$A$ coil has $L = 0.04 \, H$ and $R = 12 \, \Omega$. When it is connected to a $220 \, V$,$50 \, Hz$ supply,the current flowing through the coil in amperes is: (in $.7$)

When $100 \ V$ $d.c.$ is applied across a solenoid,a current of $1 \ A$ flows in it. When $100 \ V$ $a.c.$ is applied across it,the current drops to $0.5 \ A$. If the frequency is $50 \ Hz$,the impedance and inductance are:

The adjoining figure shows an $AC$ circuit with resistance $R$,inductance $L$ and source voltage $V_s$. Then

In an $RC$ $AC$ circuit,the maximum voltage and maximum current are $100 \ V$ and $1.1 \ A$ respectively. If $X_{C} = 60 \ \Omega$ and $R = 80 \ \Omega$,then the power consumed in the circuit will be . . . . . . . (in $W$)

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$)

When $80 \ V$ $d.c.$ is applied across a solenoid,a current of $0.8 \ A$ flows in it. When $80 \ V$ $a.c.$ is applied across the same solenoid,the current becomes $0.4 \ A$. If the frequency of $a.c.$ source is $50 \ Hz$,the impedance and inductance of the solenoid are nearly: