CHAPTER 12 - THREE-PHASE CIRCUITS

List of topics for this chapter :Balanced Three-Phase VoltagesBalanced Wye-Wye ConnectionBalanced Wye-Delta ConnectionBalanced Delta-Delta ConnectionBalanced Delta-Wye ConnectionPower in a Balanced SystemUnbalanced Three-Phase SystemsPSpice for Three-Phase CircuitsApplications

BALANCED THREE-PHASE VOLTAGES

Problem 12.1 Given = 110Van , = 0VV abab , and > 0 , determine all the phasevoltages, line voltages, and the phase sequence.

The easiest way to completely solve for all of the elements of the balanced 3-phase voltages is todo it graphically. All balanced 3-phase voltages generate the same shape. We start off bysketching Vab, which has to be a horizontal line pointing to the right. Next we know that the valueof Van has its head connected to a, either coming down (indicating its phase angle is negative)or rising up from below (indicating that its phase angle is positive.

It is now easy to determine all of the values and the rotation.

ab

c

n110

30|Vab|

30

Clearly the rotation is acb, or a negative phase sequence.

Because this is a triangle with equal sides, the interior angles are 60 and the line-to-neutralvoltages bisect the angle. Thus Van has to be at 30. This then leads to the following;

Van = 11030 V, Vbn = 110150 V, Vcn = 11090 Vand

|Vab| = 110 3 = 190.52therefore

Vab = 190.520 V, Vbc = 190.52120 V, Vca = 190.52120 V

Problem 12.2 Is it possible to generate the effect of a balanced three-phase delta-connectedsource with only two voltage sources? If so, how?

Let us start with three sources arranged in a delta as shown below.

Looking at this figure, we can see that KVL applies and going around the loop leads to a sum ofzero volts. Now, if we remove any single source, KVL still states that the voltage across the openpart of the circuit still is equal to the same voltage and phase of the source that was removed.Thus, you can generate a perfect, delta-connected voltage source configuration with just twosources. This will be expanded on later when we actually look at what happens to powerdistribution when you have two sources. (Clearly the two remaining sources must make up for thepower the third source would have supplied.) Also, you will see later how this leads to being ableto measure power delivered in a 3-phase system using only 2 wattmeters.

Problem 12.3 Is it possible to generate the effect of a balanced three-phase wye-connectedsource with only two voltage sources? Why or why not?

Again, we can look at the sources arranged in a wye and see what would happen. If we were tomeasure the line-to-line voltages, we would see a balanced 3-phase voltage system. Removingany one of the sources actually takes away two line-to-line voltages. Clearly, unlike the deltaconfiguration, you cannot deliver 3-phase voltage without 3 sources in a wye configuration.

+2000200120

200-120

+

+

BALANCED WYE-WYE CONNECTION

Problem 12.4 [12.7] Obtain the line currents in the three-phase circuit of Figure 12.1.

Figure 12.1

This is a balanced Y-Y system.

4400 V +

ZY = 6 j8

Ia

6 j8

A

N

6 j8 6 j8

Ic

Ib

+

+

440120440120+

4400

a

n

+200120 2000

+

200-120

+

Using the per-phase circuit shown above,

=

= 8j6

0440aI A53.1344

== 120-ab II A66.87-44 == 120ac II A13.73144

BALANCED WYE-DELTA CONNECTION

Problem 12.5 [12.15] In a wye-delta three-phase circuit, the source is a balanced,positive phase sequence with V0120an =V . It feeds a balanced load with += 12j9Zper phase through a balanced line with += 5.0j1LZ per phase. Calculate the phase voltagesand currents in the load.

Convert the -connected load to a Y-connected load and use per-phase analysis.

4j33Y +==ZZ

=+++

=

+= 48.37-931.19)5.0j1()4j3(

0120LY

an

a ZZV

I

But = 30-3ABa II

=

=

30-348.37-931.19

ABI A18.37-51.11

=BCI A138.4-51.11 =CAI A101.651.11

)53.1315)(18.37-51.11(ABAB == ZIV

ZY

ZL

+

Van

Ia

=ABV V76.436.172 =BCV V85.24-6.172 =CAV V8.5416.172

BALANCED DELTA-DELTA CONNECTION

Problem 12.6 [12.17] For the - circuit of Figure 12.1, calculate the phase and linecurrents.

Figure 12.1

=+= 18.4362.3110j30Z

The phase currents are

=

==

18.4362.310173ab

AB ZV

I A18.43-47.5

== 120-ABBC II A138.43-47.5 == 120ABCA II A101.5747.5

The line currents are== 30-3ABCAABa IIII

a

c

A

B

C

30

30

j10 +

173120 V

1730 V+

j10

+

173120 V

30

j10

b

== 48.43-347.5aI A48.43-474.9

== 120-ab II A168.43-474.9 == 120ac II A71.57474.9

BALANCED DELTA-WYE CONNECTION

Problem 12.7 [12.23] In a balanced three-phase -Y circuit, the source is connected inthe positive phase sequence, with V20220ab =V and += 15j20YZ . Find the linecurrents.

)15j20(310-220

330-

Y

aba

+

=

=

ZV

I

=aI A46.87-081.5 == 120-ab II A166.87-081.5

== 120ac II A73.13081.5

POWER IN A BALANCED SYSTEM

Problem 12.8 Given a delta-connected source with line-to-line voltages of 100 volts-rmsand a delta-connected load of 10-ohm resistors, calculate the power absorbed by each resistor andthe power delivered by each source.

As we can see from the circuit, there is 100 volts across each 100-ohm resistor.

+

100120

1000

+

100-120

+10

10

10

P = (100)2/10 = 1000 watts (per phase)

So a total of 3000 watts is delivered to the load and supplied by the source, with each individualvoltage source supplying 1000 watts.

Problem 12.9 Given 100v = volts for the load in Figure 12.1, determine 1R and 2R sothat the sources see a balanced load. Find 1I and 2I .

Figure 12.1

We can combine the two circuits as follows since each corresponding resistor is in parallel witheach other.

Obviously, the 20||20 is equal to 10 ohms. This means that the others need to combine into 10ohms if possible.

1

1

R10R10+

is = 10 if and only if R1 =

222

2 R10300R3010R30

R30+==

+

or R2 = 300/(30 10) = 15

Clearly, since the load appears balanced to the sources, I1 = 0.

20 R1

R2

n

20 10

30

nI2I1

20||201

1

R10R10+

2

2

R30R30+

I2 can be found by summing all of the currents flowing into it from the second circuit (see Figure12.3). It is easier if we assume that V20 = VAN = 1000, V10 = VBN = 100120, and V30= VCN = 100120.

30120100

10120100

200100

30V

10V

20VI CNBNAN2

+

+

=++=

= 5 + 10120 +3.333120 = 5 + (5 + J8.66) +(1.667 J2.887)

= 1.667 + J5.773 = 6.009106.1 A

UNBALANCED THREE-PHASE SYSTEMS

Problem 12.10 Given a balanced wye-connected source of 200 volts each, calculate thepower delivered to each of the resistors in the unbalanced wye-connected load in Figure 12.1.

Figure 12.1

We start by labeling the circuit and arbitrarily placing the sources.

Let the voltage from a to the source neutral, n, be Van = 2000

Likewise, Vbn = 200120 and Vcn = 200120

10 20

5

10 20

5

n Sourceneutral

c

b

a

N

C

B

A

This can easily be solved if we look at it as a simple circuits problem with only one unknownnode voltage, VN (measured with respect to n).

We now can write a nodal equation at node N.

020

VV5

VV10

VV cnNbnNanN=

+

+

Solving for VN we get,

++= 1207

2001207

8007

400VN

= 57.14 + (14.285 j24.74) + (57.14 + j98.98)

= 14.28 + j74.24 = 75.6100.9

Now that we have VN, we can either find the currents through the resistors or voltages acrossthem and then calculate the power. Solving for the voltages we get,

VAN = Van VN = 200 (14.28 + j74.24) = 214.28 j74.24 = 226.819.11

VBN = Vbn VN = 100 + j173.21 (14.28 + j74.24) = 85.72 + j98.97

= 130.93130.9

VCN = Vcn VN = 100 j173.21 (14.28 + j74.24) = 85.72 j247.45

= 261.88109.11

Now to calculate the power.

