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