The resistance of the meter bridge AB in given figure is 4 ohm. With a cell of emf epsilon = 0.5 V a

Q: Why doesn't the resistance of the galvanometer affect the calculations?

At the null point, the potential difference across the galvanometer is zero, meaning no current flows through that branch. Since current is zero, the resistance of the galvanometer ($V = IR$) does not cause any voltage drop and can be completely ignored in the balancing equation.

Q: Why did the potential gradient change in the second case?

The potential gradient is the voltage drop per unit length of the wire ($k = I \cdot R_{AB} / L$). Because the rheostat resistance was changed from $2\Omega$ to $6\Omega$, the total resistance of the primary circuit increased. This decreased the current $I$ flowing through the wire $AB$, which directly reduced the potential gradient.

Q: What exactly is a potential gradient?

Think of the wire like a gentle slope. The potential gradient is the 'steepness' of that slope. It tells you how much voltage drops for every centimeter (or meter) you move along the wire. A higher current makes the slope steeper (higher gradient).

Q: Could the balancing length $x$ ever be greater than the total length $L$?

No. The maximum potential difference you can measure is the total voltage drop across the entire wire $AB$. If the cell's emf is greater than this total voltage drop, you will never find a null point on the wire. The galvanometer will always deflect in one direction.

Q: Why do we use a rheostat in the primary circuit?

The rheostat acts as a control valve. By changing its resistance, we can adjust the current flowing through the main wire. This allows us to control the potential gradient, ensuring that the voltage drop across the wire is suitable for measuring the specific emfs we are testing.

Question

The resistance of the meter bridge $AB$ in given figure is $4\Omega$. With a cell of emf $\varepsilon = 0.5	ext{ V}$ and rheostat resistance $R_h = 2\Omega$. The null point is obtained at some point $J$. When the cell is replaced by another one of emf $\varepsilon = \varepsilon_2$, the same null point $J$ is found for $R_h = 6\Omega$. The emf $\varepsilon_2$ is

Options:

(A) 0.6 V

(B) 0.3 V

(C) 0.5 V

(D) 0.4 V

Sigma Percentile · Accepted Answer

This problem is a beautiful demonstration of how a potentiometer (or a meter bridge acting as one) works. It elegantly links the concepts of Ohm's law, potential gradients, and the null deflection method. Let's break down the physics step-by-step.

### Analyzing the Setup

Imagine the circuit as two distinct but interacting loops. The bottom loop is the **primary circuit**. It consists of a powerful $6	ext{ V}$ battery, a rheostat $R_h$, and the meter bridge wire $AB$ which has a resistance of $4\Omega$. This primary circuit acts like an engine; its sole job is to drive a steady current $I$ through the wire $AB$, creating a voltage drop along its length.

The top loop is the **secondary circuit**. It contains the cell whose emf $\varepsilon$ we want to measure, and a galvanometer. The jockey $J$ connects this secondary circuit to the wire $AB$ at a distance $x$ from point $A$.

When the galvanometer shows zero deflection (the null point), it means no current is flowing through the secondary circuit. This happens when the potential difference across the length $AJ$ of the wire perfectly matches the emf of the cell. Mathematically, $\varepsilon = V_{AJ}$.

### The Master Equation: Potential Gradient

The potential difference $V_{AJ}$ is proportional to the length $x$. We define the **potential gradient** $k$ as the voltage drop per unit length of the wire.

$$k = \frac{V_{AB}}{L} = \frac{I \cdot R_{AB}}{L}$$

Therefore, the balancing condition is simply:

$$\varepsilon = k \cdot x$$

### Case 1: Finding the Golden Ratio

In the first scenario, we are given $\varepsilon_1 = 0.5	ext{ V}$ and the rheostat is set to $R_h = 2\Omega$. Let's find the current $I_1$ in the primary circuit. The total resistance is the wire's resistance plus the rheostat's resistance.

$$I_1 = \frac{V}{R_{AB} + R_h} = \frac{6}{4 + 2} = 1	ext{ A}$$

Now, we can write the potential gradient $k_1$:

$$k_1 = \frac{I_1 \cdot R_{AB}}{L} = \frac{1 \cdot 4}{L} = \frac{4}{L}$$

Using the balancing condition $\varepsilon_1 = k_1 \cdot x$:

$$0.5 = \frac{4}{L} \cdot x$$

Rearranging this gives us a crucial ratio:

$$\frac{x}{L} = \frac{0.5}{4} = \frac{1}{8}$$

This ratio $\frac{x}{L}$ is our golden key because the problem states that the balancing length $x$ remains the same in the second case!

### Case 2: Unveiling the Unknown EMF

Now, the cell is replaced by an unknown $\varepsilon_2$, and the rheostat is adjusted to $R_h = 6\Omega$. This changes the current in the primary circuit, which in turn changes the potential gradient.

Let's find the new current $I_2$:

$$I_2 = \frac{6}{4 + 6} = 0.6	ext{ A}$$

The new potential gradient $k_2$ is:

$$k_2 = \frac{I_2 \cdot R_{AB}}{L} = \frac{0.6 \cdot 4}{L} = \frac{2.4}{L}$$

We apply the balancing condition again for the new cell. Remember, the balancing length $x$ is unchanged.

$$\varepsilon_2 = k_2 \cdot x = \frac{2.4}{L} \cdot x$$

### Final Calculation

We can rewrite the equation for $\varepsilon_2$ to isolate our golden ratio:

$$\varepsilon_2 = 2.4 \cdot \left(\frac{x}{L}ight)$$

Substitute the ratio $\frac{x}{L} = \frac{1}{8}$ that we found in Case 1:

$$\varepsilon_2 = 2.4 \cdot \frac{1}{8}$$

**$$\varepsilon_2 = 0.3	ext{ V}$$**

The elegance of this problem lies in realizing that you don't need to know the absolute values of $x$ or $L$; their ratio is sufficient to bridge the two scenarios.

Select Answer:

Visualized Solution

\text{Analyzing the Circuit}

\text{Principle of Potentiometer}

\text{Case 1: } \varepsilon_1 = 0.5\text{ V}, R_h = 2\Omega

\text{Evaluating Case 1}

\text{Case 2: } \varepsilon_2 = ?, R_h = 6\Omega

\text{Evaluating Case 2}

\text{Final Answer}

\text{The Way Forward}

The Sigma Insight: Measuring Instruments

Analyzing the Setup

The Master Equation

Case 1

Case 2

Final Calculation

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