Question:
Two cells of e.m.f.s $E_1$ and $E_2$ ($E_1>E_2$) are connected as shown in figure. When the potentiometer is connected between A and B, the balancing length of the potentiometer wire is 3.60 m. On connecting the potentiometer between A and C, the balancing length is 0.90 m. The ratio $E_1/E_2$ is
Two cells of e.m.f.s $E_1$ and $E_2$ ($E_1>E_2$) are connected as shown in figure. When the potentiometer is connected between A and B, the balancing length of the potentiometer wire is 3.60 m. On connecting the potentiometer between A and C, the balancing length is 0.90 m. The ratio $E_1/E_2$ is
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In the sum and difference method, $E_1/E_2 = (L_1 + L_2) / (L_1 - L_2)$ if connections are assisting and opposing.
Updated On: Apr 30, 2026
- 5 : 4
- 4 : 3
- 3 : 4
- 4 : 5
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The Correct Option is B
Solution and Explanation
Step 1: Analyze Connections
- Between A and B: Balancing $E_1$ alone. $E_1 = k(3.60)$.
- Between A and C: Based on the diagram (series opposing), balancing $E_1 - E_2$. $E_1 - E_2 = k(0.90)$.
Step 2: Form Equations
$\frac{E_1 - E_2}{E_1} = \frac{0.90}{3.60} = \frac{1}{4}$.
Step 3: Solve for Ratio
$1 - \frac{E_2}{E_1} = \frac{1}{4} \Rightarrow \frac{E_2}{E_1} = 1 - \frac{1}{4} = \frac{3}{4}$.
Step 4: Reciprocal
$E_1/E_2 = 4/3$.
Final Answer:(B)
- Between A and B: Balancing $E_1$ alone. $E_1 = k(3.60)$.
- Between A and C: Based on the diagram (series opposing), balancing $E_1 - E_2$. $E_1 - E_2 = k(0.90)$.
Step 2: Form Equations
$\frac{E_1 - E_2}{E_1} = \frac{0.90}{3.60} = \frac{1}{4}$.
Step 3: Solve for Ratio
$1 - \frac{E_2}{E_1} = \frac{1}{4} \Rightarrow \frac{E_2}{E_1} = 1 - \frac{1}{4} = \frac{3}{4}$.
Step 4: Reciprocal
$E_1/E_2 = 4/3$.
Final Answer:(B)
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