20
CIRCUIT
THEOREMS
IN AC ANALYSIS
20-1
20-2
20-3
20-4
20-5
ffi
The Superposition
Theorem
Thevenin's
Theorem
Norton'sTheorem
MaximumPowerTransfer
Theorem
Technology
Theoryinto Practice
Electronics
Workbench(EWB)and
PSpiceTutorialsat
http://www.prenhalLcom/floyd
r INTRODUCTION
Severalimportant theorems were covered in Chapter8
with emphasison their applications in the analysisof
dc circuits. This chapter is a continuation of that coverage with emphasison applications in the analysisof
ac circuits with reactive elements.
The theoremsin this chapter make analysiseasier for certain types of circuits. These methods do not
replace Ohm's law and Kirchhoff's laws, but they are
normally used in conjunction with those laws in certain situations.
Although you are already familiar with the theorems covered in this chapter, a restatementof therr
purposesmay be helpful. The superpositiontheorem
will help you to deal with circuits that have multiple
sources.Thevenin'sand Norton's theoremsprovide
methods for reducing a circuit to a simple equivalent
form for easier analysis.The maximum power transfer
theorem is used in applications where it is important
for a given circuit to provide maximum power to a
load.
In the TECH TIP assignmentin Section20-5,
you will evaluate a band-passfilter module to determine its internal component values, and you will
apply Thevenin's theorem to determine an optimum
load impedancefor maximum power transfer.
r TECHnology
Theorv
Into
Practice
Band-pass filter
module
rN c,ro ioui
r CHAPTER
OBIECTIVES
tr Apply the superpositiontheorem to ac circuit
analysis
A Apply Thevenin's theorem to simplify reactive ac
circuits for analysis
tr Apply Norlon's theorem to simplify reactive ac
circuits
tr Apply the maximum power transfer theorem
IN AC ANALYSIS
794 T CIRCUITTHEOREMS
THEOREM
2O-1 . THE SUPERPOSITION
The superposition theorem w&s introduced in Chapter I for ase in dc circait analysi*
In this section, the superposition theorem is applied to circuits with ac sources and
reuctive elements.
After completing this section, you should be sble to
r Apply the superposition theorem to ac circuit analysis
. State the superpositiontheorem
. List the stepsin applying the theorem
The superposition theorem can be stated as follows:
The current in any given branch of a multiple-source circuit can be
found by determining the currents in that particular branch produced
by each source acting alone, with all other sources replaced by their
internal impedances. The total current in the given branch is the phasor
sum of the individual source currents in that branch.
The procedure for the application of the superpositiontheorem is as follows:
Step 1.
Leave one of the sourcesin the circuit, and replace all others with their internal impedance. For ideal voltage sources,the internal impedance is zero. For
ideal current sources,the intemal impedance is infinite. We will call this procedure zeroing the source.
Step 2.
Find the cunent in the branch of interestproduced by the one remaining source.
Step 3.
Repeat Steps 1 and 2 for each source in turn. When complete, you will havea
number of current values equal to the number of sourcesin the circuit.
Step 4.
Add the individual current values as phasor quantities.
The following three examplesillustrate this procedure.
Find the current in R of Figure 20-1 using the superposition theorem. Assume the
internal source impedancesare zero.
FIGURE2O-1
c1
0.01prF
V,r
1020"v
J=10kHz
vrz
820"V
/= 10kHz
Solution
Step 1. ReplaceV,twith its internalimpedance(zero),andfind the currentin R due
to V"1,as indicatedinFigure 20-2.
FIGURE2O-2
cr
0.01pF
%r
1020"v
f = 10k}jz
c2
0.02pF
THESUPERPOSITION
THEOREM. 795
l
=
=l.59ko
X.,=,f
2tr[CI 2n(l0 kHzlt}.0| pFr
=196d)
X e 1= , ] = = - - L
2nfC2 2n(10kHz)(O.02
p.F)
Looking from V"r,the impedanceis
RXs,
= 1.591-90"ke + 0'010" kQ)(1961-90"Q)
R + X.,
1.0kO - i196A
= 1.59.1-90'kA + 6222-51.5'O
= -jt.59 kQ + 387A - j481O = 387O -j2.08 kO
Z =X., *
Convertingto polar form yields
Z = 2.122-19.5' kO
The total curent from source 1 is
1",=E= - -t9'liI. - =4.t2tte.5"mA
z
2 . 1 2 1 - 1 9 . 5 "k Q
Use the current-divider formula. The current through R due to V", is
t., = (I-4!L\t.,
- jxr,
(
a,
= !e64-e0" \+.tzz'tg.s"
' " ^o
\ R
) - ' ' \ t . ok o - j i 9 6 a ) " ' - - '
= (0.6232-51.5"
{>)(4.12279.5'
mA)= 2.94128.0"
mA
Step2.
Find the current in R due to source V,2by replacing %, with its internal
impedance (zero), as shown in Figure 20-3.
F I C U R E2 O - 3
V"z
820"V
-f = l0 kHz
LookingfromV,2,the impedance
is
RX.'
= 7961-90"{, + (1.020' kA)0.59t-90' kO)
R + xcr
l.0kfl ,li,
kf,
= 1962-90' Q + 847Z-32.2' Q,
=-j796{L+111
O - j 4 5 I C 2 = l l lA - j n 4 7 A
rq' = vA c 2-
Convertingto polar form yields
Z = 1438/-60.1'f)
The total currentfrom source2 is
_
v-.
810" v
=
l" =
5 ' 5 6 2 6 0 ' lm
' A
t
t+zal4rLf o=
Usethe current-dividerformula.The currentthroughR dueto V"2is
-l p )
=
/xctl-go" \-l I . '
\ R-.iXc' l "
l-
= ( . ! ?Yu.?"
r" mA= 4..t
0Z2j.eomA
=\^?
=)s.roroo.
\ r . 0k o- j r . s 9k o/
796 I
CIRCUITTHEOREMS
IN AC ANALYSIS
Step 3.
Convert the two individual resistor currents to rectangular form and add to
get the total current through R.
lm = 2.94228.0' mA = 2.60 mA + jl.3g mA
l a z = 4 . 7 0 / 2 ' 1 . 9 om A = 4 . 1 5 m A + j 2 . 2 0 m A
In = Inr +lR2= 6.75 mA + j3.58 mA=7.64227.9. mA
Related Prohlem
EXAMPLE
2O-2
Determine IR if V"2 = 8ll80o V in Figure 20-1.
Find the coil cur:rentin Figure 20-4. Assume the sourcesare ideal.
FICURE2O-4
C
I"z
30190" mA
Solution
step 1. Find the current through the inductor due to cunent source 1"1by replacing
source1"2with an open, as shown in Figure 20-5. As you can see,the entire
100 mA from the current source1,1is through the coil.
FIGURE2O-5
Step 2.
C
Find the current through the inductor due to current source 1"2by replacing
source 1"1with an open, as indicated in Figure 20-6. Notice that all of the
30 mA from source1"2is through the coil.
(.
F I C U R E2 0 - 6
Step 3.
To get the total inductor current, superimposethe two individualcurrentsand
add as phasor quantities.
lr=lr, + I'
= 70020' mA+ 30290o mA = 100 mA +j30 mA
= 104216.7" mA
Related Problem
Find the current through the capacitor in Figure20-4.
T H E S U P E R P O S I T IT
OHNE O R E M .
EXAMPTE 20-3
797
Find the total current in the resistor R1 in Figure 20*7. Assume the sourcesare ideal.
FIGURE20-7
c
0.22pF
V"r
520" V rms
f = IkHz
Rl
1.0ko
RL
- vsz
- 15V
2ko
Solution
Step 1. Find the current through R1 due to source V"1by zeroing (replacing with its
internal impedance) the dc source V52,as shown in Figure 20-8. Looking
from V"1,the impedanceis
z=xc.+#
|
= 7 2 3I
2n(1.0kllz)(0.22p.F)
(1'020" kflx2z0' kQ)
z = 7232-g0"O +
320'k{,
-j723
=
9 + 667dl = 9842-47.3"C,
X-. *
The total curent from source1 is
r^,,=E = - ,t(ol.-Y-mA
^ =5-082473"
z
9842.-47.3'{L
Usethe current-dividerapproach.The currentin R; dueto V"1is
*'
l'0 ko\5
I o , , " ,=, /
. 0 8 r 4 i . 3 "n A = 1 . 6 9 2 4 7 . m
3 "A
\ t '. ', - /
\Rr+RLl
\ 3 kA /
Irr,
Rr
2ko
520'V rms
f =lffiz
FIGURE2O-B
FIGURE2O-9
Step 2.
Find the current in R1 due to the dc source V52by zeroing V"1(replacing with
its internal impedance), as shown in Figure 20-9.The impedancemagnitude
as seenby Vs2is
Z=Rr*R.=3Lg2
The currentproducedby V52is
IRz(sz)=
ry=#=5mAdc
IN AC ANALYSIS
798 T CIRCUITTHEOREMS
Step3.
By superposition,the total current in R. is 1.69241.3" mA riding on a dc
level of 5 mA, as indicated in Figure 20-10.
FICURE20-10
Related Problem
sEcTloN20-1
REVIEW
Ir.
Determine the current through RLlf Vs2is changed to 9 V.
l. If two equal currentsare in opposing directionsat any instant of time in a given
branchof a circuit, what is the net current at thal"instant?
2. Why is the superpositiontheoremuseful in the analysisof multiple-source
3. Using the superposition theorem, find the magnitude of the current through R in
Figure20-tl.
FtcuRE20-11
THEOREM
2O_2. THEVENIN'S
Thevenin's theorem, as applied to ac circuits, provides a method for reducing any cir'
cuit to an equivalentform that consi.stsof an equivalent ac voltage soarce in serieswith
an equivalerutimp edance.
After completing this section' you should be able to
r Apply Thevenin's theorem to simplify reactive ac circuits for analysis
. Describe the form of a Thevenin equivalent circuit
. Obtain the Thevenin equivalent ac voltage source
. Obtain the Thevenin equivalent impedance
. List the stepsin applying Thevenin's theorem to an ac circuit
T H E V E N I N 'TSH E O R E M .
799
The form of Thevenin's equivalent circuit is shown in Figure 20-12. Regardlessof how
complex the original circuit is, it can always be reduced to this equivalent form. The
equivalent voltage source is designatedYx,; the equivalent impedance is designated2,1,
(lowercaseitalic subscript denotesac quantity). Notice that the impedanceis represented
by a block in the circuit diagram. This is becausethe equivalent impedancecan be of several forms: purely resistive, purely capacitive, purely inductive, or a combination of a
resistanceand a reactance.
F I G U R E2 O - 1 2
Thevenin'sequivalentcircuit.
