Question

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

• A. $5\sin (\omega t+{90}^{\circ })$
• B. $5\sin (\omega t+{37}^{\circ })$
• C. $5\sin (\omega t+{45}^{\circ })$
• D. $5\sin (\omega t+{53}^{\circ })$

Answer 1

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

Applying $\text{KCL}$ at node

$i_3=i_1+i_2$

$\Rightarrow i_3=3\sin \omega t+4\cos \omega t$

$\Rightarrow i_3=3\sin \omega t+4\sin (\omega t+{90}^{\circ })$


Drawing Phasor diagram,


$|i_3|=\sqrt{3^2+4^2+(2\times 3\times 4\cos {90}^{\circ })}$

$|i_3|=5\text{A}$

Also, $\tan \varphi =\left(\frac{4}{3}\right)$

$\Rightarrow \varphi ={\tan }^{-1}\left(\frac{4}{3}\right)={53}^{\circ }$

$\therefore i_3=5\sin (\omega t+{53}^{\circ })$

Hence, option $\left(\text{D}\right)$ is correct.

Alternate solution: $i_3=i_1+i_2$ $\Rightarrow i_3=3\sin \omega t+4\cos \omega t$ $\Rightarrow i_3=5\times \frac{3}{5}\sin \omega t+5\times \frac{4}{5}\cos \omega t$ $\Rightarrow i_3=5\cos {53}^{\circ }\sin \omega t+5\times \sin {53}^{\circ }\cos \omega t$ $\therefore i_3=5\sin (\omega t+{53}^{\circ })$