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 having $L = 8.0 \, H$,$C = 0.5 \, \mu F$,and $R = 100 \, \Omega$ in series,the resonance frequency in radians per second 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:

$A$ series $LCR$ circuit with $L=0.12\, H$,$C=480\, nF$,$R=23\, \Omega$ is connected to a $230\, V$ variable frequency supply.
$(a)$ What is the source frequency for which current amplitude is maximum? Obtain this maximum value.
$(b)$ What is the source frequency for which average power absorbed by the circuit is maximum? Obtain the value of this maximum power.
$(c)$ For which frequencies of the source is the power transferred to the circuit half the power at resonant frequency? What is the current amplitude at these frequencies?
$(d)$ What is the $Q$-factor of the given circuit?

In a series resonant $LCR$ circuit,the voltage across $R$ is $100 \ V$ and $R = 1 \ k\Omega$ with $C = 2 \ \mu F$. The resonant frequency $\omega$ is $200 \ rad/s$. At resonance,the voltage across $L$ is:

An alternating e.m.f. of $0.2 \, V$ is applied across an $LCR$ series circuit having $R=4 \, \Omega$, $C=80 \, \mu F$, and $L=200 \, mH$. At resonance, the voltage drop across the inductor is (in $V$)