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 \(1\) H. The peak em.f. induced in secondary coil is (Frequency of a. c. is \(50\) Hz)

• A. \(25\) V
• B. \(50\) V
• C. \(100\) V
• D. \(200\) V

Hint:

Peak emf = M times omega times peak current.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The emf induced in the secondary is \(\varepsilon_s=-M\dfrac{dI}{dt}\). For \(I=I_0\sin\omega t\), the peak emf is \(M\omega I_0\).

Step 2: Find omega:
\(\omega=2\pi f=2\pi\times50=100\pi\) rad/s.

Step 3: Compute the peak emf:
\(\varepsilon_0=M\omega I_0=1\times100\pi\times\dfrac2\pi=200\) V. Option D.

Step 4: Why the other options are wrong.
25, 50 and 100 V come from missing the factor 2 in \(\omega=2\pi f\) or from using \(f\) instead of \(\omega\).

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