Electric Circuits, 9th Edition P76 ppt

Appendix _ More on Magnetically { Coupled Coils and Ideal Transformers C.l Equivalent Circuits for Magnetically Coupled Coils At times, it is convenient to model magnetically couple

726 Complex Numbers

A complex n u m b e r in polar form can b e put in rectangular form by writing

ce je = c(cos0 + /sin 0)

= c cos 0 + jc sin 0)

= a + jb

(B.5)

T h e transition from rectangular to polar form m a k e s use of the geom-etry of the right triangle, namely,

a + jb = Va 2 + b 2 )e je

where

It is not obvious from Eq B.7 in which quadrant the angle 0 lies T h e

ambi-guity can be resolved by a graphical representation of the complex number

b

0

s^ H

J?\ C

a Figure B.l • The graphical representation of a + jb

when a and b are both positive

I l

5 36.87° 5

4

4 + / 3 = 5/36.87°

(a)

143.13° '

\ 3

1 1 1 1 I X

- 4

l

- 4 + /3 = 5,/143.13°

(b)

i i 1 ilU

- 4 ,

- 3

5 216.87°

- 4 - / 3 = 5/216,87 0

(c)

i_L 1 1 1 1 1 I 0

- 3

i_

~l 1

Z y S

— 5

1 1 1

4

s

323

1,

13°

4 - / 3 = 5/323.13°

(d)

Figure B.2 A The graphical representation of four

complex numbers

B.2 The Graphical Representation

of a Complex Number

A complex number is represented graphically on a complex-number plane, which uses the horizontal axis for plotting the real component and the vertical axis for plotting the imaginary component The angle of the complex number is measured counterclockwise from the positive real axis

The graphical plot of the complex number n = a + jb = c /0°, if we assume that a and b are both positive,is shown in Fig B.l

This plot makes very clear the relationship between the rectangular and polar forms Any point in the complex-number plane is uniquely defined by

giving either its distance from each axis (that is, a and b) or its radial dis-tance from the origin (c) and the angle of the radial measurement 0

It follows from Fig B.l that 0 is in the first quadrant when a and b are both positive, in the second quadrant when a is negative and b is positive,

in the third quadrant when a and b are both negative, and in the fourth quadrant when a is positive and b is negative These observations are

illustrated in Fig B.2, where we have plotted 4 + / 3 , - 4 + / 3 , - 4 - / 3 , and 4 - / 3

Note that we can also specify 0 as a clockwise angle from the positive

real axis Thus in Fig B.2(c) we could also designate —4 - /3 as 5/-143.13° In Fig B.2(d) we observe that 5/323.13° = 5 / - 3 6 8 7 ° It is

customary to express 0 in terms of negative values when 0 lies in the third

or fourth quadrant

The graphical interpretation of a complex number also shows the

relationship between a complex number and its conjugate The conjugate

of a complex number is formed by reversing the sign of its imaginary

component Thus the conjugate of a + jb is a - jb, and the conjugate of

—a + jb is — a - jb When we write a complex number in polar form, we form its conjugate simply by reversing the sign of the angle 0 Therefore the conjugate of c/0° is c/-0° The conjugate of a complex number is

B.3 Arithmetic Operations 727

designated with an asterisk In other words, n* is understood to be the /j, = -a+jb-c &•>

conjugate of n Figure B.3 shows two complex numbers and their

conju-gates plotted on the complex-number plane

Note that conjugation simply reflects the complex numbers about the

real axis

/j- = — a+ib—v #->

^ "

-b-112 — —u-jb=c -H2

n

^f

^ ¾ ^

« ] =

-

(i-«+

~jb =

b = c 0,

1

a

c -0 {

B.3 Arithmetic Operations

Addition (Subtraction)

