Consider the $L-C-R$ circuit given below. The circuit is driven by a $50 \,Hz, AC$ source with peak voltage $220 \,V$. If $R=400 \,\Omega, C=200 \,\mu F$ and $L=6 \,H$,the maximum current in the circuit is closest to ............ $A$.

$A$ series $LCR$ circuit containing an $AC$ source of $100 \ V$ has an inductor and a capacitor of reactances $24 \ \Omega$ and $16 \ \Omega$ respectively. If a resistance of $6 \ \Omega$ is connected in series,then the potential difference across the series combination of inductor and capacitor only is (in $V$)

In the circuit shown below, the ac source has voltage $V = 30 \cos(\omega t) \text{ V}$ with $\omega = 2000 \text{ rad/s}$. What will be the amplitude of the current?

In the series $LCR$ circuit, the impedance is: (in $\Omega$)

The figure shows an $LCR$ series $AC$ circuit. If $L$ is removed from the circuit,the current leads the voltage by $45^o$,while if $C$ is removed,the current lags the voltage by $45^o$. The current passing in the original circuit is......$A$.

In an $LCR$ circuit,the potential difference between the terminals of the inductor is $60\,V$,between the terminals of the capacitor is $30\,V$,and between the terminals of the resistor is $40\,V$. The supply voltage will be equal to ......$V$.