Question

The coefficient of mutual induction is 2 H and induced e.m.f. across secondary is 2 kV. Current in the primary is reduced from 6 A to 3 A. The time required for the change of current is

• A. \( 6 \times 10^{-3} \, \text{s} \)
• B. \( 5 \times 10^{-3} \, \text{s} \)
• C. \( 3 \times 10^{-3} \, \text{s} \)
• D. \( 4 \times 10^{-3} \, \text{s} \)

Hint:

In mutual induction, the time required for a change in current is inversely proportional to the induced e.m.f. and the mutual inductance.

Answer 1

The Correct Option is C

Step 1: Formula for induced e.m.f. due to change in current.
The induced electromotive force (e.m.f.) in the secondary coil is related to the rate of change of current in the primary coil by the equation: \[ \mathcal{E} = M \frac{dI}{dt} \] where \( M \) is the mutual inductance, \( \frac{dI}{dt} \) is the rate of change of current, and \( \mathcal{E} \) is the induced e.m.f. in the secondary.
Step 2: Rearranging the equation.
Solving for \( \frac{dI}{dt} \): \[ \frac{dI}{dt} = \frac{\mathcal{E}}{M} \] Substituting the given values: \[ \frac{dI}{dt} = \frac{2 \times 10^3}{2} = 10^3 \, \text{A/s} \] Step 3: Finding the time.
The change in current \( \Delta I = 6 - 3 = 3 \, \text{A} \). Using \( \Delta I = \frac{dI}{dt} \times \Delta t \), we can solve for \( \Delta t \): \[ \Delta t = \frac{\Delta I}{\frac{dI}{dt}} = \frac{3}{10^3} = 3 \times 10^{-3} \, \text{s} \] Step 4: Conclusion.
Thus, the time required for the change of current is \( 3 \times 10^{-3} \, \text{s} \), corresponding to option (C).