To add or subtract complex numbers, we must express the numbers in

rec-tangular form Addition involves adding the real parts of the complex

numbers to form the real part of the sum, and the imaginary parts to form

the imaginary part of the sum Thus, if we are given

Figure B.3 A The complex numbers n x and n 2 amd their

conjugates n\ and «3

and

then

«! = 8 + /16

«2 = 12 - / 3 ,

n { + n 2 = (8 + 12) + /(16 - 3) = 20 + /13

Subtraction follows the same rule Thus

n 2 - «j = (12 - 8 ) + / ( - 3 - 16) = 4 - /19

If the numbers to be added or subtracted are given in polar form, they are

first converted to rectangular form For example, if

«i = 10/53.13°

and

then

and

n 2 = 5 / - 1 3 5 ° ,

m + n 2 = 6 + /8 - 3.535 - /3.535

= (6 - 3.535) + /(8 - 3.535)

= 2.465 + /4.465 = 5.10/61.10°,

/11 - n2 = 6 + / 8 - (-3.535 - /3.535)

= 9.535 + /11.535

= 14.966 /50.42°

Multiplication (Division)

Multiplication or division of complex numbers can be carried out with the

numbers written in either rectangular or polar form However, in most

cases, the polar form is more convenient As an example, let's find the

product n x n 2 when /^ = 8 + /10 and n 2 = 5 - /4 Using the rectangular

form, we have

n x n 2 = (8 + /10)(5 - /4) = 40 - /32 + /50 + 40

= 80 + /18

= 82/12.68°

If we use the polar form, the multiplication n.\n 2 becomes

n 1 n 2 = (12.81 /51.34° )(6.40 / - 3 8 6 6 ° )

= 82/12.68°

= 80 + /18

The first step in dividing two complex numbers in rectangular form is to

multiply the numerator and denominator by the conjugate of the

denomi-nator This reduces the denominator to a real number We then divide the

real number into the new numerator As an example, let's find the value of

n\/n 2 , where rt\ = 6 + /3 and n 2 = 3 - / 1 We have

« 1

n 2

6 + /3

3 - /1

(6 + ( 3

18 + /6 + /9

-9 + 1

15 + /15

10

= 2.12 /45°

/3)(3 + /1)(3 +

- 3

= 1.5 + /1.5

/1) /1)

In polar form, the division of n x by n 2 is

n { 6.71 /26.57°

n 2 3 1 6 / - 1 8 4 3 °

= 1.5 + /1.5

2.12 / 4 5 '

B.4 Useful Identities

In working with complex numbers and quantities, the following identities

are very useful:

( - / ) ( / ) = U (B.9) / = ^ 7 , (B.10)

ff*/»/2 = ± / (B.i2)

Given that n = a + jb = c/0°, it follows that

« + n = 2«, (B.14)

n - n* = jib, (B.15)

«/w* = 1/20° (B.16)

B.5 The Integer Power

of a Complex Number

To raise a complex number to an integer power k, it is easier to first write

the complex number in polar form Thus

n k = (a + jb) k

= (cei°) k = c k e jk0

= c k (coskd + j sinkO)

For example,

(2e/12°)5 = 2V6 ( )° = 32e mr

= 16 +)27.71, and

(3 + /4)4 = (5e^y ) 4 = 5 Vm 5 2°

= 625^212,52°

= -527 - /336

B.6 The Roots of a Complex Number

To find the /cth root of a complex number, we must recognize that we are

solving the equation

which is an equation of the kth degree and therefore has k roots

To find the k roots, we first note that

ce je = ce mi*) = ce m-w = > (B 1 8 )

It follows from Eqs B.17 and B.18 that

Xj = (ce'°y /k = c yk eW\

X 2 = [ ce W+2iryil/k = c l/k e j{fi+2v)/k^

X$ = [ce' i0+47T) } ]/k = c Vk e J(0+47r)/k^

We continue the process outlined by Eqs B.19, B.20, and B.21 until the

roots start repeating This will happen when the multiple of n is equal to 2k For example, let's find the four roots of 81tfy6{)° We have

