A car battery of emf 12V and internal resistance `0.05 omega` receives a current of 60 A from an external source, then the terminal potential difference of the battery is

Options

A. 32 V
B. 10V
C. 15V
D. 9V

Explanation

To find the terminal potential difference (V) of the car battery, we can use the formula that relates the electromotive force (E), internal resistance (r), and current (I):

\[ V = E - I \times r \]

Where:
- \( V \) = terminal potential difference
- \( E \) = electromotive force (emf) of the battery
- \( I \) = current flowing through the battery
- \( r \) = internal resistance of the battery

### Step-by-Step Solution:

1. **Identify the given values:**
- Electromotive force (E) = 12 V
- Internal resistance (r) = 0.05 Ω
- Current (I) = 60 A

2. **Substitute the values into the formula:**
\[ V = E - I \times r \]
\[ V = 12 V - (60 A \times 0.05 Ω) \]

3. **Calculate the product of current and internal resistance:**
\[ I \times r = 60 A \times 0.05 Ω = 3 V \]

4. **Subtract this value from the emf:**
\[ V = 12 V - 3 V = 9 V \]

5. **Final result:**
The terminal potential difference of the battery is \( V = 9 V \).