The resonance frequency of an $L-C-R$ $AC$ circuit is $\nu_{0}$. If the capacitance is made $4$ times its initial value,then the new resonance frequency will become . . . . . . .

In a series $LR$ circuit with $X_L = R$, the power factor is $P_1$. If a capacitor of capacitance $C$ with $X_C = X_L$ is added to the circuit, the power factor becomes $P_2$. The ratio of $P_1$ to $P_2$ will be:

$A$ series $LCR$ circuit containing a resistance of $120\,\Omega$ has a resonance frequency of $4 \times 10^5\, rad\, s^{-1}$. The voltages,at resonance,across the resistance and inductance are $60\,V$ and $40\,V$ respectively. The values of $L$ and $C$ respectively are:

In the circuit shown,the power factor of the circuit is $1$ and the power factor of the box is $\frac{3}{5}$. Find the reading of the ammeter in $A$.

The $Q$ factor of a series $LCR$ circuit with $L = 2 \ H$,$C = 32 \ \mu F$,and $R = 10 \ \Omega$ 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 ... $A$.