lecture 12 - shanghaitechsist.shanghaitech.edu.cn/faculty/zhoupq/teaching/fall15/lec12...two-pin and...

Electric Circuits (Fall 2015) Pingqiang Zhou

Lecture 12

- Three-Phase Circuits/Transformers

Part I

11/26/2015

Reading: Chapter 12

1Lecture 12


Outline

• Why Three-Phase for AC supply?

• Balanced Three-Phase System

Balanced sources

Balanced loads

• Circuit analysis

Phase voltage/current

Line voltage/current

Lecture 12 2


Two-Pin and Three-Pin Sockets

3Lecture 12

https://en.wikipedia.org/wiki/AC_power_plugs_and_sockets

https://en.wikipedia.org/wiki/AC_power_plugs_and_sockets


Electrical Safety

4Lecture 12

https://cnx.org/contents/8f205833-26b8-4eb4-98bb-936aad728cc6@2/Electrical-Safety-Systems-and-


5Lecture 12


Single Phase vs. Polyphase

• Households have single-phase power supply

This typically in a three wire form, where two 120V sources with the

same phase are connected in series.

This allows for appliances to use either 120 or 240V

• Circuits that operate at the same frequency but with multiple

sources at different phases are called polyphase.

Lecture 12 6


Three-Phase System

• In power grids, three phase

power is used for a variety of

reasons.

It is easy to extract single or two

phase power from a three phase

system, satisfying the cases where

this is needed.

The instantaneous power in a three

phase system does not pulsate like

it does in a single phase system.

(refer to Ch. 12.7)

Lastly, the transmission of three

phase is more economical than

transmitting the equivalent single

phase power. (refer to Ch. 12.7)

Lecture 12 7


Balanced Three Phase

• Three phase voltages are typically

produced by a three phase AC

generator.

• The output voltages look like below.

Lecture 12 8


Connecting the Sources

• Three phase voltage sources can be connected the loads

by either three or four wire configurations.

Three-wire configuration accomplished by Delta connected source.

Four-wire system accomplished using a Y connected source.

Lecture 12 9


Balanced Source

• A wye connected source is said

to be balanced when

• Two sequences for the phases:

0an bn cnV V V

an bn cnV V V

0

120

240 120

an p

bn p

cn p p

V V

V V

V V V

0

120

240 120

an p

cn p

bn p p

V V

V V

V V V

Lecture 12 10

Positive Negative


Balanced Loads

• Similar to the source, a balanced load is one that has the

same impedance presented to all three voltage sources.

• They may also be connected in either Delta or wye

For a balanced wye connected load:

For a balanced delta connected load:1 2 3 YZ Z Z Z

a b cZ Z Z Z

Lecture 12 11


Source-Load configurations

• The load impedance per phase for the two load

configurations can be interchanged:

Lecture 12 12


Y and ∆, Which One Better?

13Lecture 12

http://www.allaboutcircuits.com/textbook/alternating-current/chpt-10/three-phase-y-delta-configurations/

http://www.allaboutcircuits.com/textbook/alternating-current/chpt-10/three-phase-y-delta-configurations/


Balanced Y-Y connection

• Any three-phase system can

be reduced to an equivalent

Y-Y system.

• We will consider an example

of a balanced four wire Y-Y

system shown here.

• The load impedances Zy will

be assumed to be balanced.

This can be the source 𝑍𝑠, line

𝑍𝑙 and load 𝑍𝐿 together.

Lecture 12 14


Line-to-Line Voltage

• We will use the positive sequence for

this circuit, meaning the voltages are:

0

120 120

an p

bn p cn p

V V

V V V V

The line to line (or line in short) voltages:Thus the magnitude of the line

voltages VL is:

3L pV V

p an bn cn

L ab bc ca

V V V V

V V V V

3 30

3 90

3 210

ab P

bc P

ca P

V V

V V

V V

Lecture 12 15


Line Currents

• If we apply KVL to each phase, we

find the line currents are:

120

240

ana b a

Y

c a

VI I I

Z

I I

0a b cI I I

• From this one can see the line currents add up to zero.

This shows that the neutral wire has zero voltage and no current.

Thus it can be removed without affecting the system.

Lecture 12 16


Per Phase Analysis

• An alternative way to analyze the

Y-Y circuit is to look at each phase

individually. Let us look at phase a:

The equivalent circuit for that phase is

shown here.

