In an $LCR$ circuit, at a particular angular frequency $\omega_0$, the capacitive reactance and inductive reactance are the same. If the angular frequency is doubled to $2\omega_0$, what will be the ratio of the reactance of the capacitor to that of the inductor?

$A$ $20 \Omega$ resistance,$10 \text{ mH}$ inductance coil,and $15 \mu \text{F}$ capacitor are joined in series. When a suitable frequency alternating current source is joined to this combination,the circuit resonates. If the resistance is made $1/3$ rd of its original value,the resonant frequency:

$A$ series $LCR$ circuit is connected to an alternating source of emf $E$. The current amplitude at resonant frequency is $I_0$. If the value of resistance $R$ becomes twice of its initial value,then the amplitude of current at resonance will be

An $AC$ circuit consists of an inductor of inductance $0.5 \, H$ and a capacitor of capacitance $8 \, \mu F$ in series. The current in the circuit is maximum when the angular frequency of the $AC$ source is

In the circuit shown in the figure, the voltmeter reads $75 \, V$. The value of $C$ is ....... $\mu F$.

An $LCR$ circuit is at resonance for a capacitor $C$,inductance $L$,and resistance $R$. If the value of the resistance is halved while keeping all other parameters the same,the current amplitude at resonance will be: