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

$A$ resonance circuit having inductance and resistance $2 \times 10^{-4} \ H$ and $6.28 \ \Omega$ respectively oscillates at $10 \ MHz$ frequency. The value of the quality factor of this resonator is .........
$[\pi = 3.14]$

What is resonance in an $LCR$ series circuit?

The frequencies at which the current amplitude in an $LCR$ series circuit becomes $\frac{1}{\sqrt{2}}$ times its maximum value are $212\,rad\,s^{-1}$ and $232\,rad\,s^{-1}$. The value of resistance in the circuit is $R = 5\,\Omega$. The self-inductance in the circuit is $.........\,mH$.

The figure shows a series $LCR$ circuit connected to a variable frequency $230 \; V$ source. Given $L = 5.0 \; H$,$C = 80 \; \mu F$,and $R = 40 \; \Omega$.
$(a)$ Determine the source frequency which drives the circuit in resonance.
$(b)$ Obtain the impedance of the circuit and the amplitude of current at the resonating frequency.
$(c)$ Determine the $rms$ potential drops across the three elements of the circuit. Show that the potential drop across the $LC$ combination is zero at the resonating frequency.

An $LCR$ circuit is at resonance for a capacitor $C$,inductance $L$,and resistance $R$. If the value of the resistance is halved while keeping all other parameters the same,the current amplitude at resonance will be: