The impedance of an $LR$ circuit with $L=\frac{60}{\pi} \text{ mH}$,$R=8 \Omega$ and frequency $f=50 \text{ Hz}$ is (in $Omega$)

An induction coil has an impedance of $10\,\Omega$. When an $AC$ signal of frequency $1000\,Hz$ is applied to the coil,the voltage leads the current by $45^o$. The inductance of the coil is

An inductive circuit contains a resistance of $10 \, \Omega$ and an inductance of $2.0 \, H$. If an $AC$ voltage of $120 \, V$ and frequency $60 \, Hz$ is applied to this circuit,the current would be nearly: ..... $A$

$A$ virtual current of $4 \, A$ and $50 \, Hz$ flows in an $AC$ circuit containing a coil. The power consumed in the coil is $240 \, W$. If the virtual voltage across the coil is $100 \, V$,its inductance will be:

The power factor of an $R-L$ circuit is $\frac{1}{\sqrt{2}}$. If the frequency of $AC$ is doubled,the power factor will now be

In the $AC$ circuit shown,$E = E_0 \sin(\omega t + \phi)$ and $i = i_0 \sin(\omega t + \phi + \frac{\pi}{4})$. Then,the box contains: