Two AC sources  ${\xi }_1=10\sin \left(\omega t\right)V$ and  ${\xi }_2=10\sin \left(\omega t-\frac{\pi }{3}\right)V$ are fed to an LR circuit. The resistance is  $R=10\Omega$ and inductive reactance is  $X_L=10\sqrt{3}\Omega$:

$E=E_1+E_2=10\sqrt{3}\sin \left(\omega t-\frac{\pi }{6}\right)V$         (by phasors) Option (a) is correct.

${\mathrm{I}}_{\text{max}}=\frac{E_{\max }}{Z}=\frac{10\sqrt{3}}{\sqrt{R^2+{\left(X_L\right)}^2}}=\frac{10\sqrt{3}}{\sqrt{100+300}}=\frac{\sqrt{3}}{2}\text{A}$   Option (b) is correct.

$\cos \varphi =\frac{R}{Z}=\frac{10}{20}\Rightarrow \varphi =\frac{\pi }{3}$

Current equation, $I=\frac{\sqrt{3}}{2}\sin \left(\omega t-\frac{\pi }{6}-\frac{\pi }{3}\right)\text{V}\Rightarrow I=-\frac{\sqrt{3}}{2}\text{cos}\omega \text{t\hspace{0.17em}A}$

${\left(V_L\right)}_{\max }={\mathrm{I}}_{\text{max}}\left(X_L\right)=\frac{\sqrt{3}}{2}\left(10\sqrt{3}\right)=15\text{V}$

$V_L=-15\text{cos}\left(\omega \text{t}+\frac{\pi }{2}\right)\text{V}=\text{15sin}\omega \text{t\hspace{0.17em}\hspace{0.17em}V}$

At  $t=\frac{\pi }{3\omega },V_L=\text{15sin}\left(\omega \frac{\pi }{3\omega }\right)=\frac{\text{15}\sqrt{3}}{2}\text{V}$.    Option (c) is correct. Option (d) is wrong.