A circuit consists of a resistance R connected to n similar cells. If the current in the circuit is the same whether the cells are connected in series or in parallel, then the internal resistancer of each cell is given by

Options

A. `r = R//n`
B. `r=nR`
C. `r=R`
D. `r= 1//R`

Answer

To solve the problem, we need to analyze the circuit with n similar cells connected in both series and parallel configurations, and then derive the expression for the internal resistance of each cell.

### Step-by-step Solution:

1. **Understanding the Circuit Configuration**:
- We have n similar cells, each with an EMF (E) and an internal resistance (r).
- The external resistance in the circuit is denoted as R.

2. **Cells Connected in Series**:
- When the cells are connected in series, the total EMF (E_eq) is:
\[ E_{eq} = nE \]
- The total internal resistance (r_eq) is:
\[ r_{eq} = nr \]
- The total resistance in the circuit when cells are in series is:
\[ R_{total} = R + nr \]
- Using Ohm's law, the current (I) in the circuit can be expressed as:
\[ I = \frac{E_{eq}}{R_{total}} = \frac{nE}{R + nr} \]

3. **Cells Connected in Parallel**:
- When the cells are connected in parallel, the total EMF (E_eq) remains the same:
\[ E_{eq} = E \]
- The total internal resistance (r_eq) for n cells in parallel is:
\[ r_{eq} = \frac{r}{n} \]
- The total resistance in the circuit when cells are in parallel is:
\[ R_{total} = R + \frac{r}{n} \]
- The current (I') in the circuit can be expressed as:
\[ I' = \frac{E_{eq}}{R_{total}} = \frac{E}{R + \frac{r}{n}} \]

4. **Setting the Currents Equal**:
- According to the problem, the current is the same in both configurations:
\[ I = I' \]
- Therefore, we can set the two expressions for current equal to each other:
\[ \frac{nE}{R + nr} = \frac{E}{R + \frac{r}{n}} \]

5. **Cross-Multiplying and Simplifying**:
- Cross-multiplying gives us:
\[ nE(R + \frac{r}{n}) = E(R + nr) \]
- Simplifying this, we get:
\[ nR + r = R + nr \]
- Rearranging terms leads to:
\[ nR - R = nr - r \]
- Factoring out common terms gives us:
\[ (n - 1)R = (n - 1)r \]

6. **Finding Internal Resistance**:
- Assuming \(n \neq 1\) (since if n=1, there is no difference between series and parallel), we can divide both sides by (n - 1):
\[ R = r \]
- Thus, the internal resistance of each cell is equal to the external resistance.

### Final Answer:
The internal resistance of each cell is given by:
\[ r = R \]