$A$ $100 \;\mu F$ capacitor in series with a $40 \;\Omega$ resistance is connected to a $110 \;V, 12 \;kHz$ supply.
$(a)$ What is the maximum current in the circuit?
$(b)$ What is the time lag between the current maximum and the voltage maximum?
Hence,explain the statement that a capacitor is a conductor at very high frequencies. Compare this behaviour with that of a capacitor in a $dc$ circuit after the steady state.

(A) Given: $C = 100 \;\mu F = 100 \times 10^{-6} \;F$,$R = 40 \;\Omega$,$V_{rms} = 110 \;V$,$f = 12 \;kHz = 12 \times 10^3 \;Hz$.
Angular frequency $\omega = 2\pi f = 2 \times \pi \times 12 \times 10^3 = 24\pi \times 10^3 \;rad/s$.
Peak voltage $V_0 = V_{rms} \sqrt{2} = 110 \sqrt{2} \;V$.
Capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{24\pi \times 10^3 \times 100 \times 10^{-6}} = \frac{1}{2.4\pi} \approx 0.1326 \;\Omega$.
Impedance $Z = \sqrt{R^2 + X_C^2} = \sqrt{40^2 + (0.1326)^2} \approx 40 \;\Omega$.
$(a)$ Maximum current $I_0 = \frac{V_0}{Z} = \frac{110 \sqrt{2}}{40} \approx 3.89 \;A \approx 3.9 \;A$.
$(b)$ Phase angle $\phi = \tan^{-1}(\frac{X_C}{R}) = \tan^{-1}(\frac{0.1326}{40}) \approx 0.19^\circ$.
Time lag $\Delta t = \frac{\phi}{\omega} = \frac{0.19 \times \pi}{180 \times 24\pi \times 10^3} \approx 4.4 \times 10^{-8} \;s$.
At very high frequencies,$X_C = \frac{1}{\omega C} \to 0$,so the capacitor acts as a conductor. In a $dc$ circuit,$f=0$,so $\omega=0$ and $X_C \to \infty$,acting as an open circuit.