Equivalency
Figure 20-13(a) shows a block diagram that representsan ac circuit of any given complexity. This circuit has two output terminals, A and B. A load impedance, 27, is connected to the terminals. The circuit produces a certain voltage, V1, and a certain current,
L. as illustrated.
zrn
tu)V,l
vr
lr,
v -
B
(b)
F I G U R E2 O - 1 3
An ac circuit of any complexity can be reduced to
purposes.
a Thevenin
equivalent for analysis
By Thevenin's theorem, the circuit in the block can be reduced to an equivalent
form, as indicated in the beige area of Figure 20-13(b). The term equivalent means that
when the same value of load is connected to both the original circuit and Thevenin's
equivalent circuit, the load voltages and currents are equal for both. Therefore, as far as
the load is concerned,there is no difference between the original circuit and Thevenin's
equivalent circuit. The load "sees" the same current and voltage regardlessof whether it
is connected to the original circuit or to the Thevenin equivalent. For ac circuits, the
equivalent circuit is for one particular frequency. When the frequency is changed, the
equivalent circuit must be recalculated.
B O O T C I R C U I T H E O R E MISN A C A N A L Y S I S
Thevenin'sEquivalentVoltage(V,6)
As you have seen,the equivalentvoltage, V,7,,is one part of the complete Thevenin equivalent circuit.
Thevenin's equivalent voltage is defined as the open circuit voltage
between two specified terminals in a circuit.
To illustrate, let's assumethat an ac circuit of some type has a resistor connected
between two specified terminals, A and B, as shown in Figure 20-14(a). We wish to find
the Thevenin equivalentcircuit for the circuit as "seen" by R. V,7,is the voltage acrossterminals A and B, with R removed, as shown in part (b) of the figure. The circuit is viewed
from the open across terminals A and B, and R is considered external to the circuit for
which the Thevenin equivalent is to be found. The following three examples show how to
Iind V,7,.
(a) Circuit
(b) With R removed
F I G U R E2 O - 1 4
How Ys, is determined.
EXAMPTE 20-4
Determine V,7,for the circuit external to R; in Figure 20-15. The beige area identifies
the porlion of the circuit to be thevenized.
F I G U R E2 O _ 1 5
100C)
v"
2520"V
Solution Remove R, and determine the voltage from A to B (Y,,).In this case,the
voltage from A to B is the same as the voltage acrossX;. This is determined using the
voltage-divider method.
I XLZ90" \,,
1 50290" Q
252.0'Y= I1.2263.4'Y
\Rr+jXt)
\112126.6"Q
Y tn= Y ea = Y L = 11.2263.4" Y
rr
"
Related Problem
Determine Vr, if R1 is changed to 47 Q in Figure 20-15.
EXAMPLE 20-5
For the circuit in Figure 20-16, determine the Thevenin voltage as seenby R1.
F t c u R E2 0 - 1 6
Solution Thevenin's voltage for the circuit between terminals A utd B is the voltage
that appearsacrossA and B with R1 removed from the circuit.
There is no voltage drop acrossR2 becausethe open across terminals A and B
preventscurrent through it. Thus, Vas is the sameas V62 and can be found by the voltage-divider formula.
l'52-90'ko
xczl-9j"
V,. = v-, = [ -.rXcr\v" = /
\tozo"v
l
Rr
iXrz)
\
\ 0 kA -i3 kO/
ka
- ( 1'51-90"
\tozo" y = 4151-18.4'v
\3.162-11.6'kQ/
Ytn= Yen= 4.752-18.4"Y
Related Problem
EXAMPLE
20-6
Determine V,7,if R1 is changedto 2.2kQ in Figure 20-16.
For Figure 20-11 , find V,a for the circuit external to R..
F T G U R2E0 - 1 7
x.
R
10ko
Solution First remove R1 and determine the voltage acrossthe resulting open terminals, which is V,7,.Find V,7,by applying the voltage-divider formula to X6and R.
v,,= vR
' = ( :t9: \v, = f;*q*
^ )sro"u
\R-jx./' \ r0ko-jl0ko/
=(
tozo' ug> \szo' Y = 3.s424s"
Y
\14.141-45' kQ I
Notice the L has no effect on the result, since the 5 V source appearsacross C and R
in combination.
Related Problem
Find V,7,if R is 22 kQ and R1 is 39 kO in Figure 20-17.
BO2 I
CIRCUITTHEOREMS
IN AC ANALYSIS
Thevenin's
Equivalent
lmpedance (Ztn)
The previous examplesillustrated how to find only one part of a Thevenin equivalentcircuit. Now, let's turn our attention to determining the Thevenin equivalent impedance,Z,1,.
As defined by Thevenin's theorem,
Thevenin's equivalent impedance is the total impedance appearing
between two specified terminals in a given circuit with all sources
replaced by their internal impedances.
Thus, when you wish to find Z,l,between any two terminals in a circuit, all the voltage
sourcesare replacedby a short (any internal impedanceremains in series).Al1 the cunent
sources are replaced by an open (any internal impedance remains in parallel). Then the
total impedancebetween the two terminals is determined.The following three examples
illustrate how to find Z,y.
EXAMPTE
2O-7
Find Zn for the part of the circuit in Figure 20-18 that is externalto R,. This is the
same circuit used in Example 204.
F I G U R E2 O - 1 8
100f,)
v,
2520'V
Solution First, replace V" with its internal impedance (zero), as shown in Figure
20-19.
FIGURE
2O-19
I
I
I
I Ztn
I
I
I
Lookingin betweenterminalsA andB, R andXl arein parallel.Thus,
ry = -(R110")(xLzg0")= Q0020" ox50zg0" o)
"'h
Rt + jxLroo (z + /--.50
o
(10020'
ox50z90' Q)
_
= 44,6263.4.dL
112126.6{t
RelatedProblem Change\ to 47 Q anddetermine2,7,.
THEVENINT
' SH E O R E M
EXAMPLE 20-B
For the circuit in Figure 20-20, determine Zs, ts saenby R;. This is the same circuit
used in Example 20-5.
FtcuRE20-20
Solution First. replace the voltage source with its internal impedance (zero), as
shown in Figure 20:21.
