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: C
Solution :
\[\tan \phi =\frac{\omega L-\frac{1}{\omega C}}{R}\] \[\phi \]being the angle by which the current leads the voltage. Given, \[\phi ={{45}^{o}}\] \[\therefore \] \[\tan {{45}^{o}}=\frac{\omega L-\frac{1}{\omega C}}{R}\] \[\Rightarrow \] \[1=\frac{\omega L-\frac{1}{\omega C}}{R}\] \[\Rightarrow \] \[R=\omega L-\frac{1}{\omega C}\] \[\Rightarrow \] \[C=\frac{1}{\omega (\omega L-R)}\] \[=\frac{1}{2\pi f(2\pi fL-R)}\] Note: In series\[L\cdot C\cdot R\]circuit,\[\frac{1}{\omega CR}\]is greater than unity.
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