For the series $LCR$ circuit connected with a $220 \ V$, $50 \ Hz$ a.c. source as shown in the figure, the power factor is $\frac{\alpha}{10}$. The value of $\alpha$ is . . . . . .

In an $AC$ circuit $E = 200 \sin(50t) \text{ V}$ and $I = 100 \sin(50t + \frac{\pi}{3}) \text{ mA}$. The power dissipated in the circuit is (Given: $\sin 30^{\circ} = \cos 60^{\circ} = 0.5$,$\sin 60^{\circ} = \cos 30^{\circ} = \frac{\sqrt{3}}{2}$) (in $watt$)

If the power factor changes from $0.5$ to $0.25$ because impedance changes from $Z_1$ to $Z_2$, then $Z_1 = x Z_2$. The value of $x$ is (Resistance remains constant).

In a series $LCR$ circuit,$R = 18 \ \Omega$ and the impedance $Z = 33 \ \Omega$. An $RMS$ voltage of $220 \ V$ is applied across the circuit. The true power consumed in the $AC$ circuit is: (in $W$)

Power dissipated in an $LCR$ series circuit connected to an $A.C.$ source of $e.m.f.$ $\varepsilon$ is

Average power in the $L-C-R$ circuit depends upon