F I G U R E2 0 - 2 1
T
Looking from terminals A and B, C2 appearsin parallel with the series combination of R1 and C1. This entire combination is in series with Rr. The calculation for
Z,yis as follows:
*V(2{F=Pi
z,n=Rzloo
jxa - jxc2
h-
=56otoon.#
kO)
(1'52-9-0'FOX1'82-56'3"
3.t61-11.6"k{t
= 56020"Q + 8541-74.7o{r= 560 O + 225 Q - j824O
= 785 Q - j824Q = 11382-46.4" O
= 56010"()+
RelatedProblem DetermineZth if Rr is changedto 2.2 kQ in Figure20-20.
B O 4 T C I R C U I T H E O R E MISN A C A N A L Y S I S
EXAMPLE
2O-9
For the circuit in Figure 20-22, determine 2,7, for the portion of the circuit externalto
R.. This is the same circuit as in Example 20-6.
F I G U R E2 0 - 2 2
Solution With the voltage source replaced by its internal impedence (zero), X. is
effectively out of the circuit. R and c appear in parallel when viewed from the open
terminals, as indicated in Figure 20-23. 2,6 is calculatedas follows:
( lozo"kftxloz- 90"ko)
= 7'07Z -45' kO
l4.rr-41.6:
,,,. =W- R jxc
FIGURE20-23
X^
I
I
I
:.!i:l
l.i:i
irl:.
I
I
Related Problem
Find 2,61f R is 22 kQ and R. is 39 kf) in Figure 20-22.
EXAMPTE
20-10
20-24
that
isexternar
to
ilTr'li]lt?n':#:'Jfil';:ffi,::'i311fi1[.i*'re
FICURE20-24
100C)
v,
2520"y
Solution From Examples 204 and 20-l , respectively,Y tn = Il .2263.4. Y and 2,7
= 44.6263.4" Q. In rectangular form, the impedanceis
Z,n=20O+j40O
This form indicates that the impedanceis a 20 Q resistor in serieswith a 40 Q inductive reactance.The Thevenin equivalent circuit is shown in Figure 20-25.
T H E V E N I N ,TSH E O R E MI
805
FIGURE20_25
1r.2163.4"
v
Related Problem
4',7Q.
Draw the Thevenin eouivalent circuit for Fieure 2O-24 with
EXAMPLE20_11 For the circuit in Figure 20*26, sketch the Thevenin equivalent circuit external to R1.
This is the circuit used in Examples 20-5 and 20-8.
FTGURE
20-26
R]
Solution From Examples 20-5 and 20-8, respectively,Y*=
Ztn = 11382-464" f). In rectangularform, the impedanceis
4.J52-18.4'V
and
Ztn= 785 A - j824 A
The Thevenin equivalent circuit is shown in Figure 20-21 .
z,n
FIGURE20-27
4.752-r8.4"
V
Related Prohlem
with Rt = 2.2 kd).
Sketch the Thevenin equivalent for the circuit in Figure 20-26
B06 !
C I R C U I T H E O R E MISN A C A N A L Y S I S
EXAMPLE20_12 For the circuit in Figure 20-28, determine the Thevenin equivalent circuit as seenby
R.. This is the circuit in Examples 20-6 and20-9.
F I G U R E2 0 - 2 8
R
10ko
Solution From Examples 20-6 and 20-9, respectively,Y tn = 354245" , and Zx, =
7 .OiZ-45" kQ. The impedancein rectangular form is
Z,n= 5 kf) - j5 kO
Thus, the Thevenin equivalent circuit is as shown in Figure 20-29.
FIGURE20-29
Ztn
3.54245"V
Related Problem Change R to 22 kQ and R, to 39 kO
the Thevenin equivalent circuit.
Thevenin's
Equivalent
Circuit
The previous examples have shown how to find the two equivalent components of a
Thevenincircuit, YTpandZx,.Keepin mind that Vr7,and2,6 can be found for any circuit.
Once these equivalentvalues are determined,they must be connectedin seriesto form the
Thevenin equivalent circuit. The following examples use the previous examples to illustrate this final step.
Summaryof Thevenin's
Theorem
Remember that the Thevenin equivalent circuit is always a voltage sourcein serieswith a
resistanceregardlessof the original circuit that it replaces.The significance of Thevenin's
theorem is that the equivalent circuit can replace the original circuit as far as any external
load is concerned.Any load connectedbetween the terminals of a Thevenin equivalent
circuit experiencesthe same current and voltage as if it were connectedto the terminals
of the original circuit.
A summary of steps for applying Thevenin's theorem follows.
Step 1.
Open the two terminals between which you want to find the Thevenin circuit,
This is done by removing the component from which the circuit is to be
viewed.
NORTON,STHEOREM .
sEcTroN20-2
REVIEW
807
Step 2.
Determine the voltage acrossthe two open terminals.
Step 3.
Determine the impedanceviewed from the two open terminals with ideal voltage sourcesreplaced with shorts and ideal current sourcesreplaced with opens
(zeroed).
Step 4.
ConnectVr7,andZ,Tinseies to producethe completeThevenin equivalentcircuit.
1".What are the two basic componentsof a Thevenin equivalent ac circuit?
2. For a certaincircuit,Zn=25 O - j50 fl. and Ytn=5/:0o V. Sketchthe Thevenin
equivalentcircuit.
3. For the circuit in Figure 20-30, find the Thevenin equivalentlooking from terminals
A and B.
FIGURE20-30
2O-3 r NORTON'STHEOREM
Like Thevenin's theorem, Norton's theorem provides a method of reducing & more
complex circuit to a simpler, more msnageableform for analysis. The basic dffirence
is that Norton's theorem gives an equivalent cument source (rather thsn a voltage
source) in parallel (rather than in series) with an equivalent impedance.
After completing this section, you should be uhle to
r Apply Norton's theorem to simplify reactive ac circuits
. Describe the form of a Norton equivalent circuit
. Obtain the Norton equivalent ac current source
. Obtain the Norton equivalent impedance
The form of Nofion's equivalent circuit is shown in Figure 20-31. Regardless of how
complex the original circuit is, it can be reduced to this equivalent form. The equivalent
current sourceis designatedI,, and the equivalent impedanceis Z, (lowercaseitalic subscript denotesac quantity).
Nortonts theorem shows you how to find I, and 2,. Once they are known, simply
connect them in parallel to get the complete Norton equivalent circuit.
FIGURE20-31
Norton equivalentcircuit.
BOB T CIRCUITTHEOREMS
IN AC ANALYSIS
Norton/sEquivalent
CurrentSource(1")
I, is one part of the Norton equivalent circuit; Z, is the other pafi.
Norton's equivalent current is defined as the short-circuit current
between two specified terminals in a given circuit.
Any load connectedbetween these two terminals effectively "sees" a cuffent sourceI, in
paralfel with 2,.
To illustrate, let's supposethat the circuit shown in Figure 20-32has a load resistor connectedto terminals A and B, as indicated in part (a). We wish to find the Norton
equivalent for the circuit external to R.. To flnd I,, calculate the current between terminals A and B with those terminals shorted, as shown in part (b). Example 20-13 shows
how to find I,.
FIGURE20-32
How Io is determined.
(a) Circuit with load resistor
EXAMPLE 20-1J
(b) Load is replaced by short and
shoft circuit current is In.
In Figure2u33, determine
r, for the circuitas ,.seen,,by
theloadresistor.Thebeige
areaidentifiesthe portionof the circuit to be nortonized.
F I C U R E2 0 - 3 3
Solution
Short the terminals A andB, as shown in Fisure 20-34.
FIGURE20-34
I, is the current through the short and is calculated as follows. First, the total
impedanceviewed from the source is
Z = Xct * =RX=!, = 50/-90" e + 6620" Q)(1002-90" sr)
R+X.,
56(.)-1100f2
= 502-90" Q + 48.91-29.3"O
= -j50 Q + 42.6A - j23.9 CL= 42.6A - il3.9 A
NORTON'S
THEOREM
Converting to polar form yields
Z = 85.32-60.0' Q
Next, the total current from the source is
I" = 5 = - ,9oto"'
z
85.32-60.0. Q
=703260.0"
mA
Finally, apply the current-divider formula to get In (the current through the short
between terminals A and B).
= (=-\-)I,
t,,
"
'
\ R + X c zI
= l,;+4q*
-
nA = 3442r2r.
^-\tozzoo.o"
mA
\ 5 6 o j t 0 0O / '
This is the valuefor the equivalentNortoncurrentsource.
Related Prohlem DetermineIn if v" is changedto 2520" v and R is chansedto
33 Q in Fisure20-33.
Norton'sEquivalentlmpedance(Zn)
Z, is defined the same as zr7: rt is the total impedance appearing between two specified
terminals of a given circuit viewed from the open terminals with all sourcesreplaced by
their internal imoedances.
''
EXAMPLE20_14 Find z, for the circuit in Figure 20-33 (Example 20-13) viewed from the open across
terminals A and B.
solution
20-35.
First, replace v" with its internal impedance (zero), as indicated in Figure
II
l
i
l
FlcuRE20-35
xct
-l#
100c)
50f,)
R
s6c)
Z,
I
Looking in between terminals A and B, C2 is in series with the parallel combi_
nation of R and C1.Thus,
= t00l_-90"e + (5620" QX50z-90' Q)
z, = xr-"* =StR+X.,
56O_i50O
= l00Z-90' Q + 31.32-48.2.Q
= - j 1 0 0{ L + 2 4 . 8 f J *j z j . 8 { L = 2 4 . 8O - i 1 2 S O
The Nortonequivalentimpedanceis a 24.8C)resistance
in serieswith a 128C)caoacitive reactance.
RelatedProblem Find Zn in Figure 20-33 if Y, = 2510. V andR = 33 C).
810 r
C I R C U I T H E O R E MlSN A C A N A L Y S I S
The previous two exampleshave shown how to find the two equivalent components
of a Norton equivalent circuit. Keep in mind that these values can be found for any given
ac circuit. Once these values are known, they are connectedin parallel to form the Norton equivalent circuit, as the following example illustrates.
EXAMPLE 20-15
Show the complete Norlon equivalent circuit for the circuit in Figure 20-33 (Example
20_13).
Solution From Examples 20-13 and 20-14, respectively,l" = 3442121' mA and
Z, = 24.8 A - jl28 Q. The Nofton equivalent circuit is shown in Figure 20-36.
FtcuRE20-36
A
24.8A
t,
3441121"mA
n8a
zn
B
Related Problem Show the Norton equivalent for the circuit in Figure 20-33
Y, = 2520'V and R = 33 Q.
Summaryof Norton'sTheorem
Any load connectedbetween the terminals of a Norton equivalent circuit will have the
same current through it and the same voltage acrossit as it would when connectedto the
terminals of the original circuit. A summary of steps for theoretically applying Norton's
theorem is as follows:
SECTION20-3
REVIEW
Step 1.
Replacethe load connectedto the two terminals between which the Norton circuit is to be determined with a short.
Step 2.
Determine the current through the short. This is I,.
Step 3.
Open the terminals and determine the impedancebetween the two open terminals with all sourcesreplaced with their intemal impedances.This is Zn.
Step 4.
Connect ln and Z, in parallel.
1. For a given circuit, I, = 510' mA, and Z, = 150 O + j100 C).Draw the Norton
equivalentcircuit.
) Find the Nortoncircuit as seenby R1in Figure20-37.
FIGURE20-37
v_
's
I
1.2kO
1.220"V
R,
Lc
1800c)
M A X I M U MP O W E RT R A N S F ETRH E O R E MT 8 1 1
2O_4I MAXIMUM POWERTRANSFER
THEOREM
When a load is connected to a circuit, rnuximum power is transferred to the load when
the loud impedance is the complex conjugate of the circait's outptut impedance.
After completing this section, you should be sble to
r Apply the maximum power transfer theorem
. Explain the theorem
. Determine the value of load impedance for which maximum power is transferred
from a siven circuit
The complex conjugate of R - jX6 is R + jX1, where the resistancesand the reactances
are equal in magnitude.The output impedanceis effectively Thevenin's equivalent impedance viewed from the output terminals. When Zl is the complex conjugate of Zou,, maximum power is transfered from the circuit to the load with a power factor of 1. An equivalent circuit with its output impedanceand load is shown in Figure 2V38.
FIGURE20-38
Equivalentcircuit with load.
Zout
)o,
K
Example 20-16 shows that maximum power occurs when the impedancesare conjugately matched.
EXAMPLE
20_16 The circuit to the left of terminals A and B in Figure 20-39 provides power to the load
27. This can be viewed as simulating a power amplifier delivering power to a complex
load. It is the Thevenin equivalent of a more complex circuit. Calculate and plot a
graph of the power delivered to the load for each of the following frequencies:
10 kHz, 30 kHz, 50 kHz, 80 kHz, and 100 kHz.
FIGURE20-39
812 r
C I R C U I T H E O R E MlSN A C A N A L Y S I S
Solution For/= 10kHz,
-- -]=l.59kQ
xr=|-=
ZnfC 2n(10kHz)(0.01pF)
Xr = 2nfL = 2n(10kHz)(l mH) = 62'8A
The magnitudeof the total impedanceis
4o,=@=@=1.53kQ
The curent is
I- = v ' =
tou, =6.5lmA
Z,o, 1.53kQ
The load power is
pr= I2Rr= (6.54mA)'z(loQ) = 428p.W
For/= 30kIIz,
I
= 5'll Q
znt:o t Hrlro.otlrn
Xr= 2n(30kHz)(l mH) = 199g
lt =
120ful\ 1342{)l' = 343o
v, - lov =29.2mA
'r*=
Z,o, 3$ A-'
pt= I2Rt= Q9.2mA)2(10Q) = 8.53mW
z,o,=f
For/= 50 kHz,
1
= 3 1 8A
prF)
2n(50kHz)(0.01
Xr=2n(50kHz)(l mH) = 314g
X'- =
Note that Xs und Xy are very close to being equal which makes the impedances
approximately complex conjugates.The exact frequency at which Xr= Xc is 50.3 kHz.
2,, = f (20 Q)2 + (4 tL)2= 2o'4 Q'
lov =-l9omA
t=!'--
Zro,
20.4 A
Pr= I2Rr= (490 mA)2(10Q) = 2.40 W
For/= 80 kHz,
xc=- ^^_l-=1994
2n(80kHz)(0.01pF1
Xr=2n(80 kHz)(l mH) = 563g
z,o,={@
e)f * 1304fll' = 305Q
j,=T;,=
=t::^^;:T =1o
8mw
For f= 100kHz.
v _
Acv _
A L 7
-
=159Q
2n(100kHz)(0.01pF)
2n(100kHzXl mH) = 623g
x@8;
@61elf= 46etL
MAXIMUM POWERTRANSFER
THEOREM T 813
i,=,',;,=#^^;,;r,=4.54
mw
As you can see from the results, the power to the load peaks at the frequency
(50 kHz) for which the load impedanceis the complex conjugate of the output impedance (when the reactancesare equal in magnitude). A graph of the load power versus
frequency is shown in Figure 2040. Since the maximum power is so much larger than
the other values, an accurateplot is difficult to achieve without intermediate values.
FIGURE20-40
pr (w)
50 kHz
RelatedProblem rf R = 47 Q and c = 0.022prFin a seriesRC circuir,what is the
complexconjugateof the impedanceat 100kHz?
EXAMPTE
20_17
(a) Determine the frequency at which maximum power is transferredfrom the ampli-
fier to the speakerin Figure 204r(a). The amplifier and coupling capacitor are the
source, and the speaker is the load, as shown in the equivalent circuit of Figure
20-4r(b).
(b) How many watts of power are delivered to the speaker at this frequency if
% =
3.8 V rms?
Amplifier
Rw
8f,}
speaker
h*t=tr
l\
n
FIGURE20-41
)_
capacitoi
l_
IN AC ANALYSIS
814 T CIRCUITTHEOREMS
Solution
(a) Whenthe powerto the speakeris maximum,the sourceimpedance(R"-7X6') and
so
the load impedance(Rw+ jX) arecomplexconjugates,
Xc=Xr
fi, ='"n
Solving forl
t
^ - ) I
4ntLC
"
2nf rc
znr,{too-rDgt rrD
(b) The power to the speakeris calculated as follows:
Z , o , =R " * R ' , ' = 8 O + 8 ( . )= 1 6 O
I=
p**=
V'Zrn,
3'8v =238rnA
16 O
I2R*= (238 mA)2(8Q) = 453 mW
Related Problem Determine the frequency at which maximum power is transferred
from the amplifler to the speakerin Figure 2041 if the coupling capacitor is 1 pF.
sEcTroN20-4
REVIEW
L. If the output impedance of a certain driving circuit is 50 O - j10 Q, what value of
load impedancewill result in the maximum power to the load?
