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

Answer: (A)

$5 \, \sin (\omega t + 53^{\circ})$

Answer 2

From Kirchhoffs current law,
$i_{3}= i_{1} + i_{2} =3 \sin\omega t + 4 \sin \left(\omega t + 90^{\circ}\right)$
$= \sqrt{3^{2} + 4^{2} + 2\left(3\right)\left(4\right)\cos90^{\circ}} \sin \left(\omega t + \phi\right)$
where $\tan \, \, \phi = \frac{ 4 \sin 90^{\circ}}{3+4 \cos 90^{\circ}} = \frac{4}{3}$
$\therefore \, \, i_{3} = 5 \sin\left(\omega t + 53^{\circ}\right)$