Solution At first glance, the node-voltage method looks appealing, because we may define the unknown voltage as a node voltage by choosing the lower ter-minal of the dependent current so
Trang 1106 Techniques of Circuit Analysis
I / A S S E S S M E N T PROBLEMS
Objective 2—Understand and be able to use the mesh-current method
4.10 Use the mesh-current method to find the power
dissipated in the 2 ft resistor in the circuit shown
30 V
Answer: 72 W
4.11 Use the mesh-current method to find the mesh
current /a in the circuit shown
75 VI
Answer: 15 A
4.12 Use the mesh-current method to find the
power dissipated in the 1 ft resistor in the
cir-cuit shown
16 A
^vw 4 Wv 4
10 A
e
2 0
'a <\t, 1 5 n
in
• A ^ V
-2V*
2A
e
i o v T * j 2 a
Answer: 36 W
NOTE: Also try Chapter Problems 4.42, 4.44, 4.48, and 4.51
6 V
4,8 The Node-Voltage Method Versus
the Mesh-Current Method
The greatest advantage of both the node-voltage and mesh-current meth-ods is that they reduce the number of simultaneous equations that must be manipulated They also require the analyst to be quite systematic in terms
of organizing and writing these equations It is natural to ask, then, "When
is the node-voltage method preferred to the mesh-current method and vice versa?" As you might suspect, there is no clear-cut answer Asking a number of questions, however, may help you identify the more efficient method before plunging into the solution process:
• Does one of the methods result in fewer simultaneous equations
to solve?
• Does the circuit contain supernodes? If so, using the node-voltage method will permit you to reduce the number of equations to
be solved
Trang 24.8 The Node-Voltage Method Versus the Mesh-Current Method 107
• Does the circuit contain supermeshes? If so, using the mesh-current
method will permit you to reduce the number of equations to
be solved
• Will solving some portion of the circuit give the requested solution?
If so, which method is most efficient for solving just the pertinent
portion of the circuit?
Perhaps the most important observation is that, for any situation, some
time spent thinking about the problem in relation to the various analytical
approaches available is time well spent Examples 4.6 and 4.7 illustrate the
process of deciding between the node-voltage and mesh-current methods
Example 4.6 Understanding the Node-Voltage Method Versus Mesh-Current Method
Find the power dissipated in the 300 Ct resistor in
the circuit shown in Fig 4.29
300 ft J 1
Wv
150 ft 100ft
- A A A ^
250 ft -AVv t
50 / A
500 ft
256 V £200 ft 400 ft k 128 V
Figure 4.29 A The circuit for Example 4.6
Solution
To find the power dissipated in the 300 H resistor,
we need to find either the current in the resistor or
the voltage across it The mesh-current method
yields the current in the resistor; this approach
requires solving five simultaneous mesh equations,
as depicted in Fig 4.30 In writing the five
equa-tions, we must include the constraint /A = — i b
Before going further, let's also look at the circuit
in terms of the node-voltage method Note that, once
we know the node voltages, we can calculate either
the current in the 300 il resistor or the voltage across
it The circuit has four essential nodes, and therefore
only three node-voltage equations are required to
describe the circuit Because of the dependent
volt-age source between two essential nodes, we have to
sum the currents at only two nodes Hence the
prob-lem is reduced to writing two node-voltage equations
and a constraint equation Because the node-voltage
method requires only three simultaneous equations,
it is the more attractive approach
Once the decision to use the node-voltage
method has been made, the next step is to select a
reference node Two essential nodes in the circuit in
Fig 4.29 merit consideration The first is the
refer-ence node in Fig 4.31 If this node is selected, one of
the unknown node voltages is the voltage across the
300 CL resistor, namely, v 2 in Fig 4.31 Once we know this voltage, we calculate the power in the
300 f! resistor by using the expression
Psaon = «1/300
300 ft -*'A
- A A A
150ft 100ft 250 ft
