Electro Magnetic Induction And Alternating Currents Question 131
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) $ [RPET 1997]
Options:
A) 0.052 H
B) 2.42 H
C) 16.2 mH
D) 1.62 mH
Show Answer
Answer:
Correct Answer: A
Solution:
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}=99.5Volt $
Also $ V_{L}=iX_{L}=i\times (2\pi \nu L) $
$ \Rightarrow ,99.5=6\times 2\times 3.14\times 50\times L $
$ \Rightarrow L=0.052H $
BETA
