elec 250: linear circuits i chapter 13 5/4/08 chapter 13 ...amiralib/courses/elec250/chapter...

Elec 250: Linear Circuits I Chapter 13 5/4/08

Page 361

anners, computers, ste-

er across the power

r transformer opera-

utual inductance, and

ill use to solve for cir-

S.W. Neville

Chapter 13Mutual Inductance

13.1 Introduction

• We are all familiar with the ac-to-dc converters that power out printers, screo’s, etc.

- These devices require a transformer for their operation.

- Transformers are also used in the transmission of electrical powgrid.

- Electromagnetic coupling is the basic phenomena responsible fotion.

• In this chapter, we study the phenomena of magnetic coupling through mthen we discuss the principle of operation of an ideal transformer.

- Mesh (loop) analysis is the primary circuit theorem which we wcuits containing mutual inductance.

- The nature of such circuits is not well suited to nodal analysis.


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e inductor magnetic

asor equation

r.

S.W. Neville

13.2 Self Inductance

• Consider the inductor shown below, when the current passes through thlines of force are produced.

- This is what gives the self inductance and we get the usual ph

- where is the operating angular frequency of the current phaso

I

I

L

L

V jωLI=

ω


63

agnetic field is created ond inductor.

as before

uctor, denoted , V21

S.W. Neville

13.3 Mutual Inductance

• Assume that we have two inductors in close proximity as shown below

- When a sinusiods current passes through the first inductor, a maround it, and part of this magnetic field cuts the coils of the sec

- The voltage on inductor is given by the same phasor equation

- But, the same current gives rise to a voltage on the second indgiven by

- Where is the mutual inductance between the two inductors.

I1

L1

I2

L2

I1

L1

V1 jωL1I1=

I1

V21 j± ωMI1=

M


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ight be opposite to the

l induced voltages in

both voltage equations.

S.W. Neville

- The sign arises because the direction of the induced voltage massumed reference voltage.

- is measured in Henries just like the self inductance .

• In general, when currents flow in two inductors at the same time, the totathe two inductors are given by

- Each inductor will have two voltage components:

1. A component due to the self inductance.

2. A component due to the mutual inductance

- Note that the mutual inductance coefficient is the same for

±

M L

V1 jωL1I1 jωMI2±=


jωM


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ith mutual coupling are the voltages devel-

ed. Assume positive

f one inductor threads

3 30° A∠=

S.W. Neville

Ex. 13.1 Assume two coupled inductors and w and an operating frequency . What

oped across the inductors when: (a) and (b) and the second inductor is open-circuitcoupling between the inductors

Solution:

13.4 Coupling Coefficient

• The coupling coefficient is a measure of how much the magnetic flux othe coils of the other inductor.

L1 1H= L2 4H=

M 0.1H= f 100Hz=I1 2 0° A∠= I2

I1 5 60° A∠=

k


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ded between

d

13.7).

and whose L2 5H=

S.W. Neville

- is defined by

- The coupling coefficient is a dimensionless constant and is boun

- When , we say that the two coils (inductors) are not couple

- When , we say that the two coils are weakly coupled.

- When , we say that the two coils are strongly coupled.

- When , we say that we have an ideal transformer (Section

Ex. 13.2 Determine the coupling coefficient for two coils mutual inductance is .

Solution:

k

k ML1L2

----------------=

0 k 1≤ ≤

k 0=

k 1«

k 1≈

k 1=

L1 1H=

M 2H=


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es are given by the

ng, we use a dot con-

dotted terminals, the as the signs of the

d terminal. the signs

other leaves the dot-he signs of the

L

L

S.W. Neville

13.5 The Dot Convention

• Above we stated that when two inductors are coupled, the induced voltagequations,

• To know which sign to give the induced voltage due to the mutual couplivention in accordance with the following rules,

1. When both the currents in the two inductors enter the signs of the terms (mutual inductance) are the sameterms (self inductance)

2. When both current in the two inductors leave the dotteof the terms are the same sign as the terms.

3. When one current enters the dotted terminal while theted terminal, the signs of the terms are opposite to tterms.



M

M L

M


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ul in some situation:

t induces a positive

r, it induces a positive

I1

L1

I2

L2

M & L

signhave opposite

S.W. Neville

• There is an alternate way to phrase the dot convention which proves usef

1. When a current enters the dotted terminal of an inductor, ivoltage on the dotted terminal of the coupled inductor.

2. When a current enters the undotted terminal of an inductovoltage on the undotted terminal of the coupled inductor.

I1

L1

I2

L2

I1

L1

I2

L2

M & Lhave same

sign

M & Lhave same

sign

I1

L1

I2

L2

M & Lhave opposite

sign


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n below.

+

-

VI

L

S.W. Neville

Ex. 13.3 Find the equivalent inductances for the two circuits show

Solution:

+

-

VI +

-

VIL1

L2 L

(a) (b)

+

-

VI L1

L2


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es coupled inductors).

al and enters a non- sign of the self induc-

I2

S.W. Neville

13.6 Coupled Circuits

• Consider the circuit shown below (where the double headed arrow denot

- Taking the dot convention into account ( enters a dotted termindotted terminal so the mutual inductances will have the oppositetances), the two loop equations are,

- Placing in matrix form we have,

+-V I2 Z2

Z1

jωL1jωL2

jωM

I1

I1

V I1 Z1 jωL1+( )– jωMI2+ 0=

I2 Z2 jωL2+( )– jωMI1+ 0=

Z1 jωL1+ j– ωMj– ωM Z2 jωL2+

I1

I2

V0

=


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lgebra techniques.

ts and what is the value

3 2– j

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- We can now solve for the mesh currents using standard matrix a

Ex. 13.4 For the circuit below what is the value of the loop currenof the mutual inductance?

