Physics | VITEEE 2016 | Discussion Page

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if i₁=3 sin ω t and i₂= 4 cos ω t, the i₃ is

$5 (\sin\omega t+53^{0})$

$5 (\sin\omega t+37^{0})$

$5 (\sin\omega t-47^{0})$

$5 (\cos\omega t+53^{0})$

Option A

From Kirchhoff's current law,

$i_{3}=i_{1}+i_{2}=3\sin\omega t+4\sin (\omega t+90^{0})$

$=\sqrt{3^{2}+4^{2}+2(3)(4)\cos 90^{0}\sin (\omega t+\phi)}$

where $\tan\phi= \frac{4\sin 90^{0}}{3+4\cos 90^{0}}=\frac{4}{3}$

$\therefore$ i₃ = $5 (\sin\omega t+53^{0})$

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