In an $AC$ circuit,current is $3 \ A$,voltage is $210 \ V$,and power is $63 \ W$. The power factor is . . . . . . .

In an $AC$ circuit $E = 50 \sin(500t)$ $V$, $I = 600 \sin(500t + \frac{\pi}{3})$ mA. What is the power dissipated in the circuit (in $\text{ W}$)? [$\cos 60^{\circ} = 0.5$]

$A$ series $LCR$ circuit is connected to an $AC$ source of $220\,V, 50\,Hz$. The circuit contains a resistance $R = 80\,\Omega$,an inductor of inductive reactance $X_L = 70\,\Omega$,and a capacitor of capacitive reactance $X_C = 130\,\Omega$. The power factor of the circuit is $\frac{x}{10}$. The value of $x$ is:

In an $ac$ circuit,the power factor

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

For an $AC$ circuit,the potential difference and current are given by $V = 10 \sqrt{2} \sin \omega t$ (in $V$) and $I = 2 \sqrt{2} \cos \omega t$ (in $A$) respectively. The power dissipated in the instrument is ............ $W$.