In a series $L-C-R$ circuit,

An electric bulb,an open coil inductor,an $AC$ source,and a key are all connected in series to form a closed circuit. The key is closed and after some time,an iron rod is inserted into the interior of the inductor. Then:

In an $LCR$ series circuit,if $V$ is the effective value of the applied voltage,$V_R$ is the voltage across $R$,and $V_L$ and $V_C$ are the effective voltages across $L$ and $C$ respectively,then:

$A$ resistor and a capacitor are connected in series with an $AC$ source of angular frequency $\omega$. If the frequency is changed to $\omega / 3$ while keeping the voltage constant,the current becomes half. What is the ratio of the reactance to the resistance at the initial frequency?

If $E = 200\,V$,$R = 25\,\Omega$,$L = 2\,H$,and $C = 2\,\mu F$ and the frequency is variable,then the current at $f = 0$ and $f = \infty$ will be respectively:

In an $LR$-circuit,the inductive reactance is equal to the resistance $R$. If the applied voltage is $E = E_0 \cos(\omega t)$,what is the power dissipated in the circuit?