Question

If $i_1=3\sin \omega t$ and $i_2=4\cos \omega t$, then $i_3$ is

• A. $5\sin (\omega t+{53}^{\circ })$
• B. $5\sin (\omega t+{37}^{\circ })$
• C. $5\sin (\omega t+{45}^{\circ })$
• D. $5\cos (\omega t+{53}^{\circ })$

Answer 1

The correct option is A $5\sin (\omega t+{53}^{\circ })$

From Kirchoff's current law,
$i_3=i_1+i_2=3\sin wt+4\sin (wt+{90}^0)$
$=\sqrt{3^2+4^2+2\left(3\right)\left(4\right)cos{90}^0}\sin (wt+\varphi )$
where $\tan \varphi =\frac{4\sin {90}^0}{3+4\cos {90}^0}=\frac{4}{3}$
$\therefore i_3=5\sin (wt+{53}^0)$