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

The equivalent emf of the AC source has a peak value of $10\sqrt{3}V$

The peak value of current is $\frac{\sqrt{3}}{2}A$

The voltage across the inductor at $t=\frac{\pi }{3\omega }$ is $\frac{15\sqrt{3}}{2}V$

The voltage across the resistor at $t=\frac{\pi }{3\omega }$ is $\frac{15}{2}V$