Five identical cells each of internal resistance `1 Omega` and emf 5V are connected in series and in parallel with an external resistance 'R' . For what value of 'R', current in series and parallel combination will remain the same ?

Options

A. `5Omega`
B. `10Omega`
C. `25Omega`
D. `1Omega`

Answer

To solve the problem, we need to find the value of the external resistance \( R \) such that the current through the series combination of cells is equal to the current through the parallel combination of cells.

### Step 1: Calculate the total EMF and internal resistance in the series combination.

1. **Total EMF in Series**:
- Each cell has an EMF of \( 5V \).
- For 5 cells in series, the total EMF \( E_s \) is:
\[ E_s = 5 \times 5 = 25V \]

2. **Total Internal Resistance in Series**:
- Each cell has an internal resistance of \( 1 \Omega \).
- For 5 cells in series, the total internal resistance \( r_s \) is:
\[ r_s = 5 \times 1 = 5 \Omega \]

### Step 2: Calculate the current in the series combination.

The total resistance in the series circuit is the sum of the internal resistance and the external resistance \( R \):
\[ R_{total, series} = r_s + R = 5 + R \]
Using Ohm's law, the current \( I_s \) in the series circuit is given by:
\[ I_s = \frac{E_s}{R_{total, series}} = \frac{25}{5 + R} \]

### Step 3: Calculate the total EMF and internal resistance in the parallel combination.

1. **Total EMF in Parallel**:
- The EMF remains the same as that of one cell, which is \( 5V \).

2. **Total Internal Resistance in Parallel**:
- For \( n \) identical resistors in parallel, the equivalent resistance \( r_p \) is given by:
\[ \frac{1}{r_p} = \frac{1}{1} + \frac{1}{1} + \frac{1}{1} + \frac{1}{1} + \frac{1}{1} = 5 \]
Thus,
\[ r_p = \frac{1}{5} \Omega \]

### Step 4: Calculate the current in the parallel combination.

The total resistance in the parallel circuit is:
\[ R_{total, parallel} = r_p + R = \frac{1}{5} + R \]
The current \( I_p \) in the parallel circuit is given by:
\[ I_p = \frac{E_p}{R_{total, parallel}} = \frac{5}{\frac{1}{5} + R} \]

### Step 5: Set the currents equal to each other.

To find the value of \( R \) for which the currents are equal, we set \( I_s = I_p \):
\[ \frac{25}{5 + R} = \frac{5}{\frac{1}{5} + R} \]

### Step 6: Cross-multiply and solve for \( R \).

Cross-multiplying gives:
\[ 25 \left(\frac{1}{5} + R\right) = 5(5 + R) \]
Expanding both sides:
\[ 5 + 25R = 25 + 5R \]
Rearranging gives:
\[ 25R - 5R = 25 - 5 \]
\[ 20R = 20 \]
\[ R = 1 \Omega \]

### Conclusion

The value of the external resistance \( R \) for which the current in the series and parallel combination will remain the same is:
\[ \boxed{1 \Omega} \]