In the adjoining $AC$ circuit,the voltmeter whose reading will be zero at resonance is

In a series $RLC$ circuit $(AC)$:
$L = 10 \, mH$
$C = 0.01 \, \mu F$
$R = 50 \, \Omega$
If the supply voltage is $V = 10 \sin \omega t$,find the power dissipated at resonance in $W$.

$A$ $50 \text{ Hz}$ $AC$ circuit has a $10 \text{ mH}$ inductor and a $2 \text{ } \Omega$ resistor in series. The value of capacitance to be placed in series in the circuit to make the circuit power factor as unity is

$A$ $100 \, \Omega$ 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 $\text{H}$)

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 $(\pi^2=10)$

$A$ $L-C-R$ series circuit is connected to a source of alternating current. At resonance,the applied voltage and the current flowing through the circuit will have a phase difference of