Answer:
Option A
Explanation:
From Kirchhoff's current law,
$i_{3}=i_{1}+i_{2}=3\sin\omega t+4\sin (\omega t+90^{0})$
$=\sqrt{3^{2}+4^{2}+2(3)(4)\cos 90^{0}\sin (\omega t+\phi)}$
where $\tan\phi= \frac{4\sin 90^{0}}{3+4\cos 90^{0}}=\frac{4}{3}$
$\therefore$ i3 = $5 (\sin\omega t+53^{0})$