The current for this phase is:

• If the circuit is balanced, only one

phase need be analyzed.

ana

Y

VI

Z

Lecture 12 17


Example

• Calculate the line currents.

18Lecture 12


Balanced Wye-Delta

• This system consists of a

balanced Y connected source and

a balanced Delta connected load.0

120 120

an p

bn p cn p

V V

V V V V

3 30 3 90 3 150ab P AB bc P BC ca P CAV V V V V V V V V

The line voltages are equal to the voltages across the

load. From this, we can calculate the phase currents:

The line voltages are:

BC CAABAB BC CA

V VVI I I

Z Z Z

Lecture 12 19


Balanced Wye-Delta II

• An alternative way to solve for the

phase currents is to apply KVL.

• For example, applying KVL

around the loop aABbna gives:

• Or

• This is the more general way to

find phase currents.

0an AB bnV Z I V

an bn ab ABAB

V V V VI

Z Z Z

Lecture 12 20


Phase to Line Currents

• The line currents can be obtained from the phase currents

by applying KCL to nodes A, B, and C

Since ICA = IAB -240˚:

Thus:

a AB CA b BC AB c CA BCI I I I I I I I I

3 30a ABI I

3L pI I

Lecture 12 21


Alternative

• An alternate way to analyze the

Wye-Delta circuit is to transform

the Delta connected load into a

wye connected load. Using the

Delta-Wye transformation:

3Y

ZZ

With this circuit now rendered as a Y-Y circuit, single phase

analysis can be done

Lecture 12 22


Balanced Delta-Delta

• Now we turn our attention to the Delta-Delta configuration.

• Once again, the goal is to get the phase and line currents.

• Note that Delta configured generators are less typical than

the wye because any imbalance in the voltage sources will

result in current flowing through the delta loop.

Lecture 12 23


Balanced Delta-Delta II

• Assuming a positive sequence,

the phase voltages are:

0

120 120

ab p

bc p ca p

V V

V V V V

ab AB bc BC ca CAV V V V V V

• If line impedances are insignificant, then the line voltages

are the same as the phase voltages.

Hence the phase currents are:

ab BC bc CA caABAB BC CA

V V V V VVI I I

Z Z Z Z Z Z

a AB CA b BC AB c CA BCI I I I I I I I I 3L pI I

Lecture 12 24


Balanced Delta-Wye

• The last configuration to consider is the Delta-Wye system.

• The phase voltages are the same as the last case.

• There are many ways to get the line currents.

• One way is to apply KVL to the loop aANNba

0Y a b ab pZ I I V V

0p

a b

Y

VI I

Z

Keeping in mind that Ib lags Ia by

120˚, we can solve for the line

current:

/ 3 30p

a

Y

VI

Z

Lecture 12 25


Convert back to Y-Y

• Another way to solve this system is to

convert both the source and load

back to a Wye-Wye system.

• The equivalent Wye connected

source voltages are:

303

150 903 3

p

an

p p

bn cn

VV

V VV V

The load conversion goes as the

standard delta-wye conversion. Once

this is done, a single phase can be

examined to find the line current.

/ 3 30p

a

Y

VI

Z

Lecture 12 26


Summary:

Lecture 12 27


Summary II:

Lecture 12 28


Power in a Balanced System

• Power in balanced Wye load

29

[Nilsson, Ch. 11.5]

Lecture 12


Power in a Balanced System

• Power in balanced Delta load

30Lecture 12

[Nilsson, Ch. 11.5]


Further Reading

• Unbalanced Three-Phase System (Ch.12.8)

• Two-wattmeter method (Ch.12.10.1)

• Residential wiring (Ch.12.10.2)

• http://www.allaboutcircuits.com/textbook/alternating-

current/#chpt-10

31Lecture 12

http://www.allaboutcircuits.com/textbook/alternating-current/#chpt-10


Lecture 12

- Three-Phase Circuits/Transformers

Part II

11/26/2015

Reading: Chapter 13

32Lecture 12


Outline

• Mutual inductance

• Transformers

33Lecture 12


34Lecture 12


Self Inductance and Mutual Inductance

• Self inductance: reaction of

the inductor to the change in

current through itself.

