Question

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

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

Answer 1

The correct option is A $5a\sin (\omega t+{53}^o)$

$I_1=3Asin\omega t,I_2=4Acos\omega t{I}_1+I_2-I_3=0I_3=I_1+I_{2}3Asin\omega t+4Acos\omega t5A(\frac{3}{5}sin\omega t+\frac{4}{5}cos\omega t)=5A(cos53sin\omega t+sin53cos\omega t)=5A\left(sin(\omega t+53)\right)$