$A$ 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 $, H$)

In an $LCR$ series resonant circuit,at resonance,the voltage across $L$ and $C$ will cancel each other because they are:

An inductor coil is connected to a capacitor and an $AC$ source of rms voltage $8 \, V$ in series. The rms current in the circuit is $16 \, A$ and is in phase with the emf. If this inductor coil is connected to a $6 \, V$ $DC$ battery, the magnitude of the steady current is (in $A$)

For an $RLC$ 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 given by:

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:

What will be the reading of the voltmeter in the given circuit (in $V$)?