A coil of inductance 0.50 H and resistance 100 Ω 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?

Given,
Inductance, L = 0.50 H
Resistance, R = 100 $\mathrm{\Omega }$
$$\begin{gathered} \mathrm{Effective} \mathrm{voltage}, {\mathrm{E}}_{\mathrm{v}} = 240 \mathrm{V} \\ \mathrm{Frequency} \mathrm{of} \mathrm{the} \mathrm{ac} \mathrm{supply}, \mathrm{f} = 50 \mathrm{Hz} \end{gathered}$$

We can calculate,
Angular frequency, $\mathrm{\omega } = 2\mathrm{\pi f} = 100 \mathrm{\pi }$
Peak voltage,  ${\mathrm{E}}_0 = \sqrt{2} {\mathrm{E}}_{\mathrm{v}} = \sqrt{2} \times 240 \mathrm{V}$

(a) Maximum amount of current in the coil is given by,
                  $$\begin{gathered} {\mathrm{I}}_0 = \frac{{\mathrm{E}}_0}{\sqrt{{\mathrm{R}}^2+{\mathrm{\omega }}^2{\mathrm{L}}^2}} \\ =\frac{\sqrt{2} \times 240}{\sqrt{{10}^4+(100\mathrm{\pi } \times 0.5{)}^2}} \\ = 1.82 \mathrm{A} \end{gathered}$$ 

(b) In LR circuit,

If , $\mathrm{E} = {\mathrm{E}}_0 \cos \mathrm{\omega t} \mathrm{then}, \mathrm{I} = {\mathrm{I}}_0 \cos (\mathrm{\omega t}-\mathrm{\varphi })$ 

Now, 

At $\mathrm{t} = 0, \mathrm{E} = {\mathrm{E}}_o \mathrm{i}.\mathrm{e}., \mathrm{voltage} \mathrm{is} \mathrm{maximum}.$ 

At                   $\mathrm{t} = \frac{\mathrm{\varphi }}{\mathrm{\omega }}, \mathrm{I} = {\mathrm{I}}_0 \cos (\mathrm{\varphi }-\mathrm{\varphi }) = {\mathrm{I}}_0 \times 1, \mathrm{current} \mathrm{is} \mathrm{maximum}$

∴  Time lag between voltage maximum and current maximum $= \frac{\mathrm{\varphi }}{\mathrm{\omega }}$ 

Using the below given relation, we can find the value of $\mathrm{\varphi }$ .
Therefore,
                  
               $$\begin{gathered} \boxed{\tan \mathrm{\varphi } = \frac{\mathrm{\omega L}}{\mathrm{R}}} \\ = \frac{2\mathrm{\pi } \times 50 \times 0.50}{100} \\ = \frac{22}{7 \times 2} \\ = 1.571 \\ \Rightarrow \mathrm{\varphi } = {\tan }^{-1} (1.571) \\ = 57.5 ° \\ = \frac{57.5 \mathrm{\pi }}{180}\mathrm{radian} \end{gathered}$$                              
Thus,

      $$\begin{gathered} \boxed{\mathrm{Time} \mathrm{lag} = \frac{\mathrm{\varphi }}{\mathrm{\omega }}} \\ = \frac{57.5 \mathrm{\pi }}{180 \times 2\mathrm{\pi f}} \\ = \frac{57.5}{180 \times 2\times 50} \\ = 3.19 \times {10}^{-3}\mathrm{s}. \end{gathered}$$        

is the required time lag between the maximum voltage and maximum current.