In an $LCR$ series resonant circuit,at resonance,the voltage across $L$ and $C$ will cancel each other because they are:

An $LCR$ series circuit with $100 \Omega$ resistance is connected to an $AC$ source of $200 V$ and angular frequency $300 \text{ rad/s}$. When only the capacitor is removed,the current lags behind the voltage by $60^{\circ}$. When only the inductor is removed,the current leads the voltage by $60^{\circ}$. The power dissipated in the $LCR$ circuit will be:

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 ............. $\,Hz$.

An $LCR$ circuit contains $R = 50 \, \Omega$,$L = 1 \, \text{mH}$,and $C = 0.1 \, \mu\text{F}$. The impedance of the circuit will be minimum for a frequency of:

In a series $LCR$ resonance circuit,if we change the resistance only,from a lower to higher value:

$A$ 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 $V$)