For circuits used for transporting electric power,a low power factor implies . . . . . . .

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

In the circuit shown in the figure,if the value of $rms$ current is $2.2\, A$,the power factor of the box is:

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 . . . . . .

The instantaneous values of alternating current and voltage in a circuit are given as $I = \frac{1}{\sqrt{2}} \sin(100\pi t)$ and $E = \frac{1}{\sqrt{2}} \sin(100\pi t + \frac{\pi}{3})$. The average power in watts consumed in the circuit is: