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

Question

A 50 Hz AC current of crest value 1 A flows through the primary coil of a transformer. If mutual induction between primary and secondary is 1.5 H, the crest voltage induced in the secondary coil is (A) 272V (B) 320V (C) 415V (D) 471V

Tardigrade Admin · Accepted Answer

i=i0 sin ω t=i0 sin 2 π n t E max =M (d i/d t)=N ⋅ (d/d t)(i0 ⋅ sin 2 π n t) =M i0 .2 π n cos 2 π n t ∴ E max =M i0 ⋅ 2 π n =1.5 × 1 × 2 π × 50 =150 π V=471 V

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

272V

320V

415V

471V

$i = i_{0} sin ω t = i_{0} sin 2 πn t$
$E_{m a x} = M \frac{d i}{d t} = N \cdot \frac{d}{d t} (i_{0} \cdot sin 2 πn t)$
$= M i_{0} .2 πn cos 2 πn t$
$∴ E_{m a x} = M i_{0} \cdot 2 πn$
$= 1.5 \times 1 \times 2 π \times 50$
$= 150 πV = 471 V$

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