Question

Alternating current of peak value \((\frac{2}{π})\) A flows through the primary coil of a transformer. The coefficient of mutual inductance between primary and secondary coils is 1H. The peak value of induced e.m.f. in the secondary coil is
(Frequency of a.c. = 50 Hz)

• A. \(100\text{V}\)
• B. \(300\text{V}\)
• C. \(200\text{V}\)
• D. \(400\text{V}\)

Hint:

The peak emf equals M times omega times the peak current.

Answer 1

The Correct Option is C

Step 1: Formula
\(e = M\frac{dI}{dt}\) with \(I = I_0\sin\omega t\), so the peak emf is \(e_0 = M\omega I_0\).

Step 2: Substitute
\(\omega = 2\pi f = 100\pi\ \text{rad/s}\), \(I_0 = \frac2\pi\ \text{A}\), \(M = 1\ \text{H}\).

Step 3: Result
\(e_0 = 1\times100\pi\times\frac2\pi = 200\ \text{V}\). Option (C).

Final Answer:
The peak induced emf is 200 V. \[ \boxed{\text{(C)}\ 200\ \text{V}} \]