In a series $LCR$ circuit,$R = 200 \, \Omega$ and the voltage and frequency of the main supply are $220 \, V$ and $50 \, Hz$ respectively. On taking out the capacitance from the circuit,the current lags behind the voltage by $30^\circ$. On taking out the inductor from the circuit,the current leads the voltage by $30^\circ$. The power dissipated in the $LCR$ circuit is......$W$.

$A$ steady current of magnitude $I$ and an $AC$ current of peak value $I$ are allowed to pass through identical resistors for the same time. The ratio of heat produced in the two resistors will be

$A$ resistor of resistance $R$ and an inductor of inductive reactance $R$ are connected in series to an $AC$ source. $A$ capacitor of capacitive reactance $2R$ is then connected in series with $L$ and $R$. The ratio of the power factors of the $LR$ and $LCR$ circuits is:

An $LCR$ circuit is equivalent to a damped oscillator. $A$ capacitor is charged to a charge $Q_0$ as shown and then connected to $L$ and $R$. If a student plots graphs of the square of the maximum charge on the capacitor $(Q^2_{max})$ versus time for two different values of inductors $L_1$ and $L_2$ $(L_1 > L_2)$, which of the following graphs will correctly represent it? (The diagram is schematic and not drawn to scale.)

Assertion : In the purely resistive element of a series $LCR$ $AC$ circuit,the maximum value of $rms$ current increases with an increase in the angular frequency of the applied $e.m.f$.
Reason : $I_{\max} = \frac{\varepsilon_{\max}}{Z}$,where $Z = \sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2}$ and $I_{\max}$ is the peak current in a cycle.

Given below are two statements :
Statement $I$ : In an $LCR$ series circuit,current is maximum at resonance.
Statement $II$ : Current in a purely resistive circuit can never be less than that in a series $LCR$ circuit when connected to the same voltage source.
In the light of the above statements,choose the correct option from the options given below :