$A$ $10 \Omega$ resistance,$5 \ mH$ coil,and $10 \ \mu F$ capacitor are joined in series. When an alternating current source of suitable frequency is joined to this combination,the circuit resonates. If the resistance is halved,the resonant frequency is

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 $\pi^2 = 10$).

$A$ telegraph line of length $100 \, km$ has a capacity of $0.01 \, \mu F/km$ and it carries an alternating current at $0.5 \, kHz$. If minimum impedance is required, then the value of the inductance that needs to be introduced in series is . . . . . . $mH$. (Take $\pi = \sqrt{10}$)

For an $L-C-R$ $AC$ circuit,the resonance frequency is $5000 \ Hz$ and the frequencies at half-power points are $4950 \ Hz$ and $5050 \ Hz$. What will be the $Q$-factor?

On which factor does the sharpness of resonance depend?

In the circuit shown,the readings of voltmeters $V_1, V_2$ and $V_3$ at resonance are given by