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)

• A. 0.052 H
• B. 2.42 H
• C. 16.2 mH
• D. 1.62 mH

Answer 1

The correct option is A 0.052 H

Current through the bulb $i=\frac{P}{V}=\frac{60}{10}=6A$



$V=\sqrt{V_R^2+V_L^2}$
$(100{)}^2=(10{)}^2+V_L^2\Rightarrow V_L=99.5$Volt
Also $V_L=i{X}_L=i\times \left(2\pi vL\right)$
$\Rightarrow 99.5=6\times 2\times 3.14\times 50\times L\Rightarrow L=0.052$ H