Xt = 8lVV6°/4 = 3e'l s°,

X2 = 81^/^+360)/4 = ^ / 1 ( ^

x3 = 8W m+mV4 = 3e' 195'\

x = g l l/4 e /(6()+ t«H0)/4 = 2e j2 ^'\

Here, x$ is the same as X\, so the roots have started to repeat Therefore we

know the four roots of 81*?; are the values given by X\, X 2 , X 3 , and X4

It is worth noting that the roots of a complex number lie on a circle in the complex-number plane The radius of the circle is c!///< The roots are uniformly distributed around the circle, the angle between adjacent roots

being equal to lir/k radians, or 360/k degrees The four roots of 81 e '6 ( r are shown plotted in Fig B.4

3 105°

/ /

1

1 1• 1 1

3 1 9 5 ° ^

"s

N

_ \

\ fc3 15°

1 i 1 1 1

- I /

~ 3 285°

Figure B.4 • The four roots of %\e m "

(B.19) (B.20) (B.21)

Appendix

_ More on Magnetically

{ Coupled Coils and Ideal

Transformers

C.l Equivalent Circuits for Magnetically

Coupled Coils

At times, it is convenient to model magnetically coupled coils with an

equivalent circuit that does not involve magnetic coupling Consider the

two magnetically coupled coils shown in Fig C.l The resistances Ri and

R 2 represent the winding resistance of each coil The goal is to replace the

magnetically coupled coils inside the shaded area with a set of inductors

that are not magnetically coupled Before deriving the equivalent circuits,

we must point out an important restriction: The voltage between terminals

b and d must be zero In other words, if terminals b and d can be shorted

together without disturbing the voltages and currents in the original

cir-cuit, the equivalent circuits derived in the material that follows can be

used to model the coils This restriction is imposed because, while the

equivalent circuits we develop both have four terminals, two of those four

terminals are shorted together Thus, the same requirement is placed on

the original circuits

We begin developing the circuit models by writing the two equations

that relate the terminal voltages i?i and v2 to the terminal currents ix and

i 2 For the given references and polarity dots,

di\ diz

i-h U-r + M~r

and

Vi

du dh

« 1

+-' l

a

+

" i

* L*

L l )

M

« 1

••2 V 2

Figure C.l • The circuit used to develop an equivalent

circuit for magnetically coupled coils

The T-Equivalent Circuit

To arrive at an equivalent circuit for these two magnetically coupled coils,

we seek an arrangement of inductors that can be described by a set of

equations equivalent to Eqs C.l and C.2 The key to finding the

arrange-ment is to regard Eqs C.l and C.2 as mesh-current equations with i y and i 2

as the mesh variables Then we need one mesh with a total inductance of

L\ H and a second mesh with a total inductance of L 2 H Furthermore, the

two meshes must have a common inductance of M H The T-arrangement

of coils shown in Fig C.2 satisfies these requirements

Figure C.2 • The T-equivalent circuit for the

magneti-cally coupled coils of Fig C.l

731

732 More on Magnetically Coupled Coils and Ideal Transformers

You should verify that the equations relating vY and v2 to /, and i2

reduce to Eqs C.l and C.2 Note the absence of magnetic coupling between the inductors and the zero voltage between b and d

The ^-Equivalent Circuit

We can derive a 7r-equivalent circuit for the magnetically coupled coils shown in Fig C.l.This derivation is based on solving Eqs C.l and C.2 for

the derivatives dijdt and di^jdt and then regarding the resulting

expres-sions as a pair of node-voltage equations Using Cramer's method for

solv-ing simultaneous equations, we obtain expressions for di\jdt and di2 /dt:

di\

dt

V\

v 2

u

M

M

L 2

M

L 2

LiL 1^2 M Vl

M

L X L 2 - M •v 2 ; (C.3)

di 2

dt

UU - M <\^2 2 UL 1^2 w Vi +

Li

L,L>, - M l

jVl (C.4)

