The induced emf in a coil of 10 H inductance in which current varies from 9 A to 4 A in 0.2 s is

Options

A. 200 V
B. 250 V
C. 300 V
D. 350 V

Explanation

To find the induced emf in a coil with a given inductance when the current changes, we can use the formula:

\[ \text{emf} = -L \frac{di}{dt} \]

Where:
- \( L \) is the inductance in henries (H),
- \( di \) is the change in current (A),
- \( dt \) is the change in time (s).

### Step-by-Step Solution:

1. **Identify the given values**:
- Inductance, \( L = 10 \, \text{H} \)
- Initial current, \( I_i = 9 \, \text{A} \)
- Final current, \( I_f = 4 \, \text{A} \)
- Time interval, \( dt = 0.2 \, \text{s} \)

2. **Calculate the change in current (\( di \))**:
\[ di = I_f - I_i = 4 \, \text{A} - 9 \, \text{A} = -5 \, \text{A} \]

3. **Substitute the values into the emf formula**:
\[ \text{emf} = -L \frac{di}{dt} \]
\[ \text{emf} = -10 \, \text{H} \cdot \frac{-5 \, \text{A}}{0.2 \, \text{s}} \]

4. **Calculate \( \frac{di}{dt} \)**:
\[ \frac{di}{dt} = \frac{-5 \, \text{A}}{0.2 \, \text{s}} = -25 \, \text{A/s} \]

5. **Substitute \( \frac{di}{dt} \) back into the emf equation**:
\[ \text{emf} = -10 \cdot (-25) = 250 \, \text{V} \]

6. **Final result**:
The induced emf in the coil is \( 250 \, \text{V} \).