$A$ series $LCR$ circuit containing a resistance $R$ has angular frequency $\omega$. At resonance,the voltage across the resistance and the inductor are $V_R$ and $V_L$ respectively. Then,the value of inductance $L$ will be:

In a series $R-L-C$ $AC$ circuit,for a particular value of $R, L$ and $C$,the power supplied by the source is $P$ at resonance. If the value of inductance is halved,then the power from the source again at resonance is $P'$. Then:

$A$ resistor $R$,an inductor $L$,and a capacitor $C$ are connected in series to an oscillator of frequency $N$. If the resonance frequency is $N_R$,then the current lags behind the voltage when:

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 $\pi = \frac{22}{7}$)

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

In the figure shown,three $AC$ voltmeters are connected. The circuit is at resonance.