Physics - EMI/AC Question with Solution | TestHub

From Kirchhoff's current law, $\begin{array}{l}{i}_3=i_1+i_2=3\sin \omega t+4\sin \left(\omega t+90^{\circ }\right)\\ =\sqrt{3^2+4^2+2(3)(4)\cos 90^{\circ }}\sin (\omega t+\varphi )\\ \text{ where }\tan \varphi =\frac{4\sin 90^{\circ }}{3+4\cos 90^{\circ }}=\frac{4}{3}\\ \therefore i_3=5\sin \left(\omega t+53^{\circ }\right)\end{array}$