In an LCR series circuit,if the angular frequ | vedclass

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(A) The capacitive reactance is given by $X_C = \frac{1}{\omega C}$. As $\omega$ increases,$X_C$ decreases continuously. Therefore,$(A)-(iv)$.The indu

Vedclass · Accepted Answer

$(A)-(iv), (B)-(i), (C)-(ii), (D)-(iii)$

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(A) The capacitive reactance is given by $X_C = \frac{1}{\omega C}$. As $\omega$ increases,$X_C$ decreases continuously. Therefore,$(A)-(iv)$.The inductive reactance is given by $X_L = \omega L$. As $\omega$ increases,$X_L$ increases continuously. Therefore,$(B)-(i)$.The resistance $R$ is independent of the angular frequency $\omega$. Therefore,$(C)-(ii)$.The total impedance $Z$ in an $LCR$ series circuit is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$. At resonance,$X_L = X_C$,so $Z$ is minimum. As $\omega$ increases from zero,$Z$ first decreases until resonance and then increases. Therefore,$(D)-(iii)$.

Column-$I$	Column-$II$
$(A)$ Capacitive reactance	$(i)$ Will continuously increase
$(B)$ Inductive reactance	(ii) Will remain constant
$(C)$ Resistance	(iii) Will first decrease and then increase
$(D)$ Total impedance	(iv) Will continuously decrease

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