Question #167082

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)


1
Expert's answer
2021-03-03T11:28:51-0500

i1=14.42sin(wt+π2);i2=70.71cos(wt);i3=50eiπ/3i_1=14.42sin(wt+\frac{\pi}{2}); i_2=-70.71cos(wt); i_3=50e^{i\pi/3}

i4=1002π/3;i5=50(cos(π3)+isins(π3))i_4=100 \angle-2\pi/3 ; i_5=50(cos(\frac{\pi}{3})+isins(\frac{\pi}{3}))

As i(t)=1msin(wt+ϕ)i(t)=1_msin(wt+\phi) can be written as i(t)=1m2ϕ    rms valuei(t)=\frac{1_m}{\sqrt{2}}\angle\phi \implies rms \space value

i1=141.422π/2i_1=\frac{141.42}{\sqrt{2}} \angle-\pi/2

i2=70.712π/2i_2=\frac{70.71}{\sqrt{2}}\angle-\pi/2

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

i3=50π/3i_3=50\angle\pi/3

i4=1002π/3i_4=100\angle-2\pi/3

i5=50π/3i_5=50\angle\pi/3


With these values, the phasor diagram is as shown below



Sum of current, iT=i1+i2+i3+i4+i5i_T=i_1+i_2+i_3+i_4+i_5

iT=100π/2+50π/2+50π/3+1002π/3+50π/3i_T=100\angle\pi/2+50\angle-\pi/2+50\angle\pi/3+100\angle-2\pi/3+50\angle\pi/3

iT=50j+100(12+j23)+100(12j23)i_T=50j+100(\frac{1}{2}+j\frac{\sqrt{2}}{3})+100(-\frac{1}{2}-j\frac{\sqrt{2}}{3})

iT=502sin(w+π2)i_T=50\sqrt{2}sin(w+\frac{\pi}{2})


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