$A$ coil of inductance $0.50 \; H$ and resistance $100 \; \Omega$ is connected to a high-frequency supply $(240 \; V, 10 \; kHz)$.
$(a)$ What is the maximum current in the coil?
$(b)$ What is the time lag between the voltage maximum and the current maximum?
Hence,explain the statement that at very high frequency,an inductor in a circuit nearly amounts to an open circuit. How does an inductor behave in a $DC$ circuit after the steady state?

(A) Given: Inductance $L = 0.50 \; H$,Resistance $R = 100 \; \Omega$,$RMS$ Voltage $V_{rms} = 240 \; V$,Frequency $\nu = 10 \; kHz = 10^4 \; Hz$.
$(a)$ Peak voltage $V_0 = V_{rms} \sqrt{2} = 240 \sqrt{2} \approx 339.4 \; V$.
Angular frequency $\omega = 2 \pi \nu = 2 \pi \times 10^4 \; rad/s$.
Inductive reactance $X_L = \omega L = 2 \pi \times 10^4 \times 0.5 = \pi \times 10^4 \approx 31416 \; \Omega$.
Impedance $Z = \sqrt{R^2 + X_L^2} = \sqrt{100^2 + (31416)^2} \approx 31416 \; \Omega$.
Maximum current $I_0 = \frac{V_0}{Z} = \frac{339.4}{31416} \approx 1.08 \times 10^{-2} \; A$.
$(b)$ Phase angle $\phi = \tan^{-1}(\frac{\omega L}{R}) = \tan^{-1}(\frac{31416}{100}) \approx \tan^{-1}(314.16) \approx 89.82^{\circ}$.
Time lag $\Delta t = \frac{\phi}{\omega} = \frac{89.82 \times \pi / 180}{2 \pi \times 10^4} \approx 2.5 \times 10^{-5} \; s = 25 \; \mu s$.
At very high frequencies,$X_L = \omega L$ becomes very large,making $Z$ very large,so $I_0 \to 0$,acting like an open circuit. In a $DC$ circuit at steady state,$\omega = 0$,so $X_L = 0$,and the inductor acts as a pure conductor (short circuit).