In a series $LCR$ circuit,$R = 18 \ \Omega$ and the impedance $Z = 30 \ \Omega$. An $rms$ voltage of $210 \ V$ is applied across the circuit. The true power consumed in the $AC$ circuit is nearly: (in $W$)

$A$ lamp consumes only $50 \%$ of maximum power in an $AC$ circuit. What is the phase difference between the applied voltage and the circuit current?

In an $A$.$C$. circuit,the potential difference $V$ and current $I$ are given respectively by $V = 100 \sin(100t) \text{ V}$ and $I = 100 \sin(100t + \frac{\pi}{3}) \text{ mA}$. The power dissipated in the circuit will be (Given: $\cos \frac{\pi}{3} = \frac{1}{2}$)

Two electric circuits $A$ and $B$ are shown in the figure. The ratio of power factor of circuit $B$ to that of circuit $A$ is

$A$ series $LCR$ circuit is connected to an a.c. source of $230 \ V, 50 \ Hz$. The circuit contains a resistance of $80 \ \Omega$,an inductor having inductive reactance $70 \ \Omega$,and a capacitor of capacitive reactance $130 \ \Omega$. The power factor of the circuit is $x$. The value of $x$ is

The current and voltage functions in an $AC$ circuit are:
$I = 100 \sin(100t) \, mA$
$V = 100 \sin(100t + \frac{\pi}{3}) \, V$
The power dissipated in the circuit is ...... $W$.