. The value of C is:
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, \[\phi ={{45}^{o}}\] \[\therefore \,\,\,\,\tan {{45}^{o}}=\frac{\omega L-\frac{1}{\omega C}}{R}\] \[\Rightarrow 1=\frac{\omega L-\frac{1}{\omega C}}{R}\] \[\Rightarrow R=\omega L-\frac{1}{\omega L}\] \[\Rightarrow \omega C=\frac{1}{(\omega L-R)}\] \[\Rightarrow C=\frac{1}{\omega (\omega L-R)}=\frac{1}{2\pi f\,(2\pi fL-R)}\] Note: In series resonance L-C-R circuit, \[\frac{1}{\omega CR}\] is greater than unity.
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