Question

For the circuit shown, current

${\overline{I}}_1=6\angle 30{}^{\circ }Aand{\overline{I}}_2=2\angle 20{}^{\circ }A$

The instantaneous current ${\overline{I}}_3$ at 60 Hz frequency is_____

• A. 5.72 sin (377t + 34.92 )A
• B. 5.72 cos(377t - 34.92°) A
• C. 4.27 sin (314t + 26.91°) A
• D. 4.27 cos (314t - 26.91°) A

Answer 1

The correct option is A 5.72 sin (377t + 34.92 )A

We have: ${\overline{I}}_3={\overline{I}}_1-{\overline{I}}_2$

$=6\angle 30{}^{\circ }-2\angle 20{}^{\circ }$

$=4.045\angle 34.92{}^{\circ }A$

${\overline{I}}_{3max}=\sqrt{2}\times I_{rms}$

$=\sqrt{2}\times 4.045=5.72A$

Here, $\varphi =34.92{}^{\circ }$
and $\omega =2\pi f=2\pi \times 60=377rad/sec$

$\therefore$ Instantaneous current

$\overline{I_3}=5.72\sin (377t+34.92{}^{\circ })A$