When the frequency of the $AC$ voltage applied to a series $LCR$ circuit is gradually increased from a low value, the impedance of the circuit

In a series $LCR$ resonant circuit,$R = 800 \ \Omega$,$C = 2 \ \mu F$,and the voltage across the resistance is $200 \ V$. The angular frequency is $250 \ rad/s$. At resonance,the voltage across the inductance is: (in $V$)

An inductor of $0.5 \ mH$,a capacitor of $20 \ \mu F$ and a resistance of $20 \ \Omega$ are connected in series with a $220 \ V$ a.c. source. If the current is in phase with the e.m.f.,the maximum current in the circuit is $\sqrt{x} \ A$. The value of '$x$' is

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:

An $AC$ voltage $V = 220 \sin(2 \times 10^3 t) \text{ V}$ is applied to a series $LCR$ circuit. The current amplitude in this circuit is: (Given: $L = 10 \text{ mH}$, $C = 25 \mu\text{F}$, $R = 100 \Omega$) (in $\text{ A}$)

$A$ series $LCR$ circuit with $L=0.5 \text{ H}$ and $R=10 \Omega$ is connected to an $AC$ supply with $rms$ voltage and frequency equal to $200 \text{ V}$ and $\frac{150}{\pi} \text{ Hz}$, respectively. The magnitude of the capacitance is varied so that the current amplitude in the circuit becomes maximum. The $rms$ voltage difference across the inductor is (in $\text{ V}$)