A current of 10A in the primary coil of a circuit is reduced to zero. If the coefficient of mutual inductance is 3 H and emf induced in the secondary coil is 30 kV, time taken for the change of current is

Options

A. `10^3 s`
B. `10^2 s`
C. `10^(-3)s`
D. `10^(-2)s`

Answer

To solve the problem, we need to find the time taken for the change of current in the primary coil, given the mutual inductance, the induced EMF in the secondary coil, and the change in current.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Current in the primary coil (I_initial) = 10 A
- Current in the primary coil (I_final) = 0 A
- Coefficient of mutual inductance (M) = 3 H
- Induced EMF in the secondary coil (E) = 30 kV = 30 × 10^3 V

2. **Calculate the Change in Current:**
\[ \Delta I = I_{\text{final}} - I_{\text{initial}} = 0 - 10 = -10 \text{ A} \]
(The change in current is -10 A, but we will use the absolute value for calculations.)

3. **Use the Formula for Induced EMF:**
The induced EMF (E) in the secondary coil is related to the mutual inductance (M), the change in current (\(\Delta I\)), and the time taken (\(\Delta t\)) by the formula:
\[ E = -M \frac{\Delta I}{\Delta t} \]
Rearranging this formula to solve for \(\Delta t\):
\[ \Delta t = -M \frac{\Delta I}{E} \]

4. **Substituting the Values:**
\[ \Delta t = -3 \frac{-10}{30 \times 10^3} \]
(We can ignore the negative signs since we are interested in the magnitude.)
\[ \Delta t = \frac{3 \times 10}{30 \times 10^3} \]
\[ \Delta t = \frac{30}{30 \times 10^3} = \frac{1}{10^3} = 10^{-3} \text{ seconds} \]

5. **Final Answer:**
The time taken for the change of current is \(10^{-3}\) seconds.