The current in a coil decreases from 5 A to 0 in 0.1 sec. If the average e.m.f induced in the coil is 50 V, then the self inductance of the coil is

Options

A. 0.25 H
B. 0.5 H
C. 1 H
D. 2 H

Explanation

To find the self-inductance of the coil, we can use the formula for the average induced electromotive force (e.m.f.) in terms of self-inductance (L) and the rate of change of current (di/dt):

\[ \text{e.m.f.} = -L \frac{di}{dt} \]

### Step 1: Identify the change in current (di)
The current decreases from 5 A to 0 A. Therefore, the change in current (di) is:

\[ di = I_f - I_i = 0 \, \text{A} - 5 \, \text{A} = -5 \, \text{A} \]

### Step 2: Calculate the rate of change of current (di/dt)
The time interval (dt) is given as 0.1 seconds. Thus, we can calculate di/dt:

\[ \frac{di}{dt} = \frac{di}{dt} = \frac{-5 \, \text{A}}{0.1 \, \text{s}} = -50 \, \text{A/s} \]

### Step 3: Use the average e.m.f. to find self-inductance (L)
We know the average e.m.f. induced in the coil is 50 V. Using the formula for e.m.f., we can rearrange it to solve for L:

\[ 50 \, \text{V} = -L \left(-50 \, \text{A/s}\right) \]

This simplifies to:

\[ 50 = L \cdot 50 \]

### Step 4: Solve for L
Now, we can isolate L:

\[ L = \frac{50}{50} = 1 \, \text{H} \]

### Conclusion
The self-inductance of the coil is:

\[ L = 1 \, \text{Henry} \]