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. `sqrt(r_(1)r_(2))`
B. `r_(1)+r_(2)`
C. `r_(1)-r_(2)`
D. `(r_(1)+r_(2))/(2)`

Answer

To solve the problem, we need to analyze the circuit with two cells connected in series, each having the same electromotive force (emf) and different internal resistances. We will derive the value of the external resistance \( R \) based on the given conditions.

### Step-by-Step Solution:

1. **Understanding the Circuit**:
- We have two cells connected in series with the same emf \( E \).
- The internal resistances of the cells are \( r_1 \) and \( r_2 \) respectively, where \( r_1 > r_2 \).
- The external resistance is denoted as \( R \).

2. **Condition Given**:
- It is stated that the potential difference across the first cell (with internal resistance \( r_1 \)) is zero when the circuit is closed. This means that the entire voltage drop across this cell is equal to its internal resistance drop.

3. **Applying Kirchhoff's Law**:
- According to Kirchhoff's loop rule, the sum of potential differences in a closed loop must equal zero. For our circuit, we can express this as:
\[ E - I \cdot r_1 - I \cdot r_2 - I \cdot R = 0 \]
- Rearranging gives:
\[ E = I \cdot (r_1 + r_2 + R) \]

4. **Finding the Current**:
- Since the potential difference across the first cell is zero, we can write:
\[ E - I \cdot r_1 = 0 \implies I = \frac{E}{r_1} \]

5. **Substituting Current in the Equation**:
- Now substitute \( I \) back into the equation derived from Kirchhoff's law:
\[ E = \frac{E}{r_1} \cdot (r_1 + r_2 + R) \]

6. **Simplifying the Equation**:
- Dividing both sides by \( E \) (assuming \( E \neq 0 \)):
\[ 1 = \frac{(r_1 + r_2 + R)}{r_1} \]
- Rearranging gives:
\[ r_1 = r_1 + r_2 + R \]

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

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