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 oscillations, $v=60 \,Hz$

$(a)$ Angular frequency, $\omega=2 \pi v=2 \pi \times 60 \,rad / s$

For a RC circuit, we have the relation for impedance as:

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

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

Maximum current is given as:

$I_{0}=\frac{V_{0}}{Z}$

$=\frac{V_{0}}{\sqrt{R^{2}+\frac{1}{\omega^{2} C^{2}}}}$

$=\frac{110 \sqrt{2}}{\sqrt{(40)^{2}+\frac{1}{(120 \pi)^{2} \times\left(10^{-4}\right)^{2}}}}$

$=\frac{110 \sqrt{2}}{\sqrt{1600+\frac{10^{2}}{(120 \pi)}}}=3.24 \,A$

$(b)$ In a capacitor circuit, the voltage lags behind the current by a phase angle of $\phi .$ This angle is given by the relation

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

$=\frac{1}{120 \pi \times 10^{-4} \times 40}=0.6635$

$\phi=\tan ^{-1}(0.6635)=33.56^{\circ}$

$=\frac{33.56 \pi}{180} \,rad$

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

$=\frac{33.56 \pi}{180 \times 120 \pi}=1.55 \times 10^{-3} \,S=1.55\, ms$

Hence, the time lag between maximum current and maximum voltage is $1.55 \,ms$.