In a series $R-L-C$ $AC$ circuit,for a particular value of $R, L$ and $C$,the power supplied by the source is $P$ at resonance. If the value of inductance is halved,then the power from the source again at resonance is $P'$. Then:

For the series $LCR$ circuit shown in the figure,what is the resonance frequency and the amplitude of the current at the resonating frequency?

In a series $LR$ circuit, $X_{L} = R$ and the power factor of the circuit is $P_{1}$. When a capacitor with capacitance $C$ such that $X_{L} = X_{C}$ is added in series, the power factor becomes $P_{2}$. The ratio $\frac{P_{1}}{P_{2}}$ is

$A$ series $LCR$ circuit with $L=0.5 \text{ H}$ and $R=10 \Omega$ is connected to an $AC$ supply with $rms$ voltage and frequency equal to $200 \text{ V}$ and $\frac{150}{\pi} \text{ Hz}$, respectively. The magnitude of the capacitance is varied so that the current amplitude in the circuit becomes maximum. The $rms$ voltage difference across the inductor is (in $\text{ V}$)

An inductor of inductance $L$,a capacitor of capacitance $C$,and a resistor of resistance $R$ are connected in series to an $ac$ source. The quality factor of the circuit is

An $L-C-R$ series circuit with $100 \, \Omega$ resistance is connected to an $ac$ source of $100 \, V$ and angular frequency $300 \, rad/s$. When the capacitance is removed, the current lags behind the voltage by $45^{\circ}$. When the inductance is removed, the current leads the voltage by $45^{\circ}$. The current flowing in the circuit will be ... $A$.