A A A — f — W V
-500 ft
: f20uftSO50*A)f
.+
Figure 4.30 A The circuit shown in Fig 4.29, with the five
mesh currents
300 ft J s
vvV
400 ft £ 128 V
Figure 4.31 A The circuit shown in Fig 4.29, with a
reference node
Note that, in addition to selecting the reference
node, we defined the three node voltages V\, v 2 , and
u3 and indicated that nodes 1 and 3 form a super-node, because they are connected by a dependent voltage source It is understood that a node voltage is
a rise from the reference node; therefore, in Fig 4.31,
we have not placed the node voltage polarity refer-ences on the circuit diagram
Trang 3108 Techniques of Circuit Analysis
The second node that merits consideration as
the reference node is the lower node in the circuit,
as shown in Fig 4.32 It is attractive because it has
the most branches connected to it, and the
node-voltage equations are thus easier to write However,
to find either the current in the 300 11 resistor or
the voltage across it requires an additional
calcula-tion once we know the node voltages v a and v c For
example, the current in the 300 H resistor is
(v c - va)/300, whereas the voltage across the
resis-tor is v r - v»
300 ft l*
128 V
Figure 4.32 A The circuit shown in Fig 4.29 with an
alternative reference node
t? 3 + 256
+ — = 0
150 Att? 2 ,
v 2 v 2 - Vi v 2 - VT VI + 128 - v-x ' + _•_ + + — — = 0
From the supernode, the constraint equation is
v 2
v 3 = vi - 50/A = v, - —
Set 2 (Fig 4.32)
A t % ,
v a v a - 256 v d - v b v a - v c _
200 150
A t vc,
We compare these two possible reference nodes
by means of the following sets of equations The first
set pertains to the circuit shown in Fig 4.31, and the
second set is based on the circuit shown in Fig 4.32
Set 1 (Fig 4.31)
At the supernode,
V\ V\ — v 2 v 3 V3 — v 2 v 3 — (v 2 + 128)
KX)+ 250 +2 0 0+ 400 500
v c v c + 128 v c — v b v c - v a
400 500 250 300
From the supernode, the constraint equation is
50(vc - v a ) v c - v a
v b = 50/A
300
You should verify that the solution of either set leads to a power calculation of 16.57 W dissipated in
the 300 O, resistor
Example 4.7 Companng the Node-Voltage and Mesh-Current Methods
Find the voltage v 0 in the circuit shown in Fig 4.33
Solution
At first glance, the node-voltage method looks
appealing, because we may define the unknown
voltage as a node voltage by choosing the lower
ter-minal of the dependent current source as the
refer-ence node The circuit has four essential nodes and
two voltage-controlled dependent sources, so the
node-voltage method requires manipulation of
three node-voltage equations and two constraint
equations
Let's now turn to the mesh-current method for
finding v 0 The circuit contains three meshes, and
we can use the leftmost one to calculate v 0 If we
let /a denote the leftmost mesh current, then
v 0 = 193 — 10/'a The presence of the two current sources reduces the problem to manipulating a sin-gle supermesh equation and two constraint equa-tions Hence the mesh-current method is the more attractive technique here
4ft
0.8¾
6 ft 7.5 ft 8ft
Figure 4.33 • The circuit for Example 4.7
Trang 44.9 Source Transformations 109
411 2.512
193 VI
6 n 7.5 n 8 a
Figure 4.34 A The circuit shown in Fig 4.33 with the three mesh currents
4 0 2.5 ft *« 2 ft
-'vw f
-vw-M l 9 3 V ^ / t > 0 4 ^ ( T ) o 5 A
and the constraint equations are
i\j — /a = 0.4i;A = 0.8/c;
v 9 = — 7.5/b; and /c — /b = 0.5
We use the constraint equations to write the super-mesh equation in terms of /a:
160 = 80*'a, or /a = 2 A,
v a = 193 - 20 = 173 V
The node-voltage equations are
0.8 v (l
7.5 ft "h
Figure 4.35 A The circuit shown in Fig 4.33 with node voltages
To help you compare the two approaches, we
summarize both methods.The mesh-current
equa-tions are based on the circuit shown in Fig 4.34,
and the node-voltage equations are based on
the circuit shown in Fig 4.35 The supermesh
equation is
193 = 104 + 1 0'b + 10/'c + 0.8v0,
v„ - 193
2.5 10
^b t 0 5 1 % + °' 8Ve ~ Va ~ 0
7.5 " 10 The constraint equations are
v 9 = -v b , v A = "«a - (v b + 0.8¾¾)1
We use the constraint equations to reduce the node-voltage equations to three simultaneous equations
involving v () , ua, and v b You should verify that the
node-voltage approach also gives v 0 = 173 V
^ASSESSMENT PROBLEMS
Objective 3—Deciding between the node-voltage and mesh-current methods
4.13 Find the power delivered by the 2 A current
source in the circuit shown
4.14 Find the power delivered by the 4 A current
source in the circuit shown
4A
20 V
128 V
Answer: 70 W
NOTE: Also try Chapter Problems 4.52 and 4.53
Answer: 40 W
4.9 Source Transformations
Even though the node-voltage and mesh-current methods are powerful
tech-niques for solving circuits, we are still interested in methods that can be used
to simplify circuits Series-parallel reductions and A-to-Y transformations are
Trang 51 1 0 Techniques of Circuit Analysis
-•a
-•b
(b)
Figure 4.36 A Source transformations
already on our list of simplifying techniques We begin expanding this list
with source transformations A source transformation, shown in Fig 4.36,
allows a voltage source in series with a resistor to be replaced by a current source in parallel with the same resistor or vice versa The double-headed arrow emphasizes that a source transformation is bilateral; that is, we can start with either configuration and derive the other
We need to find the relationship between v s and i s that guarantees the two configurations in Fig 4.36 are equivalent with respect to nodes a,b
Equivalence is achieved if any resistor R L experiences the same current flow, and thus the same voltage drop, whether connected between nodes a,b in Fig 4.36(a) or Fig 4.36(b)
Suppose R f is connected between nodes a,b in Fig 4.36(a) Using
Ohm's law, the current in R L is
R + R L
Now suppose the same resistor R L is connected between nodes a,b in
Fig 4.36(b) Using current division, the current in R, is
R
if the two circuits in Fig 4.36 are equivalent, these resistor currents must be the same Equating the right-hand sides of Eqs 4.52 and 4.53 and simplifying
(4.54)
When Eq 4.54 is satisfied for the circuits in Fig 4.36, the current in R L is
the same for both circuits in the figure for all values of R L If the current
through R L is the same in both circuits, then the voltage drop across R { is the same in both circuits, and the circuits are equivalent at nodes a,b
If the polarity of v s is reversed, the orientation of i s must be reversed
to maintain equivalence
Example 4.8 illustrates the usefulness of making source transforma-tions to simplify a circuit-analysis problem
Example 4.8 Using Source Transformations to Solve a Circuit
a) For the circuit shown in Fig 4.37, find the power
associated with the 6 V source
b) State whether the 6 V source is absorbing or
delivering the power calculated in (a)
Solution
a) If we study the circuit shown in Fig 4.37,
know-ing that the power associated with the 6 V
source is of interest, several approaches come
to mind The circuit has four essential nodes
and six essential branches where the current is
unknown Thus we can find the current in the
branch containing the 6 V source by solving
either three [ 6 - ( 4 - 1 ) ] mesh-current
equa-tions or three [ 4 - 1 ] node-voltage equaequa-tions
Choosing the mesh-current approach involves
6 V §30 n ?20£l 40 V
Figure 4.37 • The circuit for Example 4.8
solving for the mesh current that corresponds
to the branch current in the 6 V source Choosing the node-voltage approach involves solving for the voltage across the 30 O resistor, from which the branch current in the 6 V source can be calculated But by focusing on just one branch current, we can first simplify the circuit by using source transformations
Trang 64.9 Source Transformations 111
We must reduce the circuit in a way that
pre-serves the identity of the branch containing the 6 V
source We have no reason to preserve the identity of
the branch containing the 40 V source Beginning with
this branch, we can transform the 40 V source in series with the 5 ft resistor into an 8 A current source in parallel with a 5 f i resistor, as shown
in Fig 4.38(a)
4 n 6 ii
-f 'WV
32 V
(a) First step
4 11
(b) Second step
412 12 O
19.2 V
(c) Third step Figure 4.38 A Step-by-step simplification of the circuit shown in Fig 4.37
Next, we can replace the parallel combination of
the 20 ft and 5 ft resistors with a 4 ft resistor
This 4 ft resistor is in parallel with the 8 A source
and therefore can be replaced with a 32 V source
in series with a 4 ft resistor, as shown in
Fig 4.38(b).The 32 V source is in series with 20 ft
of resistance and, hence, can be replaced by a
cur-rent source of 1.6 A in parallel with 20 ft, as shown
in Fig 4.38(c) The 20 ft and 30 ft parallel
resis-tors can be reduced to a single 12 ft resistor The
parallel combination of the 1.6 A current source
(d) Fourth step
and the 12 ft resistor transforms into a voltage source of 19.2 V in series with 12 ft Figure 4.38(d) shows the result of this last transformation The current in the direction of the voltage drop across the 6 V source is (19.2 - 6)/16, or 0.825 A Therefore the power associated with the 6 V source is
p6 V = (0.825)(6) = 4.95 W
b) The voltage source is absorbing power
A question that arises from use of the source transformation depicted
in Fig 4.38 is, "What happens if there is a resistance R p in parallel with the
voltage source or a resistance R s in series with the current source?" In
both cases, the resistance has no effect on the equivalent circuit that
pre-dicts behavior with respect to terminals a,b Figure 4.39 summarizes this
observation
The two circuits depicted in Fig 4.39(a) are equivalent with respect to
terminals a,b because they produce the same voltage and current in any
resistor R L inserted between nodes a,b The same can be said for the
cir-cuits in Fig 4.39(b) Example 4.9 illustrates an application of the
equiva-lent circuits depicted in Fig 4.39
R
-wv—»a
»« \R,
-•b (a)
(b)
Figure 4.39 • Equivalent circuits containing a
resistance in parallel with a voltage source or in series with a current source
Trang 7112 Techniques of Circuit Analysis
Example 4.9 Using Special Source Transformation Techniques
a) Use source transformations to find the voltage
v () in the circuit shown in Fig 4.40
b) Find the power developed by the 250 V voltage
source
c) Find the power developed by the 8 A current
source
b) The current supplied by the 250 V source equals the current in the 125 ft resistor plus the current in the
25 ft resistor Thus
250 250 - 20 „ „
'< = l 2 5+- ^ -= 1 U A
-25 ft
250 V
Figure 4.40 A The circuit for Example 4.9
Solution
Therefore the power developed by the voltage source is
/>25()v(developed) = (250)(11.2) = 2800 W
c) To find the power developed by the 8 A current source,
we first find the voltage across the source If we let v s
represent the voltage across the source, positive at the upper terminal of the source, we obtain
a) We begin by removing the 125 ft and 10 ft
resis-tors, because the 125 ft resistor is connected across
the 250 V voltage source and the 10 ft resistor is
connected in series with the 8 A current source We
also combine the series-connected resistors into a
single resistance of 20 ft Figure 4.41 shows the
sim-plified circuit
v s + 8(10) = v 0 = 20, or v s = - 6 0 V,
and the power developed by the 8 A source is 480 W Note that the 125 ft and 10 ft resistors do not affect
the value of v 0 but do affect the power calculations
25 ft
250 V
Figure 4.41 • A simplified version of the circuit shown in Fig 4.40
Figure 4.42 • The circuit shown in Fig 4.41 after a source
transformation
We now use a source transformation to replace
the 250 V source and 25 ft resistor with a 10 A
source in parallel with the 25 ft resistor, as shown in
Fig 4.42 We can now simplify the circuit shown in
Fig 4.42 by using Kirchhoffs current law to
com-bine the parallel current sources into a single
source The parallel resistors combine into a single
resistor Figure 4.43 shows the result Hence
v„ = 20 V
+ D„5 io ft
Figure 4.43 A The circuit shown in Fig 4.42 after combining
sources and resistors
Trang 84.10 Thevenin and Norton Equivalents 113
/ " A S S E S S M E N T PROBLEM
Objective 4—Understand source transformation
4.15 a) Use a series of source transformations to
find the voltage v in the circuit shown
b) How much power does the 120 V source
deliver to the circuit?
