To an AC power supply of 220 V at 50 Hz,a r | vedclass

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(B) Given: $V_{rms} = 220 \ V$,$R = 20 \ \Omega$,$X_C = 25 \ \Omega$,$X_L = 45 \ \Omega$.The impedance $Z$ of the $LCR$ series circuit is given by $Z

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$7.8 \ A$ and $45^{\circ}$

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(B) Given: $V_{rms} = 220 \ V$,$R = 20 \ \Omega$,$X_C = 25 \ \Omega$,$X_L = 45 \ \Omega$.The impedance $Z$ of the $LCR$ series circuit is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.Substituting the values: $Z = \sqrt{20^2 + (45 - 25)^2} = \sqrt{400 + 20^2} = \sqrt{400 + 400} = \sqrt{800} = 20\sqrt{2} \ \Omega$.The $RMS$ current $I_{rms}$ is $I_{rms} = \frac{V_{rms}}{Z} = \frac{220}{20\sqrt{2}} = \frac{11}{\sqrt{2}} \approx 7.778 \ A \approx 7.8 \ A$.The phase angle $\phi$ is given by $	an \phi = \frac{X_L - X_C}{R}$.Substituting the values: $	an \phi = \frac{45 - 25}{20} = \frac{20}{20} = 1$.Therefore,$\phi = 	an^{-1}(1) = 45^{\circ}$.

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

To an $AC$ power supply of $220 \ V$ at $50 \ Hz$,a resistor of $20 \ \Omega$,a capacitor of reactance $25 \ \Omega$,and an inductor of reactance $45 \ \Omega$ are connected in series. The corresponding current in the circuit and the phase angle between the current and the voltage are,respectively:

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