A) \[\frac{\pi }{\sqrt{3}}H\]
B) \[100H\]
C) \[\frac{\sqrt{2}}{\pi }H\]
D) \[\frac{\sqrt{3}}{\pi }H\]
Correct Answer: D
Solution :
[d] From the rating of the bulb, the resistance of the bulb can be calculated. \[R=\frac{{{V}_{rms}}^{2}}{p}=100\Omega \] For the full to be operated at its rated value the rms current through it should be 1A Also, \[rms=\frac{{{V}_{rms}}}{Z}\therefore 1=\frac{200}{\sqrt{{{100}^{2}}+{{(2\pi 50L)}^{2}}}}\Rightarrow L=\frac{\sqrt{3}}{\pi }H\]You need to login to perform this action.
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