$A$ coil of inductance $0.50 \; H$ and resistance $100 \; \Omega$ is connected to a $240 \; V, 50 \; Hz$ $AC$ supply.
$(a)$ What is the maximum current in the coil?
$(b)$ What is the time lag between the voltage maximum and the current maximum?

(A) Given:
Inductance $L = 0.50 \; H$
Resistance $R = 100 \; \Omega$
$RMS$ Voltage $V_{rms} = 240 \; V$
Frequency $f = 50 \; Hz$
$(a)$ Peak voltage $V_0 = \sqrt{2} \times V_{rms} = 1.414 \times 240 = 339.41 \; V$.
Angular frequency $\omega = 2 \pi f = 2 \pi \times 50 = 100 \pi \; rad/s$.
Impedance $Z = \sqrt{R^2 + (\omega L)^2} = \sqrt{100^2 + (100 \pi \times 0.5)^2} = \sqrt{10000 + 24674} = \sqrt{34674} \approx 186.21 \; \Omega$.
Maximum current $I_0 = \frac{V_0}{Z} = \frac{339.41}{186.21} \approx 1.82 \; A$.
$(b)$ The phase angle $\phi$ is given by $\tan \phi = \frac{\omega L}{R} = \frac{100 \pi \times 0.5}{100} = \frac{\pi}{2} \approx 1.57$.
$\phi = \tan^{-1}(1.57) \approx 57.5^{\circ}$.
Converting to radians: $\phi = 57.5 \times \frac{\pi}{180} \approx 1.003 \; rad$.
The time lag $\Delta t = \frac{\phi}{\omega} = \frac{1.003}{100 \pi} \approx 3.19 \times 10^{-3} \; s = 3.2 \; ms$.