Solution:

+- I23j 2j

0.4j

I1

v t( ) 3 10t 10°–( )sin=

2 5j+


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ng coefficient .k 0.8=

S.W. Neville

Ex. 13.5 Find the voltage for the circuit below, given the coupli

Solution:

V

+-

+

-

Vj4Ω–

3Ω

5Ω

j5Ω j10Ω

50 0°∠


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j2A–V2

+

-

S.W. Neville

Ex. 13.6 Find the unknown voltages in the circuit below.

Solution:

I28jΩ j14Ω

5jΩ

I15 0∠ °AV1

-

+


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circuit shown below.

ZL

S.W. Neville

Ex. 13.7 Find the Thevenin and Norton equivalent circuits for the

Solution:

+-

I2

15j

300j

50jI1

120 15°V∠

20Ω

V1 V2

+

-

+

-


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S.W. Neville
Ex. 13.8 Find the mesh currents in the circuit below.

Solution:

+-

+

-

Vj4Ω–

3Ω j5Ω

j10Ω50 0°∠

j3Ω


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ith .

e is used to confine the g.

and a secondary output.

k 1=

+

-

V2

2

S.W. Neville

13.7 The Ideal Transformer

• The ideal transformer consists of two inductors that are tightly coupled w

- The circuit symbol for an ideal transformer is

- The vertical lines between the inductors indicate that an iron cormagnetic flux to the two coils, thus ensuring a very tight couplin

- The ideal transformer is a two-port device with a primary input

- The primary side is the side that connect to the power supply.

+

-

V1

I1 I1:n


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nce of the primary and

ns to primary coil turns

ZL

ss----

S.W. Neville

- The secondary side is the side which connects to the load.

- A transformer is used to change the voltage, current, and impedasecondary sides.

- The main parameter of a transformer is the turns ration

- This is the ration between the number of secondary coil tur

+-

I2I1

V

1:n

VL

+

-

Primary Side Secondary Side

n

n Turns Ration Number of secondary turnNumber of primary turn

-------------------------------------------------------------= =


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ency or the load con-.

plied by the source to

l be given by

S.W. Neville

- In an ideal transformer the secondary to primary voltage ratio is

- The secondary to primary current ration is

- The above two equations are independent of the operating frequnected the transformer, but they only hold for ideal transformers

- We know from the previous Chapter that the complex power supthe primary coil is given by

- where it is assumed that and represent rms values.

- We also know that the complex power for the secondary coil wil

V2V1------ n=

I2I1---- 1

n---=

S1 V1I1*=

V1 I1

S2 V2I2*=


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rrents and voltages in plex power equations.

exactly equals the load.

ivered to it; it delivers

me power will be lost

S.W. Neville

- Using the above equations relating the primary and secondary cuideal transformers we can relate the primary and secondary com

- Thus the power delivered to the transformer’s primary coil power delivered by the transformer’s secondary coil to the

- The ideal transformer does not absorb any of the power delall of the power to the load.

- Obviously, this is an ideal case and for real transformers sowithin the transformer.

S2 V2I2* nV1

I1*

n---- S1= = =


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the secondary voltage, ower delivered by the

50 30°Ω∠

S.W. Neville

Ex. 13.9 Consider the ideal transformer circuit shown below. Findthe current in the primary and secondary sides, and the psource, and the power consumed by the load.

Solution:

+-

I2I1

110V

1:10

VL

+

-


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d as it appears to the

d impedance and is

ZL

S.W. Neville

13.7.1 Reflected Impedance

• Consider the ideal transformer shown below

- We are interested in finding the equivalent impedance of the loasource on the primary circuit.

- The impedance as seen by the primary side is called the reflectegiven by

- Since we are dealing with an ideal transformer we can write

+-

I2I1

V1

1:n

V2

+

-

ZrV1I1------=

V11n---V2=

I1 nI2=


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mpedance as

rmer appears to the

ivalent to the circuit

its by replacing the impedance.

S.W. Neville

- The reflected impedance can then be given in terms of the load i

- Thus a load connected to a secondary side of an ideal transfo

source as if it is an impedance of the value .

- Hence, the circuit above, as seen from the primary source, is equshown below

- The reflected impedance will help us simplify transformer circutransformer and the load impedance by their equivalent reflected

ZrV1I1------ 1

n2-----

V2I2------ 1

n2-----ZL= = =

ZL

1n2-----ZL

+-

I1

V11n2-----ZL


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econdary voltages and umed in and .Z1 Z2

Z2 100 j75Ω–=

S.W. Neville

Ex. 13.10 Consider the circuit shown below. Find the primary and scurrents, and the power delivered by the source and cons

Solution:

+-

I2

I112 0°∠

1:5

V2

+

-

Z1 2 j10Ω+=


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hown below. Such a itor is formed through y circuits.

2Ω

S.W. Neville

Ex. 13.11 Find the power consumed in the capacitor in the circuit scircuit can occur in practice when an unintentional capacan unintentional connecting of the primary and secondar

+-

I2I1

4 0°∠

1:2

V2

+

-V1

+

-

I3 j8Ω–

j5Ω


Page 385
S.W. Neville
Assignment #13

Refer to Elec 250 course web site for assigned problems.

• Due 1 week from today @ 5pm in the Elec 250 Assignment Drop box.

elec 250: linear circuits i chapter 13 5/4/08 chapter 13 ...amiralib/courses/elec250/chapter...

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