An a.c. e.m.f. of peak value $230 \ V$ and frequency $50 \ Hz$ is connected to a circuit with $R=11.5 \ \Omega, L=2.5 \ H$ and a capacitor all in series. The value of capacitance is '$C$' for the current in the circuit to be maximum. The value of '$C$' and maximum current are respectively $(\pi^2=10)$

If the reading of the voltmeter $V$ shown in the figure at resonance is $200 \, V$,then the quality factor of the circuit is:

$A$ series $LCR$ circuit with resistance $R=500 \ \Omega$ is connected to an a.c. source of $250 \ V$. When only the capacitance is removed,the current lags behind the voltage by $60^{\circ}$. When only the inductance is removed,the current leads the voltage by $60^{\circ}$. The impedance of the circuit is

In a series resonant $R-L-C$ circuit, the voltage across $R$ is $100 \, V$ and the value of $R = 1000 \, \Omega$. The capacitance of the capacitor is $2 \times 10^{-6} \, F$; the angular frequency of the $AC$ source is $200 \, rad \, s^{-1}$. Then the potential difference across the inductance coil is: (in $V$)

In a series $L-C-R$ circuit,$C = 10^{-11} \, F$,$L = 10^{-5} \, H$,and $R = 100 \, \Omega$. When a constant $D.C.$ voltage $E$ is applied to the circuit,the capacitor acquires a charge of $10^{-9} \, C$. The $D.C.$ source is replaced by a sinusoidal voltage source in which the peak voltage $E_0$ is equal to the constant $D.C.$ voltage $E$. At resonance,the peak value of the charge acquired by the capacitor will be:

$A$ resonance circuit having inductance and resistance $2 \times 10^{-4} \ H$ and $6.28 \ \Omega$ respectively oscillates at $10 \ MHz$ frequency. The value of the quality factor of this resonator is .........
$[\pi = 3.14]$