In a series resonant L – C – R circuit, the voltage across R is 100 V and $R\ = \ 1\ k\ \Omega$ with $C = 2\mu F$. The resonant frequency $\omega$ is $200\ $rad$s^{- 1}.$ At resonance the voltage across L is

Question

Dr. Abubacker · Accepted Answer

(c) At resonance, $\omega L = \frac{1}{\omega C}$
Current flowing through the circuit,
$I = \frac{V_{R}}{R}$
$= \frac{100}{1000} = 0.1\ A$
So, voltage across L is given by
$V_{L} = IX_{L} = I\omega L$
But $\omega L = \frac{1}{\omega C}$
$	herefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ V_{L} = \frac{1}{\omega C}$
$= \frac{0.1}{200 	imes 2 	imes 10^{- 6}} = 250\ $V

Q. In a series resonant L – C – R circuit, the voltage across R is 100 V and $R\ = \ 1\ k\ \Omega$ with $C = 2\mu F$. The resonant frequency $\omega$ is $200\ $rad$s^{- 1}.$ At resonance the voltage across L is

Explanation

Reviewed by Dr. Abubacker

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