Question

A 50Hz AC current of crest value 1A flows through the primary coil of transformer. If mutual induction between primary and secondary is 1.5H, then the crest voltage induced in the secondary coil is:

• A. 272V
• B. 320V
• C. 415V
• D. 471V

Answer 1

The correct option is D

471V

$I=I_{0}sin\omega t=I_{0}sin2\pi ft{E}_{max}=M\frac{di}{dt}=M\frac{d}{dt}\left(I_{0}sin2\pi ft\right)E_{max}=M{I}_{0}2\pi ft=2\pi M{I}_0=2\times 3.14\times 1.5\times 50=471volt$