$A$ bulb and a capacitor are connected in series to a source of alternating current. If its frequency is increased,while keeping the voltage of the source constant,then

An e.m.f. $E = E_{0} \sin \omega t$ is applied to a circuit containing $L$ and $R$ in series. If $X_{L} = R$,then the power dissipated in the circuit is

$A$ light bulb connected in series with a capacitor and an $A.C.$ source is glowing with a certain brightness. On reducing the capacitance of the capacitor and the frequency of the source,the brightness of the lamp (respectively):

An $AC$ source of angular frequency $\omega$ is connected across a resistor $R$ and a capacitor $C$ in series. The current flowing in the circuit is found to be $I$. Now,the frequency of the source is changed to $\frac{\omega}{3}$ (maintaining the same voltage),and the current in the circuit is found to be halved. What is the ratio of reactance to resistance at the original frequency?

$A$ capacitor of reactance $4 \sqrt{3} \Omega$ and a resistor of resistance $4 \Omega$ are connected in series with an $AC$ source of peak value $8 \sqrt{2} \text{ V}$. The power dissipation in the circuit is . . . . . $\text{W}$.

What will be the reading in the voltmeter and ammeter of the circuit shown?