In a coil current falls from 5 A to 0 A in 0.2 s. If an average emf of 150 V is induced, then the self inductance of the coil is

Options

A. 4H
B. 2 H
C. 3 H
D. 6 H

Explanation

To solve the problem, we need to find the self-inductance of the coil using the given information about the change in current and the induced emf.

### Step-by-Step Solution:

1. **Identify the change in current (di)**:
- The current falls from 5 A to 0 A.
- Therefore, the change in current (di) is:
\[ di = 0 A - 5 A = -5 A \]

2. **Calculate the rate of change of current (di/dt)**:
- The time interval (dt) is given as 0.2 s.
- The rate of change of current can be calculated as:
\[ \frac{di}{dt} = \frac{-5 A}{0.2 s} = -25 \, \text{A/s} \]
- Since we are interested in the magnitude, we take:
\[ \left| \frac{di}{dt} \right| = 25 \, \text{A/s} \]

3. **Use the formula for induced emf (E)**:
- The induced emf (E) is related to the self-inductance (L) and the rate of change of current by the formula:
\[ E = L \left| \frac{di}{dt} \right| \]
- Rearranging this gives us the formula for self-inductance:
\[ L = \frac{E}{\left| \frac{di}{dt} \right|} \]

4. **Substitute the known values**:
- We know the average induced emf (E) is 150 V.
- Substituting the values into the formula:
\[ L = \frac{150 \, \text{V}}{25 \, \text{A/s}} = 6 \, \text{H} \]

5. **Conclusion**:
- The self-inductance of the coil is:
\[ L = 6 \, \text{H} \]

### Final Answer:
The self-inductance of the coil is **6 Henry**.