Circuit Theory I
Circuit Theorems
Assistant Professor Suna BOLAT !
Eastern Mediterranean University Department of Electric and Electronic Eng
Circuit Theorems
•
Linearity Property •
Superposi7on •
Source Transforma7ons •
Thevenin’s Theorem •
Norton’s Theorem •
Maximum Power Transfer
!2
A)large)
complex)circuits)
Simplify)
circuit)analysis)
Circuit)Theorems)
Thevenin’s)theorem
Circuit)linearity
Source)transforma7on
)
)
)
)Norton)theorem)
)Superposi7on)
)Max.)power)transfer)
3"
!3
Linearity
Homogeneity property (Scaling)
•
α!A1
αA
!
αA2
!
•
Addi7vity property !
A1
A2
B1
B2
A
A1+B1
A2+B2
A+B
B
!4
Linearity •  A!linear!circuit!is$one$whose$output$is$linearly$
related$(directly$propor4onal)$to$its$input.$
i
I0
V0
v
!5
Linearity
•  A!linear!circuit!consist(of(:(
•  linear(elements((i.e.(R=5 Ω)((
•  linear(dependent(sources((i.e. vs=6Io V)
•  independent(sources((i.e. vs=12 V)((
vs = 10 V → i = 2A
vs = 1V → i = 0.2A
vs = 5mV ← i = 1mA
2
v
p =i R =
: nonlinear
R
2
!6
Linearity
Example(4.1:(For$the$circuit$in$below$find$Io$when$vs=12V
and$vs =24V.
!7
KVL$$
12i1 − 4i2 + vs = 0
− 4i1 + 16i2 − 3vx − vs = 0
vx = 2i1
(1)
(2)
Eq(2)$becomes*
− 10i1 + 16i2 − vs = 0
(3)
Using$Eqs(1)$and$(3)$we**get*
2i1 + 12i2 = 0 → i1 = −6i2
vs
From%Eq(1)%we%get:% − 76i2 + vs = 0 ⇒ i2 =
76
when% vs = 12V
12
I 0 = i2 = A
76
when% vs = 24V
24
I 0 = i2 = A
76
!8
Linearity Example(4.2:(Assume&Io&= 1A and&use&linearity&to&find&the&&
actual&value&of&Io&in&the&following&circuit.
!9
Superposi7on
Specific superposition example:
V1
0
V1
0
y1
0
V2
0
V2
y2
y1
y2
!10
Superposi7on
!
Homogeneity
Superposition
Linearity
The output of a circuit is determined by summing the responses to each source acAng alone !
Specific superposition example:
(Independent sources only!!!) !
V1
0
V1
0
y1
0
V2
0
V2
y2
y1
y2
!11
Superposi7on
!
• The superposiAon principle states that the voltage across (or current through) an element in a linear circuit is the algebraic sum of the voltages across (or currents through) that element diue to each +
independent source acAng alone. V 0 +
v
–
• Turn off, kill, inacAve source(s): i
+
v
-
•
independent voltage source: 0 V (short circuit) •
independent current source: 0 A (open circuit) short
• Dependent sources are leN intact.
i
V
0 +
–
i
i
+
v
+
v
-
short
I
0
i
+
v
+
v
-
-
open
!12
Applica7on of superposi7on
1. Turn off all independent sources except one source. Find the output (voltage or current) due to that acAve source using nodal or mesh analysis. 2. Repeat step 1 for each of the other independent sources. 3. Find the total contribuAon by adding algebraically all the contribuAons due to the independent sources.
!13
how to turn off independent sources
•
•
Turn off voltages sources = short voltage sources; make it equal to zero voltage Turn off current sources = open current sources; make it equal to zero current !
• SuperposiAon involves more work but simpler circuits. • SuperposiAon is not applicable to the effect on power.
