Q. 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 $mH$
Explanation
(a) 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 $V_{L} = iX_{L} = i \times (2\pi vL)$
$\Rightarrow 99.5 = 6 \times 2 \times 3.14 \times 50 \times L$
$\Rightarrow L = 0.052\ H$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.