In the given series LCR circuit,if the value | vedclass

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(C) In a series $LCR$ circuit,the current $I$ is given by $I = \frac{V}{\sqrt{R^2 + (X_L - X_C)^2}}$.From the circuit diagram,the voltage across the i

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voltage across $L-C$ combination remains same

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(C) In a series $LCR$ circuit,the current $I$ is given by $I = \frac{V}{\sqrt{R^2 + (X_L - X_C)^2}}$.From the circuit diagram,the voltage across the inductor $V_L = 10 	ext{ V}$ and the voltage across the capacitor $V_C = 10 	ext{ V}$.Since $V_L = V_C$,the circuit is in resonance,meaning $X_L = X_C$.The voltage across the $L-C$ combination is $V_{LC} = |V_L - V_C|$.Since $V_L = V_C = 10 	ext{ V}$,$V_{LC} = |10 - 10| = 0 	ext{ V}$.If the resistance $R$ is changed,the current $I$ in the circuit changes.However,the condition for resonance $(X_L = X_C)$ depends only on the frequency of the source and the values of $L$ and $C$,not on $R$.Therefore,$V_L = I X_L$ and $V_C = I X_C$ will change as $I$ changes,but their difference $V_{LC} = I(X_L - X_C)$ will remain $0 	ext{ V}$ because $X_L - X_C = 0$.Thus,the voltage across the $L-C$ combination remains the same (zero).

In the given series $LCR$ circuit,if the value of $R$ is changed,then:

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