Question

If current $I_1=3Asin\omega t$ and $I_2=4Acos\omega t$, then $I_3$ is

• A. $5Asin(\omega t+{53}^{\circ })$
• B. $5Asin(\omega t+{37}^{\circ })$
• C. $5Asin(\omega t+{45}^{\circ })$
• D. $5Asin(\omega t+{30}^{\circ })$

Answer 1

Answer: (A)

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

$I_1=3A\sin \omega t$

$I_2=4A\cos \omega t$

as from kirchoff's law

$I_3=I_1+I_2$

$I_2=3A\sin \omega t+4A\cos \omega t$

Multiplying and dividing the RHS with $\sqrt{(3A{)}^2+(4A{)}^2}=5A$

$I_2=5A(\frac{3}{5}\sin \omega t+\frac{4}{5}\cos \omega t)$

as $\cos {53}^0=\frac{3}{5}$

$\sin {53}^0=\frac{4}{5}$

$\Rightarrow I_3=5A(\cos {53}^0\sin \omega t+\sin {53}^0\cot \omega t)$

$I_2=5A\sin (\cot +{53}^0)$

as $\sin (A+B)=\sin A\cos B+\cos A\sin B$

$I_3=5A\sin (\omega t+{53}^0)$