Q108 / 120 · VITEEE 2016 · Physics
If $i_{1}=3 \sin \omega t$ and $i_{2}=4 \cos \omega t$, then $i_{3}$ is
- Correct answerA$5 \sin \left(\omega t+53^{\circ}\right)$
- B$5 \sin \left(\omega t+37^{\circ}\right)$
- C$5 \sin \left(\omega t+45^{\circ}\right)$
- D$5 \cos \left(\omega t+53^{\circ}\right)$
Solution
From Kirchhoff's current law,
$\begin{array}{l}
i_{3}=i_{1}+i_{2}=3 \sin \omega t+4 \sin \left(\omega t+90^{\circ}\right) \\
=\sqrt{3^{2}+4^{2}+2(3)(4) \cos 90^{\circ}} \sin (\omega t+\phi) \\
\text { where } \tan \phi=\frac{4 \sin 90^{\circ}}{3+4 \cos 90^{\circ}}=\frac{4}{3} \\
\therefore i_{3}=5 \sin \left(\omega t+53^{\circ}\right)
\end{array}$
Hence, the correct answer is option: 'A'.