Switch $S$ is closed at $t = 0$. After a sufficiently long time,an iron rod is inserted into the inductor $L$. Then,the light bulb:

An alternating emf is applied across a parallel combination of a resistance $R$,capacitance $C$ and an inductance $L$. If $I_R$,$I_L$,and $I_C$ are the currents through $R$,$L$,and $C$ respectively,then the diagram which correctly represents the phase relationship among $I_R$,$I_L$,$I_C$,and the source emf $E$ is:

An alternating e.m.f. having voltage $V = V_0 \sin \omega t$ is applied to a series $L-C-R$ circuit. Given: $|X_L - X_C| = R$. The r.m.s. value of potential difference across the capacitor will be:

In the given $AC$ circuit,the phase difference between $I_1$ and $I_2$ is:

An a.c. source of $15 \, V, 50 \, Hz$ is connected across an inductor $(L)$ and resistance $(R)$ in series. The $R.M.S.$ current of $0.5 \, A$ flows in the circuit. The phase difference between the applied voltage and current is $\left(\frac{\pi}{3}\right)$ radian. The value of resistance $(R)$ is $\left(\tan 60^{\circ}=\sqrt{3}\right)$. (in $\Omega$)

When an $AC$ generator of $120 \, V$ is connected in series with a capacitor and a resistor of $30 \, \Omega$,the circuit carries a current of $1.5 \, A$. The potential difference across the capacitor will be ..... $V$.