The average power dissipated in an $A.C.$ circuit is $2 \ W$. If the current flowing through the circuit is $2 \ A$ and the impedance is $1 \ \Omega$,what is the power factor of the $A.C.$ circuit?

In an $A.C.$ circuit,$V$ and $I$ are given by:
$V = 100 \sin(100 t) \text{ V}$
$I = 100 \sin(100 t + \pi/3) \text{ mA}$
The power dissipated in the circuit is:

In the circuit shown in the figure,if the value of $rms$ current is $2.2\, A$,the power factor of the box 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 $\text{ W}$)? [$\cos 60^{\circ} = 0.5$]

In an $A.C.$ circuit,the $e.m.f.$ $(\varepsilon)$ and the current $(i)$ at any instant are given respectively by $E = E_o \sin \omega t$ and $i = I_o \sin(\omega t - \phi)$. The average power in the circuit over one cycle of $A.C.$ is:

If a current $I$ given by $I = I_0 \sin(\omega t - \frac{\pi}{2})$ flows in an $AC$ circuit across which an $AC$ potential of $E = E_0 \sin(\omega t)$ has been applied,then the power consumption $P$ in the circuit will be