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 will analyze the cases when the cells are connected in series and in parallel, and then find the conditions under which the same current flows in both configurations.

### Step-by-Step Solution:

**Step 1: Analyze the Series Configuration**
- When n cells, each with an EMF \( E \) and internal resistance \( r \), are connected in series, the total EMF \( E_{series} \) is:
\[ E_{series} = nE \]
- The total internal resistance \( R_{internal, series} \) is:
\[ R_{internal, series} = nr \]
- The total resistance in the circuit \( R_{total, series} \) is:
\[ R_{total, series} = R + nr \]
- The current \( I_{series} \) flowing through the circuit can be calculated using Ohm's law:
\[ I_{series} = \frac{E_{series}}{R_{total, series}} = \frac{nE}{R + nr} \]

**Step 2: Analyze the Parallel Configuration**
- When the same n cells are connected in parallel, the total EMF \( E_{parallel} \) remains the same:
\[ E_{parallel} = E \]
- The equivalent internal resistance \( R_{internal, parallel} \) is given by:
\[ R_{internal, parallel} = \frac{r}{n} \]
- The total resistance in the circuit \( R_{total, parallel} \) is:
\[ R_{total, parallel} = R + \frac{r}{n} \]
- The current \( I_{parallel} \) flowing through the circuit can be calculated as:
\[ I_{parallel} = \frac{E_{parallel}}{R_{total, parallel}} = \frac{E}{R + \frac{r}{n}} \]

**Step 3: Set the Currents Equal**
- According to the problem, the same current flows in both configurations:
\[ I_{series} = I_{parallel} \]
- Substituting the expressions for current:
\[ \frac{nE}{R + nr} = \frac{E}{R + \frac{r}{n}} \]

**Step 4: Simplify the Equation**
- Cancel \( E \) from both sides (assuming \( E \neq 0 \)):
\[ \frac{n}{R + nr} = \frac{1}{R + \frac{r}{n}} \]
- Cross-multiplying gives:
\[ n(R + \frac{r}{n}) = R + nr \]
- Expanding both sides:
\[ nR + r = R + nr \]
- Rearranging terms:
\[ nR - R = nr - r \]
- Factoring out common terms:
\[ (n - 1)R = (n - 1)r \]

**Step 5: Solve for Resistance**
- Assuming \( n - 1 \neq 0 \) (which is valid for \( n > 1 \)):
\[ R = r \]

### Conclusion
The condition for the same current to flow in both series and parallel configurations is that the resistance \( R \) connected in the circuit must equal the internal resistance \( r \) of the cells.