An e.m.f. $E = E_0 \cos \omega t$ is applied to an $L-R$ circuit. The inductive reactance is equal to the resistance $R$ of the circuit. The power consumed in the circuit is:

When an $AC$ generator of $120 \, V$ is connected in series with a capacitor and a resistor of $30 \, \Omega$,the circuit carries a current of $1.5 \, A$. The potential difference across the capacitor will be ..... $V$.

The given figure represents the phasor diagram of a series $LCR$ circuit connected to an $ac$ source. At the instant $t'$ when the source voltage is given by $V = V_0 \cos(\omega t')$,the current in the circuit will be:
Given: $V_{OL} = 3 \text{ V}$,$V_{OR} = \sqrt{3} \text{ V}$,$V_{OC} = 2 \text{ V}$.

An $AC$ source of angular frequency $\omega$ is connected across a resistor $R$ and a capacitor $C$ in series. The current registered is $I$. If the frequency of the source is changed to $\omega/3$ (maintaining the same voltage),the current in the circuit is found to be halved. Calculate the ratio of reactance to resistance at the original frequency $\omega$.

An $a.c.$ circuit contains a resistance of $12 \ \Omega$ and an inductor of inductive reactance $5 \ \Omega$. The phase angle between current and potential difference will be

In an $L-R$ circuit,the inductive reactance is equal to $\sqrt{3}$ times the resistance $R$ of the circuit. An e.m.f. $E = E_0 \sin(\omega t)$ is applied to the circuit. The power consumed in the circuit is: