The tuning circuit of a radio receiver has a resistance of $50\,\Omega$,an inductor of $10\,mH$ and a variable capacitor. $A$ $1\,MHz$ radio wave produces a potential difference of $0.1\,mV$. The value of the capacitor to produce resonance is.......$pF$ (Take $\pi^2 = 10$).

$A$ 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 $\text{Watt}$	$(4)$ $2250$
	$(5)$ $500$

An inductance of $2 \text{ mH}$,a capacitor of $20 \mu\text{F}$,and a resistance of $50 \Omega$ are connected in series to an $A.C.$ source. The reactance of the inductor and the capacitor are the same. The reactance of either of them will be: (in $\Omega$)

$A$ series $LCR$ circuit with $R = 20 \ \Omega$, $L = 1.6 \ \text{H}$ and $C = 40 \ \mu\text{F}$ is connected to a variable frequency a.c. source. The inductive reactance at resonant frequency is . . . . . . $\Omega$.

At resonance frequency,the impedance in a series $LCR$ circuit is

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: