Electronic Circuits I
Instructor:
Dr. Alaa Mahmoud
alaa_y_emam@hotmail.com
Chapter 3
METHODS OF ANALYSIS
Outline:
— Introduction
— Nodal analysis
— Nodal analysis with voltage sources
— Mesh analysis
— Mesh analysis with current sources
INTRODUCTION
— Two powerful techniques for circuit analysis:
— Nodal analysis à based on a systematic application of Kirchhoff’s current
law (KCL)
— Mesh analysis à based on a systematic application of Kirchhoff’s voltage
law (KVL)
— We can analyze almost any circuit by obtaining a set of simultaneous
equations that are then solved to obtain the required values of current or
voltage
— One method of solving simultaneous equations involves Cramer’s rule,
which allows us to calculate circuit variables as a quotient of determinants
node voltages as the circuit variables. Choosing node voltages instead
of element voltages as circuit variables is convenient and reduces the
NODAL
ANALYSIS method (node-voltage method)
number of equations one must solve simultaneously.
To simplify
matters,
wefor
shall
assume
in thisusing
section
circuits
do
Nodal analysis:
general
procedure
analyzing
circuits
nodethat
voltages
as the
circuit
variablesvoltage
instead of
element Circuits
voltages that contain voltage sources will be
not contain
sources.
àanalyzed
reduces the
of section.
equations must solved
innumber
the next
In nodal analysis, we are interested in finding the node voltages.
— In this analysis: we assume that circuits do not contain voltage sources
Given
a circuitthewith
n nodes
voltage
sources,
thenode
nodal
We determine
voltage
at eachwithout
node relative
to the
reference
(or analysis
ground)
of the circuit involves taking the following three steps.
— For circuit of n nodes without voltage sources, the nodal analysis involves 3 steps:
Steps to Determine Node Voltages:
1. Select a node as the reference node. Assign voltages
v1 , v2 , . . . , vn−1 to the remaining n − 1 nodes. The voltages are
referenced with respect to the reference node.
2. Apply KCL to each of the n − 1 nonreference nodes. Use Ohm’s
law to express the branch currents in terms of node voltages.
3. Solve the resulting simultaneous equations to obtain the unknown
node voltages.
.
the number of independent equations that we
e — Select a node aswill
derive.or datum node or ground node (has zero potential)
the reference
— A reference node is indicated by any of the three symbols
e
n
Earth potential is
Earth potential is
used as reference
used
as
reference
e
eethods of Analysis
77
n Fig. 3.1(b) is
ase, enclosure,
he potential of
Fig. 3.1(a) or
(a)
(b)
(c)
e
chassis ground chassis acts as
a reference point for all circuits
e
Figure
3.1
Common symbols for
,voltage desig- indicating
The number of nonreference nodes is equal to
reference
circuit in Fig.
the number of a
independent
equations that wenode.
2 are
will derive.
testhat1 and
the node
(c) for
(a)
(b)
3.1 Common
symbols
ions
to nonreference
nodes.
Consider,
the
circuit
in
As
second
step,
apply
KCL
towhere
eachAs
nonreference
node
in the
the number
of independent
equations
voltages
are
defined
with
respect
toexample,
the
reference
node.add
As
illustrated
inFigure
with
respect
tothe
the
node.
circuit
in Fig.
3.2(a)
isreference
redrawn
in
Fig.for
3.2(b),
we
now
iFig.
es
are
defined
with
respect
to
thewe
reference
node.
illustrated
in
1 , i2 ,
indicating a(a)
reference node.
(b)
(a).and
Node
0
is
the
reference
node
(v
=
0),
while
nodes
1
and
2
are
circuit.
To
avoid
putting
too
much
information
on
the
same
circuit,
the
will
derive.
Fig.
3.2(a),
each
node
voltage
is
the
voltage
rise
from
the
reference
node
As
the
second
step,
we
apply
KCL
to
each
nonreference
node
in
the
i
as
the
currents
through
resistors
R
,
R
,
and
R
,
respectively.
At
3
1
2
3
Figure 3.1 Common symbols for
2(a), each circuit
node voltage
is theis voltage
rise
from
thewhere
reference
node
in
Fig.
3.2(a)
redrawn
in
Fig.
3.2(b),
we
now
add
i
,
i
,
igned
voltages
v
and
v
,
respectively.
Keep
in
mind
that
the
node
1
2
2
1
to
the
corresponding
nonreference
node
or
simply
the
voltage
of
that
node
circuit.
Tononreference
avoid
putting
tooormuch
information
on
the
same
circuit, the indicating a reference node.
node
applying
KCL
gives
— 1,Assign
voltage
to
nonreference
nodes
orresponding
node
simply
the
voltage
of
that
node
andrespect
i3 aswith
thetocurrents
through
resistors
R1node.
, R2 , and
At
Figure 3.1I2 Common s
3 , respectively.
tagescircuit
arewith
defined
respect
to the node.
reference
AsRillustrated
inadd
theisreference
in
Fig.
3.2(a)
redrawn
in
Fig.
3.2(b),
where
we
now
i
,
i
,
indicating
a (b)
reference n
espect
to—each
the
reference
node.
node
1, applying
KCL
Node
0voltage
issecond
the
reference
node
(v
0),
while
nodes 1node
and
Ithe
= voltage
I2 +
i1 +
i2 =from
(3.1)2 are1 2
(a)
1gives
. 3.2(a),
node
is
rise
the
reference
As
the
step,
we
apply
KCL
to
each
nonreference
node
in
the
I2 (c)
(a)
(b)
and
i3 asassigned
the
currents
through
resistors
R1 , R2 , andnode
R3 , in
respectively.
At
As
the
second
step,
we
apply
KCL
to
each
nonreference
the
voltages
v
and
v
,
respectively
1 too
he At
corresponding
node
or
that
nodecircuit,
I1 =
I2simply
i2 voltage
(3.1)
2 + iinformation
1 +the
circuit.
Tononreference
avoid
putting
much
onofthe
same
the
node
2,applying
R2 Common
node
1,
KCL
gives
Figure
3.1
sy
1
2
.h To
avoid
putting
too
much
information
on
the
same
circuit,
the
respectcircuit
to
the
reference
node.
— AtThe
voltages
are defined
with3.2(b),
respect
to the
I
innode
Fig.
is redrawn
in Fig.
where
wereference
now add i1Figure
, i2 , 3.1 Common
2
node
2, 3.2(a)
symbols
for
indicating
no
i2 = i3where we now add i , i ,(3.2)
I2 +
Ra2 reference
(c)
(a)
(b)
1
in As
Fig.theand
3.2(a)
is
redrawn
in
Fig.
3.2(b),
2
second
we apply
KCL
in2the
I1 =toIresistors
+ i1nonreference
+Ri12, R2 , andnode
(3.1)
node
i3 as step,
the currents
through
R3 ,1respectively.
At
2each
indicating
a reference
node. +
+
+
i
=
i
(3.2)
I
2
2
3
v2
v1 + R1
as We
theTonow
currents
through
resistors
R1 , R
andon
R3the
, respectively.
R3
apply
Ohm’stoo
law
to express
the2 , unknown
currents
i1 , i2 , At
and
cuit.
avoid
much
information
same circuit,
thei3
I1
node
1,putting
applying
KCL
gives
+
—
Each
node
voltage
is
the
voltage
rise
from
the
reference
node
Figure
3.1
Common
symbols
At
node
I2
!v
! R for
v
inin
terms
of2,
node
The
key
idea to
bear
inwe
mind
isadd
that,
We
now
apply
Ohm’s in
lawFig.
to
express
the
unknown
currents
i1 ,i1i2since
cuit
Fig.
3.2(a)
isvoltages.
redrawn
3.2(b),
where
now
,,iand
R
I
, applying
KCL
gives
2
1
2 , i3
3 R
2
1
1
1
indicating
a
reference
node.
to
the
corresponding
nonreference
node
0 I
I
=
I
+
i
+
i
(3.1)
1
2
1
2
!
!
2
in terms
ofthrough
nodeelement,
voltages.
to
bear
in mind is that,
is a passive
byThe
the
sign
current
d i3resistance
as the currents
resistors
R+
,passive
R2 2idea
, and
R
respectively.
Atsince (3.2)
1key
3 ,convention,
i
=
i
I
2
3
0
Ia1passive
=
I2 +
i1potential
+ i2by the
(3.1)current
element,
convention,
must
always
flow
from
a higher
to apassive
lower sign
potential.
+
de 1,
applying
KCL
gives
Atresistance
node
2, is
R
1
2
— Apply
KCLflow
to each
node
in potential.
the
circuit
must always
from
higher
potential
a lower
lawanonreference
to
express
the to
unknown
currents
i1 , i2 , and i3
e 2,We now apply Ohm’s
I1 = I2 +The
i1 +
(3.1)
(3.2)
Ikey
2 iidea
2 i+
2 = ito
3 bear in mind is
At node
1: voltages.
in terms
of node
that, since
Currentisflows
fromIa2 higher
potential
to athe
lowerpassive
potential sign
in a resistor.
+
i
=
i
(3.2)i , i current
a
passive
element,
by
2
3
noderesistance
2, We
At node
2:
now apply
Ohm’s
law
to
express
the
unknown
currents
Current flows from a higher potential to a lower potential inconvention,
a resistor.
1 2 , and i3
mustinOhm’s
always
flow
a higherunknown
potentialcurrents
to atolower
terms law
of
node
voltages.
bear
mind is that, since
w apply
tofrom
express
i1 , in
ipotential.
I1
i2 =The
i3 key idea
I2 +the
2 , and i(3.2)
3
We
can
express
this
principle
as
resistance
is
a
passive
element,
by
the
passive
sign
convention,
current
Applying
Ohm’s
lawidea
to express
i-smind
in terms
of vsince
of the node
ms of—node
voltages.
The
to bear in
is that,
We
can express
thiskey
principle
as
now apply
Ohm’s
law
tofrom
express
the unknown
currents
i1 ,potential.
i2 , and i3
I1
must
always
flow
apassive
higher
potential
to a lower
since
resistance
is
a
element
nce
is
a
passive
element,
by
the
passive
sign
convention,
current
vhigher
Current
flows from
a=higher
potential
to a lower
potential
in a resistor.
erms of node
voltages.
The
key
idea
tovlower
bear
in mind
is that,
since
v−
higher − vlower
i
(3.3)
I1 I
lways
flow
from
a
higher
potential
to
a
lower
potential.
i
=
1
R
istance is a passive element, by the passive
sign convention, current (3.3)
R
I1 (a)
(a)R2
1
I2v11
!
I2 +
R1
2
2
0
I
R2v21 + R12
!
v2i R3
R2 i22+0 v
I
+ 11
v1
Ri12
! v+1v i2 R2 ! (a)
v22
v
v1 01 R
2
i3R3
i11
i
i
1
!
! 3 I2
0RR11
(a) R
R33
I2
à from
Currenta flows
frompotential
a higher potential
to a lower
potential in a resistor.
(a)
st always flow
higher
to a lower
potential.
i2 R
Note
this
principle
isprinciple
in agreement
we defined
resistance
2
Note
that
thisthis
principle
is in agreement
withway
the way
we defined
resistance
We that
can
express
as with the
v1
(a) I
2
Current
flows
from
potential
a lower
potential
inwea resistor.
in Chapter
2 (seea higher
Fig.
2.1).
inthis
mind,
we obtain
fromfrom
Fig.Fig.
3.2(b),
in Chapter
2 (see
Fig.With
2.1).tothis
With
in mind,
obtain
3.2(b),
i2 i1 R
(b)
(b)
2
We
can
express
this
principle
as
I
vlower in a resistor.
2 v1
Current flows from a higher
lower−potential
v1potential
− 0v1 −to0vahigher
(3.3)
I1
i1 = i1 = i = or ori1 = iG
1 =
1 v1G1 v1
iR11
i
i
R1 R1 v R − v
2
2
Figure3.23.2
Typical
circuit for
for nodal
R2circuit
Figure
Typical
lower
n express this principle as v1 − v2i = higher
v
v2nodal
1
analysis.
i
i
(3.3)
I
v1i −=v2
2
1analysis.
R2 2 R1v
or
i
=
G
(v
−
v
)
(3.4)
canNote
express
this
principle
as
2
2
2
1
2
v
R
=
or
i2 =with
G2 (v
v2 ) we defined(3.4)
1 i1
i32
1−
that thisi2principle
in2 agreement
the
way
resistance
Rv2 is R
−
vlower
i3
i1
higher
in Chapter
2
(see
Fig.
2.1).
With
this
in
mind,
we
obtain
from
Fig.
3.2(b),
v
−
0
2
i
=
(3.3)
I
R
R
Note that this principle
agreement
vi23−
0is in−
v=higher
vloweror with
1
3
1
i3 =the
G3way
v2 we defined resistance
R
(a)
Note that this principle is in agreement with the way we defined resistance
Noteinthat
this principle
in agreement
with
the we
wayobtain
we defined
resistance
Chapter
2 (see Fig.is2.1).
With this in
mind,
from Fig.
3.2(b),
in
Chapterto
2 (see
Fig.potential
2.1). With
in mind, we obtain from Fig. 3.2(b),
r potential
a lower
in athis
resistor.
v1 − 0
i1 =
or
i1 = G1 v1
v1 −R0
1
i1 =
or
i1 = G1 v1
R
v1 − v12
i
=
or
i2 = G2 (v1 − v2 )
2 v −v
e as
1
2
R2
i2 =
or
i2 = G2 (v1 − v2 )
R2 v2 − 0
i
or
i3 = G3 v2
vhigher − vlower 3 =v −R0
2
3
=
(3.3)
(3.4)
i3 = G3 v2
or
With this in mind, we obtain
Fig.
+
+ 3.2(b),
I1 = I2 from
0
0
or
▲
|
|
▲
▲
or
vR
Rv22
1−
1 v2
=
v −Rv2
vR3
I2 +
i1 = G1 v1
(b)
(b)
(3.4) v
I1
R Eq. (3.4) inREqs.
Substituting
(3.1) and (3.2) results, respectively, in
3
— Substitute
v1 (3.2)
v1 −
v2
Substituting
Eq. (3.4) in Eqs. (3.1) and
results,
respectively, in
greement with
the way we
resistance
I2 +
+
(3.5)
I1 =defined
R2v2
v1R1 v1 −
i3 =
I2
i2
Figure 3.2
Typical circ
analysis.
Typical c
i23.2
RFigure
2
v
2 analysis.
1
i1
i3
R1
R3
(3.5)
(3.6)
(b)
n − 1 simultaneous equations
(3.6)
Figure
3.2
Typical circuit for nodal
R2
R3
Solving Workbook C
e-Text Main Menu| Textbook Table of Contents |Problem
analysis.
i2 = G2 (v1 − v2 )
(3.4)
I2 +
1
2
=
2
Table ofvContents
Solving Workbook
Menu
| Textbook
|Problem
— e-Text
SolveMain
Eqs. to
obtain
the node voltages
any standard
1 and v2 using
or method,
i3 =such
G3 vas2 the substitution method, the elimination method,
Cramer’s rule, or matrix inversion
(3.1) and (3.2) results, respectively, in
v1
v1 − v2
I2 +
+
(3.5)
R
R
calculators
as HP48
or3 with
−Gsuch
GPro.
v2 softwareI2 packages such as Matlab
2Quattro
2+G
Mathcad,
Maple, and
calculators
such
as
HP48
or
with
software packages such as Matlab,
Mathcad, Maple, and Quattro Pro.
whichMathcad,
can be solved
to
get
v
and
v
.
Equation
3.9 will be generalized
1
2Pro.
Maple, and Quattro
in Section 3.6. The simultaneous equations may also be solved using
calculators such as HP48 or with software packages such as Matlab,
Mathcad,the
Maple,
Quattro
Pro.circuit shown in Fig. 3.3(a).
Calculate
node and
voltages
in
the
Calculate
the
node
voltages
in the circuit shown in Fig. 3.3(a).
5A
5A
Calculate
the
node
voltages
in
the
circuit shown in Fig. 3.3(a).
Solution:Solution:
5
A
E X A M P L E 3 . 1
Solution:
Consider
Fig.
3.3(b),
where
thewhere
circuitthe
incircuit
Fig. 3.3(a)
has
beenhas
prepared
for
Consider
Fig.
3.3(b),
in Fig.
3.3(a)
been prepared
for
4"
Consider
Fig.
3.3(b),
where
the
circuit
in
Fig.
3.3(a)
has
been
prepared
for
24 "
nodal
analysis.
Notice
how
the
currents
are
selected
for
the
application
Calculate
the
node
voltages
in
the
circuit
shown
in
Fig.
3.3(a).
2
nodal analysis. Notice how the currents are selected for the application
1
51 A 4 "
2
nodal
analysis.
how
the current
currents
are selected
for the
the labeling
application
ofSolution:
KCL. Except
forExcept
theNotice
branches
with
the labeling
of the of the
of KCL.
for the
branches
withsources,
current
sources,
1
ofisKCL.
Except
for
the but
branches
with
current
sources,
the
labeling
of the
currents
arbitrary
consistent.
(By
consistent,
webeen
mean
if,forfor
currents
is but
arbitrary
consistent.
(By consistent,
wethat
mean
that
if, for
2"
Consider
Fig.
3.3(b),
where
the circuit
in Fig.
3.3(a)
has
prepared
2 "6 "
6 "10 A
10 A
currents
is
arbitrary
but
consistent.
(By
consistent,
we
mean
that
if,
for
2" 4"
62 "
10 Aexample,
we
assume
that
i
enters
the
4
!
resistor
from
the
left-hand
side,
2 the
example,
we assume
thatcurrents
i2 entersare
theselected
4 ! resistor
from
the left-hand side
nodal analysis.
Notice
how
for the
application
1
example,
we
assume
that
i
enters
the
4
!
resistor
from
the
left-hand
side,
2
i2 of
must
leave
the for
resistor
from
thewith
right-hand
side.) The
reference
node
iExcept
leave
resistor
from
the right-hand
side.)
The
reference
node
KCL.
thethe
branches
current
sources,
the
labeling
of
the
2 must
i
must
leave
the
resistor
from
the
right-hand
side.)
The
reference
node
2
is currents
selected,
the node
voltages
v1 (By
andconsistent,
v2v1are
now
to mean
be
determined.
isisand
selected,
and
the node voltages
and
v2we
are
now
to
arbitrary
but consistent.
thatbeif,determined.
for
2"
6"
10 A
is
selected,
and
the
node
voltages
v
and
v
are
now
to
be
determined.
1
2
At node
1,Atapplying
and
Ohm’s
law
gives
node
1,iapplying
KCL
Ohm’s
example,
we assume
that
enters
the
4 ! and
resistor
fromlaw
thegives
left-hand side,
(a)
2KCL
(a)
At
node
1,
applying
KCL
and
Ohm’s
law
gives
(a)
i2 must leave the resistor from the right-hand
node
v1 side.)
− v2 vThe
v1reference
v−
0
1−
2 0 v1 −
5A
v
−
v
−
0
v
=
i
+
i
"⇒
5
=
+
i
5A
=
i
+
i
"⇒
5
=
+
i
1
2
1
1 and 2the node
13
2
3
is
selected,
voltages
v
and
v
are
now
to
be
determined.
1
2 5 =
5A
"⇒
i1 = i2 + i3
4
2
4 +
2
4
2
At
node
1,
applying
KCL
and
Ohm’s
law
gives
(a)
Multiplying
term
the last equation
4, we obtain
each termeach
in the
lastinequation
by 4, weby
obtain
i1 = 5
i1 = 5
i1 = 5 Multiplying
i1 = 5
Multiplying
each
term
in
the
last
equation
by
4,
we
obtain
i1 = 5
v
−
v
v
i1 = 5
1
2
1−0
5 Ai
5+
=
i1 = i2 + i3 20 "⇒
i4 = 10
− v2 + +
2v
20v=
v 2v
i = 10
= v1 −
i2
4 2v1 1 2
4 "i2 2 v2 4 " 4
20 = v21 −1 v21+
v2 i4 = 10
4"
v2
v1
v1
or each term in the last equation by 4, we obtain
orMultiplying
iv11= 5
i =5
or
E X A M PE LX EA 3M . P 1L E 3 . 1
E X A M P L E 3 . 1
i3
i2i3 i
5
i2 1 i
i2 i i4 =5 10
5
−2vv12 = 20
−
203v
=1 v−
1 20
v2v3v
=
(3.1.1) (3.1.1
1 3v
2+
(3.1.1)
1 − v2 = 20
6
"
2
"
v
10
A
6"
2"
1
10 A
node
2, we
do the
same
thing
and get
2,At
we
do
the
same
thing
and
get
6"
2"
10or
A At node
At
node
2,
we
do
the
same
thing
and
get
i2 i
i3
5
v1 − v2
v2 − 0
=
3v
v1 20
− vv21 −
0(3.1.1)
v=
1 − v2"⇒
2−
v
−
0
v
2
2
+
i
=
i
+
i
+
10
5
+
i
2
4
1
5
i1i+
i5 i1 + i5"⇒ "⇒
+ 10
= 10
5+
i2 + i4 i=
+
=
+
=
5
+
2
4
4
6"
2"
10 A
6 6 6
At node 2, we do the same thing and4 get 4
Multiplying
each12
term by 12inresults in
Multiplying
each term
Multiplying
eachby
term results
by 12 results
v1 −inv2
v2 − 0
+
i
=
i
+
i
"⇒
+
10
=
5
+
i
2
4
1
5
(b)
−
3v=
= 60 + 2v2 6
2+
(b)
(b)
460120
3v123v−
+1 3v
120
+ 60
2v
3v1 − 3v
120
=
2+ 2v2
2+
Multiplying
or each term by 12 results in
Figure
3.3
For Example 3.1: (a) original
or
or
Figure 3.3 Figure
For 3.3
Example
(a) original
For3.1:
Example
3.1: (a) original
circuit,
(b) circuit for analysis.
(b)
=1 60
+ 22v=2 60
3v1 − 3v2 + 120
+605v
(3.1.2
−3v
circuit, (b)circuit,
circuit (b)
for circuit
analysis.
for analysis.
5v
=5v
(3.1.2)(3.1.2)
−3v1 +
−3v
12+
2 = 60
or
Figure 3.3 For Example 3.1: (a) original
i2
i3
4"
v2
4
2Eqs. (3.1.1) and (3.1.2). We can solve
Now we have two simultaneous
M
E
T
H
O
D
1
Using
the
elimination
technique,
add
and
anywe
method
and obtain
the
v1Eqs.
and (3.1.1)
v2 .
but
(Byusing
consistent,
we mean
that
if, values
forweof
in consistent.
the the
lastequations
equation
by 4,
obtain
Now
we have two simultaneous Eqs. (3.1.1) and (3.1.2). We can solve
(3.1.2).
hat 20
i2 enters
the
resistor
from
the
M Evequations
T1H−
O Dv42 1!
Using
themethod
elimination
technique,
we add
(3.1.1)
the
andleft-hand
obtain
theside,
values
of Eqs.
v and
v2 . and
+using
2v1 any
=
!⇒
v2 = 20 V 1
4v
2 = 80
stor from
the
right-hand
side.)
The
reference
node
(3.1.2).
MSubstituting
E T HvO1Dand
1 vvUsing
technique, we add Eqs. (3.1.1) and
ode voltages
are20the
now
to(3.1.1)
be!⇒
determined.
in elimination
Eq.
gives
22 =
=
80
v2 = 20 V
4v
2
(3.1.2).
ying KCL
and
v2 Ohm’s
= 20 law gives
(3.1.1)40
3v1 −
20
=
20
!⇒
v
= 20 V= 13.33 V
=−20
in
Eq.
(3.1.1)
gives
Substituting v3v
2 1
!⇒
v21 =
4v2 = 80
3
v
−
v
−
0
v
2
1
o the same thing and get1
40
i3
"⇒
5 2=
in20
Eq.+
(3.1.1)
Substituting
13.33 V
−
!⇒
vwe
1 =need =
2 vgives
M E T H Ov3vD1v1−
2=v20220
To=4use
Cramer’s
rule,
2−0
3 to put Eqs. (3.1.1) and
"⇒
+
10
5+
40
(3.1.2)
in3vmatrix
as=
m in the
last
equation
4, 20
we
obtain
13.33 V
− by
20form
=
!⇒
v
=
6
14
1
! 3to"=put
!
" ! we" need
M E T H O D 2 To use Cramer’s
rule,
Eqs. (3.1.1) and
20
3
−1
v
1
by 12
results
in
v2 + form
2v1 as
20(3.1.2)
= v1in−matrix
=
(3.1.3)
−3 "rule,
5! "vwe
2 need
M E T H O D 2 To use !Cramer’s
! 60
"to put Eqs. (3.1.1) and
= 60form
+ 2vas2 3 −1 v1
3v1 − 3v
20
2 + 120
(3.1.2)
in matrix
=
(3.1.3)
The determinant
of the matrix
is
v
v
!−31# 52" !v#2 " !60"
20
3## −1
(3.1.1)
3v1 − v2 = 20
3 −1v##1
=
(3.1.3)
!
=
=
15
−
3 = 12
The determinant of the matrix
is
#−3 5 # 5v#2
60
−3
#
5v2 =
60get #
(3.1.2)
1 +thing
do the−3v
same
and
#
3
−1
#
!and
=matrix
The
v##−3
Wedeterminant
now obtainofv1the
2 as is 5# = 15 − 3 = 12
#
# #v2 − 0
v1 − v2
#
#
##
3
−1
#
"⇒
+
10
=
5
+
20
−1
5
#
# #= 15 − 3 = 12
!
=
#
We now obtain 4
v1 and v2 #as
# 6
−3
5
#
!1 # 60 # 5#
100 + 60
#20 −1# =
=
= 13.33 V
v1 =
m by 12We
results
in
now obtain v1 and!v2 ##as ! ##
12
!1
#60 # 5# #100 + 60
#220 # −1
# =#
=
=
= 13.33 V
60
+
2v
3v1 − 3v2 + 120v=
1
# 3 # 20
# 12
# !
!
# #−35# 60
# #100 180
+ 60
!1 !2 #60
#
#= = + 60 = 13.33
= 20VV
v1 =v2 = = =
3
20
! ! ## ! ! ##
12 12
180 + 60
#−3 60#
60!2result
(3.1.2)method.
−3v
giving
us 2the
elimination
#=
v=2same
=
= # as3did20the
= 20 V
1 + 5v
# ! #
!
12
#−3 60#
!2
180 + 60
giving usIfthe
result
as
thecan
elimination
method.
we same
need
the
currents,
them
v2 =
= did we
=easily calculate
= 20
V from the values
!
!
12
of the nodal voltages.
wethe
need
theresult
currents,
wethe
canelimination
easily calculate
them from the values
givingIfus
same
as did
method.
v1 − v2
v1
of thei1nodal
voltages.i2 =
= 5 A,
= −1.6667 A,
i3 =
= 6.666
4we can easily calculate them from
2 the values
If we need the currents,
v1 − v2
v1
5 A, voltages.
i2 =
= −1.6667v2A,
i3 =
= 6.666
i1 =nodal
of the
4 A,
i4 = 10
i5 =
= 3.333 A 2
v1 − v2
v1
6
v2 A,
i2 =
= −1.6667
i3 =
= 6.666
i1 = 5 A,
10 A,
i5 that
= the
= current
3.333 Aflows2in the direction
negative
The fact that i2 iis4 =
4 shows
tents |Problem Solving Workbook Contents
ntents |Problem Solving Workbook Contents
80
3A
Figure
2 " 3.5
4"
4"
20 "
2
he voltages at the nodes
1 in Fig. 3.5(a).
v1
3A
ix
At node
1,
2ix
8"
de 1,
ix
i3
4"
4"
3A
2i
v1 −x v3
v 1 − v2
3=
+
3 = i1 + ix i1 !⇒
i1
i
i2
2
2"
v2 2 8 "4
v
v
3
1
3
Multiplying
by 4 and
rearranging terms, we get
(a)
in this example has
nodes,
3A
4 " unlike the pre- 2ix
3 Athree nonreference
ple which has two nonreference nodes. We assign voltages
to
At node 2,
3.5 in Fig.
For Example
3.2: label
(a) original
circuit,
(b)
circuit
for analysis.
des asFigure
shown
3.5(b) and
the currents.
0
ix
3v1
(b)
ix
i3
− 2v2 − v3 = 12
4"
2ix
(3.2.1)
v 1 − v2
v2 − v3
v2 − 0
=
+
2
8
4
4"
At node 1,
(a)
(b) we get
Multiplying by 8 and rearranging
v1 − vterms,
v 1 − v2
3
i1 3 = i + i
!⇒
3=
+
i2
i2
1
x
2"
8"
4 7v2 − v32= 0
(3.2.2)
2
Figure
3.5 vFor
Example 3.2: (a) voriginal
circuit, (b) circuit for analysis. −4v1 +
3
Multiplying by 4 and rearranging terms, we get
i
At node 3,
ix = i2 + i3
ix
!⇒
3
4"
3A
circuit for analysis.
2ix
v3
3A
ix
i3
4"
(a)
ix
3A
v3
(b)
The circuit in this example has three nonreference
nodes, unlike the previous example iwhich
has two nonreference nodes.
i1 We assign voltages to
1
i2
i2
For Example
3.2:
(a)
original
circuit,
(b)
circuit
for
analysis.
2"
8 " and label the currents.
8"
v2 3.5(b)
the three nodes as shown
in Fig.
2
3A
v1
2
2ix
3 A nonreference nodes,4unlike
"
"
The4circuit
in this example
has three
the previous example which has two nonreference nodes. We assign voltages to
the 0three nodes
as shown
Fig. 3.5(b)
labelinthe
currents.
Determine
the in
voltages
at theand
nodes
Fig.
3.5(a).
3
i1
v
Solution:
4"
ix
DC Circuits
3 at the nodes1 in Fig. 3.5(a).
Determine the voltages
3A
ix
ix
Solution:
E X A M P L E 3 . 2
DC Circuits 1
PART 1
2
1
(b)
At node 3v
1, 1 − 2v2 − v3 = 12 v1 − v3
v2 − v3(3.2.1)2(v1 − v2 )
!⇒
+
i1 + i2 = 2ix
v1 − v3 8 v1 −=v2
4
2
At node 2,
!⇒
3=
+
3 = i1 + ix
4v − 0 by23, we get
Multiplying by 8, rearranging
v1 − v2 terms,
v3 dividing
v2 − and
2
i2 + i3
=terms, we+
ix =Multiplying
by!⇒
4 and rearranging
get
2 2v − 3v 8 + v = 04
(3.2.3)
1
2
3
−
v
=
12
(3.2.1)
3v1 −
Multiplying by 8 and rearranging terms,
we2v
get
2
3
We have three simultaneous equations to solve to get the node voltages
Atvnode
2, −4v1 + shall
7v2 −solve
v3 =the
0 equations in two ways.
(3.2.2)
1 , v2 , and v3 . We
v 1 − v2
v2 − v3
v2 − 0
At node 3,
!⇒
=
M E TiHx O=Di2 1+ iUsing
the elimination technique,
we+add Eqs. (3.2.1) and
3
2
8
4
2ix
▲
▲
▲
▲
▲
2
3,
, 1 v2 =
v3 =
+
7v
−
v
=
0
(3.2.2)
−4v
!
!
vV
V,
v2 1= 2.4 2V, 3 v1v3−=v3−2.4!
v1 = 4.8 V,
v2 = 2.4 V,
v3 = −2.4 V
1 − v2
At node 2, 3v1==i14.8
+ ix
!⇒
3=
+
4to
At node
where
!1 , !2 , and !3 are the determinants
as follows.
v1we
− put
v2 Eqs.
−calculated
v3 to2v(3.2.3)
v2be
2 − 0 in M E T H O D 2 To use Cramer’s rule, we put Eqs. (3.2.1) to (3.2
M E T H!,
OixD3,
2
To
use
Cramer’s
rule,
(3.2.1)
= iin
iand
!⇒
=
+
2+
3
Multiplying
4Appendix
rearranging
wethe
get
As
explained
A, to calculate
by 3 form.
vterms,
v2 determinant
−8v3
2(v14of
− va23) matrix
1 −2v3
matrix
form. by
v1 v2 v3
+
i
=
2i
!⇒
+
=
i
2
xthe first two rows and cross multiply.
matrix,1we repeat
⎡
⎤
⎡
⎤
⎡
⎤
⎡
⎤⎡ ⎤ ⎡ ⎤
4 we get 8
2
Multiplying by 8 and rearranging
terms,
(3.2.1)
3v1 −
3 −2
−12v2 −v1v3 = 1212
3 −2 −1
v1
12
Multiplying by 8,⎣rearranging
terms,
and
dividing
by
3,
we
get
⎦
−4 −4v
7 1−1
==
à ⎣−4 7 −1⎦ ⎣v2 ⎦ = ⎣ 0⎦
+⎦7v⎣2v−
(3.2.2)
2 ⎦v3
At node 2,
−1
3⎣ 00−2
2 −3
1
0
v3
2 −3
0
v3 v −4
2v
3 −2
−11 −13vv2 1+
− v3 2= 0 v2 7− v−1
v2 − 0 (3.2.3)
At node 3,
3
+ i3 7 !⇒
= −3
+
ix!==i2 −4
=
−1
2
1
2
8
From
this,
we
obtain
We have three simultaneous
to vget
node
v1 − vto3 solve
v2 −
2(v
v ) From this, we obtain
PART 1
DCequations
Circuits
3 the
14− voltages
+ 2
+3 −2 −1
=
i1 + i2 = 2ix2 −3!⇒ 1 3 − −2 −1
v1 , v2 , and vby
shall
solve the terms,
equations
in two
Multiplying
8 and
get
3 . We
8!73ways.
2
!2
!3
!rearranging
!24 we
−1 +
−4 −1
−4 7 !1
1
−4
−1
7
∆=
=
3
−
(−2)
−1
−
−4
7
−1
v
=
=
=
,
v
,
v
v
=
=
=
,
v
,
v
1
2
3
1
2
3
−3
1 get
2
1
2 −3 !
Multiplying
rearranging
terms,
by 3, we
!
!
!the−4v
!and
!
−
v3dividing
1= 0 we
M E T H O D by
1 8,
Using
elimination
technique,
Eqs. +
(3.2.1)(3.2.2)
and
1 +2 7v
12
−1 = 3 7add
3−2−3
− 3 + 2 −4 + 2 − 12 − 14
(3.2.3).
3v
+−
v38=
(3.2.3)
where
!,3,
!1=
, !21
!3+
are42v
the
determinants
to0be
calculated
follows.
1−
29
2 , and
−
12
+
14
−
=
10
At
node
0
−4
−1
=
3
4
+ 2 as
−2
− −2where !, !1 , !2 , and !3 are the determinants to be calculated as fo
PART
1
DC
Circuits
in Appendix
A, to calculate the determinant of a 3
As explained in Appendix
to calculate
the determinant
of2+a=0310
by 24
3 As
−=
4the
+
!2 = A, 5v
0
−
− explained
0 − 0 + 48
= 24
2v5v
1v2=−
We have three simultaneous
equations
to3 12
solve
to12
get
node
voltages
1−
2 0=
−
v
v
2(v
−
v
)
1
3
1
2
matrix, we repeat the first two rows and cross multiply.
matrix,
wewe
first two rows and cross+multiply. =
Similarly,
irepeat
=obtain
2ithe
2v
x shall!⇒
+
12
−1
v1 , vi21, +
and
in
two
ways.
3 . We
−solve the3equations
4
8
2
+
0
−4
−1
12
−1
3
−−1
3 −2 −1
12
−1
−2 the
3dividing
−2 we
Multiplying
8,
rearranging
terms, and
byadd
3, we
M E1T H O DDC1by
Using
elimination
technique,
Eqs.get
(3.2.1) and
PART
Circuits
+
3 −2 −1
−4
7 −1
3 −2
−1−1
−4
7 −1
0−4
7 0−−1
(3.2.3).
Textbook
Table
of
Contents
−
3v
+
v
=
0
(3.2.3)
2v
Problem
Solving
Workbook
Contents
|
1
2
3
2 −3
1
= ! = 0−42−37 0 −11 1=
=20++
240−−036
−−
0+
4848
= 24 ! = −4 7 −1 =
−3
!!
= 84
0 0+−0 1−
0=
1 2=
−
5v
=
12
5v
12
+
31 − −2
2 −3
1
3 −2 −1
2 −312 −11−1
−2 to−1
−
+ +2 to3 solve
We have
equations
get the+node voltages
12 −1−4
3 3−2
−−three12simultaneous
+
−4
7 −1
−1ways.+
−
−+equations
+7 −4 0 in7 two
−1the
v1 , v2 ,−and
.0−4
We shall
−1
70 0 solve
− v3−4
−1
+
!3 =
= 0 ++144 + 0 − 168 − 0 − 0 = −24
0
2 − −3
−
+
+
−
−
!M2E=
= 012
+ 0 − we
24 add
− 0Eqs.
− 0(3.2.1)
+ 48 =
24
0the elimination
2
1
= 21 − 12 + 4 + 14 − 9 − 8 = 10
T H O D =1 21Using
and
− 12−
+ 4 + 143− 9−2
−technique,
8 = 10 +
+
12
−1
3
Textbook
Contents
Problem 0Solving
+ Workbook Contents
(3.2.3).−Table of
− 12 −4| + 7
3
−2
Similarly, we obtain
Similarly,
0 −1
−4
− we obtain
+
−4
7− 05v1 −+
5v2 = 12
−
12 −2Workbook
−1
122 Main
−2
−10 |
!
= 0 + 144
+ 0of−Contents
168 − 0 − 0|Problem
= −24 Solving
−3we
Table
Contents
Menu
| 3 = e-Text
Thus,
find Textbook
0
7
−1
03 −2
7 −1
+
1212
3
−2
−
= 84 + 0 + 0 − 0 − 36 − 0 =
1
0 −3
!1 =
= 8448
+ 0 + 0 − 0 − 36 − 0 =!
482 !24
1
0 −3
1=
!
+
1
0
−4
7
− −4
7 −1
+
v01 = + =
= 4.8 V,
v2 =
=
=−2.4 V12 −2 −1
−2
−Table12of Contents
Textbook
Problem
Solving
Workbook
Contents
+
|!
10
!
10
!3 = −−
=
0
+
144
+
0
−
168
−
0
−
0
=
−24
0
20 −37 −1
0
+
7 −1
−
+
+
+
3 −2 12
−
+
−
− we
!3
−24
Thus,
find
+
v3 =
=
= −2.4 V
0
−4
7
−
!
10
+
!1
!2
48
24
−
v1 = as we
= obtained
= 4.8with
V, Method
v2 =1.
=
= 2.4 V
!
10
!
10
e-Text
Main Menu
| Solving
| Textbook Table of Contents
Textbook Table of Contents | Problem
Workbook
Contents
▲
v1 =
flows 2in the
The fact that i2 is negative shows that the!current
= 0 + 144 + 0 − 168 − 0 − 0 = −24
−3direction
0
3 =
opposite to the one assumed.
+
3 −2 12
−
+
0
−4
7
−
PRACTICE PROBLEM 3.1
+
−
Obtain the node voltages in the circuit in Fig.
3.4.
Thus,
we find
Answer: v1 = −2 V, v2 = −14 V.
v1 =
1
!1
48
=
= 4.8 V,
1A
!
10
v3 =
v2 =
6"
2
!2
24
=
= 2.4 V
!2 " 10 7 "
4A
!3
−24
=
= −2.4 V
!
10
as we obtained with Method 1.
Figure 3.4
For Practice Prob. 3.1.
|
▲
▲
PRACTICE PROBLEM 3.2
2"
Table ofvContents
e-Text Main Menu| TextbookAnswer:
|Problem Solving Workbook Contents
1 = 80 V, v2 = −64 V, v3 = 156 V.
3"
1
10 A
Figure 3.6
Find the voltages at the three nonreference nodes in the circuit of Fig. 3.6.
4ix
2
ix
4"
3
6"
For Practice Prob. 3.2.
Electronic Testing Tutorials
3.3 NODAL ANALYSIS WITH VOLTAGE SOURCES
We now consider how voltage sources affect nodal analysis. We use the
i2
i3
NODAL
ANALYSIS
WITH VOLTAGE
SOURCES
CHAPTER 3
NODAL
VOLTAGE
SOURCES
"
6"
10 V +ANALYSIS8 WITH
!
w consider how voltage sources affect nodal analysis. We use the
If a voltage sources affect nodal analysis à two cases:
n Fig. 3.7 for illustration. Consider the following two possibilities.
The voltage
source
is connected
1 If 1.
a voltage
source is
connected
betweenbetween
the reference node
nonreference
nodesat the nonreference
v1
onreferencereference
node, weand
simply
set the voltage
à nonreference
voltage
= voltage
ual to the voltage
of the voltage
source.
In Fig.source
3.7, for example,
Figure 3.7
A circuit with a supernode.
v1 = 10 V
(3.10)
10 V +
!
Me
4"
Super
i4
2"
i1
5V
v2
+!
i2
v3
i3
8"
6"
r analysis is somewhat simplified by this knowledge of the voltage
2.S E 2If the
voltage
source
(dependent
ororindependent)
C
A
If
the
voltage
source
(dependent
independent) is connected
ode.
connected
between
two
nodes
betweenistwo
nonreference
nodes,
thenonreference
two nonreference
nodes form a gennode orwesupernode)
eralized (generalized
node or supernode;
apply both KCL and KVL to determine
A supernode may b
Figure
3.7
A
circuit
with
a
supernode.
apply KCL & KVL to determine the node voltages
the nodeà
voltages.
enclosing the volta
Table of Contents |Problem Solving Workbook Contents
CASE 2
If the voltage source (dependent or independent)
between two nonreference nodes, the two nonreference node
eralized node or supernode; we apply both KCL and KVL
the node voltages.
A supernode is formed by enclosing a (dependent or independent) voltage
source connected between two nonreference nodes and any
elements connected in parallel with it.
A supernode is formed by enclosing a (dependent or independent
source connected between two nonreference nodes and an
In Fig. 3.7, nodes 2 and 3 form a supernode. (We could have moreelements
thanconnected in parallel with it.
ernode
isconnected
formed bybetween
enclosing
a nonreference
(dependent ornodes
independent)
voltage
source
two
and
any
A supernode
is formed
by enclosing a nodes
(dependent
or
independent) voltage
nected
between
two
nonreference
and
any
source
connected
between
two
nonreference
nodes
elements
connected
in two
parallel
with it. nodesandandanyany
source
connected
between
nonreference
lementsProperties
connected
inconnected
parallel
with
it. with it.
elements
in
parallel
of
a
supernode:
elements connected in parallel with it.
nodes1.2 and
3 form
a supernode.
(We could
have more than
The
voltage
source inside
the supernode
nd
3 form
a23supernode.
have
more
than
forming
single
supernode.
Forcould
example,
see
the
circuit
nodes
2 aand
form
a constraint
supernode.
(We
could
have
more
than
g.
3.7,
nodes
and
3 form
a (We
supernode.
(We
could
have
morein
than
provides
equation
needed
to
We analyze
aacircuit
with
supernodes
using
the
same
three in
single
supernode.
For
example,
see
the see
circuit
nodes
forming
single
supernode.
For
example,
seethe
thein
circuit
in
forming
a single
supernode.
For
example,
circuit
solve
for
the
node
voltages
ioned
in
previous
section
except
that
theusing
supernodes
3.14.)
Wethe
analyze
asupernodes
circuit
supernodes
the
sameare
three
yze
circuit
with
using
the
same
three
Weaanalyze
a circuit
withwith
supernodes
using
the
same
three
mentioned
inprevious
the Because
previous
section
except
that
supernodes
are
erently.
Why?
an that
essential
component
of
he
previous
section
except
the that
supernodes
arenodalare
oned
in the
section
except
thethe
2.
A
supernode
has
no
voltage
of
itssupernodes
own
ed
differently.
Because
an
essential
component
nodal
applying
KCL,
which
requires
knowing
the
current
through
v1
Why?
Because
an essential
component
of nodal
erently.
Why?Why?
Because
an essential
component
ofof nodal
ysis
is applying
KCL,
requires
knowingthrough
the current
through
nt. There
isAnosupernode
way
ofwhich
knowing
the
current
aof
voltage
KCL,
which
requires
knowing
the
current
through
applying
KCL,
which
requires
knowing
the current
through
3.
requires
the
application
element. However,
There is noKCL
way must
of knowing
the current
through a voltage
dvance.
be
satisfied
at
a
supernode
like
is
no
way
of
knowing
the
current
through
a
voltage
nt.
There
is
no
way
of
knowing
the
current
through
a
voltage
both
KCL
and
KVL
ce
in advance.
KCL must
be 3.7,
satisfied at a supernode like
ode.
Hence, atHowever,
the supernode
in Fig.
10 V +
owever,
KCL
must
besupernode
satisfied
atFig.
a supernode
like like!
dvance.
However,
must be in
satisfied
other
node.
Hence,
atKCL
the
3.7,at a supernode
Applying
supernode:
ce,
the
supernode
ode.atHence,
at the
supernode
in
iKCL
iin
=Fig.
i2 +3.7,
i3 Fig. 3.7,
1+
4at
i1 + i4 = i2 + i3
i1 + i4 =i1i2++i4i3= i2 + i3
CHAPTER 3
Methods of A
4"
Supernode
i4
2"
i1
5V
v2
+!
i2
v3
i3
8"
6"
(3.11a)
(3.11a)
(3.11a) (3.11a)
v1 − vv2 − vv1 − vv3 − v v2 −v0 − 0v3 −v 0 − 0
1 + 2
1 = 3
2
3
(3.11b)
5V
(3.11b)
+4
=8 + +6
2
Figure
3.7 A circuit with a supernode.
5V
2
4
8
6
v
−
v
−
0
−
0
v
v
v
v2 v1v−
−
v
−
0
−
0
v
v
1 2 3
2 3
3
+ +! !
1 2 3
(3.11b)
+
=
+
irchhoff’s
voltage
law
to
the
supernode
in
Fig.
3.7,
we
redraw
(3.11b)
+
=
+
+
pply Kirchhoff’s
voltage
law to the8supernode6in Fig. 3.7, we redraw
+ 5 V 5 V ++
2
4
4
8
6
as
shown
in Fig.in3.8.
around
the loop
in the
clockwise
ircuit
as shown
Fig. Going
3.8. Going
around
the loop
in the
clockwise
+ ! + ! is
C
A
S
E redraw
2 If the voltage source (dependentv2orv2independent)
v3vconn
rchhoff’s
voltage
law
to
the
supernode
in
Fig.
3.7,
we
3
+
+
Applying
KVL
at
super
node:
voltage
law
to
the
supernode
in
Fig.
3.7,
we
redraw
ives
+
+
tion gives
between two nonreference nodes, the two nonreference nodes form a
as
in Fig.
3.8.around
Goingthe
around
loopclockwise
in the
clockwise
n shown
Fig. 3.8.
Going
loopthe
in the
eralized
node or supernode; we apply both!KCL
! v and KVL to!!deter
v
−v2 +−v
5 2++v53 +
= v03 = 0 "⇒ "⇒ v2 −v2v3−=v35= 5the node(3.12)
(3.12)
v
v3
2
2
voltages.
ves
Figure
Applying
KVL
a supernode.
m(3.10),
Eqs. (3.10),
(3.11b),
and (3.12),
we obtain
the node
voltages.
Figure
3.8 3.8 Applying
(3.11b),
and (3.12),
we obtain
the node
voltages.
!to toa supernode.
!
! KVL
!
−v
+
5
+
v
=
0
"⇒
v
−
v
=
5
(3.12)
+ v23 = 0 3 "⇒
v2 − v3 =2 5 3
(3.12)
A supernode is formed by enclosing a (dependent or independent) voltage
2=
in terms of the node voltages
Example 3.3.
2
+
4
supernode
application of both KCL and KVL.
+ 73. A "⇒
8 =requires
2v1 + v2the
+ 28
or
v2 − 0
+
v2 = −20 − 2v1
E X7 A M P"⇒
L E 3 . 83 = 2v1 + v2 + 28
4E X A M P L E 3 . 3
(3.3.1)
To get the relationship between v1 and v2 , we apply KVL to the circuit
For the circuit shown in Fig. 3.9, find the node voltages.
10 "
thecircuit
loop, we
obtainin Fig. 3.9, find the node voltages.
For the
shown
10 in
" Fig. 3.10(b). Going around
Solution:
v2 = v1 + 2
(3.3.2)
The supernode contains the 2-V source, nodes 1 and 2, and the 10-! rev1
v2
2V
The supernode
contains the 2-V source, nodes 1 and 2, and the 10-! rev2 (3.3.1) and (3.3.2),
we write
Applying KCL to the supernode as shown in Fig. 3.10(a) gives
hip between vv11 and v2 From
, we Eqs.
apply
KVL to thesistor.
circuit
sistor. Applying KCL to the supernode as shown in Fig. 3.10(a) gives
ng around the loop, we obtain
v = v1 + 2 = −20 − 2v1
2 = i1 + i 2 + 7
2A
7 A2
2"
4"
2
=
i1 + i 2 + 7
2A
7A
2"
4"
or
2 + v2 = 0
"⇒
v2 = v1 + 2
(3.3.2) i1 and i2 in terms of the node voltages
Expressing
Expressing i1 and i2 in terms of the node voltages
"⇒ v v−
= −7.333
3v1 = −22
v2 −V0
1 1 0
nd (3.3.2), we write
"⇒
8 = 2v1 + v2 + 28
2=
v2 − 0 + 7
v1 − 0 +
4 +does
and v2 = v1 + 2 = −5.333 V. Note
+ resistor
7 not make
"⇒
8 = 2v1 + v2 + 28
2 = that the2 10-!
Example
3.3.
2
4
v1 +3.92 =For
−20
− difference
2v
v2 = Figure
any
because it is connected
across
the supernode.
or
1
Figure 3.9
For Example 3.3.
or
v2 = −20 − 2v1
(3.3.1)
v2 = −20 − 2v1
(3.3.1)
1 v1
2 v2
2 relationship
V
To
get
the
between
v
and
v
,
we
apply
KVL
to
the
circuit
1
2
= −22
"⇒
v1 = −7.333 V
2
1
To in
getFig.
the3.10(b).
relationship
v1 and
, we
apply KVL to the circuit
Goingbetween
around the
loop,v2we
obtain
i2 7 A
i1
2A
+
+
Fig.make
3.10(b). Going around the loop, we obtain
−5.333
V. Note 2that
the
10-! resistor
doesinnot
7A
2A
"
4"
−vv12 − 2 + v2 = 0
"⇒
v2 = v1 + 2
(3.3.2)
v1
use it is connected across the supernode.
−v1 − 2 + v2 = 0
"⇒
v2 = v1 + 2
(3.3.2)
From!Eqs. (3.3.1)! and (3.3.2), we write
From Eqs. (3.3.1) and (3.3.2), we write
(b)
v2 = v1 + 2 = −20 − 2v1
(a)
v2 = v1 + 2 = −20 − 2v1
2V
or
2
Figure 3.10 1Applying: (a) KCL to the supernode,
(b) KVLor
to the loop.
"⇒
v1 = −7.333 V
3v1 = −22
+
+
"⇒
v1 = −7.333 V
3v = −22
10-! resistor does not make
and v2 = v1 + 21 = −5.333 V. Note that the
v2
v1
difference
because
it V.
is connected
theresistor
supernode.
Workbook
Contents
e-Text Main Menu| Textbook Table of Contents
= v1 + 2Solving
=
−5.333
Note
that across
the 10-!
does not make
andany
v|2Problem
!
!
any difference because it is connected across the supernode.
v2 = −20 − 2v21 V
(3.3.1)"⇒
−v1 − 2 + vSolution:
2 =0
+!
+!
+!
+!
10 A
E X A M P L E 3 . 4
4"
E X A M P L E 3 . 4
1"
Find the node voltages in the circuit of Fig. 3.12.
3"
Find the node voltages in the circuit of Fig. 3.12.
Figure 3.12
+ vx !
20 V
For Example 3.4.
1
3"
Solution:
+ v !
Nodes 1 and 2 form a supernode; so do nodes 3x and 4. We apply KCL to
3vx
V 3.13(a). At supernode
the two supernodes as in20Fig.
1-2,
6
"
3
2
+!
+!
1
i3 + 10 = i1 + i2
Expressing this in terms of the node voltages,
2"
v3 − v210 A
v1 − v4
v1
+ 10 =
+
6
3
2
or
At supernode 3-4,
Figure 3.12
i1 = i3 + i4 + i5
For Example 3.4.
v1 − v4
v3 − v2
v4
"⇒
=
+
+
3
6
1
2
+!
2"
+!
4
1"
4
Figure 3.12
For Example 3.4.
PART 1
1"
Solution:
DC
Nodes 1 and 2 form a supernode; so do nodes 3 and 4. We apply KCL to
the two supernodes as in Fig. 3.13(a).
3 "At supernode 1-2,
i3 + 10+ =vxi1 !+ i2
i1
Expressing this in terms ofv the node6voltages,
"
v3
2
v1
v3 − v 2
v1 − v4
v1
+ 10
=
+
i3
v3
3i3
2i5
i2 6
4 or
10 A
2"
4"
5v1 + v2 − v3 − 2v4 = 60
i1
(3.4.1)
i1 = i3 + i4 + i5
"⇒
v4
i4
1"
(3.4.1
v1 − v4
v3 − v2
v4
v3
=
+
+
3
6
1
4
(a)
or
i3 + 10Table
= i1 of+Contents
i2
Solving Workbook Contents
e-Text Main Menu| Textbook
|Problem
+ 2v2 − 5v
16vto
4 =
Figure 3.13 4v1Applying:
(a)3 −
KCL
the0 two supernodes, (3.4.2
(b) K
Expressing this in terms of the node voltages,
▲
▲
▲
3vx
3
4"
10 A
Solution:
or
Nodes 1 and 2 form
a supernode;
so4 do
apply3-4,KCL to
2v2 − 5v3 − 16v
= 0nodes 3 and 4.
(3.4.2)
4v1 +
AtWe
supernode
the two supernodes as in Fig. 3.13(a). At supernode 1-2,
|
6"
86
4"
5v1 + v2 − v3 − 2v4 = 60
▲
|
2"
We now apply KVL to the86branches involving the voltage sources
PART 1
as shown in Fig. 3.13(b). For loop 1,
e two supernodes, (b) KVL to the loops.
DC Circuits
3"
−v + 20 + v = 0
"⇒
v1 − v23 =
20
(3.4.3)
We now1apply KVL2 to the branches involving
the" voltage sources
+ vx !
as shown
For
loop 2,in Fig. 3.13(b). For loop 1,
i1
i1
2 −
−v1 + 20 + v−v
0 3vx +"⇒
v1v4 = 0 vv
2 =
1
3+
Forvloop
2,1 − v4 so that
But
x =v
i3
i2
2"
−v1 3−+v3v
3v
2vv4 4==00
x +
3−
6"
v2 =
20
i3
v3
Loop 3
i5
i4
(3.4.4)
1"
Butloop
vx =3,v1 − v4 so that
For
+
v1
+!
Loop 1
!
3v3v
v36i
−3 2v
vx −
− 420==0 0
1−
x +
For6i
loop
But
v3 − v2 and vx = v1 − v4 . Hence
3 =3,
Figure 3.13
3vx
i3
20 V
(3.4.3)
v4
4"
10 A
+ vx !
+
6"
+
v2
v3
!
!
+!
Loop 2
+
v4
!
(3.4.4)
(a)
(b)
Applying: (a) KCL to the two supernodes, (b) KVL to the loops.
vx1−−3v
+v6i
20
= 20
0
−2v
v2x +
3 3+−2v
4 =
(3.4.5)
But 6iWe
v3 −four
v2 and
vx voltages,
= v1 − v4v.1Hence
node
, v2 , v3 , and v4 , and it requires We now apply KVL to the branches involving the voltage sources
3 =need
as shown in Fig. 3.13(b). For loop 1,
only four out of the five Eqs. (3.4.1) to (3.4.5) to find them. Although
−2v1 − v2 + v3 + 2v4 = 20
(3.4.5)
the fifth equation is redundant, it can be used to check results. We can
−v1 + 20 + v2 = 0
"⇒
v1 − v2 = 20
(3.4.3)
eliminate
node
voltage
that wevsolve
three
simultaneous
equations
Weone
need
four
node so
voltages,
,
v
,
v
,
and
v
,
and
it
requires
1
2
3
4
For loop 2,
instead
of
four.
From
Eq.
(3.4.3),
v
=
v
−
20.
Substituting
this into
only four out of the five Eqs. (3.4.1)
to find them. Although
2 to (3.4.5)
1
Eqs.
(3.4.1)
and (3.4.2),
respectively,
−v3 + 3vx + v4 = 0
the fifth
equation
is redundant,
it cangives
be used to check results. We can
eliminate one node voltage so that we solve three simultaneous equations
But vx = v1 − v4 so that
6v1 − v3 − 2v4 = 80
(3.4.6)
instead of four. From Eq. (3.4.3),
v2 = v1 − 20. Substituting this into
3v1 − v3 − 2v4 = 0
(3.4.4)
Eqs. (3.4.1) and (3.4.2), respectively, gives
and
2v44 =
6v6v
5vv33−−16v
= 80
40
1−
1−
For loop 3,
(3.4.6)
(3.4.7)
vx − 3vx + 6i3 − 20 = 0
and
Equations
(3.4.4), (3.4.6), and (3.4.7) can be cast in matrix form as
But 6i3 = v3 − v2 and vx = v1 − v4 . Hence
⎡
⎤⎡ ⎤ ⎡ ⎤
6v1 − 5v
(3.4.7)
3 −1
−23 − 16v
v1 4 = 40 0
−2v1 − v2 + v3 + 2v4 = 20
(3.4.5)
⎣6 −1
−2⎦ ⎣v3 ⎦ = ⎣80⎦
Equations (3.4.4), (3.4.6),
(3.4.7) can
6 −5and−16
40in matrix form as
v4 be cast
We need four node voltages, v1 , v2 , v3 , and v4 , and it requires
⎡
⎤⎡ ⎤ ⎡ ⎤
only four out of the five Eqs. (3.4.1) to (3.4.5) to find them. Although
3 −1 −2
v1
0
the fifth equation is redundant, it can be used to check results. We can
⎣6 −1 −2⎦ ⎣v ⎦ = ⎣80⎦
CHAPTER 3
Methods of Analysis
87
Using Cramer’s rule,
!
!
!
!
! 0 −1 −2!
!3 −1 −2!
!
!
!
!
! = !!6 −1 −2!! = −18,
!1 = !!80 −1 −2!! = −480
!40 −5 −16!
!6 −5 −16!
!
!
!
!
!3 0 −2!
!3 −1 0!
!
!
!
!
!3 = !!6 80 −2!! = −3120,
!4 = !!6 −1 80!! = 840
!6 40 −16!
!6 −5 40!
Thus, we arrive at the node voltages as
!1
−480
=
= 26.667 V,
!
−18
v4 =
v3 =
!3
−3120
=
= 173.333 V
!
−18
!4
840
=
= −46.667 V
!
−18
and v2 = v1 − 20 = 6.667 V. We have not used Eq. (3.4.5); it can be used
to cross check results.
@
PRACTICE PROBLEM 3.4
Find v1 , v2 , and v3 in the circuit in Fig. 3.14 using nodal analysis.
Answer: v1 = 3.043 V, v2 = −6.956 V, v3 = 0.6522 V.
6"
10 V
v1
+!
5i
v2
+!
v1 =
i
2"
4"
v3
3"
!40 −5 −16!
!6 −5 −16!
!
!
!
!
!3 0 −2!
!3 −1 0!
P R!A3 =
C !!T6 I C80E P−2
R !!O=B −3120,
L E M 3 . 3!4 = !!6 −1 80!! = 840
!
!
!
!
!6 40 −16!
!6 −5 40!
Find v and i in the circuit in Fig. 3.11.
Thus, we arrive
Answer:
−0.2atV,the
1.4node
A. voltages as
v4 =
v3 =
!3
−3120
=
= 173.333 V
!
−18
!4
840
=
= −46.667 V
!
−18
and v2 = v1 − 20 = 6.667 V. We have not used Eq. (3.4.5); it can be used
to cross check results.
3"
Figure 3.11
For Practice Prob. 3.3.
+ vx !
20 V
1
2"
+!
2
10 A
6"
+!
v1
4"
+!
2"
For Example 3.4.
4"
1"
Figure 3.14
Figure 3.12
5i
v2
i
4
6"
6"
10 V
3vx
3
2"
@
Find the
voltages
in circuit
the circuit
of Fig.
v1 , vnode
v3 in the
in Fig.
3.143.12.
using nodal analysis.
2 , and
3"
i
7V +
!
PE RX AA CM TPI LC EE 3P .R 4O B L E M 3 . 4
Answer: v1 = 3.043 V, v2 = −6.956 V, v3 = 0.6522 V.
+
v
!
+!
!1
−480
=
= 26.667 V,
!
−18
+!
v1 =
3V
4"
For Practice Prob. 3.4.
v3
3"
8"
11 "
nown voltages in a given
unknown currents. Mesh
Mesh analysis is also known as loop analysis or the
MESH
ANALYSIS
10 "
sis because it is only apmesh-current method.
circuit —is In
one
thatanalysis,
can
3.16 be
A nonplanar circuit.
mesh Figure
mesh
currents is used as the circuit variables instead of
6
"
e another;element
otherwise
it is
currents
understand
hes and still be To
planar
if itmesh analysis, we should first explain more about
what weloop
meanisbya aclosed
mesh. path with no node passed more than once
—
Recall:
anches. For example, the
with crossing
awn with no
A mesh is a loop which does not contain any other loops within it.
ble of Contents |Problem Solving Workbook Contents
mesh,
osely
sis to
In Fig. 3.17, for example, paths abefa and bcdeb are meshes, but path
— Nodal
KCL
to find
unknown
voltages
abcdefaanalysis
is not a applies
mesh. The
current
through
a mesh
is known as mesh
Mesh
analysis
find unknown
currents
current.
In mesh applies
analysis,KVL
we aretointerested
in applying
KVL to find the
mesh currents in a given circuit.
— Mesh analysis is only applicable to a circuit that is planar
A planar circuit (≠I nonplanar): is one
that can be drawn in a plane with
I2
1
R1
R2
b
c
a
no branches crossing
one another
I3 and still be planar if it can be redrawn
— A circuit may have crossing branches
such that it has no crossing
branches
i1
+
!
V1
f
R3
e
i2
+ V
2
!
d
the circuit in Fig. 3.16 is nonplanar, because there is no way to redraw
the branches crossing. Nonplanar circuits can be handled
PARTit1 and avoid
DC Circuits
using nodal analysis, but they will not be considered in this text.
2"
88
88
PART 1
Planar circuit
5" 6"
1A
1"
1A
3"
4"
2"
2 8""
7"
DC Circuits
Nonplanar circuit
circuit in Fig. 3.15(a) has two crossing branches,
but it can be red
1"
circuit
Fig. 3.15(b).
3.15(a) has
two crossing
branches,
but3.15(a)
it can beisredrawn
as ininFig.
Hence,
the circuit
in Fig.
planar. How
as the
in Fig.
3.15(b).
Hence,
theiscircuit
in Fig. 3.15(a)
is planar.
circuit
in Fig.
3.16
nonplanar,
because
there isHowever,
no way to re
5"
theitcircuit
in Fig.the
3.16branches
is nonplanar,
becauseNonplanar
there is no way
to redraw
and avoid
crossing.
circuits
be ha
2 can
"
4"
7"
it and avoid the branches crossing. Nonplanar circuits can be handled
using nodal analysis, but they will
6 " not be considered in this text.
using nodal analysis, but they will not be considered in3 "
this text.
1"
1"
55"
"
" 6
6"
44 "
(a)
13 "
3 3""
1A
5"
2"
1"
3"
— Two crossing(a)branches
4 " but it
5"
(a)
can be redrawn
1A
8"
2"
Figure 3.15
6"
7"
4"
8"
2"
7"
2"
A nonplanar circuit.
3"
"
12 " mesh analysis, we 9should
To understand
first explain more about
8"
11 "
A mean by a mesh.
9"
what5we
12 "
8"
11
"
— Nonplanar circuits can be
5A
handled10using
nodal analysis
"
2"
3"
5"
1"
3"
way
to redraw it and avoid
13 "
13 " crossing
the branches
(b)
(a) A planar circuit with crossing
1 branches,
"
3 " circuit redrawn with no
(b) the same
4"
crossing branches.
1"
Figure
— 3.16
No
6"
7"
1A
11 "
4"
6 " 10 "
77""
8 8""
5"
12 "
5A
1"
9"
6"
A mesh is a loop which does not contain any other loops within it.
Figure 3.16
A nonplanar circuit. 10 "
In Fig. 3.17, for example, paths abefa and bcdeb are meshes, but path
Figureis 3.16
A nonplanar
circuit.through a mesh is known as mesh
abcdefa
not a mesh.
The current
4"
To understand mesh analysis, we should first explain more about
8"
7"
6"
5"
current. In mesh analysis, we are interested in applying KVL to find the
what we mean by a mesh.
mesh currents in a given circuit.
To understand mesh analysis, we should first explain more
Although
path
abcdefa
is
a
loop
and
not
a
mesh,
8"
7"
(b)
KVL still holds. This is the reason for loosely
what we mean by a mesh.
using the terms loop analysis and mesh analysis to
2
R1 contain any Iother
R2 within it.
A mesh is a loopawhichI1does not
loops
b
c
Figure 3.15 mean(a)theAsame
planar
thing.circuit with crossing
branches, (b) the same (b)
circuit redrawn with no
I
ot a mesh,
or loosely
analysis to
I1
a
R1
b
I2
R2
c
I3
+
!
V1
f
i2
i1
R3
e
+ V
2
!
d
— Paths abefa and bcdeb
are meshes,
buttwo
path
abcdefa is not a mesh
Figure 3.17
A circuit with
meshes.
— The current through a mesh is known as mesh current
—
In this section, we will apply mesh analysis to planar circuits that
do not contain current sources. In the next sections, we will consider
In meshcircuits
analysis,
KVL is applied to find the mesh currents
in awith
given
circuit
with current sources. In the mesh analysis
n Analysis
CHAPTERof3a circuit
Methods of
meshes, we take the following three steps.
— We will apply mesh analysis to planar circuits that do not contain current sources
S t e p s Table
t o D eoft Contents
e r m i n e M|eProblem
s h C u rSolving
r e n t s :Workbook Contents
Main Menu| Textbook
1. Assign mesh currents i1 , i2 , . . . , in to the n meshes.
2. Apply KVL to each of the n meshes. Use Ohm’s law to express
the voltages in terms of the mesh currents.
3. Solve the resulting n simultaneous equations to get the mesh
currents.
3.17, to
forget
example,
paths abefa and bcdeb are meshe
of
meshcurrents.
currents.
Solve
simultaneous
equations
the mesh
the
voltages
in
of
S t e p 3.
s tthe
o Rvoltages
D1 e+the
t Re resulting
minterms
i terms
n−R
e n3M
sthe
hMethods
eAnalysis
nFig.
CHAPTER
ofIn
89
imesh
Vt1 s :
3 ethe
1C u r r
3r
abcdefa
is
not
a
mesh.
The
current
through
a
mesh
is know
=
(3.15)
currents.
3.
Solve
the
resulting
n
simultaneous
equations
to
get
the
mesh
−R
R
+
R
i
−V
3
2 n simultaneous
3i1 , i22, . . . , icurrent.
1. Assign
mesh
currents
to 2the In
n meshes.
3. Solve
the resulting
to get
the mesh
nequations
mesh
analysis,
we are interested in applying KVL
currents.
mesh
currents
in1 a&
given
currents.
—
Mesh
currents
ithe
i2ofare
toUse
meshes
2 to circuit.
2.
Apply
KVL to
each
theassigned
n meshes.
Ohm’s
law
express
1 and
h can be
solved
to
obtain
mesh
currents
i
and
i
.
We
are
at
liberty
1
2
Although
path
abcdefa
is
a
loop
and
not
a
mesh,
m i n e M e sthe
h Cvoltages
u r r e n in
t s :terms of the mesh currents.
any technique
theloosely
simultaneous
equations.
According
KVL
holds. for
This issolving
the reason
for
Tostillillustrate
the
steps,
consider
the circuit
in Fig.
3.17. The first
—
We
assume
mesh
current
flows
clockwise
I1 mesh
sh
currents
i1terms
,ai2circuit
,loop
. . the
.analysis
, inresulting
to
nanalysis
meshes.
using3.the
andthe
mesh
to ib and
R1 1 and I2
R2
Solve
n simultaneous
equations
toaget
the
. (2.12),
if
has
n
nodes,
branches,
and
l
independent
step
requires
that
mesh
currents
i
are
assigned
to
meshes
b
c
1
2
To
illustrate
the
steps,
consider
the
circuit
in
Fig.
3.17.
The
first
mean
the
same
thing.
LortoTo
each
of the
n meshes.
Use
lawbetolthe
express
currents.
illustrate
consider
circuit
in
Fig.mesh
3.17.
Thearbifirst
meshes,
then
lthe
= bsteps,
−
n +Ohm’s
1.
Hence,
independent
simultaneous
Note:
2.
Although
a
mesh
current
may
assigned
to
each
in
an
The direction
tep
requires
that
mesh
currents i1 and i2 are assigned to meshes 1 and I3
s
in
terms
of
the
mesh
currents.
ions
are
required
toissolve
the circuit
using
analysis.
requires
that itmesh
currents
i1toand
i2mesh
are
assigned
to meshes
1 and
rary
direction,
conventional
assume
that
each
mesh
current
flows
(clockwise
or
.esulting
Although
mesh
current
may
be
assigned
to
each
mesh
in
an
arbii2
—
i1 aand
i2 are
mesh current
i
The
directio
1
simultaneous
equations
toare
getdifferent
theassigned
meshfrom the
+theV vali
Notice nthat
the
branch
currents
mesh
V1 +currents
Although
a
mesh
current
may
be
to
each
mesh
in
an
arbiclockwise.
affect
2
R
The
!
! direc
(imaginary,
not
measurable
directly)
3
ary direction,
it is conventional
to assume
that each
mesh
current
(clockwise
o
To illustrate
the steps, consider
the circuit
in Fig.
3.17.
Theflows
first
!
"! "
!
"
the mesh
isolated.
Towe
distinguish
between
the
twomesh
types
of
As theisitsecond
step,
apply
KVL
tothat
each
mesh.
Applying
KVL
yslockwise.
direction,
is conventional
to assume
each
current
flows
(clockwis
affect
the va
stepwe
requires
that
mesh
currents
i
and
i
are
assigned
to
meshes
1
and
1
2
nts,
use
i
for
a
current
and
I
for
a
branch
current.
The
—
I
,
I
and
I
are
branches
current,
1 obtain
2
3
okwise.
mesh
1,
we
affect the
As
the
second
step,
we
apply
KVL
to
each
mesh.
Applying
KVL
2.
Although
a
mesh
current
may
be
assigned
to
each
mesh
in
an
arbint elements algebraic
I1 , I2 , andsums
I3 areofalgebraic
of the meshf currents.
the meshsums
currents
e
d
The direction
As
the
second
step,
we
apply
KVL
to
each
mesh.
Applying
KVL
he
steps,
the
circuit
Fig.to3.17.
The ifirst
ovident
mesh
1,consider
we
obtain
trary
direction,
it3.17
is conventional
assume
each
from
Fig.
that1 +inR
−V
R
0 mesh current flows
(clockwise or
(real,
measurable
directly)
1 i1 +
3 (i1 − that
2) =
esh
currents
i1 and i2 are assigned to meshes 1 and Figure 3.17 A circuit with two meshes. affect the vali
mesh
1, we obtain
clockwise.
+
Ri21,i1 mesh
+ R33in
(i
i2i)2 = 0
horcurrent
be iassigned
each
arbiI1 =
I12to
=
=1an
i−
−
(3.16)
—may
1 , −V
1to
The
direction ofKVL
the mesh current is arbitrary—
As
the
second
step,
we
apply IKVL
each mesh. Applying
−Vthat
R1 i1mesh
+ Rcurrent
i2 ) = 0
conventional
to assume
1 + each
3 (i1 −flows
(clockwise
orapply
counterclockwise)—and
does
not
to
mesh
1,
we
obtain
In
this
section,
we
will
mesh
analysis
to
planar c
r — Apply KVL to each
(R1 +
R3 )i1 − R3 i2 = V1
(3.13)
mesh
affect
the
validity
of
the
solution.
do not contain current sources. In the next sections, we wi
▲
▲
step, weFor
apply
KVL
each
mesh.
Applying
mesh
1: to−V
RR
R3R
(i1circuits
−KVL
)Vwith
= 0 current sources. In(3.13)
1 ++
1 i1)i+
the mesh Workbook
analysis of a Co
circ
(R
−
i
=i2Contents
Textbook
Table
of
1
3
1
3
2
1
For mesh
2,
applying
KVL
gives
Problem
Solving
e-Text
Main
Menu
|
|
n
meshes, we take the following three steps.
(R
+
R
)i
−
R
i
(3.13)
or
1
3 1
3 2 = V1
or mesh 2, applying R
KVL
gives
−V1 + R1 i1 + R3 (i1 − i22)i2=+0V2 + R3 (i2 − i1 ) = 0
(Rgives
(3.13)
mesh
applying
KVL
1 + R3 )i1 − R3 i2 = V1
—| 2,For
mesh
2:
R
i
+
V
+
R
(i
−
i
)
=
0
or
2
2
2
3
2
1
e-Text Main Menu| Textbook Table of Contents |Problem Solving Workboo
|
For mesh 2, applying
KVL
gives
i
+
V
+ R3 (iR23−
2
2
r (R1 + R3 )i1 − R3R
−R
i
)i2i1(3.13)
=) =
−V02
(3.14)
i2 = V311 +2 (R2 +
R2 i2 + V2 + R3 (i2 − i1 ) = 0
Note
ingives
Eq. (3.13) that−R
the3 icoefficient
the−V
sum
in
R3i)i
(3.14)
g KVL
1 2is=
1 + (R2 + of
2 of the resistances
or −V + R i + R (i − i ) = 0
1 ofgives
1 1 resistances
3 1
2in
2,i1applying
KVL
is the sum
the
−R
(3.14)
3 i1 + (R2 + R3 )i2 = −V2
of
the
resistance
2 is the negative
R2 i2 + V2 + R3 (i2 − i1 ) = 0
erve Note
that in
theEq.same
true
Eq. of i1 is the sum of the resistances in
(3.13)isthat
thein
coefficient
(Rmesh,
+R
(3.13)
1the
3 )i1 −
3 i2 = V1 of i is the negative
y of the
writing
mesh
equations.
first
while
theRcoefficient
of the
2
The shortcut way
will resistance
not apply if one mesh curcommon
meshes 1 and 2. Now observe that
sameclockwise
is trueand
in the
Eq.other assumed
−R3toigives
(3.14)
rent isthe
assumed
1 + (R2 + R3 )i2 = −V2
applying
KVL
(3.14). This
can serve
as (3.13)
a shortcut way of writing
the mesh
The sh
mesh currents.
Putting
Eqs.
anticlockwise,
althoughequations.
this is permissible.
q. (3.13)
of iin
the sum
of the resistances in
1 is
Wethat
will
rent is
R2the
i2exploit
+coefficient
V2 this
+ Ridea
iand
—
Note
the
coefficient
i1Section
i2 03.6.
3 (iof
2−
1) =
esh, while the
oftoi2 solve
is thefor
negative
of the
resistance
Thecoefficient
third step is
the mesh
currents.
Putting Eqs. (3.13)
anticlo
!
"
!
"
o meshes
1 andfor
2.the
Now
observe
that
the shortcut
same is way
true of
in writing
Eq. the mesh
andSolve
(3.14)
in
matrix
form
yields
—
mesh
currents
using
i
V
1 serve as 1a shortcut way of writing the mesh equations.
his can
"! " !
"
=equations !
(3.15)
The shortcut way will not ap
i
−V
+
−R32
i1
V1 (3.14)
2 this−R
21 +
(R2 R+13.6.
R3R)i32 = −V
xploit
idea3 iin
Section
rent is assumed clockwise and
=
(3.15)
−R
R2 + R3 Putting
i2 Eqs.
−V
3
2
third step is to solve for the
mesh
currents.
(3.13)
anticlockwise, although this i
urrents
i1 form
andcoefficient
iyields
atiliberty
3.13)
that
the
2 . We areof
1 is the sum of the resistances in
in matrix
which
can
be solved of
to
obtain
thenegative
mesh currents
i1 resistance
and i2 . We are at liberty
h, while
the
coefficient
i
is
the
of
the
multaneous
equations.
According
" !b branches,
"
— ! Note: if a circuit has"2n!nodes,
and l independent loops or
to use
any
technique
for
solving
the
simultaneous
equations. According
R
+ R2.
−R
i+1 1that the
V1same is true
1and
3observe
3 then
meshes
1
Now
in Eq.
meshes,
l
=
b
−
n
b branches,
andifl aindependent
=n nodes, b branches,(3.15)
to Eq.
(2.12),
circuit
has
and l independent
−R
R
+
R
i
−V
3
2
3
2
2
à as
l independent
simultaneous
equations
are required
to solve the circuit
can serve
a shortcut
way of writing
the mesh
equations.
ence,
l independent
simultaneous
loops
or meshes, then
l = b − n + 1. Hence, l independent simultaneousThe shortcut way will
using
analysis
oit
this
idea
inmesh
Section
3.6.to
be using
solved
to
obtain
the
mesh
currents
i1 and
i2 . We
aremesh
at liberty
rent is assumed clockw
uit
mesh
analysis.
equations
are
required
solve the
circuit
using
analysis.
step is to
solve
formesh
thebranch
mesh
currents.
Eqs.
(3.13)
technique
for
solving
the
simultaneous
equations.
According
Notice
that
the
currents
arePutting
different
from
the mesh currentsanticlockwise, although
eird
different
from
the
currents
12),
ifunless
a form
circuit
has n isnodes,
b branches,
and l independent
matrix
yields
the
mesh
isolated.
To
distinguish
between the two types of
guish
between
the
two
types
of
eshes,
then l =we
b −use
n +i 1.
!currents,
"a !mesh
" l independent
!
" Isimultaneous
forHence,
current
and
for a branch current. The
and
I
for
a
branch
current.
The
R
+
R
−R
i
V
1
3solve the 3circuit 1using mesh1analysis.
are required
to
current
elements
I1 , I2 , and I3=are algebraic sums of the
mesh currents.
(3.15)
R2mesh
+Fig.
R3are
i2that from
−V2 the mesh currents
ce that
the
branch
currents
different
braic
currents.
3of the
Itsums
is−R
evident
from
3.17
90
PART 1
DC Circuits
E X A M P L E 3 . 5
I1
5"
I2
For the circuit in Fig. 3.18, find the branch currents I1 , I2 , and I3 using
mesh analysis.
6"
I3
Solution:
We first obtain the mesh currents using KVL. For mesh 1,
10 "
15 V +
!
i1
i2
+ 10 V
!
−15 + 5i1 + 10(i1 − i2 ) + 10 = 0
4"
or
3i1 − 2i2 = 1
For mesh 2,
Figure 3.18
(3.5.1)
6i2 + 4i2 + 10(i2 − i1 ) − 10 = 0
For Example 3.5.
or
i1 = 2i2 − 1
METHOD 1
(3.5.2)
Using the substitution method, we substitute Eq. (3.5.2)
into Eq. (3.5.1), and write
6i2 − 3 − 2i2 = 1
"⇒
i2 = 1 A
From Eq. (3.5.2), i1 = 2i2 − 1 = 2 − 1 = 1 A. Thus,
I1 = i1 = 1 A,
METHOD 2
I2 = i2 = 1 A,
I3 = i1 − i2 = 0
To use Cramer’s rule, we cast Eqs. (3.5.1) and (3.5.2) in
matrix form as
!
3
−1
We obtain the determinants
"! " ! "
i1
1
=
i2
1
−2
2
From Eq. (3.5.2), i1 = 2i2 − 1 = 2 − 1 = 1 A. Thus,
I1 = i1 = 1 A,
METHOD 2
I2 = i2 = 1 A,
I3 = i1 − i2 = 0
To use Cramer’s rule, we cast Eqs. (3.5.1) and (3.5.2) in
matrix form as
!
"! " ! "
3 −2 i1
1
=
−1
2 i2
1
We obtain the determinants
#
#
# 3 −2#
#=6−2=4
! = ##
−1
2#
#
#
#
#
#1 −2#
# 3 1#
# = 2 + 2 = 4,
#=3+1=4
!1 = ##
!2 = ##
#
1
2
−1 1#
Thus,
i1 =
!1
= 1 A,
!
i2 =
!2
=1A
!
as before.
M 3.5
+
!
Calculate the mesh currents i1 and i2 in the circuit of Fig. 3.19.
Answer: i1 = 23 A, i2 = 0 A.
8V
E X A M P L E 3 . 6
Use mesh analysis to find the current io in the circuit in Fig. 3.20.
i1
Solution:
We apply KVL to the three meshes in turn. For mesh 1,
For mesh 2,
11i1 − 5i2 − 6i3 = 12
For mesh 3,
−5i1 + 19i2 − 2i3 = 0
24 V
(3.6.2)
4(i1 − i2 ) + 12(i3 − i1 ) + 4(i3 − i2 ) = 0
In matrix form, Eqs. (3.6.1) to (3.6.3) become
⎡
⎤⎡ ⎤ ⎡ ⎤
11 −5 −6
i1
12
⎣−5 19 −2⎦ ⎣i2 ⎦ = ⎣ 0⎦
−1 −1
2
i3
0
We obtain the determinants as
11 −5
−6
(3.6.3)
i1
12 "
Figure 3.20
But at node A, io = i1 − i2 , so that
−i1 − i2 + 2i3 = 0
+
!
(3.6.1)
4io + 12(i3 − i1 ) + 4(i3 − i2 ) = 0
or
i2
10 "
24i2 + 4(i2 − i3 ) + 10(i2 − i1 ) = 0
or
i2
io
−24 + 10(i1 − i2 ) + 12(i1 − i3 ) = 0
or
A
24 "
4"
i3
For Example 3.6.
+
!
4io
−6
11 −5
−1
−1
2
i3−6
11
−5
+0
−
11 −5 −6 !5 +19 −2
+
− as!5 19 −2
n the determinants
!
=
!1
−1
2
+
!5 19 −2
+
− 11 11
−5+−5−6 −6
+
−
=−
418!5−!5
30
10
114 −+22 − 50 = 192
−2−−2
19−19
− 30!−
22 −−6
+
= 10 −
− 114
!1
−1
250 = 192
12 −
−5
−5−
−2
1910−6
41811
−0 30
− 114+− 22 − 50 = 192
12 −5 −=−6
−1 −2
1 =−−2 !5
0 !19
−62 + = 456 − 24 = 432
120 19
−5
++= 432
−6− 24
−5
−2
0 19
= 456
0 −1 −− 2 12
−2
19− 114
−−6− 3000−+−1
=50
456
24 = 432
=418
2 − 22+−
1=
10
=−
192
12 !−5
12 −5 −6
++
0 19 −−−2
−6
12 −5+
PART
1 −2 DC Circuits
+
19
0
−
−2 −6
0 11
19+ 12
+ PART 1
92
DC Circuits
−
!1 =
0 !5
−1 0 2 −2 = 456 − 24 = 432
!5 =12144
11 + 120
−6
11
1 !12
DC
−62 +
11
12−6
=
24
!1
0
2 =Circuits
12
−5
−
!5 19
0
11 !5 12
0−2 −2
!5 −
0 −−20 !5
−6 + +
11
19 12
! =
= 6019
+ 2280= 288
!1= 144
0
!5
= 144
24!1+ 120
00 24−2
2+3+
+
!5
=
120
=
!1 !2−
0= − 2 !1
11 !5 12 −
11 !5!3 =
12
+!1 !1
= 60 + 228 = 288
0
−6
11
12
+
−
+
11 12 −−6 !5+ 19
0 +
0− + 11 !5 12
!5 19
−
+
− 11 !512 0−6 −2
+
!5 !30= −2
!1
0 +
− = 60 + 228 =−288+!5 19
+
0
− !5 !10 −2
++ 120
+
!2 =
− !1 110+
= 24
= 144
2 12 the
We !5
calculate
mesh
currents
using−Cramer’s rule as
Table of Contents |Problem Solving Workbook Contents
e-Text− Main
Menu
| Textbook
19
+
0
!5
−6
11
12
+
We calculate the mesh currents
using Cramer’s rule as
−
!1
!2
432
144
= + =+Table
= of
2.25
A,
i2 =
= Solving
= 0.75Workbook
A
0 |−2i1Textbook
!5 Menu
Contents
Problem
e-Text
− −Main
|
!
!2
432
144Contents
!
192
!
192
1
=
= 2.25 A,
i2 =
=
= 0.75 A
−
alculate
| the mesh currents using+ Cramer’s rulei1!=|as3 ! 288
192
!
192
n Menu Textbook Table of Contents Problem Solving Workbook Contents
i =
=
= 1.5 A
!3
288
!1
!23 144
432
!
192
=
= 1.5 A
i1 =
=
= 2.25 A,
i2 =
=
= 0.75 Ai3 =
!
192
!
192 Thus, io = i1 − i2 = 1.5
! A. 192
of Contents
Problem Solving Workbook Contents
e-Text Main Menu| Textbook TableThus,
io = i1 − i2| = 1.5 A.
P R O B L E M 3 . i63 = !3 = 288 = 1.5 A
!1
!2
= 1 A,
i2 =
=1A
! !5 12
!
11
as before.
!5 19
0
!3 =
= 60 + 228 = 288
!1 !1
0
5
11 !5 12
+
−
+
0
!5 19
−
Calculate the mesh currents i1 and i2 in the circuit of Fig. 3.19.
+
−
Answer: i1 = 23 A, i2 = 0 A.
We calculate the mesh currents using Cramer’s rule as
i1 =
PRACTICE PROBLEM 3.
2"
9"
12 "
i1
12 V +
!
i2
4"
Figure 3.19
+
!
8V
i1 =
!1
432
=
= 2.25 A,
!
192
3"
i3 =
For Practice Prob. 3.5.
Thus, io = i1 − i2 = 1.5 A.
i2 =
!2
144
=
= 0.75 A
!
192
!3
288
=
= 1.5 A
!
192
|
Tableanalysis,
of Contents
Problem
Workbook Contents
e-Text
| Textbook
6 " Main Menu
Using mesh
find io in |the
circuit Solving
in Fig. 3.21.
▲
|
▲
PRACTICE PROBLEM 3.6
io
20 V +
!
Figure 3.21
4"
i1
Answer: −5 A.
i3
8"
2"
i2
–
+
10io
For Practice Prob. 3.6.
Electronic Testing Tutorials
3.5 MESH ANALYSIS WITH CURRENT SOURCES
Applying mesh analysis to circuits containing current sources (dependent
i
4"
6i "
5 3.5iMESH
ANALYSIS
WITH
CURRENT
SOURCES
+
20
V
!
6 " ANALYSIS WITH
10 "
MESH
CURRENT SOURCES
6 A2 "
1
i210 "
2
1
3.5 M
4"
Electronic Testing Tutorials
MESH ANALYSIS WITH CURRENT
SOURCES (dependent or independent)4 "
pplying
mesh analysis to circuits
containing
current sources (dependent
+ Applying
6"
10 "
i
i
4
"
mesh
analysis
to
circuits
containing
current sources (dependent
1
2
!
+
i1 easier i2
independent)
may
appear
complicated.
But
it
is
actually
much
20
V
2
"
!
i
i
0 may2 appear complicated. But it is actually
or independent)
much
easier
1
6A
(b)
Exclude
these
an what
we
encountered
in
the
previous
section,
because
the
presence
+ what
in the iprevious
because the presence
i1 we encountered elements
V than
4 section,
"
2
!
(a)
theofcurrent
sources
reduces
the the
number
of equations.
+ Consider
i1the
i2
20 V Consider
the cases:
current
sources
reduces
number
of equations.
the
!
Two
60 A cases.
i2 cases.
lowing
two i1possible
following
two possible
4"
Applyin
or indep
than wh
of the c
followin
4"
CASE
3"
(b)
(a) Two meshes having a current source
in common,
(b) a supermesh, created by excluding
the current source.
Exclude
these
elements
1. When
aa(a)current
source
exists
only
one
mesh:
When
current
source
exists
only
inin
one
mesh:
Consider
thethe
When
a
current
source
exists
only
in
one
mesh:
Consider
i
i
0
2 and as
1 i a=supermesh
10 V +
6"
i2
i1
g.
3.23(b),
we
create
thethese
periphery
of and write a mesh
!
à
We
set
−5
A
write
a
mesh
equation
cuit
in
Fig.
3.22,
for
example.
We
set
i
=
−5
A
2
(b)
Exclude
2 i2 = −5 A and write a mesh
circuit in Fig. 3.22, for example. We set
23and (a)
Twoitmeshes
having a current
in
common,
(b)
supermesh, created by excluding the current source.
treat
differently.
(Ifinathe
circuit
has
two
ora more
elements
forthe
the
other
insource
usual
way
uation
for
other
mesh
usual
way,
that
is, is,
(a)mesh
equation
for
the
other
mesh
inthe
the
usual
way,
that
circuit i
equatio
A S EC A1S E 1
5A
CASE
the circ
at intersect, they should be combined to form a larger
CHAPTER 3
Methods of
cluding
Figure 3.22 A circuit with a current source.
Fig.
3.23(b),
we
create
a
supermesh
as
the
periphery
of
hy
treat
the
supermesh
differently?
Because
mesh
analy−10
+
4i
+
6(i
−
i
)
=
0
"⇒
i
=
−2
A
(3.17)
−10
+
4i
+
6(i
−
i
)
=
0
"⇒
i
=
−2
A
(3.17)
1
1
2
1 having
2 source in common, (b) a1 supermesh,
1
ure 3.23 (a) Two meshes
a1 current
created by excluding the current source.
as show
6"
10 "
—which
requires
that we know
voltage
es
and treat
it differently.
(If a the
circuit
has across
two oreach
more
A2S E3.23(b),
2 When
a current
source
exists
between
two
meshes:
Consider
Ado
S ECnot
When
a voltage
current
source
between
two
meshes:
Consider
the
across
a exists
current
source
adintersect,
they
should
bea combined
to
form
ain
larger
nthat
in2.
Fig.know
we
create
supermesh
as
the
periphery
of
A
2"
When
aFig.
current
source
exists
between
two
the
circuit
inmust
3.23(a),
for example.
We
create
a meshes:
supermesh
byexexe
circuit
in
Fig.
3.23(a),
for
example.
We
create
a
supermesh
by
r,
a
supermesh
satisfy
KVL
like
any
other
mesh.
Why
treat
the
supermesh
differently?
Because
mesh
analymeshes and treat it differently. (If a circuit has two or more
i
i
20 V +
4"
!
We
create
a
supermesh
by
excluding
the
current
cluding
the
current
source
and
any
elements
connected
in
series
with
it,
uding
the
current
source
and
any
elements
connected
in
series
with
it,
ying
KVL
to
the
supermesh
in
Fig.
3.23(b)
gives
VL—which
requires
that
we
know
the
voltage
across
each
hes that intersect, they should be combined to form a larger
6A
as
shown
in
Fig.
3.23(b).
Thus,
source
and
any
elements
connected
in
series
with
it
shown
in
Fig.
3.23(b).
Thus,
we
do
not
know
the
voltage
across
a
current
source
in
adh.) Why
supermesh differently? Because mesh analy−20treat
+ 6ithe
1 + 10i2 + 4i2 = 0
CHAPTER 3
Methods of Analysis
i
i
0
a supermeshrequires
must satisfy
KVL
like
other across
mesh.
sver,
KVL—which
that we
know
theany
voltage
each
Exclude these
e-Text
Main
Menu
|
|
| Textbook
elements
(a)
plying
to the
supermesh
in Fig.
3.23(b)
givessource in adand weKVL
do not
know
the voltage
across
a current
▲
A supermesh results when two meshes6 "have a (dependent10 "or independent)
▲
1
2
1
2
A supermesh
results when
meshes
havelike
a (dependent
or mesh.
independent) Figure 3.23 (a) Two meshes having a current source in common, (b) a
owever,
a supermesh
musttwo
satisfy
KVL
any other
−206i+1 6i
+ 210i
+ current
4i2source
= 0source
in3.23(b)
common.gives
114i
2 20
+to
(3.18)
6"
10 "
current
in common.
, applying KVL
the =
supermesh
in Fig.
As shown in Fig. 3.23(b), we create a supermesh as the periph
2"
the two meshes and treat it differently. (If a circuit has two o
to a node in −20
the branch
where
the
two
meshes
intersect.
+
that intersect, they should be combined to form a
i2
+ 6i1 + 10i202 V+ 4i2 =i1 0
4supermeshes
"
+ the supermesh
i1
i2
supermesh.) 20
Why
differently?
Because
o node 0 in Fig. 3.23(a) gives !
V treat
4 " mesh
!
(3.18)
6i1 + 14i2 = 20
6A
sis applies KVL—which requires that we know the voltage acros
branch—and we do not know the voltage across a current source
(3.19)
i 2 = i1 + 6
vance. However, a supermesh must satisfy KVL like any other
L to a node in the branch where the|Problem
two meshes
intersect.
Contents
i2 Workbook
i1 Solving
0
Therefore, applying KVL to the supermesh
in Fig. 3.23(b) gives
(3.18)
6i1 + 14i2 =| 20
(b)
Exclude these
L18)
to and
node(3.19),
0 in Fig.
3.23(a)
gives
we get
Textbook
of ContentsProblem Solving Workbook Contents
book
TableTable
of Contents
the two meshes and Figure
treat3.23
it differently.
a circuit
hasthese
two
or more
Exclude
(a) Two meshes(If
having
a current
source
in common,
(b) a supermesh,
the twosupermeshes
meshes and treat
it differently.
(If a circuit hasbe
twocombined
or more elements
intersect,
to form a larger
6that
"
10 " they should (a)
6
"
10
"
supermeshes that intersect, they should be combined to form a larger
supermesh.) Why
treat the
supermesh
differently?
Because
Assupermesh
shown
indifferently?
Fig.