===

10)8.226(

10V

P22

AN10 5.144 kwatts

===

5)93.130(

5V

P22

BN5 3.429 kwatts

===

20)88.261(

20V

P22

CN20 3.429 kwatts

Problem 12.11 Given a balanced wye-connected source of 200 volts each, calculate thepower delivered to each of the resistors in the unbalanced wye-connected load in Figure 12.1.Explain the difference between this problem and Problem 12.10.

Figure 12.1

First of all, this problem is a lot easier to solve since the neutral connection means that each phaseof the wye-connected source is across only its corresponding resistor, unlike the previousproblem. In fact, there are no unknown node voltages in this circuit.

Stated mathematically, VAN = Van, VBN = Vbn, and VCN = Vcn

Thus, P10 = (200)2/10 =2 kwatts

P5 = (200)2/5 = 4 kwatts

P20 = (200)2/20 = 1 kwatt

PSPICE FOR THREE-PHASE CIRCUITS

Problem 12.12 Solve Problem 12.8 using PSpice. The circuit was set up as the followingschematic, calculating the current through each resistor and the voltage across each resistor.Since this is a resistance circuit, there is no phase angle between the voltage and the current. Soall we need to do to calculate power is to multiply the magnitudes of the currents and voltages.Since it is a balanced circuit, each source supplies the same power equal to the power beingabsorbed by each resistor. After storing the circuit, we run Simulate and then examined theoutput file. The results yield the following:

P10 = 100x10 = 1 kwatt, for a total of 3 kwatts (both delivered and absorbed)

It is interesting to note that a small resistor needed to be placed in the voltage loop to allowPSpice to run properly.

10 20

5

n

Problem 12.13 Solve Problem 12.10 using PSpice. We develop the following schematic inPSpice. Note, we are making more measurements than necessary to solve this problem.However, these extra measurements will allow us to check the results of the voltage calculationsin 12.10.

After saving the schematic, we run Simulate and open the output file. The results allow us todetermine the voltages and currents in the circuit. We then can calculate the power delivered toeach resistor.

V10 = 226.819.11 and I10 = 22.6819.11

Therefore P10 = (226.8)(22.68) = 5.144 kwatts (please note that cos= 1 since these are resistors)

V5 = 130.9130.9 and I5 = 26.19130.9

Therefore P20 = (130.9)(26.19) = 3.428 kwatts

V20 = 261.9109.1 and I20 = 13.09109.1

Therefore P20 = (261.9)(13.09) = 3.428 kwattsGoing back to the solution obtained by hand, we see that we have virtually the same values ofbranch voltages. PSpice also gave us the value of VNn = 79.59100.9, which agrees. Thisshows the value of using PSpice to check results obtained by hand.

Problem 12.14 [12.49] Given the circuit in Figure 12.1, use PSpice to determinecurrents aAI and voltage BNV .

Figure 12.1The schematic is shown below.

A

C

a j3 4 2400 V

+

b j3 4 240120 V

+

c j3 4 +

240120 Vj36

j36 j36

n NB j15 10

j15 10

j15 10

Pseudocomponents IPRINT and PRINT are inserted to measure IaA and VBN. In the AC Sweepbox, we set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. Once the circuit issimulated, we get an output file which includes

FREQ VM(2) VP(2)1.592 E01 2.308 E+02 1.334 E+02

FREQ IM(V_PRINT2) IP(V_PRINT2)1.592 E01 1.115 E+01 3.699 E+01

from which=aAI A3715.11 , =BNV V4.133-8.230

APPLICATIONS

Problem 12.15 [12.59] For the circuit displayed in Figure 12.1, find the wattmeterreadings.

Figure 12.1

240120 V +

24060 V +

Z

Z

Z = 10 + j30

W1

W2

Consider the circuit as shown below.

