In the given potentiometer circuit arrangement, the balancing length AC is measured to be 250 cm. Wh

Q: Why does the balancing length increase when we switch to point (2)?

When you switch to point (2), both cells $\varepsilon_1$ and $\varepsilon_2$ are connected in series. Their total EMF is $\varepsilon_1 + \varepsilon_2$, which is greater than just $\varepsilon_1$. A larger EMF requires a larger potential drop to balance it, which corresponds to a longer length on the potentiometer wire.

Q: What would happen if the cells were connected with opposite polarities?

If the cells were connected in opposition (e.g., positive to positive), the net EMF would be $\varepsilon_1 - \varepsilon_2$ (assuming $\varepsilon_1 > \varepsilon_2$). In that case, the balancing length would be shorter than the initial $250\text{ cm}$.

Q: Why don't we consider the internal resistance of the cells in our equations?

This is the beauty of the null-point method! At the balancing point, the galvanometer shows zero deflection, meaning absolutely no current flows through the secondary circuit. Since current $I = 0$, the potential drop across the internal resistance ($Ir$) is zero. Therefore, the terminal voltage is exactly equal to the EMF.

Q: Can we find the exact values of $\varepsilon_1$ and $\varepsilon_2$ from this data?

No, we cannot. We can only find their ratio. To find the exact values, we would need to know the potential gradient $k$ of the potentiometer wire, which requires knowing the EMF of the primary battery $V$ and the total resistance of the primary circuit.

Q: What is the purpose of the primary circuit with battery V?

The primary circuit's job is to establish a steady, constant current through the potentiometer wire $AB$. This constant current creates the uniform potential gradient $k$ along the wire, which acts as our 'measuring scale' for the unknown EMFs.

Question

In the given potentiometer circuit arrangement, the balancing length AC is measured to be 250 cm. When the galvanometer connection is shifted from point (1) to point (2) in the given diagram, the balancing length becomes 400 cm. The ratio of the EMF of two cells, $\frac{\varepsilon_1}{\varepsilon_2}$ is

Options:

(A) $\frac{5}{3}$

(B) $\frac{8}{5}$

(C) $\frac{4}{3}$

(D) $\frac{3}{2}$

Sigma Percentile · Accepted Answer

### The Magic of the Potentiometer

Imagine you have a magical wire where the electrical pressure drops perfectly evenly as you walk along it. This is exactly what a potentiometer wire is! The core principle of a potentiometer is beautifully simple: the potential drop $V$ across any portion of the wire is directly proportional to its length $l$.

Mathematically, we write this as:
$$V = k l$$
where $k$ is the potential gradient (the voltage drop per unit length).

### Analyzing the First Setup

Let's look at the first scenario. When the galvanometer is connected to point (1), the circuit only includes the first cell, $\varepsilon_1$. The jockey slides along the wire until the galvanometer shows zero deflection. This means the potential drop across the balancing length $l_1$ perfectly matches the EMF of the cell.

We are given that the balancing length $l_1$ is $250	ext{ cm}$. Using our principle, we can write our first master equation:
$$\varepsilon_1 = k 	imes 250$$

### The Series Combination

Now, things get interesting. We shift the connection to point (2). Notice how the current path now forces it to travel through both cell $\varepsilon_1$ and cell $\varepsilon_2$. Because their polarities are aligned (negative to positive), they are in a series-aiding combination. The total EMF in the circuit is now $\varepsilon_1 + \varepsilon_2$.

To balance this larger EMF, we naturally need a longer section of the potentiometer wire. The new balancing length $l_2$ is given as $400	ext{ cm}$. This gives us our second master equation:
$$\varepsilon_1 + \varepsilon_2 = k 	imes 400$$

### The Final Calculation

We now have a system of two equations. The most elegant way to solve for the ratio is to divide the first equation by the second. This brilliantly eliminates the unknown potential gradient $k$:
$$\frac{\varepsilon_1}{\varepsilon_1 + \varepsilon_2} = \frac{k 	imes 250}{k 	imes 400}$$

Simplifying the fraction on the right side:
$$\frac{\varepsilon_1}{\varepsilon_1 + \varepsilon_2} = \frac{250}{400} = \frac{5}{8}$$

Now, it's just a matter of simple algebra. Let's cross-multiply to isolate our variables:
$$8\varepsilon_1 = 5(\varepsilon_1 + \varepsilon_2)$$
$$8\varepsilon_1 = 5\varepsilon_1 + 5\varepsilon_2$$

Subtracting $5\varepsilon_1$ from both sides, we get:
$$3\varepsilon_1 = 5\varepsilon_2$$

Finally, rearranging to find the ratio $\frac{\varepsilon_1}{\varepsilon_2}$:
$$\frac{\varepsilon_1}{\varepsilon_2} = \frac{5}{3}$$

**And there we have it! The ratio of the EMFs of the two cells is $\frac{5}{3}$.**

Select Answer:

Visualized Solution

$V = k l$

$ε_{1} = k l_{1}$

$ε_{1} + ε_{2} = k l_{2}$

$\frac{ε _{1}}{ε _{1} + ε _{2}} = \frac{5}{8}$

$\frac{ε _{1}}{ε _{2}} = \frac{5}{3}$

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

The Sigma Insight: Electrical Instruments

The Magic of the Potentiometer

Analyzing the First Setup

The Series Combination

The Final Calculation

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } V=kl

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ε1​=kl1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ε1​+ε2​=kl2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ε1​+ε2​ε1​​=85​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ε2​ε1​​=35​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ratio=35​