!14
Superposi7on
•  Example(4.3:(Use$the$superposi,on$theorem$to$find$v$$$
in$the$circuit$below.(
!15
Superposi7on
•  Example(4.4:(Find%Io%in%the%circuit%below%using%
superposi3on.%
!16
Superposi7on
Turn%off%20V$voltage$source:$
!17
Superposi7on
Turn%off%4A#current#source:#
!18
Source transforma7on
•  A!source!transforma-on!is$the$process$of$replacing$a$
voltage!source!vs!in!series!with!a!resistor!R!by$a$$$
current!source!is!in!parallel!with!a!resistor!R,$or$vice$
versa$
vs
vs = is R or is =
R
!19
Source transforma7on
•  Equivalent+Circuits:!
i
i
+
+
v = iR + vs
v
i=
v
v vs
−
R R
-
i
-is
vs
v
!20
Source transforma7on
Important points 1. Arrow of the current source is directed toward the posiAve terminal of the voltage source. 2. TransformaAon is not possible when: • ideal voltage source (R = 0) • ideal current source (R = ∞)
!21
Source transforma7on
•  Example(4.6:(Use$source$transforma.on$to$find$vo in$
the$circuit$below.$
!22
Source transforma7on
•  Example(4.6:(Use$source$transforma.on$to$find$vo in$
the$circuit$below.$
!23
Source transforma7on
•  Example(4.7:(Find%vx in%the%circuit%below%using%source%
transforma5on.%
!24
Thevenin’s theorem Thevenin’s theorem states that !
a linear two-­‐terminal circuit can be replaced by an equivalent circuit consisAng of a voltage source VTh in series with a resistor RTh !
where VTh is the open circuit voltage at the terminals and RTh is the input or equivalent resistance at the terminals when the independent sources are turned off. !
!
!
!
!25
Thevenin’s theorem •  Property(of(Linear(Circuits(
i
i
Any two-terminal
Linear Circuits
+
v
Slope=1/RTh
-
v
Vth
Isc
!26
Thevenin’s theorem •  Replacing*a*linear'two+terminal'circuit'(a)''by*its*Thevenin'equivalent'
circuit'(b).'
!27
•  How$to$find$Thevenin$Voltage?$$
!  Equivalent+circuit:+same+voltage3current+rela4on+at+the+
terminals.+
!  V++Th = voc : open circuit voltage at a − b
!28
•  How$to$find$Thevenin$Resistance?$$
RTh = Rin : input − resistance of the dead circuit at a − b.
• a − b open circuited
• Turn off all independent sources
!29
•  How$to$find$Thevenin$Resistance?$
!  There%are%two$cases$in%finding%Thevenin%Resistance RTh.%%
CASE$1$
•  $If%the%network%has%no$dependent$sources:$
●  Turn$off$all$independent$sources.$
●  RTh:$can$be$obtained$via$simplificaAon$of$either$parallel$or$
series$connecAon$seen$from$aCb"
!30
•  How$to$find$Thevenin$Resistance?$
!  There%are%two$cases$in%finding%Thevenin%Resistance RTh.%%
CASE$2$
•  $If%the%network%has%dependent$sources:$
●  Turn$off$all$independent$sources.$
●  Apply$a$voltage$source$vo$at$$a-b
vo
RTh =
io
●  AlternaAvely,$apply$a$current$source$io$$
vo
$$$$at$aCb$
RTh =
io
!31
•  How$to$find$Thevenin$Resistance?$
!  The$Thevenin$resistance,$RTh ,$may$be$nega9ve,$
indica9ng$that$the$circuit$has$ability$providing$
(supplying)$power.$
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•  A"er%the%Thevenin'Equivalent'is%obtained,%the'simplified'
circuit'can%be%used%to%calculate%IL!and'VL!easily.'
Simplified'circuit
%%
VTh
IL =
RTh + RL
RL
VL = RL I L =
VTh
RTh + RL
Voltage'divider
%%
!33
•  Example(4.8:(!Find!the!Thevenin!equivalent!circuit!of!the!circuit!
shown(below,!to!the!le7!of!the!terminals!a-b.!Then!find!the!current!
