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

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(C) The current $i$ in the series $R-L-C$ circuit is given by $i = \frac{V_R}{R}$.Substituting the given values: $i = \frac{100 \, V}{1000 \, \Omega}

Vedclass · Accepted Answer

$250$

Vedclass · Answer

(C) The current $i$ in the series $R-L-C$ circuit is given by $i = \frac{V_R}{R}$.Substituting the given values: $i = \frac{100 \, V}{1000 \, \Omega} = 0.1 \, A$.At resonance, the inductive reactance $X_L$ is equal to the capacitive reactance $X_C$, and the potential difference across the inductor $V_L$ is equal to the potential difference across the capacitor $V_C$.The potential difference across the capacitor is $V_C = i X_C = i \left( \frac{1}{\omega C} ight)$.Substituting the values: $V_C = 0.1 	imes \left( \frac{1}{200 	imes 2 	imes 10^{-6}} ight)$.$V_C = 0.1 	imes \left( \frac{1}{400 	imes 10^{-6}} ight) = 0.1 	imes \left( \frac{10^6}{400} ight) = 0.1 	imes 2500 = 250 \, V$.Since $V_L = V_C$ at resonance, the potential difference across the inductance coil is $250 \, V$.

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