In a series resonant L-C-R circuit,the voltag | vedclass

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Question

(C) At resonance,the inductive reactance equals the capacitive reactance: $\omega L = \frac{1}{\omega C}$.The current $I$ flowing through the circuit

Vedclass · Accepted Answer

$250$

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(C) At resonance,the inductive reactance equals the capacitive reactance: $\omega L = \frac{1}{\omega C}$.The current $I$ flowing through the circuit is given by $I = \frac{V_R}{R} = \frac{100 \, V}{1000 \, \Omega} = 0.1 \, A$.The voltage across the inductor $L$ is $V_L = I X_L = I \omega L$.Since at resonance $\omega L = \frac{1}{\omega C}$,we can write $V_L = \frac{I}{\omega C}$.Substituting the given values: $V_L = \frac{0.1}{200 	imes 2 	imes 10^{-6}} = \frac{0.1}{400 	imes 10^{-6}} = \frac{0.1}{4 	imes 10^{-4}} = \frac{1000}{4} = 250 \, V$.

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.....$V$

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