2. For the circuit in Question 1, how much power is delivered to the load when the
load impedance is the complex conjugate of the output impedance and when the
load current is 2 A?
20-5 r TECHnologyTheory Into Practice
In this TECH TIP, you are given a seuled band-pass filter module that hqs been
removed from a system and two schemutics. Both schematics indicate that the bqndpassfilter is implemented with a low-pass/high-passcomhinution, It is uncertain which
schemutic corresponds to the filter module, but one of them does, By certain measurements, you will determine which schematic representsthe filter so thut the filter circuit
can be reproduced. Also, you will determine the proper load for maximum power
transfer.
The fllter circuit contained in a sealedmodule and two schematics,one of which conespondsto the filter circuit, are shown in Figure 2042.
FilterMeasurement
andAnalysis
r
Based on the oscilloscope measurementof the filter output shown in Figure 20-43,
determine which schematicin Figure 2042 representsthe component values of the filter circuit in the module. A 10 V peak-to-peakvoltage is applied to the input.
TECHNOLOCY
THEORYINTO PRACTICET 815
Schematic A
Schematic B
FIGURE20-42
Filter module and schematics.
10 V peak+o-peak signal
from function
FIGURE20-43
816 I
C I R C U I T H E O R E MISN A C A N A L Y S I S
Based on the oscilloscopemeasurementin Figure 20-43, determine if the filter is operating at its approximatecenterfrequency.
Using Thevenin's theorem, determine the load impedance that will provide for maximum power transfer at the center frequency when connectedto the output of the filter.
Assumethe sourceimpedanceis zero.
sEcTtoN20-5
REVIEW
L. Determine the peak-to-peakoutput voltage at the frequency shown in Figure 20-43
of the circuit in Figure 2042 that was determined not to be in the module.
2. Find the center frequency of the circuit in Figure 20-42 that was determined not to
be in the module.
r SUMMARY
r The superyosition theorem is useful for the analysis of both ac and dc multiple-source circuits.
r Thevenin's theorem provides a method for the reduction of any ac circuit to an equivalent form
consisting of an equivalent voltage source in series with an equivalent impedance.
r The term equivalency, as used in Thevenin's and Norton's theorems, means that when a given
load impedance is connectedto the equivaient circuit, it will have the same voltage acrossit and
the same cuffent through it as when it is connectedto the original circuit.
r Norton's theorem provides a method for the reduction of any ac circuit to an equivalent form
consisting of an equivalent current source in parallel with an equivalent impedance.
r Maximum power is transferred to a load when the load impedance is the complex conjugateof
the impedance of the driving circuit.
r GLOSSARY
Theseterms are also in the end-of-book glossary.
Complex conjugate An impedance containing the same resistance and a reactance oppositein
phase but equal in magnitude to that of a given impedance.
Equivalent circuit A circuit that produces the same voltage and current to a given load as the
original circuit that it replaces.
Norton's theorem A method for simplifying a two-terminal circuit to an equivalent circuit with
only a current source in parallel with an impedance.
Superposition theorem
A method for the analysis of circuits with more than one source.
Thevenin's theorem A method for simplifying a two-terminal circuit to an equivalent circuit with
onlv a voltase source in series with an impedance.
r SELF-TEST
L. In applying the supetposition theorem,
(a) all sourcesare consideredsimultaneously
(b) all voltage sourcesare considered simultaneously
(c) the sourcesare consideredone at a time with all others replaced by a short
(d) the sources are considered one at a time with all others replaced by their internal impedances
2. A Thevenin ac equivalent circuit always consists of an equivalent ac voltage source
(a) and an equivalentcapacitance
(b) and an equivaient inductive reactance
(c) and an equivalent impedance
(d) in series with an equivalent capacitive reactance
PROBLEMS .
817
3. One circuit is equivalent to another when
(a) the same load has the same voltage and current when connectedto either circuit
(b) different loads have the same voltage and cument when connectedto either circuit
(c) the circuits have equal voltage sourcesand equal series impedances
(d) the circuits produce the same output voltage
4. The Thevenin equivalent voltage is
(a) the open circuit voltage
(b) the short circuit voltage
(c) the voltage acrossan equivalent load
(d) none of the above
The Thevenin equivalent impedanceis the impedance looking from
(a) the source with the output shorted
(b) the source with the output open
(c) any two specified open terminals with all sourcesreplaced by their internal impedances
(d) any two specified open terminals with all sourcesreplaced by a short
6. A Norton ac equivalent circuit always consists of
(a) an equivalent ac cuffent source in series with an equivalent impedance
(b) an equivalent ac cuffent source in parallel with an equivalent reactance
(c) an equivalent ac current source in parallel with an equivalent impedance
(d) an equivalent ac voltage source in parallel with an equivalent impedance
1
The Norton equivalent current is
(a) the total current from the source
(c) the current to an equivalent load
(b) the short circuit current
(d) none of the above
8. The complex conjugateof 50 Q +j100 Q is
(a)50f2-j50O
(b) 100O+js0O
(c) 100O -j50 C)
(d) s0 O -1100o
9. In order to get maximum power transfer from a capacitive source, the load must
(a) have a capacitanceequal to the source capacitance
(b) have an impedance equai in magnitude to the source impedance
(c) be inductive
(d) have an impedance that is the complex conjugate of the source impedance
(e) answers (a) and (d)
r PROBLEMS
More dfficult problems are indicated by an asterisk (*).
SECTION20-1 The Superposition
Theorem
1. Using the superpositionmethod, calculate the current through R3 in Figure 20-44.
2. Use the supetposition theorem to find the cument in and the voltage across the R2 branch of
Figlure 2044.
FIGURE 20-44
V
8 1 B T C I R C U I T H E O R E MISN A C A N A L Y S I S
FIGURE20-45
3. Using the superpositiontheorem, solve for the current through R1 in Figure 20-45.
4. Using the superpositiontheorem, find the current through Rr in each circuit ofFigure 20-46.
c3
1 0 0 pF
R2
1.0Mo
40t60" v
Rr
5MO
v2
20Z30'y
(b)
(aJ
FIGURE20-46
*5. Determine the voltage at each point (A, B, C, D) in Figure 20-47. Assume Xc = 0 for all
capacitors.Sketch the voltage waveforms at each of the points.
FtcuRE20-47
Ll
Rz
R1
vj
R^
R3
3.9 kO
10ko
9V
(peak)
Rs
5.1ko
I
l
R6
10ko
*6. Use the superpositiontheorem to find the capacitor current in Figure 20-48.
FTGURE20-48
R
20a
%
I)230.y
|
_J_ Xc
T18o
|
30f)
I.'
0.52t20. A
Theorem
SECTION20-2 Thevenin's
7. For each circuit in Figwe 2049, determine the Thevenin equivalent circuit for the portion of
the circuit viewed by Rr.
HGURE20-49
1000
|
2 70
320"y
r.
l:.
-
X.
J_
-l-7sf)
Rr
1.0kO
Rr
861o
X.^
s8off
vl
1520. y
R2
100ko
Y2
10230.v
8. using Thevenin's theorem, determine the curent through the load Rl in Figure 20-50.
FIGURE20-50
Rz
R3
*9. Using Thevenin's
FIGURE 20_51
3.3kO ..
Lc
5ko
Rr
1.0ko
*10. Simplify the circuit external to R, in Figwe
20-52 to its Thevenin equivalent.
FIGURE20-52
B 2 O T C I R C U I T H E O R E MISN A C A N A L Y S I S
SECTION20-3 Norton'sTheorem
11. For each circuit in Figure 2049, determinethe Norton equivalent as seenby Rr.
12. Using Norton's theorem, find the current through the load resistor R. in Figure 20-50.
*13. Using Norton's theorem, flnd the voltage acrossR4 in Figure 20-51.
SECTION20-4 Maximum PowerTransferTheorem
14. For each circuit in Figure 20-53, maximum power is to be transferred to the load R . Determine the appropriate value for the load impedance in each case.
I,
v"
$20"v
f = 3kHz
F s.zto
:
(al
F I G U R E2 0 - 5 3
*15. Determine Zl for maximum power in Figure 20-54.
FlcuRE20-54
*16. Find the load impedancerequired for maximum power transfer to Zlin Figure 20-55. Determine the maximum true Dower.
FIGURE20-55
r2Q
12o'
41,
*17, A load is to be connected in the place of R2 in Figure 20-52 to achieve maximum power
transfer. Determine the type of load, and expressit in rectangular form.
. 821
ANSWERS
TO SECTION
REVIEWS
EWBTroubleshooting
and Analysis
Theseproblems require your EWB compact disk.
18. Open file PRO20-18.EWB and determine if there is a fault. If so, find the fault.
19. Open file PRO20-19.EWB and determine if there is a fault. If so, find the fault.
20. Open file PRO20-20.EWB and determine if there is a fault. If so, find the fault.
21. Open file PRO20-21.EWB and deterrnineif there is a fault. If so, find the fault.
22, Open fiIe PRO20-22.EWB and determine the Thevenin equivalent circuit by measuremenl
looking from Point A.
23. Open file PRO20-23.EWB and determine the Norton equivalent circuit by measurementlookins from Point A.
I ANSWERS
TO SECTION
REVIEWS
Section20-1
1. The net current is zero.
2. The circuit can be analyzed one source at a time using superposition.
3. 1^=12*4
Section20-2
1. The componentsof a Thevenin equivalent ac circuit are equivalent voltage source and equivalent
series impedance.
2. See Fieure 20-56.
z*
FIGURE20-56
3. Z,h=21.5A * j15.7Q; V,o= 4.14253.8'y
Section20-3
1. SeeFigure20-57.
FIGURE20_57
ln
520" mA
2. Z,= RZ| = 1.220'kdl;1,=
1020" mA
Section20-4
l . Z L =5 0Q + i l 0 O
2. Pt=2ggl^r
822 r
lN AC ANALYSIS
CIRCUITTHEOREMS
Section20-5
mYPP
l. Yo,,=1662-66.7'
=
2. fo 4.76kHz
r ANSWERS
TO RETATED
PROBLEMS
FOR
EXAMPLES
20-l l.'772-752"mA
20-2 30290" mA
20-3 1.69147.3"mA ridingon a dc levelof 3 mA
V
20-4 18.2243.2"
20-5 4.032-36.3"V
204 4.552244'V
20-7 34.3243.2"g>
20-8 1.3'7t-47.8"kQ
2cF9 9J02-65s'kfJ
20-10 SeeFigure20-58.
F I G U R E2 0 - 5 8
v
18.2143.2"
20-L1 SeeFigure20-59
F I G U R E2 0 - 5 9
V
4.032-36.3"
20-12 SeeFigure20-60.
FIGURE20-60
/0 (r'
A
l--------.
8.28ko
4.55224.4"v
20-13 7172135" mA
20-14 r172-78.7 A
20-15 SeeFigure20-61
. 823
ANSWERS
TO SELF-TEST
FIGURE20-61
A
R
22.9A
l l 7 2 1 3 5 "m A
^c
1 1 5f )
2Ul6 47 O + j723 A
20-17 503 Hz
1. (d)
I ANSWERS
e.
TO SELF-TEST (d)
2.(c)
3. (a)
4.(a)
5. (c)
6. (c)
7. (b)
8. (d)