When current in a coil changes from 25 A to 12 A in 0.1 s , average voltage of 50 V is produced .The self inductance of the coil is

Options

A. a) 2/ 13 H
B. b) 3 / 13 H
C. c) 4 / 13 H
D. d) 5 / 13 H

Answer

To solve the problem, we need to find the self-inductance \( L \) of a coil given the change in current, the time taken for that change, and the average induced voltage.

### Step-by-step Solution:

1. **Identify the given values:**
- Initial current \( I_1 = 25 \, \text{A} \)
- Final current \( I_2 = 12 \, \text{A} \)
- Time interval \( \Delta t = 0.1 \, \text{s} \)
- Average induced voltage (EMF) \( e = 50 \, \text{V} \)

2. **Calculate the change in current (\( \Delta I \)):**
\[ \Delta I = I_1 - I_2 = 25 \, \text{A} - 12 \, \text{A} = 13 \, \text{A} \]

3. **Use the formula for induced EMF:**
The formula relating induced EMF, self-inductance, and the rate of change of current is:
\[ e = -L \frac{\Delta I}{\Delta t} \]
Since we are interested in magnitude, we can drop the negative sign:
\[ e = L \frac{\Delta I}{\Delta t} \]

4. **Rearrange the formula to solve for self-inductance \( L \):**
\[ L = \frac{e \cdot \Delta t}{\Delta I} \]

5. **Substitute the known values into the equation:**
\[ L = \frac{50 \, \text{V} \cdot 0.1 \, \text{s}}{13 \, \text{A}} \]

6. **Calculate the numerator:**
\[ 50 \cdot 0.1 = 5 \]

7. **Now, calculate \( L \):**
\[ L = \frac{5}{13} \, \text{H} \]

8. **Final answer:**
\[ L \approx 0.3846 \, \text{H} \quad \text{(or } \frac{5}{13} \text{ H)} \]