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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