The output voltage (taken across the resistance) of an $L-C-R$ series resonant circuit falls to half its peak value at a frequency of $200 \,Hz$ and again reaches the same value at $800 \,Hz$. The bandwidth of this circuit is ............. $\,Hz$.

In a series $R-L-C$ circuit,the frequency of the source is half of the resonance frequency. The nature of the circuit will be

$A$ resistor of $2 \ \Omega$,an inductor of $100 \ \mu H$,and a capacitor of $400 \ pF$ are connected in series across an $A$.$C$. source of $e_{rms} = 0.1 \ V$. At resonance,the voltage drop across the inductor is: (in $V$)

An a.c. e.m.f. of peak value $230 \ V$ and frequency $50 \ Hz$ is connected to a circuit with $R=11.5 \ \Omega, L=2.5 \ H$ and a capacitor all in series. The value of capacitance is '$C$' for the current in the circuit to be maximum. The value of '$C$' and maximum current are respectively $(\pi^2=10)$

An $AC$ circuit consists of an inductor of inductance $0.5 \, H$ and a capacitor of capacitance $8 \, \mu F$ in series. The current in the circuit is maximum when the angular frequency of the $AC$ source is

Obtain the resonant frequency and $Q$-factor of a series $LCR$ circuit with $L=3.0\; H$,$C=27\; \mu F$,and $R=7.4\; \Omega$. It is desired to improve the sharpness of the resonance of the circuit by reducing its 'full width at half maximum' by a factor of $2$. Suggest a suitable way.