In the given circuit of potentiometer, the potential difference E across AB (10 m length) is larger

Q: Why do we take the length from the right side for the 4th and 8th wires?

The potentiometer wire is folded back and forth. The 1st wire goes from left to right, the 2nd from right to left, the 3rd from left to right, and so on. Therefore, all even-numbered wires (like the 4th and 8th) run from right to left. To find the continuous length from the starting point A (on the left), we must measure the distance along the wire's path, which means measuring from the right edge for even-numbered wires.

Q: What is the potential gradient $k$?

The potential gradient $k$ is the drop in electrical potential per unit length of the potentiometer wire. It is constant throughout the wire if the primary circuit maintains a steady current and the wire has a uniform cross-section.

Q: Why does the galvanometer show zero deflection at the balance point?

At the balance point, the potential difference across the balancing length of the potentiometer wire is exactly equal and opposite to the EMF of the cell being tested. Because the potential difference is zero, no current flows through the galvanometer branch.

Q: Can we measure an EMF greater than the driver cell $E$?

No, we cannot. The maximum potential difference available across the potentiometer wire is slightly less than the EMF of the driver cell $E$. If the test cell has a higher EMF, the balance point will lie outside the wire, making it impossible to find a null deflection.

Q: Does the internal resistance of the cells $E_1$ and $E_2$ affect the balancing length?

No, the internal resistance of the test cells does not affect the balancing length. At the null point, no current is drawn from the test cells, so there is no voltage drop across their internal resistances. The potentiometer measures the true EMF of the cells.

Question

In the given circuit of potentiometer, the potential difference $E$ across $AB$ (10 m length) is larger than $E_1$ and $E_2$ as well. For key $K_1$ (closed), the jockey is adjusted to touch the wire at point $J_1$, so that there is no deflection in the galvanometer. Now, the first battery ($E_1$) is replaced by second battery ($E_2$) for working by making $K_1$ open and $K_2$ closed. The galvanometer gives then null deflection at $J_2$. The value of $\frac{E_1}{E_2}$ is $\frac{a}{b}$, where $a = \dots\dots$ .

Sigma Percentile · Accepted Answer

### Introduction to the Potentiometer

A potentiometer is a versatile and highly accurate instrument used to measure the electromotive force (EMF) of a cell. Unlike a standard voltmeter, it draws absolutely zero current from the test cell at the balance point. This means it measures the true EMF, unaffected by the cell's internal resistance.

In this problem, we are presented with a classic 10-wire potentiometer setup. The total length of the wire $AB$ is $10 	ext{ m}$, which is equivalent to $1000 	ext{ cm}$.

### The Principle of the Potentiometer

The core principle of a potentiometer is elegantly simple. When a constant current flows through a wire of uniform cross-section, the potential drop across any segment of the wire is directly proportional to the length of that segment.

Mathematically, we can express this as $E = k \cdot l$. Here, $E$ is the potential difference, $l$ is the balancing length, and $k$ is the potential gradient (voltage drop per unit length).

### Analyzing the First Case

Let's dive into the first scenario. Key $K_1$ is closed, bringing cell $E_1$ into the circuit, while $K_2$ remains open. The jockey finds a null point at $J_1$.

To find the exact balancing length $l_1$, we must carefully trace the wire from point $A$. The wire is folded into 10 segments, each $100 	ext{ cm}$ long. The point $J_1$ is located on the 4th segment, $20 	ext{ cm}$ from the left edge.

The first three segments contribute a full $300 	ext{ cm}$. The 4th segment runs from right to left. Since $J_1$ is $20 	ext{ cm}$ from the left edge, its distance from the right edge (where the 4th segment begins) is $100 - 20 = 80 	ext{ cm}$.

Therefore, the total balancing length is $l_1 = 300 + 80 = 380 	ext{ cm}$. This gives us our first crucial equation: $E_1 = k 	imes 380$.

### Analyzing the Second Case

Now, we switch the setup. Key $K_1$ is opened and $K_2$ is closed, connecting cell $E_2$. The new balance point is found at $J_2$.

Point $J_2$ is situated on the 8th segment, $60 	ext{ cm}$ from the right edge. The first seven segments provide a continuous length of $700 	ext{ cm}$.

The 8th segment also runs from right to left. The distance from the right edge is given directly as $60 	ext{ cm}$. Thus, the total balancing length is $l_2 = 700 + 60 = 760 	ext{ cm}$.

This leads to our second equation: $E_2 = k 	imes 760$.

### The Final Calculation

We are asked to find the ratio of $E_1$ to $E_2$. By dividing our two equations, the potential gradient $k$ beautifully cancels out.

$$ \frac{E_1}{E_2} = \frac{l_1}{l_2} = \frac{380}{760} $$

Simplifying this fraction, we get exactly $\frac{1}{2}$. The problem states that this ratio is equal to $\frac{a}{b}$.

Comparing $\frac{1}{2}$ with $\frac{a}{b}$, it is crystal clear that the value of $a$ is $1$.

### Key Takeaways

This problem perfectly illustrates how to navigate a multi-wire potentiometer. Always pay close attention to the direction of the wire segments to calculate the true balancing length from the starting point.

Furthermore, remember that a null point can only be achieved if the driver cell's EMF is strictly greater than the EMF of the cells being tested. If this condition is not met, the potential drop across the entire wire will be insufficient to balance the test cell!

Enter Numerical Value:

Visualized Solution

Analyzing the Setup

Principle of Potentiometer

Balancing Cell $E_{1}$

Balancing Cell $E_{2}$

Ratio of EMFs

Final Answer

Key Takeaway

The Sigma Insight: Electrical Instruments

Introduction to the Potentiometer

The Principle of the Potentiometer

Analyzing the First Case

Analyzing the Second Case

The Final Calculation

Key Takeaways

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