Question

The coefficient of mutual induction is \(3\)H and induced e.m.f. across secondary is \(4\) kV. Current in primary is reduced from \(7\)A to \(2\)A. The time required for the change of current is

• A. \(3.75\times 10^{-3}\) s
• B. \(2.5\times 10^{-3}\) s
• C. \(4.5\times 10^{-3}\) s
• D. \(3.5\times 10^{-3}\) s

Hint:

Use e = M dI/dt with a 5 A change.

Answer 1

The Correct Option is A

Step 1: Rate of change:
$\dfrac{dI}{dt}=\dfrac{e}{M}=\dfrac{4000}{3}\ \text{A/s}$.

Step 2: Time:
$t=\dfrac{\Delta I}{dI/dt}=\dfrac{5\times3}{4000}=3.75\times10^{-3}$ s.

Step 3: Option:
(A).

Final Answer:
Rate of current change is 4000/3 A/s. \[ \boxed{A} \]