In an $L-C-R$ circuit,the capacitance is changed from $C$ to $4 C$. For the same resonant frequency,the inductance should be changed from $L$ to

In the circuit shown in the figure,the $AC$ source provides a voltage $V = 20 \cos(2000t)$. Neglecting source resistance,the voltmeter and ammeter readings will be:

$A$ sinusoidal voltage of peak value $283 \;V$ and frequency $50 \;Hz$ is applied to a series $LCR$ circuit in which $R = 3 \;\Omega, L = 25.48 \;mH,$ and $C = 796 \;\mu F.$ Find
$(a)$ the impedance of the circuit;
$(b)$ the phase difference between the voltage across the source and the current;
$(c)$ the power dissipated in the circuit; and
$(d)$ the power factor.

When a capacitor is connected in series with an $LR$ circuit, the alternating current flowing in the circuit:

In an $LCR$ circuit,$R = 100 \ \Omega$. When the capacitor $C$ is removed,the current lags behind the voltage by a phase of $\pi / 3$. When the inductor $L$ is removed,the current leads the voltage by a phase of $\pi / 3$. What is the impedance of the circuit in $\Omega$?

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 ............ $A$.