The internal resistance of a cell of e.m.f. 12 V is `5 xx 10^(-2) Omega`. It is connected across an unknown resistance. Voltage across the cell, when a current of 60 A is drawn from it, is

Options

A. 15V
B. 12V
C. 9V
D. 6V

Answer

To find the voltage across the cell when a current of 60 A is drawn from it, we can use the formula that relates the electromotive force (e.m.f.), internal resistance, current, and terminal voltage.

### Step-by-Step Solution:

1. **Identify the given values:**
- Electromotive force (e.m.f.), \( E = 12 \, \text{V} \)
- Internal resistance, \( r = 5 \times 10^{-2} \, \Omega \)
- Current drawn, \( I = 60 \, \text{A} \)

2. **Calculate the voltage drop across the internal resistance:**
The voltage drop across the internal resistance can be calculated using Ohm's law:
\[ V_{\text{drop}} = I \times r \]
Substituting the values:
\[ V_{\text{drop}} = 60 \, \text{A} \times 5 \times 10^{-2} \, \Omega = 60 \times 0.05 = 3 \, \text{V} \]

3. **Calculate the terminal voltage across the cell:**
The terminal voltage \( V \) can be found by subtracting the voltage drop from the e.m.f.:
\[ V = E - V_{\text{drop}} \]
Substituting the values:
\[ V = 12 \, \text{V} - 3 \, \text{V} = 9 \, \text{V} \]

4. **Conclusion:**
The voltage across the cell when a current of 60 A is drawn from it is \( 9 \, \text{V} \).