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

The resistance of the resistor, $\mathrm{R}=40 \Omega$

Supply voltage, $\mathrm{V}=110 \mathrm{~V}$

Frequency, $v=12 \times 10^{3} \mathrm{~Hz}$

Angular frequency, $\omega=2 \pi v=2 \pi \times 12 \times 10^{3}=24 \pi \times 10^{3} \mathrm{rad} / \mathrm{s}$

Maximum current in the circuit

${{I}_{0}}={{V}_{0}}/Z$

${{V}_{0}}=V\sqrt{2}=110\sqrt{2}=155.6~\text{V}$

Here, $Z=\sqrt{R^{2}+\frac{1}{\omega^{2} C^{2}}}$

$Z=\sqrt{40^{2}+\frac{1}{\left(24 \pi \times 10^{3}\right)^{2}\left(10^{-4}\right)^{2}}}$

$Z=\sqrt{1600+\frac{1}{\left(24 \pi \times 10^{3} \times 10^{-4}\right)^{2}}}$

$Z=\sqrt{1600+0.178}$

$Z=\sqrt{1600.178}$

$\text{Z}=40$

${{\text{I}}_{0}}=110\sqrt{2}/40$

${{\text{I}}_{0}}=155.6/40=3.89~\text{A}$

(b) In the capacitor circuit, the voltage will lag behind the current by a phase angle $\Phi$.

Therefore, $\tan \Phi=1 / \omega \mathrm{CR}$

$=1/24\pi \times {{10}^{3}}\times {{10}^{-4}}\times 40$

$\tan \Phi =1/96\pi $

$\Phi =\tan -1(0.0033)={{0.2}^{0}}$

$=0.2\pi /180\text{rad}$

Time lag, $\mathrm{t}=\Phi / \omega=0.2 \pi /\left(180 \times 24 \pi \times 10^{3}\right)$

$=0.2/\left( 4320\times {{10}^{3}} \right)$

$=4.6\times {{10}^{-8}}~\text{s}=0.04\mu \text{s}$

Thus, at high frequency, the capacitor acts like a conductor. For a dc circuit, after steady state, $\omega=0$, the capacitor amounts to an open circuit.