Match List-I with List-II :List-IList-II(a) o | vedclass

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(C) In an $LCR$ series circuit, the phase difference $\phi$ is given by $	an \phi = \frac{X_L - X_C}{R}$.$(a)$ When $\omega L > \frac{1}{\omega C}$,

Vedclass · Accepted Answer

$(a) - (ii); (b) - (i); (c) - (iv); (d) - (iii)$

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(C) In an $LCR$ series circuit, the phase difference $\phi$ is given by $	an \phi = \frac{X_L - X_C}{R}$.$(a)$ When $\omega L > \frac{1}{\omega C}$, i.e., $X_L > X_C$, the circuit is inductive. The voltage leads the current, which means the current lags behind the applied $emf$. Thus, $(a) - (ii)$.$(b)$ When $\omega L = \frac{1}{\omega C}$, i.e., $X_L = X_C$, the circuit is at resonance. The impedance is minimum $(Z = R)$, and the current is in phase with the applied $emf$. Thus, $(b) - (i)$.$(c)$ When $\omega L $(d)$ At resonant frequency, $X_L = X_C$, the impedance is minimum, resulting in maximum current. Thus, $(d) - (iii)$.Therefore, the correct matching is $(a) - (ii), (b) - (i), (c) - (iv), (d) - (iii)$.

List-$I$	List-$II$
$(a)$ $\omega L > \frac{1}{\omega C}$	$(i)$ Current is in phase with $emf$
$(b)$ $\omega L = \frac{1}{\omega C}$	$(ii)$ Current lags behind the applied $emf$
$(c)$ $\omega L < \frac{1}{\omega C}$	$(iii)$ Maximum current occurs
$(d)$ Resonant frequency	$(iv)$ Current leads the $emf$

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