Review
ofB
asic
Concep
ts:Volta
ge
10 V- +
- ++
-+
-
2kΩ
2kΩ
5kΩ
7kΩ
v1
v2
v3
v4
•T
he
voltagedrop
fromon
enode
toan
other
isth
esam
e,no
matter
what
path
isch
osen
•K
irchhoff
’svoltage
law
Portla
nd
Sta
teU
niv
ersity
EC
E221
Analy
sisM
eth
ods
Ver.
1.6
63
Analysis
Meth
ods
Overview
•Solvin
gLin
earEquation
s
•N
odal
Analysis
•Supern
odes
(Nodal
Analysis
with
Voltage
Sou
rces)
•M
eshA
nalysis
•Superm
eshes
(Mesh
Analysis
with
Curren
tSou
rces)
•In
troduction
toB
JTTran
sistors
This
isa
veryim
portan
tch
apter.
Portla
nd
Sta
teU
niv
ersity
EC
E221
Analy
sisM
eth
ods
Ver.
1.6
61
Resisto
rsin
Para
llelw
ithVolta
ge
Sources
Circuit
RV
sv
o- +
Circuit
Vs
vo- +
•W
hat
isv
oin
eachcase?
•W
hat
effect
does
the
resistorhave
onth
ecu
rrent
pum
ped
into
the
circuit?
Portla
nd
Sta
teU
niv
ersity
EC
E221
Analy
sisM
eth
ods
Ver.
1.6
64
Review
ofB
asic
Concep
ts:Curren
t
i4i5
i3i2
i1
•W
hat
goes
in,has
tocom
eou
t
•K
irchhoff
’scu
rrent
law
•Sim
ilarto
conservation
ofm
ass
•Con
servationof
electrons
Portla
nd
Sta
teU
niv
ersity
EC
E221
Analy
sisM
eth
ods
Ver.
1.6
62
Exa
mple
1:
Term
inolo
gy
20 V2 A
R1
R2
R3
R4
R4
R6
R7
R8
35ip
ip
Iden
tifyth
efollow
ing
inform
ationN
odes:
Essen
tialN
odes:
Bran
ches:
Essen
tialB
ranch
es:EB
’sw
ithU
nkn
own
Curren
t:M
eshes:
Portla
nd
Sta
teU
niv
ersity
EC
E221
Analy
sisM
eth
ods
Ver.
1.6
67
Resisto
rsin
Series
with
Curren
tSources
Circuit
IsC
ircuitIs
Rio
io
•W
hat
isio
ineach
case?
•W
hat
effect
does
the
resistorhave
onth
evoltage
seenby
the
circuit?
Portla
nd
Sta
teU
niv
ersity
EC
E221
Analy
sisM
eth
ods
Ver.
1.6
65
Exa
mple
2:
Circu
itA
nalysis
The
Hard
Way
10 V
i1i3
i2i4
2 mA
1kΩ
2kΩ
5kΩ
10
kΩ
Can
solvew
ithK
CL
&K
VL.Fou
runkn
owns.
Portla
nd
Sta
teU
niv
ersity
EC
E221
Analy
sisM
eth
ods
Ver.
1.6
68
Netw
ork
Term
inolo
gy
Plan
arC
ircuit
Acircu
itth
atcan
be
draw
non
aplan
ew
ithno
crossing
branch
es
Node
Poin
tor
portion
ofa
circuit
where
2or
more
elemen
tsare
joined
Essen
tialN
ode
Poin
tor
portion
ofa
circuit
where
3or
more
elemen
tsare
joined
Bran
chPath
that
connects
2nodes
Essen
tialB
ranch
Path
that
connects
2essen
tialnodes
w/o
passin
gth
rough
anessen
tialnode
Loop
Path
with
lastnode
same
asstartin
gnode
that
does
not
crossitself
Mesh
Loop
that
does
not
enclose
any
other
loop
s
Note:
this
isn’t
inth
etext.
Portla
nd
Sta
teU
niv
ersity
EC
E221
Analy
sisM
eth
ods
Ver.
1.6
66
Exa
mple
2:
Contin
ued
(1)
i1−
i2−
i3=
0−
i3+
i4=
2m1k
i1+
5ki2
=10
+5k
i2−
2ki3
−10k
i4=
0
InM
atrixform
this
becom
es
1−
1−
10
00
−1
11k
5k0
00
5k−
2k−
10k i1i2i3i4
= 0
2m100
or
Ai=
b
Portla
nd
Sta
teU
niv
ersity
EC
E221
Analy
sisM
eth
ods
Ver.
