The current in a series $LCR$ circuit will be maximum when $\omega$ is

With a gradual increase in the frequency of an $A.C.$ supply,the impedance of an $LCR$ series circuit

$A$ tuned amplifier circuit is used to generate a carrier frequency of $ 2 \text{ MHz} $ for amplitude modulation. The value of $ \sqrt{LC} $ is

$A$ series $LCR$ circuit is connected to an ac voltage source. When $L$ is removed from the circuit,the phase difference between current and voltage is $\frac{\pi}{3}$. If instead $C$ is removed from the circuit,the phase difference is again $\frac{\pi}{3}$ between current and voltage. The power factor of the circuit is:

On which factor does the sharpness of resonance depend?

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