Question

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

• A. $4 \times 10^{-3}\text{ s}$
• B. $6 \times 10^{-3}\text{ s}$
• C. $2 \times 10^{-3}\text{ s}$
• D. $3 \times 10^{-3}\text{ s}$

Hint:

Remember to convert units! $1\text{ kV} = 1000\text{ V}$.

Answer 1

The Correct Option is D

Step 1: Concept
The induced e.m.f. in the secondary coil is related to the rate of change of current in the primary coil by the formula $e = M \frac{dI}{dt}$.

Step 2: Meaning
$M = 2\text{ H}$, $e = 2\text{ kV} = 2000\text{ V}$, and $dI = 6 - 3 = 3\text{ A}$.

Step 3: Analysis
$2000 = 2 \times \frac{3}{dt}$
$1000 = \frac{3}{dt}$
$dt = \frac{3}{1000} = 3 \times 10^{-3}\text{ s}$.

Step 4: Conclusion
The time required for the change is $3 \times 10^{-3}\text{ s}$. Final Answer: (D)