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

In a series $LCR$ circuit,the frequency of the source is half of the resonance frequency. The nature of the circuit will be:

In an $LCR$ circuit,the resonance frequency of the circuit increases to two times its initial value by changing the capacitance from $C$ to $C^{\prime}$ and the resistance from $100 \ \Omega$ to $400 \ \Omega$,while the inductance $L$ is kept constant. The ratio $C / C^{\prime}$ is:

$A$ series $LCR$ circuit is connected to an ac voltage source. When $L$ is removed from the circuit,the phase difference between current and voltage is $\frac{\pi}{3}$. If instead $C$ is removed from the circuit,the phase difference is again $\frac{\pi}{3}$ between current and voltage. The power factor of the circuit is:

If the reading of the voltmeter $V$ shown in the figure at resonance is $200 \, V$,then the quality factor of the circuit is:

In the circuit shown in the figure,neglecting source resistance,the voltmeter and ammeter readings will respectively be: