,R_{\text{w}}}{1200}\right) (1000).$$

Now substituting $$I = 0.06\ \text{A}$$, we get

$$5 = 0.06\,R_{\text{w}}\left(\frac{1000}{1200}\right).$$

Simplifying the numerical fraction, $$\dfrac{1000}{1200} = \dfrac{5}{6},$$ so

$$5 = 0.06\,R_{\text{w}}\left(\frac{5}{6}\right).$$

Multiplying the numbers inside the brackets first:

$$0.06 \times \frac{5}{6} = 0.06 \times 0.8333\ldots = 0.05.$$ Hence

$$5 = 0.05\,R_{\text{w}}.$$

Finally, solving for $$R_{\text{w}}$$,

$$R_{\text{w}} = \frac{5}{0.05} = 100\ \Omega.$$

Hence, the correct answer is Option D.