$A$ $100 \;\mu\,F$ capacitor in series with a $40\; \Omega$ resistance is connected to a $110\; V, 60\; Hz$ supply.
$(a)$ What is the maximum current in the circuit?
$(b)$ What is the time lag between the current maximum and the voltage maximum?

(A) Capacitance of the capacitor,$C = 100 \;\mu\,F = 100 \times 10^{-6} \;F$.
Resistance of the resistor,$R = 40 \;\Omega$.
Supply voltage $(RMS)$,$V_{rms} = 110 \;V$.
Frequency,$f = 60 \;Hz$.
$(a)$ Angular frequency,$\omega = 2\pi f = 120\pi \;rad/s$.
Impedance $Z = \sqrt{R^2 + X_C^2}$,where $X_C = \frac{1}{\omega C} = \frac{1}{120\pi \times 100 \times 10^{-6}} \approx 26.53 \;\Omega$.
$Z = \sqrt{40^2 + 26.53^2} = \sqrt{1600 + 703.84} = \sqrt{2303.84} \approx 48 \;\Omega$.
Peak voltage $V_0 = V_{rms} \sqrt{2} = 110 \times 1.414 = 155.56 \;V$.
Maximum current $I_0 = \frac{V_0}{Z} = \frac{155.56}{48} \approx 3.24 \;A$.
$(b)$ In an $RC$ circuit,current leads voltage by phase angle $\phi$,where $\tan \phi = \frac{X_C}{R} = \frac{26.53}{40} = 0.66325$.
$\phi = \tan^{-1}(0.66325) \approx 33.55^{\circ} = 0.5856 \;rad$.
The time lag $\Delta t = \frac{\phi}{\omega} = \frac{0.5856}{120\pi} \approx 1.55 \times 10^{-3} \;s = 1.55 \;ms$.