In a series LCR circuit, the voltage across R | vedclass

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(C) In a series $LCR$ circuit, the current $I$ is given by $I = V_R / R$. Given $V_R = 100 	ext{ V}$ and $R = 1000 \Omega$, we have $I = 100 / 1000 =

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$250$

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(C) In a series $LCR$ circuit, the current $I$ is given by $I = V_R / R$. Given $V_R = 100 	ext{ V}$ and $R = 1000 \Omega$, we have $I = 100 / 1000 = 0.1 	ext{ A}$.At resonance, the inductive reactance $X_L$ is equal to the capacitive reactance $X_C$.The capacitive reactance is $X_C = 1 / (\omega C) = 1 / (200 	imes 2 	imes 10^{-6}) = 1 / (400 	imes 10^{-6}) = 10^6 / 400 = 2500 \Omega$.Since the circuit is at resonance, $X_L = X_C = 2500 \Omega$.The voltage across the inductor $L$ is $V_L = I 	imes X_L = 0.1 	ext{ A} 	imes 2500 \Omega = 250 	ext{ V}$.

In a series $LCR$ circuit, the voltage across $R$ is $100 \text{ V}$, $R = 1 \text{ k}\Omega$ and $C = 2 \mu\text{F}$. The angular frequency $\omega$ is $200 \text{ rad s}^{-1}$. At resonance, the voltage across '$L$' is (in $V$)

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