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

• A. 5 : 4
• B. 4 : 3
• C. 3 : 4
• D. 4 : 5

Hint:

In the sum and difference method, $E_1/E_2 = (L_1 + L_2) / (L_1 - L_2)$ if connections are assisting and opposing.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
Points A and B enclose cell $E_1$. Points A and C enclose both $E_1$ and $E_2$. From the circuit diagram, the cells are in series opposing combination when considering points A and C (negative of $E_1$ to negative of $E_2$).
Step 2: Key Formula or Approach:
Potential balancing principle: $E \propto l$.
Step 3: Detailed Explanation:
Case 1: Potential between A and B.
Only $E_1$ is in the circuit branch.
\[ E_1 = k \cdot l_1 = k \cdot (3.60) \quad \text{--- (i)} \]
Case 2: Potential between A and C.
Both cells are connected in series. Looking at the polarities (positive terminals at A and C, negative terminals connected together at B), the net EMF is $E_1 - E_2$.
\[ E_1 - E_2 = k \cdot l_2 = k \cdot (0.90) \quad \text{--- (ii)} \]
Divide equation (i) by (ii):
\[ \frac{E_1}{E_1 - E_2} = \frac{3.60}{0.90} = 4 \]
Cross multiply:
\[ E_1 = 4E_1 - 4E_2 \]
\[ 4E_2 = 3E_1 \]
\[ \frac{E_1}{E_2} = \frac{4}{3} \]
Step 4: Final Answer:
The ratio $E_1/E_2$ is 4 : 3.