Capacitance of the capacitor, $C =100 \,\mu \,F =100 \times 10^{-6} \,F$

Resistance of the resistor, $R =40\, \Omega$

Supply voltage, $V=110\, V$

Frequency of the supply, $v=12 \,k\,Hz =12 \times 10^{3}\, Hz$

Angular Frequency, $\omega=2 \pi v=2 \times \pi \times 12 \times 10^{3}\, rad / s$

$=24 \pi \times 10^{3} \,rad / s$

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

Maximum current, $I_{0}=\frac{V_{o}}{\sqrt{R^{2}+\frac{1}{\omega^{2} C^{2}}}}$

$=\frac{110 \sqrt{2}}{\sqrt{(40)^{2}+\frac{1}{\left(24 \pi \times 10^{3} \times 100 \times 10^{-6}\right)^{2}}}}$

$=\frac{110 \sqrt{2}}{\sqrt{1600+\left(\frac{10}{24 \pi}\right)^{2}}}=3.9 \,A$

For an $RC$ circuit, the voltage lags behind the current by a phase angle of $\phi$ given as:

$\tan \phi=\frac{\frac{1}{\omega C}}{R}=\frac{1}{\omega C R}$

$=\frac{1}{24 \pi \times 10^{3} \times 100 \times 10^{-6} \times 40}$

$\tan \phi=\frac{1}{96 \pi}$

$\therefore \phi=0.2^{\circ}$

Timelag $=\frac{\phi}{\omega}$

$=\frac{0.2 \pi}{180 \times 24 \pi \times 10^{3}}=1.55 \times 10^{-3}\, s =0.04 \,\mu \,s$

Hence, $\phi$ tends to become zero at high frequencies. At a high frequency, capacitor $C$ acts as a conductor.

In a dc circuit, after the steady state is achieved, $\omega=0 .$ Hence, capacitor $C$ amounts to an open circuit.