$A$ 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 $V_0 = 230 \ V$ and frequency $f = 50 \ Hz$ is connected to a circuit with $R = 11.5 \ \Omega$,$L = 2.5 \ H$ and a capacitor $C$ all in series. For the current in the circuit to be maximum,the circuit must be in resonance. Find the value of $C$ and the maximum current $I_{max}$ (take $\pi^2 = 10$).

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 $(\pi^2=10)$

$A$ resistor of $100 \Omega$,an inductor of self-inductance $(\frac{4}{\pi^2}) \text{ H}$,and a capacitor of unknown capacitance are connected in series to an $A$.$C$. source of $200 \text{ V}, 50 \text{ Hz}$. When the current and voltage are in phase,the capacitance and the power dissipated are respectively:

$A$ resistor of resistance $R$, an inductor of inductive reactance $2R$, and a capacitor of capacitive reactance $X_C$ are connected in series to an $A.C.$ source. If the series $LCR$ circuit is in resonance, then the power factor of the circuit and the value $X_C$ are respectively:

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: