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If $i_{1}=3 \sin \omega t$ and $i_{2}=4 \cos \omega t$, then $i_{3}$ is
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Verified Answer
The correct answer is:
$5 \sin \left(\omega t+53^{\circ}\right)$
From Kirchhoff's current law,
$\begin{array}{l}
i_{3}=i_{1}+i_{2}=3 \sin \omega t+4 \sin \left(\omega t+90^{\circ}\right) \\
=\sqrt{3^{2}+4^{2}+2(3)(4) \cos 90^{\circ}} \sin (\omega t+\phi) \\
\text { where } \tan \phi=\frac{4 \sin 90^{\circ}}{3+4 \cos 90^{\circ}}=\frac{4}{3} \\
\therefore i_{3}=5 \sin \left(\omega t+53^{\circ}\right)
\end{array}$
$\begin{array}{l}
i_{3}=i_{1}+i_{2}=3 \sin \omega t+4 \sin \left(\omega t+90^{\circ}\right) \\
=\sqrt{3^{2}+4^{2}+2(3)(4) \cos 90^{\circ}} \sin (\omega t+\phi) \\
\text { where } \tan \phi=\frac{4 \sin 90^{\circ}}{3+4 \cos 90^{\circ}}=\frac{4}{3} \\
\therefore i_{3}=5 \sin \left(\omega t+53^{\circ}\right)
\end{array}$
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