First slide

AC Circuits

Question

One 10 V, 60 W bulb is to be connected to 100 V line. The required induction coil has self inductance of value $(f = 50 Hz)$

Moderate

A

0.052 H
B

2.42 H
C

16.2 mH
D

1.62 mH

Solution

Current through the bulb $i = \frac{P}{V} = \frac{60}{10} = 6 A$
$V = \sqrt{V_{R}^{2} + V_{L}^{2}}$
${(100)}^{2} = {(10)}^{2} + V_{L}^{2} \Rightarrow V_{L} = 99 .5 Volt$
Also $V_{L} = {iX}_{L} = i \times (2 πνL)$
$\Rightarrow 99 .5 = 6 \times 2 \times 3 .14 \times 50 \times L \Rightarrow L = 0 .052 H$

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