Question

In potentiometer experiment, cells of e.m.f. $E_1$ and $E_2$ are connected in series ($E_1 > E_2$) the balancing length is 64 cm of the wire. If the polarity of $E_2$ is reversed, the balancing length becomes 32 cm. The ratio $\frac{E_1}{E_2}$ is

• A. 1 : 1
• B. 6 : 1
• C. 3 : 1
• D. 2 : 1

Hint:

To solve sum-and-difference potentiometer ratios quickly without full equations, use the relationship: $\frac{E_1}{E_2} = \frac{\text{Sum of lengths}}{\text{Difference of lengths}}$. Here, $\frac{64+32}{64-32} = \frac{96}{32} = 3$, giving an immediate answer of 3 : 1.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
A potentiometer is used to analyze two cells using the sum and difference method. We are given the balancing lengths for both configurations—when the cells assist each other and when they oppose each other—and we need to determine the ratio of their electromotive forces ($\frac{E_1}{E_2}$).

Step 2: Key Formula or Approach:
Let $k$ be the potential gradient of the potentiometer wire. 1. When the cells assist each other in series (same polarity orientation), their net voltage is $E_1 + E_2$, balancing at length $l_1$:
$$E_1 + E_2 = k \cdot l_1$$ 2. When the polarity of $E_2$ is reversed, the cells oppose each other, yielding a net voltage of $E_1 - E_2$ that balances at length $l_2$:
$$E_1 - E_2 = k \cdot l_2$$ Dividing these two expressions gives the property ratio:
$$\frac{E_1 + E_2}{E_1 - E_2} = \frac{l_1}{l_2}$$ Using componendo and dividendo rules simplifies this directly to:
$$\frac{E_1}{E_2} = \frac{l_1 + l_2}{l_1 - l_2}$$

Step 3: Detailed Explanation:
Let's collect the balancing lengths provided in the problem statement:
Assisting combination length, $l_1 = 64\ \text{cm}$ Opposing combination length, $l_2 = 32\ \text{cm}$ Substitute these lengths directly into our simplified ratio formula:
$$\frac{E_1}{E_2} = \frac{64 + 32}{64 - 32}$$ Calculate the sums and differences in the fraction:
$$\frac{E_1}{E_2} = \frac{96}{32}$$ Reduce the fraction by dividing both terms by 32:
$$\frac{E_1}{E_2} = \frac{3}{1}$$ This calculation gives an exact ratio of 3 : 1 between the two electromotive forces.

Step 4: Final Answer:
The ratio $\frac{E_1}{E_2}$ is equal to 3 : 1, matching option (C).