A) \[\frac{1}{\pi v(2\pi vL-R)}\]
B) \[\frac{1}{2\pi v(2\pi vL-R)}\]
C) \[\frac{1}{\pi v(2\pi vL+R)}\]
D) \[\frac{1}{2\pi v(2\pi vL+R)}\]
Correct Answer: D
Solution :
(d) \[\frac{1}{2\pi v(2\pi vL+R)}\] As the current leads the voltage by \[45{}^\circ \], therefore, \[{{X}_{C}}>{{X}_{L}},\tan \phi =\frac{{{X}_{C}}-{{X}_{L}}}{R}=\tan \,45{}^\circ =1\] \[\Rightarrow {{X}_{C}}-{{X}_{L}}=R\] or \[{{X}_{C}}={{X}_{L}}+R=\omega L+R\] \[\frac{1}{\omega C}=\omega L+R,C=\frac{1}{\omega (\omega L+R)}\] \[=\frac{1}{2\pi v(2\pi vL+R)}\]
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