In a series $LCR$ circuit,resonance occurs at frequency $f = f_0$. If at this moment the amplitude of current is $I_0$,then calculate the wattless current at the same moment.

$A$ series $LCR$ circuit driven by $300 \, V$ at a frequency of $50 \, Hz$ contains a resistance $R = 3 \, k\Omega$, an inductor of inductive reactance $X_{L} = 250 \pi \, \Omega$ and an unknown capacitor. The value of capacitance to maximize the average power should be: (Take $\pi^{2} = 10$) (in $\mu F$)

The figure shows a series $LCR$ circuit connected to a variable frequency $230 \; V$ source. Given $L = 5.0 \; H$,$C = 80 \; \mu F$,and $R = 40 \; \Omega$.
$(a)$ Determine the source frequency which drives the circuit in resonance.
$(b)$ Obtain the impedance of the circuit and the amplitude of current at the resonating frequency.
$(c)$ Determine the $rms$ potential drops across the three elements of the circuit. Show that the potential drop across the $LC$ combination is zero at the resonating frequency.

In the given circuit,the voltmeter reads $75 \ V$. The value of $C$ is $.... \mu F$. (Given: $\pi^2 = 10$)

In the given circuit,the readings of voltmeters $V_1$ and $V_2$ are $300 \, V$ each. The readings of voltmeter $V_3$ and ammeter $A$ are respectively:

$A$ series $LCR$ circuit of $R=5 \, \Omega, L=20 \, \text{mH}$ and $C=0.5 \, \mu \text{F}$ is connected across an $AC$ supply of $250 \, \text{V}$,having variable frequency. The power dissipated at resonance condition is $..... \times 10^{2} \, \text{W}$.