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 $



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