First slide

Alternating current and voltage

Question

One 10 V, 60 W bulb is to be connected to 100 V line. The required induction coil has self inductance of value

Easy

A

0.052 H
B

2.42 H
C

16.2 mH
D

1.62 mH

Solution

Current through the bulb $i = \frac{P}{V} = \frac{60}{10} = 6 A$
$V = \sqrt{V_{R}^{2} + V_{L}^{2}}$ = 99.5 Volt
$V_{L} = i X_{L} \Rightarrow L = 0.052 H$

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