Now we solve for /, and i2 by multiplying both sides of Eqs C.3 and C.4 by

dt and then integrating:

k = *i(0) +

L X L 2 - M 2J{)

L X L 2 - M l h v

and

' 2 (0)

UU Mx / v\dr + r / 2J{) LXL2-M2k

If we regard vx and v2 as node voltages, Eqs C.5 and C.6 describe a circuit

of the form shown in Fig C.3

All that remains to be done in deriving the 7r-equivalent circuit is to find LA, LB, and Lc as functions of Lh L 2 , and M We easily do so by

writ-ing the equations for /t and i2 in Fig C.3 and then comparing them with

Eqs C.5 and C.6 Thus

Figure C.3 • The circuit used to derive the 7r-equivalent circuit for

magnetically coupled coils

C.l Equivalent Circuits for Magnetically Coupled Coils 733

1 f 1 /*'

ii = /j(0) + — / Vidr + — I («! - v 2 )dr

and

1 /"' 1 / '

* 2 = «2(0) + -^- v 2 dT + — I (v 2 - v x )dr

I-C /0 L 3 JO

= «' 2 (0) + 7" / M * + ( 7 - + 7 - ]

Then

M

LA

L 2 -M

L X L 2 - M2 '

(CIO)

Lc

L i /V/

When we incorporate Eqs C.9-C.11 into the circuit shown in Fig C.3, the

^-equivalent circuit for the magnetically coupled coils shown in Fig C.l is

as shown in Fig C.4

Note that the initial values of iy and i 2 are explicit in the ^-equivalent

circuit but implicit in the T-equivalent circuit We are focusing on the

sinu-soidal steady-state behavior of circuits containing mutual inductance, so

we can assume that the initial values of ij and i2 are zero We can thus

eliminate the current sources in the ^-equivalent circuit, and the circuit

shown in Fig C.4 simplifies to the one shown in Fig C.5

The mutual inductance carries its own algebraic sign in the T- and

^-equivalent circuits In other words, if the magnetic polarity of the

cou-pled coils is reversed from that given in Fig C.l, the algebraic sign of M

Figure C.4 A The 7r-equivalent circuit for the magnetically coupled coils of Fig C.l Figure C.5 • The ^-equivalent circuit used for

sinusoidal steady-state analysis

734 More on Magnetically Coupled Coils and Ideal Transformers

reverses A reversal in magnetic polarity requires moving one polarity dot without changing the reference polarities of the terminal currents and voltages

Example C.l illustrates the application of theT-equivalent circuit

Example C.l

a) Use the T-equivalent circuit for the magnetically

coupled coils shown in Fig C.6 to find the phasor

currents I| and I2 The source frequency is

400 rad/s

b) Repeat (a), but with the polarity dot on the

sec-ondary winding moved to the lower terminal

Solution

a) For the polarity dots shown in Fig C.6, M carries

a value of +3 H in the T-equivalent circuit

Therefore the three inductances in the

equiva-lent circuit are

L { - M = 9 - 3 = 6 H;

L2 - M = 4 - 3 = 1 H;

M = 3 H

Figure C.7 shows the T-equivalent circuit, and

Fig C.8 shows the frequency-domain equivalent

circuit at a frequency of 400 rad/s

Figure C.9 shows the frequency-domain

circuit for the original system

Here the magnetically coupled coils are

modeled by the circuit shown in Fig C.8 To find

the phasor currents I] and I2, we first find the

node voltage across the 1200 O inductive

reac-tance If we use the lower node as the reference,

the single node-voltage equation is

300

700 + y'2500 /1200

Solving for V yields

V = 136 - /8 = 136.24/-3.37° V(rms)

Then

300 - ( 1 3 6 - /8)

700 + /2500 63.25 / - 7 1 5 7 ° mA (rms)