3.23(b),
we create
a6 "supermesh
as theanalyperiphery of
10 " mesh
supermesh.) Why treat the
Because
mesh
analy6 a"current
10 " in common, (b) a supermesh, cr
Figure
3.23
(a)
Two
meshes
having
source
2"
sis applies
KVL—which
thatvoltage
know
across
each
the
and treat
itwe
differently.
(Ifvoltage
a circuit
has two
or more
2requires
" two meshes
sis applies
KVL—which
that requires
we know
the
acrossthe
each
i supermeshes that
i intersect,
20 +V +
4"
!do
they
should
combined
to4 periphery
form
a larger
branch—and
do
not
know
the
across
aadcurrent
source
in adi notwe
i
20 V we
4 " voltage
branch—and
know
the
voltage
across
a current
source
+ a in
ibe
i as
!
20
V
As shown in Fig. 3.23(b),
we
create
supermesh
the
of
"
+
i
i
!
20 V !
4"
6
A
supermesh.)
Why
treat
the
supermesh
differently?
Because
mesh
analyvance. vance.
However,However,
a supermesh
satisfy
any other
mesh.
a must
supermesh
mustlike
KVL
any other
mesh.
6 A two
the
meshes
andKVL
treat
itsatisfy
differently.
(Iflike
a circuit
has two
or more
Therefore,
applying KVL
the KVL
supermesh
in Fig.
3.23(b)
gives
sistoapplies
KVL—which
requires
that
we
know
thegives
voltage
across
each
Therefore,
applying
to
the
supermesh
in
Fig.
3.23(b)
supermeshes
that
intersect,
they
should
be
combined
to
form
a
larger
i
i
0
1
1
2
2
1
1
2
2
0
2
1branch—and
(b)
we
do these
not
know
the voltage
a currentmesh
source
in adExclude
(b) across
supermesh.)
Why
treat
the
supermesh
differently?
Because
analyExclude
these
−20
+ 6i(a)1 + 10i
+
4i
=
0
2elements
(a)
vance.
However,
satisfy
like anyacross
other each
mesh.
−20
+ elements
6ia21 supermesh
+ 10i
4i
0knowKVL
2 + must
2 =
sis
applies
KVL—which
requires
that
we
the voltage
or we Figure
Therefore,
supermesh
intheFig.
3.23(b)
gives in adFigure
3.23(a) Two
(a) Two
meshes
having
a applying
current
inKVL
common,
(b)the
a supermesh,
by excluding
current
branch—and
we
do
not
know
voltage
across
a the
current
source
3.23 the
meshes
having
a current
sourcesource
in common,
(b) a to
supermesh,
createdcreated
by excluding
current
source. source.
Why
treat
supermesh
differently?
or
vance.
However,
a supermesh
mustknowing
satisfy KVL
like any across
other mesh.
Because
mesh
analysis
the
voltage
shown
in Fig.
3.23(b),
we
aKVL,
as−20
the requires
periphery
of
+a 14i
= 20aswhich
(3.18)
6i1create
As As
shown
in Fig.
3.23(b),
we applies
create
supermesh
the
periphery
of
2supermesh
+
6isupermesh
4i2Fig.
= 03.23(b) gives
1 + 10i2 +in
Therefore,
applying
KVL
to
the
the
two
meshes
and
treat
it
differently.
(If
a
circuit
has
two
or
more
the
two
meshes
and
treat
it
differently.
(If
a
circuit
has
two
or
more
each branch, but the voltage across 6i
a current
source
is not known
+
14i
=
20
(3.18)
1
2
supermeshes
that
intersect,
they
should
be
combined
to
form
a
larger
supermeshes
that
intersect,
they
should
be
combined
to
form
a
larger
We apply KCL to a nodeorin the branch where the two meshes intersect.
−20
+mesh
6i1analy+ 10i2 + 4i2 = 0
Why
treat
the
differently?
Because
supermesh.)
Why
treat
the 0supermesh
differently?
Because
mesh
analy1.supermesh.)
Applying
KVL
to
the supermesh
Applying
KCL
to
node
insupermesh
Fig.
3.23(a)
gives
sis applies KVL—which requires that we know the voltage across each
i
1
i
2
sis applies
KVL—which
requires
we know
voltage
across each
We apply
KCL
to that
a node
inthethe
branch
where the two meshes intersect.
branch—and
we not
do not
the voltage
across
a current
source
branch—and
we do
know
the voltage
across
a current
source
in6i
ad14i2 =
20
(3.18)
1in+adorknow
=
i
+
6
(3.19)
i
Applying
KCL
to
node
0
in
Fig.
3.23(a)
gives
2
1
vance.
However,
a supermesh
mustmust
satisfy
KVLKVL
like any
mesh.mesh.
vance.
However,
a supermesh
satisfy
like other
any other
2.
Apply
KCL
to
a
node
in
the
branch
where
the two meshes intersect
Therefore,
applying
KVL
to the
supermesh
in Fig.
3.23(b)
givesgives
Therefore,
applying
KVL
to
the
supermesh
in
Fig.
3.23(b)
We
apply
KCL to a node in
the two meshes intersect.
+ branch
14i2 = where
20
(3.18)
6i1the
Solving Eqs.
(3.18) KCL
and (3.19),
we0get
Applying
to node
=
i
+
6
(3.19)
i
−20−20
+Applying
6i+
+KCL
4i
6i10i
4i02to
= node
02
1+
2 =
01in Fig. 3.23(a) gives
1 +2 10i
2+
3.
Solving
applyA,KCL ito
a node
= −3.2
2.8 Ain the branch where
(3.20) the two meshes intersect.
i1 We
2 =
Solving Eqs. Applying
(3.18) and
(3.19),
we0 get
KCL
to node
in Fig.i23.23(a)
= i1 +gives
6
(3.19)
14i
=
20
(3.18)
6iproperties
1+
2
Note
the
following
of
a
supermesh:
20supermesh:
(3.18)
6i1 + 14iof
2 =a
Note the following properties
+
6A
ii22 =
=where
−3.2
A,not
=i1get
2.8
(3.20)(3.19)
iEqs.
WeWe
apply
KCL
to ato
node
inSolving
the
where
the
two
intersect.
(3.18)
and
(3.19),
we
1supermesh
1.
The
current
in the
is
completely
ignored;
apply
KCL
asource
node
in branch
the
branch
the meshes
two
meshes
intersect.
1.Applying
TheKCL
current
source
in
the
supermesh
is not completely ignored; it provides the
Applying
to node
0 in 0Fig.
3.23(a)
gives
KCL
to node
in Fig.
3.23(a)
gives necessary to solve for the
it provides
the
constraint
equation
Solving
Eqs.
(3.18)
andi for
(3.19),
weA,get
constraint
equation
necessary
to solve
the(3.19)
mesh
currents
the
following
a
supermesh:
i2 = 2.8 A
(3.20)
+i6 + 6properties
i2 =i i1=
1 =of−3.2
meshNote
currents.
(3.19)
2
1
2. 2.AAEqs.
supermesh
has
no
of its
itsinown.
own
Solving
(3.18)
and and
(3.19),
wecurrent
get
−3.2 A, is inot
=completely
2.8 A
i1 =supermesh
has
no
current
of
1.(3.18)
The
current
source
the
ignored;(3.20)
Solvingsupermesh
Eqs.
(3.19),
we
get
Note the following
properties of a2 supermesh:
3. 3.AAsupermesh
ofequation
both(3.20)
KVLand
andKCL.
KCL to solve for the
=requires
−3.2 A, the
i2application
=constraint
2.8 A
iti1provides
the
necessary
supermesh
the
application
of
both
KVL
=
−3.2
A,
i
=
2.8
A
(3.20)
irequires
1
Note
the2 current
following
properties
a supermesh:
1. The
source
in the of
supermesh
is not completely ignored;
or or
2i
2
5A
5A
6"
6"
i2
i2
o
E X A M P L E 3 . 7
i
3io
3io 3 i3
94
8"
8"
i4
i4
io
+ 10 V
! + 10 V PART 1
!
DC Circuits
2"
For the circuit in Fig. 3.24,
Q find i1 to i4 using mesh analysis.
Q i3
i3
Solution:
i1
i1
Figure
For1Example
3.7. a supermesh since they have an independent
4"
2"
Note
that3.24
meshes
and 2 form
P
Figure
3.24
For Example 3.7.
current source in common. Also, meshes 2 and 3 form another supermesh i2
io
5A
because they have a dependent current source in common. The two
+ 10 V
i2
i3
i4
3io
6"
8"
!
because
they
have
a
dependent
current
source
in
common.
The
two
supermeshes intersect and form a larger supermesh as shown. Applying
supermeshes
intersect
and form a larger supermesh as shown. Applying
KVL
to the larger
supermesh,
KVL
largerMain
supermesh,
Q
Table of Contents |Problem Solvingi2Workbook
Contents
e-Text
| to the
2i1 +Menu
4i3 |+ 8(iTextbook
i3
3 − i4 ) + 6i2 = 0
2i1 + 4i3 + 8(i3 − i4 ) + 6i2 = 0
or
Figure 3.24 For Example 3.7.
or
i1 + 3i2 + 6i3 − 4i4 = 0
(3.7.1)
because they have a dependent current source in common. The two
i1 +
3i2 + we
6i3 apply
− 4i4KCL
= 0 to node P :
(3.7.1)
supermeshes intersect and form a larger supermesh as shown. Applying
For the independent current
source,
For the independent current source,
we5 apply KCL to node P : (3.7.2) KVL to the larger supermesh,
i2 = i1 +
2i1 + 4i3 + 8(i3 − i4 ) + 6i2 = 0
i
=
i
+
5
(3.7.2)
2
1
For the dependent current source, we apply KCL to node Q:
or
i1 + 3i2 + 6i3 − 4i4 = 0
(3.7.1)
For the dependent current source,
apply
KCL to node Q:
i2 = i3we
+ 3i
o
For the independent current source, we apply KCL to node P :
i2 = i3 + 3io
But io = −i4 , hence,
i2 = i1 + 5
(3.7.2)
But io = −i4 , hence,
i2 = i3 − 3i4
(3.7.3) For the dependent current source, we apply KCL to node Q:
i2 = i3 + 3io
i2 = i3 − 3i4
(3.7.3)
Applying KVL in mesh 4,
But io = −i4 , hence,
Applying KVL in mesh
2i44,+ 8(i4 − i3 ) + 10 = 0
i2 = i3 − 3i4
(3.7.3)
Applying KVL in mesh 4,
2i4 + 8(i4 − i3 ) + 10 = 0
or
2i4 + 8(i4 − i3 ) + 10 = 0
5i4 − 4i3 = −5
(3.7.4)
or
or
From Eqs. (3.7.1) to (3.7.4), 5i4 − 4i3 = −5
5i4 − 4i3 = −5
(3.7.4)
(3.7.4)
i1 = −7.5
A,
i2 (3.7.4),
= −2.5 A,
i3 = 3.93 A,
i4 = 2.143 A From Eqs. (3.7.1) to (3.7.4),
From
Eqs. (3.7.1)
to
i2
▲
▲
i2
. 7 i1 = −7.5 A,
i2 = −2.5 A,
i1 = −7.5 A,
i3P R=A 3.93
A,
i = 2.143 A
C T I C E P R O 4B L E M 3 . 7
i2 = −2.5 A,
i3 = 3.93 A,
i4 = 2.143 A
PRACTICE PROBLEM 3.7
2"
6V +
!
i1
3A
1"
|
2"
(3.7.4)
i4 = 2.143 A
Use mesh analysis to determine i1 , i2 , and i3 in Fig. 3.25.
Answer: i1 = 3.474 A, i2 = 0.4737 A, i3 = 1.1052 A.
4"
i2
For Practice Prob. 3.7.
▲
|
▲
Figure 3.25
i3
5i4 − 4i3 = −5
From Eqs. (3.7.1) to (3.7.4),
i1 = −7.5 A,
i2 = −2.5 A,
i3 = 3.93 A,
8"
@
e-Text Main Menu| Textbook Table of Contents |Problem Solving Workbook Contents
NODAL VERSUS MESH ANALYSIS
Nodal & mesh analyses provide a systematic way of analyzing a network
How do we know which method is better or more efficient?
The choice of the better method is dictated by two factors:
1. The first factor is the nature the network
— If networks contain series-connected elements, voltage sources, or supermeshes à
mesh analysis
— If networks have parallel- connected elements, current sources, or supernodes à
nodal analysis
— A circuit with fewer nodes than meshes à nodal analysis
— A circuit with fewer meshes than nodes à mesh analysis
— The second factor is the information required
— If node voltages are required à nodal analysis
— If branch or mesh currents are required à mesh analysis
— It is helpful to be familiar with both methods of analysis
— one method can be used to check the results from the other method
— since each method has its limitations, only one method may be suitable for a particular
problem
— For example
— Mesh analysis is the only method to use in analyzing transistor circuits
— Mesh analysis cannot easily be used to solve an op amp circuit because there is no direct
way to obtain the voltage across the op amp itself
— For nonplanar networks, nodal analysis is the only option, because mesh analysis only
applies to planar networks
Summary
1. Nodal analysis is the application of Kirchhoff’s current law at the nonreference
nodes. (It is applicable to both planar and nonplanar circuits.) We express the result
in terms of the node voltages. Solving the simultaneous equations yields the node
voltages.
2. A supernode consists of two nonreference nodes connected by a (dependent or
independent) voltage source.
3. Mesh analysis is the application of Kirchhoff’s voltage law around meshes in a
planar circuit. We express the result in terms of mesh currents. Solving the
simultaneous equations yields the mesh currents.
4. A supermesh consists of two meshes that have a (dependent or independent)
current source in common.
5. Nodal analysis is normally used when a circuit has fewer node equations than mesh
equations. Mesh analysis is normally used when a circuit has fewer mesh equations
than node equations.