Question

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

• A. 400 V
• B. 200 V
• C. 300 V
• D. 100 V

Hint:

In transformers, the induced e.m.f. depends on the mutual inductance, frequency, and the peak current.

Answer 1

The Correct Option is B

Step 1: Using the formula for e.m.f. in a transformer.
The induced e.m.f. in the secondary coil is given by: \[ E_2 = M \frac{dI}{dt} \] where \( M \) is the mutual inductance and \( dI/dt \) is the rate of change of current. For alternating current, the peak value of induced e.m.f. is given by: \[ E_2 = M \cdot \omega I_{\text{peak}} = M \cdot 2 \pi f I_{\text{peak}} = 1 \cdot 2 \pi \cdot 50 \cdot \frac{I}{\sqrt{2}} \] Substituting the given values: \[ E_2 = 200 \, \text{V} \] Step 2: Conclusion.
The correct answer is (B), 200 V.