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 containing $5.0 \, H$ inductor,$80 \, \mu F$ capacitor,and $40 \, \Omega$ resistor is connected to a $230 \, V$ variable frequency $AC$ source. The angular frequencies of the source at which the power transferred to the circuit is half the power at the resonant angular frequency are likely to be:

For the $LCR$ circuit shown here,the current is observed to lead the applied voltage. An additional capacitor $C'$,when joined with the capacitor $C$ present in the circuit,makes the power factor of the circuit unity. The capacitor $C'$ must have been connected in

$A$ series $LCR$ circuit contains inductance $5 \ mH$,capacitance $2 \ \mu F$,and resistance $10 \ \Omega$. If the frequency of the $A.C.$ source is varied,what is the frequency at which maximum power is dissipated?

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

An $LCR$ series circuit is connected to an external $emf$,$e = 200 \sin(100 \pi t) \ V$. The values of capacitance and resistance in the circuit are $1 \ \mu F$ and $100 \ \Omega$ respectively. The amplitude of current in the circuit is maximum when the inductance is (in henry):