A) \[\frac{1}{2\pi f(2\pi fL+R)}\]
B) \[\frac{1}{\pi f(2\pi fL+R)}\]
C) \[\frac{1}{2\pi f(2\pi fL-R)}\]
D) \[\frac{1}{\pi f(2\pi fL-R)}\]
Correct Answer: A
Solution :
\[\tan \varphi =\frac{{{X}_{C}}-{{X}_{L}}}{R}\] Þ \[\tan {{45}^{o}}=\frac{\frac{1}{2\pi fC}-2\pi fL}{R}\] Þ \[C=\frac{1}{2\pi \,f(2\pi fL+R)}\]
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