There is a $5\,\Omega$ resistance in an $AC$ circuit. An inductance of $0.1\,H$ is connected in series with it. If the equation of the $AC$ $EMF$ is $E = 5\sin(50t)$,then the phase difference between the current and the $EMF$ is:

$A$ circuit is made of $1 \Omega$ resistance and $2.5 \text{ mH}$ inductance in series. If an alternating source of $200 \text{ V}$ and $50 \text{ Hz}$ is connected to the circuit,then the phase difference between current and voltage is . . . . . . .

When a d.c. voltage of $200 \, V$ is applied to a coil of self-inductance $\left(\frac{2 \sqrt{3}}{\pi}\right) \, H$, a current of $1 \, A$ flows through it. But by replacing the d.c. source with an a.c. source of $200 \, V$, the current in the coil is reduced to $0.5 \, A$. Then the frequency of the a.c. supply is: (in $ \, Hz$)

$A$ series combination of resistor $R$ and capacitor $C$ is connected to an $A$.$C$. source of angular frequency $\omega$. Keeping the voltage same,if the frequency is changed to $\frac{\omega}{3}$,the current becomes half of the original current. Then the ratio of capacitive reactance and resistance at the former frequency is

$A$ series $LCR$ circuit containing an $a.c.$ source of $100 \text{ V}$ has an inductor and a capacitor of reactance $24 \text{ } \Omega$ and $16 \text{ } \Omega$ respectively. If a resistance of $6 \text{ } \Omega$ is connected in series, then the potential difference across the series combination of inductor and capacitor only is: (in $\text{ V}$)

The current $I$,potential difference $V_L$ across the inductor,and potential difference $V_C$ across the capacitor in the circuit as shown in the figure are best represented vectorially as: