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 \[{{V}_{L}}=i{{X}_{L}}=i\times (2\pi \nu L)\] \[\Rightarrow \,99.5=6\times 2\times 3.14\times 50\times L\]\[\Rightarrow \,\,L=0.052\,\,H\]You need to login to perform this action.
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