1.6
611
Solvin
gLin
earEquatio
ns
•M
uch
ofou
rcircu
itan
alysisw
illfo
cus
onfindin
ga
setof
linear
equation
san
dsolvin
gth
eseeq
uation
s
•N
eedas
man
yeq
uation
sas
there
areunkn
owns
•T
hree
possib
leap
proaches
–A
lgebra(elim
ination
,su
bstitu
tion,etc.)
–Cram
er’sru
le
–Lin
earalgebra
•Last
iseasiest
and
leastsu
sceptib
leto
errors
•Req
uires
use
your
scientifi
ccalcu
lators
Portla
nd
Sta
teU
niv
ersity
EC
E221
Analy
sisM
eth
ods
Ver.
1.6
69
Exa
mple
2:
Contin
ued
(2)
Ai=
bw
here
A=
1−
1−
10
00
−1
11k
5k0
00
+5k
−2k
−10k
i=
i1i2i3i4 b
= 0
2m100
•You
rcalcu
latorsh
ould
be
able
tosolve
this
directly
•You
shou
ldon
lyneed
toen
terA
and
b
•You
rcalcu
latorw
illretu
rna
vectori
•Sim
ultan
eously
solvesfor
allth
eunkn
own
variables
•M
uch
fasterth
anCram
er’sru
leor
brute-force
algrebra
•Read
the
man
uals
foryou
rcalcu
lators
•T
his
will
saveyou
time
(hom
ework
&exam
s)an
dred
uce
errors
Portla
nd
Sta
teU
niv
ersity
EC
E221
Analy
sisM
eth
ods
Ver.
1.6
612
Exa
mple
2:
Solvin
gLin
earEquatio
ns
i1=
i2+
i3
i4=
i3+
2m10
=1k
i1+
5ki2
5ki2
=2k
i3+
10ki4
Rew
riteso
variables
arein
consisten
tord
eron
leftsid
ean
dcon
stants
areon
the
right
side
i1−
i2−
i3=
0−
i3+
i4=
2m1k
i1+
5ki2
=10
+5k
i2−
2ki3
−10k
i4=
0
Portla
nd
Sta
teU
niv
ersity
EC
E221
Analy
sisM
eth
ods
Ver.
1.6
610
NodalAnalysis:
Step
1–
Iden
tifyEssen
tialN
odes
10 V2 m
A
1kΩ
2kΩ
5kΩ
10
kΩ
•Som
eessen
tialnodes
may
inclu
de
portion
sof
the
circuit
(pieces
ofw
ire)
•Circle
the
entire
node
topreven
terrors
Portla
nd
Sta
teU
niv
ersity
EC
E221
Analy
sisM
eth
ods
Ver.
1.6
615
Exa
mple
2:
Contin
ued
(3)
Lin
earEquation
s:
1−
1−
10
00
−1
11k
5k0
00
5k−
2k−
10k i1i2i3i4
= 02m100
Calcu
latorsh
ould
return
:i1i2i3i4
= +
0.909+
1.818−
0.909+
1.091 m
A
Portla
nd
Sta
teU
niv
ersity
EC
E221
Analy
sisM
eth
ods
Ver.
1.6
613
NodalAnalysis:
Step
2–
Pick
aReferen
ce
10 V2 m
A
1kΩ
2kΩ
5kΩ
10
kΩ
•Secon
dstep
isto
pick
areferen
cenode
•Is
ofteneasiest
toch
oose
the
node
that
intercon
nects
the
most
branch
es
•M
ust
be
anessen
tialnode
•U
sually
isat
bottom
ofcircu
it
•Lab
elw
ithth
esam
esym
bol
used
forgrou
nd
Portla
nd
Sta
teU
niv
ersity
EC
E221
Analy
sisM
eth
ods
Ver.
1.6
616
NodalAnalysis:
Intro
ductio
n
•T
here
isan
anoth
erway
tosolve
forcu
rrents
and
voltages
–Easier
–M
orem
ethodical
–Still
based
onO
hm
’slaw
,K
VL,&
KCL
•N
odal
analysis
ison
eof
two
keym
ethods
•M
eshan
alysisis
the
other
•W
ew
illdiscu
ssnodal
analysis
first
•B
asedon
KCL
•M
ust
understan
dterm
inology
intro
duced
earlier
•U
seto
solvefor
voltages
•A
llvoltages
have
acom
mon
reference
poin
t
Portla
nd
Sta
teU
niv
ersity
EC
E221
Analy
sisM
eth
ods
Ver.
1.6
614
NodalAnalysis:
Step
5–
Solve
Lin
earEquatio
ns
Lin
earEquation
s:
Solu
tion(from
calculator):
Portla
nd
Sta
teU
niv
ersity
EC
E221
Analy
sisM
eth
ods
Ver.
1.6
619
NodalAnalysis:
Step
3–
Label
Oth
erEssen
tialN
odes
10 V2 m
A
1kΩ
2kΩ
5kΩ
10
kΩ
•A
lsoa
bit
easierif
voltagesare
labeled
•A
llvoltages
arem
easured
relativeto
the
reference
node
Portla
nd
Sta
teU
niv
ersity
EC
E221
Analy
sisM
eth
ods
Ver.
1.6
617
NodalAnalysis:
Step
6–
Solve
for
Varia
bles
ofIn
terest
10 V2 m
A
12
- +v2
- +v1
i1i3
i2i4
1kΩ
2kΩ
5kΩ
10
kΩ
i1=
i2=
i3=
i4=
Portla
nd
Sta
teU
niv
ersity
EC
E221
Analy
sisM
eth
ods
Ver.
1.6
620
NodalAnalysis:
Step
4–
Apply
KCL
All
Labeled
Nodes
10 V2 m
A
12
- +v2
- +v1
1kΩ
2kΩ
5kΩ
10
kΩ
Portla
nd
Sta
teU
niv
ersity
EC
E221
Analy
sisM
eth
ods
Ver.
1.6
618
Exa
mple
3:
NodalA
nalysis
144 V
- +v2
- +v1
3 A
4Ω
5Ω
10
Ω
80
Ω
Solve
forv1
and
v2 .
Portla
nd
Sta
teU
niv
ersity
EC
E221
Analy
sisM
eth
ods
Ver.
1.6
623
NodalAnalysis:
Review
ofStep
s
•Step
1:Id
entify
essential
nodes
•Step
2:Pick
areferen
ce
–M
ust
be
anessen
tialnode
–A
lways
label
with
the
ground
symbol
–B
estto
pick
essential
node
with
most
branch
es
–O
ftenat
the
bottom
ofth
ecircu
itdiagram
•Step
3:Lab
eloth
eressen
tialnodes
•Step
4:A
pply
KCL
toall
labelled
nodes
except
reference
node
•Step
5:Solve
linear
equation
s
–G
enerates
voltageat
eachnode
(relativeto
reference
node)
•Step
6:Solve
forvariab
lesof
interest
–U
sually
easyafter
Step
5
Portla
nd
Sta
teU
niv
ersity
EC
E221
Analy
sisM
eth
ods
Ver.
1.6
621
Exa
mple
3:
Work
space
Portla
nd
Sta
teU
niv
ersity
EC
E221
Analy
sisM
eth
ods
Ver.
1.6
624
NodalAnalysis:
Use
ofLaw
s
•A
llth
reelaw
sare
used
•K
CL
isap
plied
ateach
labelled
node
except
the
reference
node
•O
hm
’slaw
isused
todeterm
ine
the
curren
tin
branch
esth
atcon
tainresistors
•K
VL
isused
todeterm
ine
the
voltagedrop
acrossth
eresistors
Portla
nd
Sta
teU
niv
ersity
EC
E221
Analy
sisM
eth
ods
Ver.
1.6
622
Exa
mple
5:
Dep
enden
tVolta
ge
Source
50 V- +
10
Ω
10
Ω
30
Ω39
Ω78
Ω
v/5
v
Solve
forv
.
•W
hat
effect
does
the
10Ω
resistorhave
onth
ecircu
it?
•W
hat
isth
ecu
rrent
flow
ing
throu
ghth
edep
enden
tsou
rce?
•H
owcan
we
apply
KCL
atth
eessen
tialnodes
with
out
this
inform
ation?
•A
ns:
One
extravariab
le
•Im
plies
we
need
anextra
equation
Portla
nd
Sta
teU
niv
ersity
EC
E221
Analy
sisM
eth
ods
Ver.
1.6
627
Exa
mple
4:
NodalA
nalysis
20 mA
- +v2
- +v1
- +v3
5 V2
kΩ
2.7
kΩ
2.7
kΩ
3.3
kΩ
4.7
kΩ
10
kΩ
Solve
forv1 ,
v2 ,
and
v3 .
Portla
nd
Sta
teU
niv
ersity
EC
E221
Analy
sisM
eth
ods
Ver.
1.6
625
Exa
mple
5:
Contin
ued
50 V- +
10
Ω
10
Ω
30
Ω39
Ω78
Ω
v/5
v
Solve
forv
.
Portla
nd
Sta
teU
niv
ersity
EC
E221
Analy
sisM
eth
ods
Ver.
1.6
628
Exa
mple
4:
Work
space
Portla
nd
Sta
teU
niv
ersity
EC
E221
Analy
sisM
eth
ods
Ver.
1.6
626
Exa
mple
6:
Dep
enden
tSource
Contin
ued
50 V- +
10
Ω
10
Ω
30
Ω39
Ω78
Ω
v/5
v
Solve
forv
.U
sea
supern
ode.
Portla
nd
Sta
teU
niv
ersity
EC
E221
Analy
sisM
eth
ods
Ver.
1.6
631
Exa
mple
5:
Work
space
Portla
nd
Sta
teU
niv
ersity
EC
E221
Analy
sisM
eth
ods
Ver.
1.6
629
Exa
mple
6:
Work
space
Portla
nd
Sta
teU
niv
ersity
EC
E221
Analy
sisM
eth
ods
Ver.
1.6
632
NodalAnalysis
and
Supern
odes
•Supern
odes
elimin
ateth
eneed
toin
troduce
anextra
variable
(unkn
own
curren
t)
•N
ecessaryw
hen
avoltage
source
isbetw
eentw
olab
elednodes
(excludin
greferen
cenode)
•Still
need
touse
voltagesou
rceto
generate
one
ofth
eeq
uation
s
Portla
nd
Sta
teU
niv
ersity
EC
E221
Analy
sisM
eth
ods
Ver.
1.6
630
Exa
mple
8:
NodalA
nalysis
11 mA
i1
20 Vi2
10 Vi3
250
Ω
500
Ω1kΩ
25
kΩ
Solve
fori1 ,
i2 ,an
di3 .
Portla
nd
Sta
teU
niv
ersity
EC
E221
Analy
sisM
eth
ods
Ver.
1.6
635
Exa
mple
7:
Dep
enden
tVolta
ge
Source
20 V
+-
1Ω
2Ω
4Ω
20
Ω40
Ω80
Ω3.1
25v
v
35iφ
iφ
Fin
dth
epow
erdevelop
edby
the
20V
source.
Portla
nd
Sta
teU
niv
ersity
EC
E221
Analy
sisM
eth
ods
Ver.
1.6
633
Exa
mple
8:
Work
space
Portla
nd
Sta
teU
niv
ersity
EC
E221
Analy
sisM
eth
ods
Ver.
1.6
636
Exa
mple
7:
Work
space
Portla
nd
Sta
teU
niv
ersity
EC
E221
Analy
sisM
eth
ods
Ver.
1.6
634
Mesh
Analysis:
Intro
ductio
n
•Recall:
There
isan
easierway
tosolve
forcu
rrents
and
voltagesth
anap
plyin
gK
VL
and
KCL
directly
•N
odal
analysis
ison
eof
two
keym
ethods
•M
eshan
alysisis
the
other
–A
pplies
KV
Lto
solvefor
curren
ts
–M
oreab
stract
–W
orkw
ithim
aginary
curren
ts
–O
nly
applies
toplan
arcircu
its
Portla
nd
Sta
teU
niv
ersity
EC
E221
Analy
sisM
eth
ods
Ver.
1.6
639
Exa
mple
9:
NodalA
nalysis
1 A
3i
i
- +v
1Ω
1Ω
2Ω
2Ω
4Ω
Solve
forv.
Portla
nd
Sta
teU
niv
ersity
EC
E221
Analy
sisM
eth
ods
Ver.
1.6
637
Mesh
Analysis:
Step
1–
Label
Mesh
es
40 V64 V
iaic
ib
1.5
Ω2
Ω
3Ω
4Ω
45
Ω
Fin
dth
ebran
chcu
rrents
ia ,ib ,
and
ic .
•Recall:
Am
eshis
alo
opth
atdoes
not
enclose
any
other
loop
s
Portla
nd
Sta
teU
niv
ersity
EC
E221
Analy
sisM
eth
ods
Ver.
1.6
640
Exa
mple
9:
Work
space
Portla
nd
Sta
teU
niv
ersity
EC
E221
Analy
sisM
eth
ods
Ver.
1.6
638
Mesh
Analysis:
Step
4–
Solve
for
Varia
bles
ofIn
terest
40 V64 V
iaic
ib
1.5
Ω2
Ω
3Ω
4Ω
45
Ω
ia=
ib=
ic=
Portla
nd
Sta
teU
niv
ersity
EC
E221
Analy
sisM
eth
ods
Ver.
