Answer to Question #167082 in Electric Circuits for Vinit Kumawat

Question

Draw the phasor diagram and find the sum of the currents: i1=141.42 Sin(t + /2) , i2 = -70.71 Cos t , i3= 50 ej/3 , i4=100 -2/3 and i5= 50 (Cos /3 + j Sin/3)

Answer 1

$i_1=14.42sin(wt+\frac{\pi}{2}); i_2=-70.71cos(wt); i_3=50e^{i\pi/3}$

$i_4=100 \angle-2\pi/3 ; i_5=50(cos(\frac{\pi}{3})+isins(\frac{\pi}{3}))$

As $i(t)=1_msin(wt+\phi)$ can be written as $i(t)=\frac{1_m}{\sqrt{2}}\angle\phi \implies rms \space value$

$i_1=\frac{141.42}{\sqrt{2}} \angle-\pi/2$

$i_2=\frac{70.71}{\sqrt{2}}\angle-\pi/2$

The polar form of representation i(t) is $r.e^{i\theta} \implies r\angle\theta$

$i_3=50\angle\pi/3$

$i_4=100\angle-2\pi/3$

$i_5=50\angle\pi/3$

With these values, the phasor diagram is as shown below

Sum of current, $i_T=i_1+i_2+i_3+i_4+i_5$

$i_T=100\angle\pi/2+50\angle-\pi/2+50\angle\pi/3+100\angle-2\pi/3+50\angle\pi/3$

$i_T=50j+100(\frac{1}{2}+j\frac{\sqrt{2}}{3})+100(-\frac{1}{2}-j\frac{\sqrt{2}}{3})$

$i_T=50\sqrt{2}sin(w+\frac{\pi}{2})$