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 $\text{ W}$)? [$\cos 60^{\circ} = 0.5$]

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 $watt$)

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 ............ $W$.

If the voltage and current in an ac circuit are respectively $50 \sin (50 t) \text{ V}$ and $50 \sin (50 t + \frac{\pi}{4}) \text{ mA}$,then the power dissipated in the circuit is nearly (in $W$)

Power factor of an ideal choke coil (i.e.,$R=0$) is

When a voltage $V_{s} = 200 \sqrt{2} \sin(\omega t + 15^{\circ})$ is applied to an $AC$ circuit,the current in the circuit is found to be $I = 2 \sin(\omega t + 45^{\circ})$. The average power consumed in the circuit is: