Average emf of a circuit 'is 200 V when current falls from 5 A to OA in 0.1 second. Self-inductance of the circuit is:

Options

A. 1 H
B. 2 H
C. 3 H
D. 4 H

Explanation

To find the self-inductance (L) of the circuit given the average EMF (E) and the change in current (I) over a specific time (Δt), we can use the formula for induced EMF due to self-inductance:

\[ E = -L \frac{\Delta I}{\Delta t} \]

### Step-by-Step Solution:

1. **Identify the given values**:
- Average EMF, \( E = 200 \, V \)
- Initial current, \( I_1 = 5 \, A \)
- Final current, \( I_2 = 0 \, A \)
- Time interval, \( \Delta t = 0.1 \, s \)

2. **Calculate the change in current (\( \Delta I \))**:
\[ \Delta I = I_2 - I_1 = 0 \, A - 5 \, A = -5 \, A \]

3. **Substitute the values into the EMF formula**:
\[ 200 = -L \frac{-5}{0.1} \]

4. **Simplify the equation**:
\[ 200 = L \frac{5}{0.1} \]

5. **Calculate \( \frac{5}{0.1} \)**:
\[ \frac{5}{0.1} = 50 \]
So the equation becomes:
\[ 200 = 50L \]

6. **Solve for \( L \)**:
\[ L = \frac{200}{50} = 4 \, H \]

### Final Answer:
The self-inductance of the circuit is \( L = 4 \, H \).

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