$A$ resistor and a pure inductor are connected to an $AC$ supply of $120\,V$ and $50\,Hz$. The current in the circuit is $3\,A$. If the power consumed in the circuit is $108\,W$,then the resistance in the circuit is.....$\Omega $

In a series $LCR$ circuit, the $rms$ voltage across the resistor and the capacitor are $30 \ V$ and $90 \ V$ respectively. If the applied voltage is $V = 50 \sqrt{2} \sin \omega t$, then the peak voltage across the inductor is

An a.c. voltage source $V=V_0 \sin \omega t$ is connected across resistance $R$ and capacitance $C$ in series. It is given that $R=\frac{1}{\omega C}$ and the peak current is $I_0$. If the angular frequency of the voltage source is changed to $\frac{\omega}{\sqrt{3}}$,then the new peak current in the circuit is

For the $AC$ circuit shown below,the phase difference between emf and current is $\frac{\pi}{4}$ radian as shown in the graph. If the impedance of the circuit is $1414 \Omega$,then the values of $P$ and $Q$ are

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 a given series $LCR$ circuit,$R = 4\,\Omega$,$X_L = 5\,\Omega$,and $X_C = 8\,\Omega$. The current: