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A 10 V, 60 W bulb is to be connected to a 100 V line. The required induction coil has self-inductance of value \((f=50\ \text{Hz})\)

1. 0.052 H
2. 2.42 H
3. 16.2 mH
4. 1.62 mH

Correct answer: 0.052 H

Solution

The bulb needs current \(I=P/V=60/10=6\) A. The series coil must drop the remaining \(100-10=90\) V, so its reactance is \(X_L=90/6=15\ \Omega\), giving \(L=X_L/(2\pi f)\approx0.052\) H.

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