Question

If current $I_1=3A$ sin $\omega t$ and $I_2=4A$ cos $\omega t$, then $I_3$ is :

• A. $5Asin(\omega t+{53}^0)$
• B. $5Asin(\omega t+{37}^0)$
• C. $5Asin(\omega t+{45}^0)$
• D. $5Asin(\omega t+{30}^0)$

Answer 1

The correct option is A $5Asin(\omega t+{53}^0)$

Given,

$I_1=3Asin\omega t$ and $I_2=4Acos\omega t$

$I_3=I_1+I_2$

$\Rightarrow I_3=5\times (\frac{3}{5}sin\omega t+\frac{4}{5}cos\omega t)=5Asin(\omega t+{53}^0)$.

Hence the correct option is A