Answer: (a) 48 V;
(b) 374.4 W
NOTE: Also try Chapter Problems 4.59 and 4.60
- r ( t ) 3 6 A
1120 V L
1.6 a
4.10 Thevenin and Norton Equivalents
At times in circuit analysis, we want to concentrate on what happens at
a specific pair of terminals For example, when we plug a toaster into an
outlet, we are interested primarily in the voltage and current at the
ter-minals of the toaster We have little or no interest in the effect that
con-necting the toaster has on voltages or currents elsewhere in the circuit
supplying the outlet We can expand this interest in terminal behavior
to a set of appliances, each requiring a different amount of power
We then are interested in how the voltage and current delivered at the
outlet change as we change appliances In other words, we want to focus
on the behavior of the circuit supplying the outlet, but only at the
out-let terminals
Thevenin and Norton equivalents are circuit simplification techniques
that focus on terminal behavior and thus are extremely valuable aids in
analysis Although here we discuss them as they pertain to resistive
cir-cuits, Thevenin and Norton equivalent circuits may be used to represent
any circuit made up of linear elements
We can best describe a Thevenin equivalent circuit by reference to
Fig 4.44, which represents any circuit made up of sources (both
inde-pendent and deinde-pendent) and resistors The letters a and b denote the
pair of terminals of interest Figure 4.44(b) shows the Thevenin
equiva-lent Thus, a Thevenin equivalent circuit is an independent voltage
source V Th in series with a resistor R Th , which replaces an
interconnec-tion of sources and resistors This series combinainterconnec-tion of VTh and R T h is
equivalent to the original circuit in the sense that, if we connect the
same load across the terminals a,b of each circuit, we get the same
volt-age and current at the terminals of the load This equivalence holds for
all possible values of load resistance
To represent the original circuit by its Thevenin equivalent, we must
be able to determine the Thevenin voltage V Jh and the Thevenin
resist-ance R lh First, we note that if the load resistance is infinitely large, we
have an open-circuit condition The open-circuit voltage at the terminals
a,b in the circuit shown in Fig 4.44(b) is Vj h By hypothesis, this must be
• a
A resistive network containing independent and dependent sources
Figure 4.44 A (a) A general circuit, (b) The Thevenin
equivalent circuit
Trang 9114 Techniques of Circuit Analysis
the same as the open-circuit voltage at the terminals a,b in the original circuit Therefore, to calculate the Thevenin voltage VTh, we simply calcu-late the open-circuit voltage in the original circuit
Reducing the load resistance to zero gives us a short-circuit condition
If we place a short circuit across the terminals a,b of the Thevenin equiva-lent circuit, the short-circuit current directed from a to b is
/„- =
Tli
Th
(4.55)
By hypothesis, this short-circuit current must be identical to the short-circuit current that exists in a short circuit placed across the terminals a,b of the original network From Eq 4.55,
Thus the Thevenin resistance is the ratio of the open-circuit voltage to the short-circuit current
4 0
^vw—• a + +
2sy Cz) 2 °^f 3 A 0 ) V] ^
Finding a Thevenin Equivalent
To find the Tlievenin equivalent of the circuit shown in Fig 4.45, we first
calculate the open-circuit voltage of v db Note that when the terminals a,b
are open, there is no current in the 4 O resistor Therefore the open-circuit
