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: Understanding the Concept:
The emf induced in the secondary due to a changing primary current is \(|e|=M\left|\dfrac{dI}{dt}\right|\).

Step 2: Substitute:
The current changes by \(7-2=5\) A in time \(t\).
\[ 4000=3\times\frac5t \]

Step 3: Solve:
\[ t=\frac{15}{4000}=3.75\times10^{-3}\ \text{s} \]

Step 4: Choose:
Option (A).

Final Answer:
The time is 3.75 ms. \[ \boxed{3.75\times10^{-3}\ \text{s}} \]