In an $LCR$ circuit at resonance, the $a.c.$ source current is

For an $R-L-C$ circuit,driven with voltage of amplitude $V_m$ and frequency $\omega_0 = \frac{1}{\sqrt{LC}}$,the current exhibits resonance. The quality factor $Q$ is

An inductor of inductance $2 \text{ } \mu\text{H}$ is connected in series with a resistance, a variable capacitor and an a.c. source of $10 \text{ kHz}$. The value of capacitance for which maximum current is drawn in the circuit is $\frac{1}{x} \text{ F}$, where the value of $x$ is (Take $\pi^2 = 10$).

An alternating $e.m.f.$ of frequency $v = \frac{1}{2\pi \sqrt{LC}}$ is applied to a series $LCR$ circuit. For this frequency of the applied $e.m.f.$:

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:

$A$ series resonant $LCR$ circuit has a quality factor ($Q$-factor) $= 0.4$. If $R = 2 \, k\Omega$ and $C = 0.1 \, \mu F$,then the value of inductance is: (in $, H$)