through!RL(= 6,16, and 36 Ω.(
!34
•  Example(4.9:(Find%the%Thevenin%equivalent%of%the%circuit%below%at%
terminals%a-b.(
!35
•  Example(4.10:(Determine(the(Thevenin(equivalent(circuit(in(Fig((a).(
Find(VTh:(
VTh = 0
Find(RTh:( Use(Fig (b):
(dependent(source(only(case)((
Apply(a(current(source(io(at((a-b
vo
RTh =
io
Nodal anaysis :
io + ix = 2ix + vo / 4
!36
•  Example(4.10:(Determine(the(Thevenin(equivalent(circuit(in(Fig((a).(
Find(RTh:( Use(Fig (b):
But
0 − vo
vo
ix =
=−
2
2
vo
vo vo
vo
io = ix + = − + = −
4
2 4
4
or vo = −4io
vo
Thus RTh = = −4Ω (Supplying power)
io
!37
•  Example(4.10:(Determine(the(Thevenin(equivalent(circuit(in(Fig((a).(
Input circuit:
Thevenin Equivalent circuit:
•  When(RTh is(nega4ve,(you(must(evaluate((check)(if(the(two(circuits(are(equivalent(
or(not!(
!38
•  Example(4.10:(Determine(the(Thevenin(equivalent(circuit(in(Fig((a).(
Input circuit:
Check with RL (9Ω) and voltage source (10V)
Thevenin Equivalent circuit:
Check with RL (9Ω) and voltage source (10V)
!39
•  Example(4.10:(Determine(the(Thevenin(equivalent(circuit(in(Fig((a).(
Input circuit:
8ix + 4i1 + 2(i1 − i2 ) = 0
− 2i1 + 6i2 = 0
, ix = i2 − i1
, i1 = 3i2 (1)
2(i2 − i1 ) + 9i2 + 10 = 0
5i2 = −10 => i2 = −2A
Check with RL (9Ω) and voltage source (10V)
Thevenin Equivalent circuit:
− 4i + 9i + 10 = 0
5i = −10 => i = −2A
Load current in both circuits are equal.
So the Thevenin Equivalent circuit is OK.
Check with RL (9Ω) and voltage source (10V)
!40
Norton’s theorem
•  Norton’s(theorem!states!that!a!linear!two-terminal!
circuit!can!be!replaced!by!an!equivalent!circuit!
consis7ng!of!a(current(source(IN(in(parallel(with!a!
resistor(RN .!!
•  IN(is!the!short-circuit!current!through!the!terminals!and!
•  RN(is!the!input!or!equivalent!resistance!at!the!terminals!
when!the!independent(source(are(turn(off.(
!41
Norton’s theorem
•  Property(of(Linear(Circuits(
i
Slope=1/RN
v
Vth
-IN
!42
Norton’s theorem
•  How$to$find$Norton$Current$
!  Thevenin!and!Norton$resistances!are!equal:$
RN = RTh
!  Short$circuit$current$from!a"to!b!!!
gives!the!Norton!current:$
VTh
I N = isc =
RTh
!43
Norton’s theorem
•  Thevenin'or'Norton'equivalent'circuit'
!  The'open'circuit'voltage'voc'across)terminals'a'and'b''
!  The'short'circuit'current'isc'at)terminals)a'and'b#''
!  The'equivalent'or'input'resistance'Rin'at)terminals)a#and#b'when)
V
all)independent'source'are'turn'off.'
VTh = voc
I N = isc
RTh
VTh
=
= RN
IN
!44
•  Example(4.11:(Find%the%Norton%equivalent%circuit%of%the%circuit%in%Fig%4.39.%
!45
•  Example(4.12:(Using&Norton’s&theorem,&find&RN#and&IN&of&the&circuit&in&Fig&
4.43&at&terminals&a;b.&
!46
Maximum power transfer
•  The$Thevenin$equivalent$is$useful$in$finding$the$
maximum&power&a$linear$circuit$can$deliver$to$a$load.$&
2
& VTH #
p = i RL = $
! RL
% RTH + RL "
2
!47
•  Maximum'power'is#transferred#to#the#load#when#the#
load#resistance#equals#the#Thevenin'resistance'as#seen#
from#the#load'(RL = RTH).'
2
& VTH #
p = i RL = $
! RL
% RTH + RL "
2
!48
dp
=0
•  Pmax!can!be!obtained!when:!!
dRL
2
(
dp
(
R
+
R
)
− 2 RL ( RTH + RL ) %
2
TH
L
= VTH &
#=0
2 2
dRL
(( RTH + RL ) )
'
$
2
(
( RTH + RL ) − 2 RL ( RTH + RL ) %
2
2 ( ( RTH + RL − 2 RL ) %
= VTH &
# = VTH &
#=0
4
3
( RTH + RL )
'
$
' ( RTH + RL )
$
⇒ ( RTH + RL − 2 RL ) = 0
⇒ ( RTH − RL ) = 0
RL = RTH
pmax
2
TH
V
=
4 RTH
!49
•  Example(4.13:(Find%the%value%of%RL%for%maximum%power%transfer%in%
the%circuit%of%Fig.%4.50.%Find%the%maximum%power.%
!50