voltage v db is identical to the voltage across the 3 A current source, labeled
V\ We find the voltage by solving a single node-voltage equation
Choosing the lower node as the reference node, we get
Figure 4.45 • A circuit used to illustrate a Thevenin
equivalent
Vi - 2 5 V]
— + — - 3 = 0
Solving for V\ yields
25 V
Figure 4.46 • The circuit shown in Fig 4.45 with
terminals a and b short-circuited
Hence the Thevenin voltage for the circuit is 32 V
The next step is to place a short circuit across the terminals and calcu-late the resulting short-circuit current Figure 4.46 shows the circuit with the short in place Note that the short-circuit current is in the direction of the open-circuit voltage drop across the terminals a,b If the short-circuit current is in the direction of the open-circuit voltage rise across the termi-nals, a minus sign must be inserted in Eq 4.56
The short-circuit current (/sc) is found easily once v 2 is known Therefore
the problem reduces to finding v 2 with the short in place Again, if we use the
lower node as the reference node, the equation for v 2 becomes
v 2 - 25 ih v-y
(4.59)
Trang 104.10 Thevenin and Norton Equivalents 115
Solving E q 4.59 for v2 gives
Hence, the short-circuit current is
1 6 A A
<ac = "J" = 4 A- (4.61)
We now find the Thevenin resistance by substituting the numerical results
from Eqs 4.58 and 4.61 into Eq 4.56:
R Th = V Th 32
Figure 4.47 shows the Thevenin equivalent for the circuit shown in Fig 4.45
You should verify that, if a 24 Cl resistor is connected across the
ter-minals a,b in Fig 4.45, the voltage across the resistor will be 24 V and
the current in the resistor will be 1 A, as would be the case with the
Thevenin circuit in Fig 4.47 This same equivalence between the circuit
in Figs 4.45 and 4.47 holds for any resistor value connected between
nodes a.b
The Norton Equivalent
A Norton equivalent circuit consists of an independent current source
in parallel with the Norton equivalent resistance, We can derive it from
a Thevenin equivalent circuit simply by making a source
transforma-tion Thus the Norton current equals the short-circuit current at the
terminals of interest, and the Norton resistance is identical to the
Thevenin resistance
32 V
SO
'VW • a
Figure 4.47 • The Thevenin equivalent of the circuit
shown in Fig 4.45
4 O
- w v — • a
J25V J 2 0 I 1 ( f J 3 A
- • b
Step 1:
Source transformation t
Step 2:
Parallel sources and parallel resistors combined •
4ft
•—-Wv—e a
8A 4 0
- • b
Step 3:
Source transformation; series resistors combined, producing the Thevenin equivalent circuit
8ft
-VW-32 V
Using Source Transformations
Sometimes we can make effective use of source transformations to
derive a Thevenin or Norton equivalent circuit For example, we can
derive the Thevenin and Norton equivalents of the circuit shown
in Fig 4.45 by making the series of source transformations shown in
Fig 4.48 This technique is most useful when the network contains only
independent sources The presence of dependent sources requires
retaining the identity of the controlling voltages and/or currents, and
this constraint usually prohibits continued reduction of the circuit
by source transformations We discuss the problem of finding the
Thevenin equivalent when a circuit contains dependent sources in
Example 4.10
Step 4:
Source transformation, producing the Norton equivalent circuit
4A 8 O
Figure 4.48 • Step-by-step derivation of the Thevenin and Norton equivalents of the circuit shown in Fig 4.45