Two cells, having the same emf, are connected in series through an external resistance `R`. Cells have internal resistance `r_(1)` and `r_(2) (r_(1) gt r_(2))` respectively. When the circuit is closed, the potentail difference across the first cell is zero the value of `R` is

Options

A. `R_(1)-R_(2)`
B. `(R_(1)+R_(2))/2`
C. `(R_(1)-R_(2))/2`
D. `R_(1)+R_(2)`

Explanation

To solve the problem, we need to analyze the circuit with two cells connected in series, taking into account their internal resistances and the external resistance \( R \). The key point is that the potential difference across the first cell is zero.

### Step-by-Step Solution:

1. **Understanding the Circuit**:
- We have two cells with the same EMF, denoted as \( E \).
- The internal resistances of the cells are \( r_1 \) (for the first cell) and \( r_2 \) (for the second cell), with the condition \( r_1 > r_2 \).
- The cells are connected in series with an external resistance \( R \).

2. **Total EMF in the Circuit**:
- Since the cells are in series, the total EMF of the circuit is:
\[ E_{\text{total}} = E + E = 2E \]

3. **Total Resistance in the Circuit**:
- The total resistance in the circuit is the sum of the internal resistances of the cells and the external resistance:
\[ R_{\text{total}} = r_1 + r_2 + R \]

4. **Current in the Circuit**:
- The current \( I \) flowing through the circuit can be expressed using Ohm's law:
\[ I = \frac{E_{\text{total}}}{R_{\text{total}}} = \frac{2E}{r_1 + r_2 + R} \]

5. **Condition for the First Cell**:
- The potential difference across the first cell is zero, which means that the voltage drop across its internal resistance \( r_1 \) must equal the EMF of the cell:
\[ E - I \cdot r_1 = 0 \implies I \cdot r_1 = E \]
- Therefore, we can express \( I \) as:
\[ I = \frac{E}{r_1} \]

6. **Equating the Two Expressions for Current**:
- From the two expressions for current, we have:
\[ \frac{E}{r_1} = \frac{2E}{r_1 + r_2 + R} \]

7. **Simplifying the Equation**:
- Cancel \( E \) from both sides (assuming \( E \neq 0 \)):
\[ \frac{1}{r_1} = \frac{2}{r_1 + r_2 + R} \]
- Cross-multiplying gives:
\[ r_1 + r_2 + R = 2r_1 \]

8. **Solving for \( R \)**:
- Rearranging the equation:
\[ R = 2r_1 - r_1 - r_2 = r_1 - r_2 \]

### Final Answer:
The value of \( R \) is:
\[ R = r_1 - r_2 \]