The tuning circuit of a radio receiver has a resistance of $50\,\Omega$,an inductor of $10\,mH$ and a variable capacitor. $A$ $1\,MHz$ radio wave produces a potential difference of $0.1\,mV$. The value of the capacitor to produce resonance is.......$pF$ (Take $\pi^2 = 10$).

$A$ series $LCR$ circuit with $L = \frac{100}{\pi} \text{ mH}$,$C = \frac{10^{-3}}{\pi} \text{ F}$,and $R = 10 \ \Omega$ is connected across an $AC$ source of $220 \text{ V}, 50 \text{ Hz}$ supply. The power factor of the circuit would be . . . . . . .

In a series $LCR$ resonant circuit,$R = 800 \ \Omega$,$C = 2 \ \mu F$,and the voltage across the resistance is $200 \ V$. The angular frequency is $250 \ rad/s$. At resonance,the voltage across the inductance is: (in $V$)

$A$ resistor of $50 \Omega$,an inductor of self-inductance $\left(\frac{3}{\pi^2}\right) H$,and a capacitor of unknown capacity are connected in series to an a.c. source of $100 \ V$ and $50 \ Hz$. When the voltage and current are in phase,the value of capacitance is (nearly):

$A$ $100 \, \Omega$ resistance,a $0.1 \, \mu \text{F}$ capacitor,and an inductor are connected in series across a $250 \, \text{V}$ supply at variable frequency. Calculate the value of inductance of the inductor at which resonance will occur. Given that the resonant frequency is $60 \, \text{Hz}$. (In $\text{H}$)

At resonance,the value of current in a series $L-C-R$ circuit is: (Symbols have their usual meanings.)