$A$ resistance $R$ draws power $P$ when connected to an $AC$ source. If an inductance is now placed in series with the resistance,such that the impedance of the circuit becomes $Z$,the power drawn will be

$A$ capacitor of $\frac{2.5}{\pi} \mu F$ and a resistor of $3000 \, \Omega$ are connected in series with a $200 \, V, 50 \, Hz$ $AC$ source. What are the power factor and the power dissipated,respectively?

For a series $LCR$ circuit,$R = X_L = 2X_C$. The impedance of the circuit and the phase difference between $V$ and $i$ will be:

An $AC$ voltage source of variable angular frequency $\omega$ and fixed amplitude $V_0$ is connected in series with a capacitance $C$ and an electric bulb of resistance $R$ (inductance zero). When $\omega$ is increased,

In an $L-C-R$ series $AC$ circuit,the voltage across each of the components $L, C$,and $R$ is $50\,V$. The voltage across the $L-R$ combination will be

The given figure represents the phasor diagram of a series $LCR$ circuit connected to an $ac$ source. At the instant $t'$ when the source voltage is given by $V = V_0 \cos(\omega t')$,the current in the circuit will be:
Given: $V_{OL} = 3 \text{ V}$,$V_{OR} = \sqrt{3} \text{ V}$,$V_{OC} = 2 \text{ V}$.