$A$ resistance of $300\,\Omega$ and an inductance of $\frac{1}{\pi}\,H$ are connected in series to an $AC$ voltage source of $20\,V$ and $200\,Hz$ frequency. The phase angle between the voltage and current is:

In a given series $LCR$ circuit,$R = 4\, \Omega, X_L = 5\, \Omega$ and $X_C = 8\, \Omega$. The current :-

An $a.c.$ 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 $\frac{\omega}{3}$ (while maintaining the same voltage),the current in the circuit is found to be halved. The ratio of reactance to resistance at the original frequency $\omega$ is:

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: $V_{OL} = 3 \text{ V}$,$V_{OR} = \sqrt{3} \text{ V}$,$V_{OC} = 2 \text{ V}$.

In the circuit shown,if the $emf$ of the source at an instant is $5 \, V$,and the potential difference across the capacitor at the same instant is $4 \, V$,then the potential difference across $R$ at that instant may be ..... $V$.

The power factor of a $CR$ circuit is $\frac{1}{\sqrt{2}}$. If the frequency of an $AC$ signal is halved,then the power factor of the circuit will become: