Inductance of the inductor, $L=0.5 \,Hz$

Resistance of the resistor, $R =100\, \Omega$

Potential of the supply voltages, $V=240 \,V$

Frequency of the supply, $v=10 \,k\,Hz =10^{4} \,Hz$

Angular frequency, $\omega=2 \pi v=2 \pi \times 10^{4} \,rad / s$

Peak Voltage $V_{0}=V \sqrt{2}=110 \sqrt{2}\, V$

$\text { Maximum current, } I_{0}=\frac{V_{0}}{\sqrt{R^{2}+\omega^{2} C^{2}}}$

$=\frac{240 \sqrt{2}}{\sqrt{(100)^{2}+\left(2 \pi \times 10^{4}\right)^{2} \times(0.50)^{2}}}=1.1 \times 10^{-2} \,A$

$(b)$ For phase difference $\phi,$ we have the relation:

$\tan \phi=\frac{\omega L}{R}$

$=\frac{2 \pi \times 10^{4} \times 0.5}{100}=100 \pi$

$\phi= 89.82^{\circ}=\frac{89.82 \pi}{180}\, rad$

$\omega t =\frac{89.82 \pi}{180}$

$t=\frac{89.82 \pi}{180 \times 2 \pi \times 10^{4}}=25\, \mu\, s$

It can be observed that $I_{0}$ is very small in this case. Hence, at high frequencies, the inductor amounts to an open circuit.

In a $dc$ circuit, affer a steady state is achieved, $\omega=0 .$ Hence, inductor $L$ behaves like a pure conducting object.