In the circuit current through source will be $\left[\text{ Given }\left({\cos }^{-1}(0.6)={53}^{\circ }\right)\right]$

Combination of DC and AC:

Impedence, 

$$\begin{gathered} \mathrm{Z} = \sqrt{{\mathrm{R}}^2 + {{\mathrm{X}}_{\mathrm{L}}}^2} \\ \Rightarrow \mathrm{Z} = \sqrt{{30}^2 + {40}^2} = 50 \mathrm{\Omega } \end{gathered}$$

DC component of current is given by: ${\mathrm{I}}_{\mathrm{dc}} = \frac{{\mathrm{V}}_{\mathrm{dc}}}{\mathrm{z}}=\frac{10}{50}=\frac{1}{5}\mathrm{A}$

AC component of current is given by: ${\mathrm{I}}_{\mathrm{ac}} = \frac{10\sqrt{2} \sin (100\mathrm{\pi t} -8°) }{50}\mathrm{A}$. As phase difference between current and voltage will be:$\varphi = {\cos }^{-1}(0.6)={53}^{\circ }$ and voltage leads the current by ${53}^{0.}$