=+= 71.5762.3130j10Z

=

= 131.57-59.7

71.5762.3160-240

aI

=

= 191.57-59.771.5762.31120-240

bI

0120-24060-240c =+ZI

=

= 108.4359.771.5731.62

240-cI

== 101.57-146.13ca1 III=+= 138.43146.13cb2 III

[ ] [ ] === ).57101146.13)(60-240(ReReP *111 IV W2360[ ] [ ] === )138.43-146.13)(120-240(ReReP *222 IV W8.632-=1P W2360 =2P W8.632-

Problem 12.16 Given two three-phase sources with line-to-line voltages of 110 volts rms, isit possible to use standard 100 watt light bulb(s) to see if the sources are in phase with each other?Is it possible to see if they have the same rotation? More than one light bulb can be used.

Actually, only one light bulb is necessary. We need to make two assumptions before we start.The first is that both sources have exactly the same frequency. This is possible if both sources aredriven by the same mechanical system. The second is that they share the same ground.

24060 V +

Ib

Ia

I1

+240120 V

Ic

Z

Z

Z

I2

Consider the following representation of the two sources.

Now we connect the light to terminal a. Then we connect it to A, B, and C of the other source. Ifthe sources are out of phase with each other, the light will light at least dimly when connected toeach of the terminals. If the sources are reasonably close to being in phase with each other, therewill be one terminal where the light will not light. Let us assume that it is the connection betweena and A.

If we now connect the light to b and B and the light does not light, then the two sources have thesame rotation. It the light lights, then they have opposite rotations. Once the rotations have beendetermined, the two sources can be connected in parallel.

Problem 12.17 Given an unbalanced-delta connected load, show how you can use twowattmeters to determine the power delivered to the load. Calculate the voltages and currents eachmeter sees and determine the power delivered to the load. The circuit is shown below.

+

100120

1000

+

100-120

+10

5

20

+

b

a

+

c

+

+

A

C

+

B

+

This can be a difficult problem to solve unless we remember what we found as an answer toProblem 12.2. It is possible to represent a balanced delta source with only two sources. That tellsus how to hook up the meters.

We can solve this using either hand calculations or PSpice. Let us use PSpice since we will needto make calculations twice in order to check our results, although the hand calculations are notthat difficult.

First, we design the schematic and save it. We then simulate and get the results shown here.

P10 = 100x10 = 1 kwatt

P5 = 100x20 = 2 kwatts

P20 = 100x5 = 500 watts

PTotal = 1000 + 2000 + 500 = 3.5 kwatts

Now look at the following circuit. If we assume that V3 is not in the circuit for the moment, wesee that V1 and V2 would supply all the power. So we place an IPRINT in series with V1 and anIPRINT in series with V2. Then all we need to do is measure the Voltages V1 and V2. (Actuallyin practice there are line losses and these voltages would be measured at the load.)

After we get the results from PSpice, we can make the following calculations.

P1 = 100x13.23cos(120 + 160.9) = 1 kwatt

P2 = 100x26.46cos(120 139.1) = 2.5 kwatts

PTotal = P1 + P2 = 3.5 kwatts (the answer checks!!)

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PrefaceChapter 1 Basic ConceptsChapter 2 Basic LawsChapter 3 Methods of AnalysisChapter 4 Circuit TheoremsChapter 5 Operational AmplifierChapter 6 Capacitors and InductorsChapter 7 First-Order CircuitsChapter 8 Second-Order CircuitsChapter 9 Sinusoids and PhasorsChapter 10 Sinusoidal Steady-State AnalysisChapter 11 AC Power AnalysisChapter 12 Three-Phase CircuitsBalanced Three-Phase VoltagesBalanced Wye-Wye ConnectionBalanced Wye-Delta ConnectionBalanced Delta-Delta ConnectionBalanced Delta-Wye ConnectionPower in a Balanced SystemUnbalanced Three-Phase Systems PSpice for Three-Phase CircuitsApplications

Chapter 13 Magnetically Coupled CircuitsChapter 14 Frequency ResponseChapter 15 Laplace TransformChapter 16 Fourier SeriesChapter 17 Fourier TransformChapter 18 Two-Port Networks

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