1.6
643
Mesh
Analysis:
Step
2–
Apply
KV
Lto
Each
Mesh
40 V64 V
iaic
ib
1.5
Ω2
Ω
3Ω
4Ω
45
Ω
Portla
nd
Sta
teU
niv
ersity
EC
E221
Analy
sisM
eth
ods
Ver.
1.6
641
Mesh
Analysis:
Review
ofStep
s
•Step
1–
Lab
elM
eshes
•Step
2–
Apply
KV
Lto
Each
Mesh
•Step
3–
Solve
Lin
earEquation
s
•Step
4–
Solve
forVariab
lesof
Interest
–U
sually
easyafter
Step
3
•Lim
itation:
Only
works
with
plan
arcircu
its
Portla
nd
Sta
teU
niv
ersity
EC
E221
Analy
sisM
eth
ods
Ver.
1.6
644
Mesh
Analysis:
Step
3–
Solve
Lin
earEquatio
ns
[50
−45
−45
50.5 ][i1i2 ]
= [4064 ]
i1=
9.8A
i2=
10A
Portla
nd
Sta
teU
niv
ersity
EC
E221
Analy
sisM
eth
ods
Ver.
1.6
642
Exa
mple
11:
Mesh
Analysis
18 V15 V
3 A
2Ω
3Ω
6Ω
9Ω
Fin
dth
etotal
pow
erdissip
ated.
•Prob
lem:
What
isth
evoltage
acrossth
e3
Asou
rce?
•Solu
tions
1A
dd
itas
avariab
le
2U
sea
superm
esh
•Secon
dop
tionreq
uires
lesswork
Portla
nd
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ods
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1.6
647
Exa
mple
10:
Mesh
Analysis
12 V
110 V70V
2Ω
3Ω
4Ω
6Ω
10
Ω12
Ω
Fin
dth
etotal
pow
erdevelop
edin
the
circuit.
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ods
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Exa
mple
11:
Mesh
Analysis
18 V15 V
3 A
2Ω
3Ω
6Ω
9Ω
Fin
dth
etotal
pow
erdissip
ated.
Add
avariab
le.
Portla
nd
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teU
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ersity
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E221
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eth
ods
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648
Exa
mple
10:
Work
space
Portla
nd
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teU
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ersity
EC
E221
Analy
sisM
eth
ods
Ver.
1.6
646
Exa
mple
12:
Work
space
Portla
nd
Sta
teU
niv
ersity
EC
E221
Analy
sisM
eth
ods
Ver.
1.6
651
Exa
mple
11:
Work
space
Portla
nd
Sta
teU
niv
ersity
EC
E221
Analy
sisM
eth
ods
Ver.
1.6
649
Exa
mple
13:
Mesh
Analysis
200 V
4.3 id
ieib
id
ia
ic
10
Ω
10
Ω
25
Ω
50
Ω
100
Ω
Fin
dth
ebran
chcu
rrents
ia–
ie .
Portla
nd
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ersity
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E221
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eth
ods
Ver.
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652
Exa
mple
12:
Mesh
Analysis
18 V15 V
3 A
2Ω
3Ω
6Ω
9Ω
Fin
dth
etotal
pow
erdissip
ated.
Use
asu
perm
esh.
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nd
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teU
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ersity
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eth
ods
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650
Exa
mple
14:
Work
space
Portla
nd
Sta
teU
niv
ersity
EC
E221
Analy
sisM
eth
ods
Ver.
1.6
655
Exa
mple
13:
Work
space
Portla
nd
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teU
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ersity
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ods
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1.6
653
Nodalversu
sM
eshA
nalysis
•You
shou
ldkn
owhow
todo
both
•W
hich
ism
oreeffi
cient
dep
ends
onth
eprob
lem
•W
illlearn
which
touse
with
experien
ce
•N
odal
analysis
used
more
often
•O
nexam
s,Iw
illsp
ecifyw
hich
meth
od
touse
Con
ciseSum
mary:
Nodal
Analysis
Mesh
Analysis
Meth
od
KCL
KV
LSolve
For
Node
Voltages
Mesh
Curren
ts“S
uper”
Con
dition
sVoltage
Sou
rcesCurren
tSou
rces
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ods
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Exa
mple
14:
Mesh
Analysis
1.5 mA
8 V
2kΩ
3kΩ
4kΩ
4kΩ
4kΩ
5kΩ
7kΩ
3iα
iα
Solve
foriα
Portla
nd
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ersity
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ods
Ver.
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654