A) \[\frac{1}{\pi f(2\pi fL+R)}\]
B) \[\frac{1}{\pi f(2\pi fL-R)}\]
C) \[\frac{1}{2\pi f(2\pi fL-R)}\]
D) \[\frac{1}{2\pi f(2\pi fL+R)}\]
Correct Answer: D
Solution :
The phase difference \[\phi \] between current and voltage is given by \[\tan \phi =\frac{{{X}_{L}}-{{X}_{C}}}{R}\] \[\frac{{{X}_{L}}-{{X}_{C}}}{R}=\tan {{45}^{\circ }}=1\] \[{{X}_{C}}={{X}_{C}}+R\] \[\frac{1}{2\pi fC}=2\pi fL+R\] \[C=\frac{1}{2\pi (2\pi fL+R)}\]
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