𝑣 = 𝑁𝑑𝜙

𝑑𝑡= 𝑁

𝑑𝜙

𝑑𝑖

𝑑𝑖

𝑑𝑡= 𝐿

𝑑𝑖

𝑑𝑡

𝐿 = 𝑁𝑑𝜙

𝑑𝑖

• Mutual inductance: reaction of

the inductor to change in current

through another inductor.

𝑣1 = 𝐿1𝑑𝑖1𝑑𝑡

𝐿1 = 𝑁1𝑑𝜙1𝑑𝑖1

𝑣2 = 𝑁2𝑑𝜙12𝑑𝑡

= 𝑁2𝑑𝜙12𝑑𝑖1

𝑑𝑖1𝑑𝑡

= 𝑀21

𝑑𝑖1𝑑𝑡

𝑀21 = 𝑁2𝑑𝜙12𝑑𝑖1

Lecture 12 35


Example

• Knowing the dot convention, we can analyze the series

aiding connection.

Applying KVL to coil 1:

For coil 2:

In the frequency domain:

1 21 1 1 1

di div i R L M

dt dt

2 12 2 2 2

di div i R L M

dt dt

𝐕𝟏 = 𝑅1 + 𝑗𝜔𝐿1 𝐈𝟏 + 𝑗𝜔𝑀𝐈𝟐

𝐕𝟐 = 𝑗𝜔𝑀𝐈𝟏 + 𝑅2 + 𝑗𝜔𝐿2 𝐈𝟐

Lecture 12 36


Energy in a Coupled Circuit

• The energy stored in an inductor is:

• For coupled inductors, the total

energy stored depends on the

individual inductance and on the

mutual inductance.

• The positive sign is selected when

the currents both enter or leave the

dotted terminals.

21

2w Li

2 2

1 1 2 2 1 2

1 1

2 2w L i L i Mi i

Lecture 12 37


Coupling Coefficient 𝒌

• With the total energy established for the mutual inductors,

we can establish an upper limit on M.

• The system cannot have negative energy because the

system is passive.

2 2

1 1 2 2 1 2

1 10

2 2L i L i Mi i

1 2M L L

• Define a parameter describes how

closely M approaches upper limit.

• Coupling coefficient, 0 ≤ 𝑘 ≤ 1.

• determined by the physical

configuration of the coils.

1 2

Mk

L L

Lecture 12 38


Linear Transformers

• A transformer is a magnetic device that takes advantage

of mutual inductance.

Generally a four terminal device comprised of two or more

magnetically coupled coils.

They are called linear if the coils are wound on a magnetically

linear material.

Lecture 12 39


Transformer Impedance

• An important parameter to know for a transformer is how

the input impedance Zin is seen from the source.

Zin is important because it governs the behavior of the primary

circuit.

Reflected impedance from

secondary to primary

Lecture 12 40


Equivalent circuits

• We already know that coupled inductors can be tricky to

work with.

• One approach is to use a transformation to create an

equivalent circuit.

The goal is to remove the mutual inductance.

This can be accomplished by using a T or a network.

The goal is to match the terminal voltages and currents from the

original network to the new network.

Lecture 12 41


Equivalent Circuits (T or )

1 1 1

2 2 2

V j L j M I

V j M j L I

Lecture 12 42


Ideal Transformers

• The ideal transformer has:

Coils with very large reactance

(L1, L2, M →)

Coupling coefficient k is equal to

unity.

Primary and secondary coils are

lossless

Lecture 12 43


Ideal Transformers II

• The voltages are related to each

other by the turns ratio n:

• The current is related as:

• Reflected impedance

1 2

2 1

V Nn

V N

2 1

1 2

1I N

I N n

Lecture 12 44


Ideal Autotransformer

• Autotransformer uses one winding for primary & secondary

It does not offer isolation!

Lecture 12 45


Three Phase Transformer

• Three-phase power has two choices for transformers:

A transformer bank, with one transformer per phase

A three phase transformer (smaller and less expensive)

Lecture 12 46

http://www.allaboutcircuits.com/textbook/alternating-current/chpt-10/three-phase-transformer-circuits/

http://www.allaboutcircuits.com/textbook/alternating-current/chpt-10/three-phase-transformer-circuits/


Further Reading (Ch. 13.9)

• Transformer as isolation/matching device

47Lecture 12

• Power distribution

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