Three-phase circuits

Solution

The textbook does not say so, but you only get its answers if you take the values as RMS.

There are three impedances in each line here: one inside the generator, one for the transmission line, and one for the load. The node names below say which is which — ag in the generator, as on the supply side, ad on the demand side.

ea0,ag,0,(120∠30°)
eb0,bg,0,(120∠-90°)
ec0,cg,0,(120∠150°)
rag,ag,as,.4+.3j
rbg,bg,bs,.4+.3j
rcg,cg,cs,.4+.3j
rat,as,ad,.6+.7j
rbt,bs,bd,.6+.7j
rct,cs,cd,.6+.7j
ra0,ad,0,24+19j
rb0,bd,0,24+19j
rc0,cd,0,24+19j

AC, with RMS phasors ticked. Every value is already an impedance, so the frequency is asked for but never used.

The question asks for the line voltages — the differences between the lines. For the answers the textbook wants, ask for these:

In Mini-Tools with aa:

• aa(vag-vbg) = 207.8∠60.00° V
• aa(vbg-vcg) = 207.8∠-60.00° V
• aa(vcg-vag) = 207.8∠180.0° V

The line currents are the currents through the transmission-line impedances:

With polar phasors on, each transmission-line impedance's current through reads:

• irat = 3.748∠-8.660° A
• irbt = 3.748∠-128.66° A
• irct = 3.748∠111.34° A

Finally the complex power. At the source:

Evaluate sea0+seb0+sec0, which gives -1054 − 842.9𝐢 VA.

Note this does not include the power lost in the source's own internal impedances, which you could add if you wanted them. At the load:

Evaluate sra0+srb0+src0, which gives 1012 + 800.8𝐢 VA.

All correct.

Solution

The centre of the generator's wye is ground, and the centre of the load's wye is a node of its own, called n.

ea,a,0,(120∠0°)
eb,b,0,(120∠-120°)
ec,c,0,(120∠120°)
ra,a,n,5j
rb,b,n,10
rc,c,n,-10j

AC, with RMS phasors ticked.

The three line currents read 56.78∠0°, 25.46∠135.0° and 42.76∠-155.1° A. The complex powers in the load are 16120𝐢, 6480 and -18280𝐢 VA, and in the three sources -6814, 790.6 − 2951𝐢 and -456.5 + 5111𝐢 VA.

All correct. Two things are worth checking for yourself. Complex power is conserved, so sea+seb+sec+sra+srb+src evaluates to zero. And the voltage at the centre of the load is not zero, as it was in the balanced case — vn is 120.0 − 283.9𝐢 V.

Solution

Two things differ from the last one. The source values are not RMS this time. And there is a neutral line joining the centre of the load to the centre of the source, which you describe as a short circuit — an s element, which takes just a name and two nodes.

Note also the angles: this is the acb sequence, so they run the other way round.

ea,a,0,(100∠0°)
eb,b,0,(100∠120°)
ec,c,0,(100∠-120°)
sn,0,n
ra,a,n,15
rb,b,n,10+5j
rc,c,n,6-8j

AC, with RMS phasors left unticked this time.

The three line currents read 6.667∠0°, 8.944∠93.43° and 10.00∠-66.87° A, and the neutral current 10.06∠178.5° A.

All four are correct. The neutral carries a current precisely because the load is unbalanced; in the balanced case it would be zero, which is why a balanced three-wire system needs no neutral at all.

Solution

e0a,0,ag,(440∠120°)
eb0,bg,0,(440∠-120°)
sla,ag,ad
slb,bg,bd
slc,0,cd
rab,ad,bd,10-5j
rbc,bd,cd,16
rca,cd,ad,8+6j

aa(isla), aa(islb) and aa(islc) read 39.71∠-41.07°, 64.12∠-139.77° and 70.13∠74.27° A.

Correct — and unlike the balanced case, all three differ in magnitude as well as angle.

Solution

These values are RMS, so tick RMS phasors in Settings — this one does ask for power.

e0a,0,ag,(220∠-120°)
eb0,bg,0,(220∠120°)
sla,ag,ad
slb,bg,bd
slc,0,cd
rab,ad,bd,-5j
rbc,bd,cd,10j
rca,cd,ad,10

The three line currents read 64.00∠80.1°, 38.11∠-60° and 42.50∠-135° A.

For the power, add the three loads' consumption in Evaluate:

prca+prab+prbc

It gives 4840 W.

Correct.

Solution

Two of the three are ordinary element currents. The first — the current inside the generator — is the awkward one, and it is dealt with after them.

e0a,0,ag,(208∠130°)
eb0,bg,0,(208∠-110°)
rla,ag,ad,2+5j
rlb,bg,bd,2+5j
rlc,0,cd,2+5j
rab,ad,bd,50
rbc,bd,cd,30j
rca,cd,ad,-40j

Find equivalent is not needed here — a plain AC solve gives both. aa(irlb) reads 9.106∠168.48° A and aa(irbc) reads 5.500∠172.47° A, matching the printed answers.

The remaining answer, the generator current, is trickier. They ask for the current in the source we did not simulate — and even for one of the two we did have, we could not trust it, since the real circuit uses three.

Now, I tried something, and I think I got lucky, because I got the answer the book gives. This is what I tried:

-(ie0a+ieb0)/3 in Mini-Tools with aa, which takes an expression as readily as a name and answers with the magnitude and the angle.

What I thought was: the current coming out of the two sources in my simulation would, in reality, come out of three. So adding those two currents and dividing by three may approximate the current out of one source. I am confident that holds in a balanced circuit. This one is not balanced, but it was the best I had. So I tried it.

5.959∠-177.18° A

And it worked. That is the answer in the book.