At very high frequency, XL increases to infinitely large, hence circuit behaves as open circuit. $X_L$ = 2πf L = 2π(10 × $10^3$) × 0.5 = 31400 Ω (a) Current in the coil, $I_{\mathrm{rms}}$ = $\frac{\varepsilon_{\mathrm{rms}}}{Z}$ Maximum current in the coil, $I_0$ = $\sqrt{2} I_{\mathrm{rms}}$ = $\frac{\sqrt{2} \times \varepsilon_{\mathrm{rms}}}{Z}$ = 1.414 × $\frac{240\,\mathrm{V}}{3.14 \times 10^4\,\Omega}$ A = 1.10 × $10^{-2}$ A This current is extremely small. Thus, at high frequencies, the inductive reactance of an inductor is so large that it behaves as an open circuit. (b) As tan ϕ = $\frac{X_L}{R}$ = $\frac{3.14 \times 10^4}{100}$ = 314 , ϕ = 90° Clearly, time lag = $\frac{90^{\circ}}{360^{\circ}}$ × $\frac{1}{10^4}$ s = 25 × $10^{-6}$ s In dc circuit (after steady state), v = 0 and as such XL = 0. In this case, the inductor behaves like a pure resistor as it has no inductive reactance.