question_answer

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]

A)            0.052 H                                   

B)            2.42 H

C)            16.2 mH                                  

D)            1.62 mH

Correct Answer: A

Solution :

                   Current through the bulb \[i=\frac{P}{V}=\frac{60}{10}=6A\]                                       \[V=\sqrt{V_{R}^{2}+V_{L}^{2}}\]                    \[{{(100)}^{2}}={{(10)}^{2}}+V_{L}^{2}\]\[\Rightarrow \,\,{{V}_{L}}=99.5\,\,Volt\]                    Also \[{{V}_{L}}=i{{X}_{L}}=i\times (2\pi \nu L)\]            \[\Rightarrow \,99.5=6\times 2\times 3.14\times 50\times L\]\[\Rightarrow \,\,L=0.052\,\,H\]