Question

In potentiometer experiment, the balancing length is $8 \text{ m}$ when two cells $E_{1}$ and $E_{2}$ are joined in series. When two cells are connected in opposition the balancing length is $4 \text{ m}$. The ratio of the e.m.f. of the two cells $\left( \frac{E_{1{E_{2 \right)$ is

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

Hint:

Use the sum and difference rule directly: $\frac{E_{1{E_{2 = \frac{L_{1} + L_{2{L_{1} - L_{2 = \frac{8 + 4}{8 - 4} = \frac{12}{4} = 3$.

Answer 1

The Correct Option is A

Step 1: In a potentiometer, the e.m.f. of a cell or combination of cells is directly proportional to the balancing length $L$. Let $k$ be the potential gradient of the wire. When cells are in series assisting each other, the total e.m.f. is $E_{1} + E_{2}$. Given $L_{1} = 8 \text{ m}$:\n\[ E_{1} + E_{2} = k \times 8 \]\n\n
Step 2: When the cells are connected in opposition, the net e.m.f. is $E_{1} - E_{2}$ (assuming $E_{1}>E_{2}$). Given $L_{2} = 4 \text{ m}$:\n\[ E_{1} - E_{2} = k \times 4 \]\n\n
Step 3: Divide the first equation by the second equation to eliminate the potential gradient $k$:\n\[ \frac{E_{1} + E_{2{E_{1} - E_{2 = \frac{8}{4} = 2 \]\n\n
Step 4: Cross-multiply to solve for the ratio $\frac{E_{1{E_{2$:\n\[ E_{1} + E_{2} = 2(E_{1} - E_{2}) \]\n\[ E_{1} + E_{2} = 2E_{1} - 2E_{2} \]\n\[ 3E_{2} = E_{1} \]\n\[ \frac{E_{1{E_{2 = 3 \]