500 a /loo a

_ T V Y Y > _

II

300/0QV

a 200 a /1200 a

4 o I •

loo a 800 a

A/W

|3H

Figure C.7 A The T-equivalent circuit for the magnetically

coupled coils in Example C.l

/2400 /400

:/1200

Figure C.8 • The frequency-domain model of the equivalent

circuit at 400 rad/s

500 a / loo a 200 a /2400 a /400 a 100 a

6 3()0,()° V / I 2 0 0 a

Figure C.9 A The circuit of Fig C.6, with the magnetically

coupled coils replaced by their T-equivalent circuit

and

I , = 136 - /8

900 - /2100 59.63 /63.43° mA (rms)

Vi /3600 a

b) When the polarity dot is moved to the lower

ter-minal of the secondary coil, M carries a value of

- 3 H in the T-equivalent circuit Before carrying out the solution with the new T-equivalent cir-cuit, we note that reversing the algebraic sign of

M has no effect on the solution for Ij and shifts

I2 by 180°.Therefore we anticipate that

/2500 a

Figure C.6 A The frequency-domain equivalent circuit for Example C.l

and

Ij = 63.25/-71.57° mA (rms)

I2 = 59.63 / - 1 1 6 5 7 ° mA (rms)

We now proceed to find these solutions

by using the new T-equivalent circuit With

M = - 3 H, the three inductances in the

equiv-alent circuit are

Lj - M = 9 - ( - 3 ) = 12 H;

L 2 - M = 4 - ( - 3 ) = 7 H ;

M = - 3 H

At an operating frequency of 400 rad/s, the

frequency-domain equivalent circuit requires two

inductors and a capacitor, as shown in Fig CIO

The resulting frequency-domain circuit for

the original system appears in Fig C.ll

As before, we first find the node voltage

across the center branch, which in this case is a

capacitive reactance of — /'1200 H If we use the

lower node as reference, the node-voltage

equation is

V - 300

+ +

700 + /4900 -/1200 900 + /300

Solving for V gives

V = - 8 - /56

= 56.57 / - 9 8 1 3 ° V (rms)

C.2 The Need for Ideal Transformers in the Equivalent Circuits 735

Then

300 - ( - 8 - /56)

h =

and

700 + /4900

= 63.25 / - 7 1 5 7 ° mA (rms)

- 8 - /56

900 + /300

= 59.63 / - 1 1 6 5 7 ° mA (rms)

/4800 fl /2800 0

-/1200O

Figure CIO • The frequency-domain equivalent circuit for

M = - 3 H and a> = 400 rad/s

500 n /loo n 200 n /48ooa /28oon 1000

r^/3°(Mjc

I

V -/120012: 800 O

-/25()0 i l

Figure C.ll • The frequency-domain equivalent circuit for

Example C.l(b)

C.2 The Need for Ideal Transformers in

the Equivalent Circuits

The inductors in the T- and 77-equivalent circuits of magnetically

cou-pled coils can have negative values For example, if L\ = 3 mH,

L 2 = 12 mH, and M = 5 mH, the T-equivalent circuit requires an

induc-tor of —2 mH, and the 7r-equivalent circuit requires an inducinduc-tor of

- 5 5 mH These negative inductance values are not troublesome when

you are using the equivalent circuits in computations However, if you

are to build the equivalent circuits with circuit components, the negative

inductors can be bothersome The reason is that whenever the frequency

of the sinusoidal source changes, you must change the capacitor used to

simulate the negative reactance For example, at a frequency of

50 krad/s, a - 2 mH inductor has an impedance of - / 1 0 0 fi.This

imped-ance can be modeled with a capacitor having a capacitimped-ance of 0.2 /xF If

the frequency changes to 25 krad/s, the - 2 mH inductor impedance

changes to - / 5 0 il At 25 krad/s, this requires a capacitor with a

capaci-tance of 0.8 /xF Obviously, in a situation where the frequency is varied