In an $A.C.$ circuit containing $L$, $C$ and $R$ in series, the ratio of apparent power to the true power is ($Z=$ impedance of the circuit and $R=$ resistance).

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:

$A$ 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 $e.m.f.$ $(\varepsilon)$ 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:

$A$ sinusoidal $AC$ current passes through a resistor of resistance $R$ in an $LCR$ series circuit. If the phase difference between the supplied voltage and the supplied current is $\theta$ and the peak value of the supplied current is $I_0$,then the power dissipated in the circuit is:

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