In an $LCR$ series circuit, if the potential differences across the inductor, capacitor, and resistor are $60 \,V$, $30 \,V$, and $40 \,V$ respectively, then the $AC$ voltage applied to the circuit is: (in $\,V$)

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......$A$.

$A$ capacitance of $\left(\frac{10^{-3}}{2 \pi}\right) F$,an inductance of $\left(\frac{100}{\pi}\right) mH$,and a resistance of $10 \Omega$ are connected in series with an $AC$ voltage source of $220 V, 50 Hz$. The phase angle of the circuit is (in $^{\circ}$)

The power factor of an $L-R$ series circuit is $0.6$ and that of a $C-R$ series circuit is $0.5$. If the elements ($L, C,$ and $R$) of the two circuits are joined in series,the power factor of this circuit is found to be $1$. The ratio of the resistance in the $L-R$ circuit to the resistance in the $C-R$ circuit is:

$A$ 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.

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