Inductance of the inductor, $L=0.50\, H$

Resistance of the resistor, $R =100 \,\Omega$

Potential of the supply voltage, $V=240\, V$

Frequency of the supply, $v=50 \,Hz$

$(a)$ Peak voltage is given as:

$V_{0}=\sqrt{2 V}$

$=\sqrt{2} \times 240=339.41 \,V$

Angular frequency of the supply, $\omega=2 \pi v$ $=2 \pi \times 50=100 \pi\, rad / s$

Maximum current in the circuit is given as:

$I_{0}=\frac{V_{0}}{\sqrt{R^{2}+\omega^{2} L^{2}}}$

$=\frac{339.41}{\sqrt{(100)^{2}+(100 \pi)^{2}(0.50)^{2}}}=1.82\, A$

$(b)$ Equation for voltage is given as: $V = V _{0} \cos \omega t$

Equation for current is given as: $I=I_{0} \cos (\omega t-\phi)$ Where.

$\phi=$ Phase difference between voltage and current At time, $t=0 . V = V _{0}($ voltage is maximum)

For $\omega t-\phi=0$ i.e., at $t=\frac{\phi}{\omega}$

$I=I_{0}$ (current is maximum) Hence, the time lag between maximum voltage and maximum current is $\frac{\phi}{\omega}$

$\tan \phi=\frac{\omega L}{R}$

$=\frac{2 \pi \times 50 \times 0.5}{100}=1.57$

$\phi=57.5^{\circ}=\frac{57.5 \pi}{180} \,rad$

$\omega t=\frac{57.5}{180 \times 2 \pi \times 50}$

$=3.19 \times 10^{-3}\, s$

$=3.2 \,ms$

Now, phase angle $\phi$ is given by the relation, Hence, the time lag between maximum voltage and maximum current is $3.2 \,ms$.