Given, $E_{rms}$ = 110 V, u = 12 kHz = 12 × $10^{3}$ Hz (a) $X_C$ = $\frac{1}{2 \times 3.14 \times (12 \times 10^{3}) \times 10^{-4}}$ = 0.1326 Ω As, R = 40 Ω, $X_c$ ⋘ R, Z ≈ R = 40 Ω $I_{rms}$ = $\frac{110}{40}$ = 2.75 A , $I_0$ = $\sqrt{2} I_{rms}$ = 1.414 (2.75) = 3.9 A This value of current is same as that without capacitor in the circuit. So, at high frequency, a capacitor offer negligible resistance (0.1326 W in this case), it behave like a conductor. (b) As tan ϕ = $\frac{X_C}{R}$ = $\frac{0.1326 \Omega}{40 \Omega}$ = 0.0033 ϕ = 0.189° = 0° Time lag = $\frac{0.189^{\circ}}{360^{\circ}}$ × $\frac{1}{12 \times 10^{3}}$ = 43.8 × $10^{-9}$ s In dc circuit, after steady state, v = 0 and accordingly, $X_C$ = ∞, i.e., a capacitor amounts to an open circuit, i.e., it is a perfect insulator of current.