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 $\pi^2 = 10$).

$A$ 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 $, H$)

An a.c. e.m.f. of peak value $V_0 = 230 \ V$ and frequency $f = 50 \ Hz$ is connected to a circuit with $R = 11.5 \ \Omega$,$L = 2.5 \ H$ and a capacitor $C$ all in series. For the current in the circuit to be maximum,the circuit must be in resonance. Find the value of $C$ and the maximum current $I_{max}$ (take $\pi^2 = 10$).

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

The self-inductance of the motor of an electric fan is $10\;H$. In order to impart maximum power at $50\;Hz$, it should be connected to a capacitance (in $\mu F$) of:

$A$ series $LCR$ circuit is connected to a $45 \sin (\omega t) \text{ V}$ source. The resonant angular frequency of the circuit is $10^5 \text{ rad s}^{-1}$ and current amplitude at resonance is $I_0$. When the angular frequency of the source is $\omega = 8 \times 10^4 \text{ rad s}^{-1}$, the current amplitude in the circuit is $0.05 I_0$. If $L = 50 \text{ mH}$, match each entry in List-$I$ with an appropriate value from List-$II$ and choose the correct option.

List-$I$	List-$II$
$(P)$ $I_0$ in $\text{mA}$	$(1)$ $44.4$
$(Q)$ The quality factor of the circuit	$(2)$ $18$
$(R)$ The bandwidth of the circuit in $\text{rad s}^{-1}$	$(3)$ $400$
$(S)$ The peak power dissipated at resonance in $\text{Watt}$	$(4)$ $2250$
	$(5)$ $500$