n equal cell having emf E and internal resistance r, are connected in circuit of a resistance R. Same current flows in circuit either they connected in series or parallel, if:

Options

A. R=nr
B. `R=r/n`
C. `R=n^(2)r`
D. R=r

Explanation

To solve the problem, we need to analyze the conditions under which the same current flows in a circuit when cells are connected in series and parallel. Let's break this down step by step.

### Step 1: Analyze the Series Connection
When \( n \) equal cells, each with an EMF \( E \) and internal resistance \( r \), are connected in series with an external resistance \( R \):

- **Effective EMF in Series**:
\[ E_{\text{effective}} = nE \]

- **Effective Resistance in Series**:
\[ R_{\text{effective}} = nr + R \]

- **Current in Series**:
Using Ohm's law, the current \( I_S \) in the series circuit is given by:
\[ I_S = \frac{E_{\text{effective}}}{R_{\text{effective}}} = \frac{nE}{nr + R} \]

### Step 2: Analyze the Parallel Connection
Now, consider the same cells connected in parallel:

- **Effective EMF in Parallel**:
\[ E_{\text{effective}} = E \]

- **Effective Resistance in Parallel**:
For \( n \) cells in parallel, the internal resistance becomes:
\[ R_{\text{internal}} = \frac{r}{n} \]
Thus, the total effective resistance when connected in parallel with an external resistance \( R \) is:
\[ R_{\text{effective}} = \frac{r}{n} + R \]

- **Current in Parallel**:
The current \( I_P \) in the parallel circuit is:
\[ I_P = \frac{E_{\text{effective}}}{R_{\text{effective}}} = \frac{E}{\frac{r}{n} + R} \]

### Step 3: Set the Currents Equal
According to the problem, the same current flows in both configurations:
\[ I_S = I_P \]
Substituting the expressions we derived:
\[ \frac{nE}{nr + R} = \frac{E}{\frac{r}{n} + R} \]

### Step 4: Simplify the Equation
Cross-multiplying gives:
\[ nE \left( \frac{r}{n} + R \right) = E(nr + R) \]
Since \( E \) is common and non-zero, we can cancel it out:
\[ n \left( \frac{r}{n} + R \right) = nr + R \]
Expanding the left side:
\[ r + nR = nr + R \]

### Step 5: Rearrange the Equation
Rearranging gives:
\[ r + nR - nr - R = 0 \]
This simplifies to:
\[ (n - 1)R = (n - 1)r \]

### Step 6: Conclusion
Dividing both sides by \( (n - 1) \) (assuming \( n \neq 1 \)):
\[ R = r \]

Thus, the condition for the same current to flow in both series and parallel configurations is:
\[ R = r \]

### Final Answer
The correct condition is \( R = r \).

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