Resonance frequency of $L-C-R$ series $AC$ circuit is $f_0$. Now,if the inductance is reduced to $\frac{1}{4}$ times and the capacitance is increased to $16$ times,then the new resonance frequency becomes:

$A$ capacitor of $250\,\mu F$ is connected in parallel to a coil of inductance $0.16\,mH$, while its effective resistance is $20\,\Omega$. Determine the resonant frequency.

To increase the resonant frequency in a series $LCR$ circuit,

For the $LCR$ circuit shown here,the current is observed to lead the applied voltage. An additional capacitor $C'$,when joined with the capacitor $C$ present in the circuit,makes the power factor of the circuit unity. The capacitor $C'$ must have been connected in

$A$ sinusoidal voltage having a maximum value of $283 \ V$ and a frequency of $50 \ Hz$ is applied to an $LCR$ series circuit where $R = 3 \ \Omega$,$L = 25.48 \ mH$,and $C = 796 \ \mu F$. What is the impedance at resonance condition (in $Omega$)?

The inductance $L$,capacitance $C$,and resistance $R$ are the values of the components connected in series to an $AC$ source of angular frequency $\omega$. The inductive and capacitive reactances are $X_L$ and $X_C$ respectively. If the circuit is purely resistive,then