$A$ coil has self-inductance $L = 0.04 \ H$ and resistance $R = 12 \ \Omega$. When it is connected to a $220 \ V, 50 \ Hz$ supply,what will be the current flowing through the coil (in $A$)?

In a series $LCR$ circuit,$R=300 \Omega$,$L=0.9 \text{ H}$,$C=2 \mu\text{F}$,and $\omega=1000 \text{ rad/s}$. The impedance of the circuit is: (in $\Omega$)

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:

$A$ 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 $\omega/3$, the current becomes half of the original current. Then the ratio of capacitive reactance and resistance 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$)

In the given circuit,the potential difference across the resistance is $54 \, V$ and the power consumed by it is $16 \, W$. If the $AC$ frequency is $60 \, Hz$,find the value of $L$ in $H$.