Three-phase circuits

Solution

Although not specified in the textbook, to get the answers they give we need to assume the values are RMS. One must describe the circuit carefully in the case of three-phase circuits, because it is very easy to make a mistake. Below is how I describe this one:

true→s\rms
"ea0,ag,0,(110.∠0°):eb0,bg,0,(110.∠-120°):ec0,cg,0,(110.∠120°):rat,ag,ad,5.–𝐢2.:rbt,bg,bd,5.–𝐢2.:rct,cg,cd,5.–𝐢2.:ra0,ad,0,10.+8.𝐢:rc0,cd,0,10.+8.𝐢:rb0,bd,0,10.+8.𝐢"→cir
s\ac(cir,ω)

Some observations about my description:

• Notice that I have specified the node at the centre of both Y's as node 0. This is something you can do only in the case of balanced Y-Y systems, not for other configurations and not for unbalanced systems.
• Since Symbulator does not differentiate between lower and upper case variables, nodes called a and A would be considered the same node. Instead, we use the names ag and ad, where the g reminds us a node is on the generation side and the d reminds us it is on the demand side.

Once the simulation is completed, we ask for the currents on the transmission lines and get:

{s\aa(irat),s\aa(irbt),s\aa(irct)}

{"6.809ᴇ0∠-21.8°","6.809ᴇ0∠-141.8°","6.809ᴇ0∠98.2°"}

The complex power at the source is sea0+seb0+sec0, which gives −2086.2 − 834.5𝐢 VA. The real part is the average power absorbed by the source; since it is negative, the source is delivering an average power of 2086 W. The imaginary part is the reactive power: 834 var.

The complex power at the load is sra0+srb0+src0, which gives 1390.8 + 1112.6𝐢 VA. The load is consuming an average power of 1391 W and a reactive power of 1113 var.

Before we move on, try simulating this circuit with a different node as the centre of the load's Y array — call it d0, for the demand side. The voltage in that node evaluates to 0 V, because the system is balanced. That is why we could use 0 for both. In an unbalanced system we cannot.

Solution

Nothing says these values are RMS, so leave the convention alone — the question asks for no powers, so it makes no difference either way. The three sources get the angles of the abc sequence.

"ea0,a,0,(100.∠10°):eb0,b,0,(100.∠-110°):ec0,c,0,(100.∠130°):rab,a,b,8.+4.𝐢:rca,c,a,8.+4.𝐢:rbc,b,c,8.+4.𝐢"→cir:s\ac(cir,ω)

The phase currents are the currents in the three load impedances:

{s\aa(irab),s\aa(irbc),s\aa(irca)}

{"19.36ᴇ0∠13.43°","19.36ᴇ0∠-106.57°","19.36ᴇ0∠133.43°"}

The line currents are the currents the sources deliver, which is the opposite of the current through each source element:

{s\aa(-iea0),s\aa(-ieb0),s\aa(-iec0)}

{"33.54ᴇ0∠-16.57°","33.54ᴇ0∠-136.57°","33.54ᴇ0∠103.43°"}

Both sets are correct, and they show the relationship you would expect of a delta load: the line current is √3 times the phase current, and lags it by 30°.

Solution

This one has line impedances, so each source reaches the load through a resistor, and the load's three nodes are separate from the source's three.

"ea1,na1,0,(100.∠0°):eb1,nb1,0,(100.∠-120°):ec1,nc1,0,(100.∠120°):raa,na1,na2,1:rbb,nb1,nb2,1:rcc,nc1,nc2,1:rac,na2,nc2,100.+24.*π*𝐢:rcb,nc2,nb2,100.+24.*π*𝐢:rba,nb2,na2,100.+24.*π*𝐢"→cir:s\ac(cir,ω)

The line current is the current in one of the line resistors, and the phase voltage is the difference between two load nodes:

{s\aa(iraa),s\aa(vna2-vnb2),s\aa(irac)}

{"2.35∠-36.2°","169.94∠30.8°","1.36∠-66.2°"}

Solution

Node c becomes ground, so the source left out is the one opposite it. To read the line currents we add three shorts to act as the lines.

"e0a,0,ag,(330.∠120°):eb0,bg,0,(330.∠-120°):sat,ag,ad:sbt,bg,bd:sct,0,cd:rab,ad,bd,20.-15.𝐢:rbc,bd,cd,20.-15.𝐢:rca,cd,ad,20.-15.𝐢"→cir:s\ac(cir,ω)

The line currents are the currents through the shorts:

{s\aa(isat),s\aa(isbt),s\aa(isct)}

{"22.86ᴇ0∠6.87°","22.86ᴇ0∠-113.13°","22.86ᴇ0∠126.87°"}

Correct.

Solution

false→s\rms
"eca,c,a,(210.∠120°):ebc,b,c,(210.∠-120°):ra,a,0,40.+𝐢25.:rb,b,0,40.+𝐢25.:rcc,c,0,40.+𝐢25."→cir:s\ac(cir,ω)

Node 0 here is the centre of the load's wye, not a node of the source at all.

{s\aa(ira),s\aa(irb),s\aa(ircc)}

{"2.570ᴇ0∠-62.01°","2.570ᴇ0∠177.99°","2.570ᴇ0∠57.99°"}

Correct — balanced, so three equal magnitudes 120° apart.