Coefficient of mutual induction between 2 coils is 0.4 H. If a current of 4A in the primary is cut off in 1/15000 second, the emf induced in the secondary is

Options

A. 24 KV
B. 15KV
C. 30 KV
D. 10 KV

Answer

To solve the problem, we need to calculate the induced electromotive force (emf) in the secondary coil when the current in the primary coil is cut off. We will use the formula for mutual induction.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Coefficient of mutual induction (M) = 0.4 H
- Initial current in the primary coil (I) = 4 A
- Change in time (dt) = 1/15000 seconds

2. **Calculate Change in Current (di):**
- Since the current is cut off, the final current is 0 A.
- Therefore, the change in current (di) = Final current - Initial current = 0 A - 4 A = -4 A.

3. **Use the Formula for Induced emf:**
- The induced emf (E) in the secondary coil can be calculated using the formula:
\[ E = -M \frac{di}{dt} \]
- Here, M is the mutual inductance, di is the change in current, and dt is the change in time.

4. **Substitute the Values:**
- Substitute M = 0.4 H, di = -4 A, and dt = 1/15000 s into the formula:
\[ E = -0.4 \times \frac{-4}{\frac{1}{15000}} \]

5. **Calculate the Induced emf:**
- First, calculate \(\frac{-4}{\frac{1}{15000}} = -4 \times 15000 = -60000\).
- Now substitute this back into the equation:
\[ E = -0.4 \times -60000 = 24000 \text{ volts} \]
- Therefore, the induced emf (E) = 24000 volts = 24 kV.

6. **Conclusion:**
- The induced emf in the secondary coil is **24 kV**.