An inductor of inductance $2\,\mu\text{H}$ is connected in series with a resistance,a variable capacitor,and an $AC$ source of frequency $7\,\text{kHz}$. The value of capacitance for which maximum current is drawn into the circuit is $\frac{1}{x}\text{ F}$,where the value of $x$ is $.........$. (Take $\pi = \frac{22}{7}$)

Resonance frequency of $L-C-R$ series $AC$ circuit is $f_0$. Now,if the inductance is reduced to $\frac{1}{4}$ times and the capacitance is increased to $16$ times,then the new resonance frequency becomes:

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

If the inductance and capacitance are both doubled in an $L-C-R$ circuit,the resonant frequency of the circuit will

An $LCR$ circuit contains $R = 50 \, \Omega$,$L = 1 \, \text{mH}$,and $C = 0.1 \, \mu\text{F}$. The impedance of the circuit will be minimum for a frequency of:

$L$,$C$,and $R$ represent the physical quantities inductance,capacitance,and resistance,respectively. The combination representing the dimension of frequency is: