Question

Two cells $E_1$ and $E_2$ having equal e.m.f ' $E$ ' and internal resistances $r_1$ and $r_2 (r_1 > r_2)$ respectively are connected in series. This combination is connected to an external resistance ' R '. It is observed that the potential difference across the cell $E_1$ becomes zero. The value of $R$ will be

• A. $\text{r}_1 - \text{r}_2$
• B. $r_1 + r_2$
• C. $\frac{r_1-r_2}{2}$
• D. $\frac{r_1+r_2}{2}$

Hint:

Potential difference across a cell is zero when the external resistance equals the difference in internal resistances.

Answer 1

The Correct Option is A

Step 1: Concept
Total current in series is $I = \frac{\sum E}{\sum r + R}$. Terminal voltage $V = E - Ir$.

Step 2: Meaning
For two cells in series, $I = \frac{2E}{r_1 + r_2 + R}$. For cell $E_1$, $V_1 = E - Ir_1 = 0$.

Step 3: Analysis
$E = I r_1 \implies E = \left(\frac{2E}{r_1 + r_2 + R}\right)r_1$.
$r_1 + r_2 + R = 2r_1$.

Step 4: Conclusion
$R = 2r_1 - r_1 - r_2 = r_1 - r_2$. Final Answer: (A)