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4.10 Thevenin and Norton Equivalents
Thevenin and Norton equivalents are circuit
simplifications techniques that focus on terminal
behavior. We can best describe a Thevenin
equivalent circuit by reference to Fig. 4.44, which
represents any circuit made up of sources
(both independent and dependent) and resistors.
The letters a and b denote the pair of terminals of interest.
Figure 4.44(b) shows the Thevenin equivalent. Thus, a Thevenin equivalent circuit
is an independent voltage source VTh in series with a resistor RTh, which replaces an
interconnection of sources and resistors. This series combination of VTh and RTh 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 voltage and current at the terminals
of the load.
Finding a Thevenin Equivalent
1)Calculate the open-circuit voltage as inFig. 4.45 which is equal to VTh.
.
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2)Calculate the short-circuit current as inFig. 4.46.
3)Calculate the Thevenin resistance which is the
ratio of the open-circuit voltage to the short-
circuit current as in Fig. 4.47.
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 transformation. Thus the Norton current
equals the short-circuit current at the terminals of interest, and the Norton resistance
is identical to the Thevenin resistance.
.
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Using Source Transformations (independent sources):
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.
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Finding the Thevenin Equivalent of a Circuit with a Dependent Source
Example 4.10:
Find the Thevenin equivalent for the
circuit containing dependent sources
shown in Fig. 4.49.
Solution:
1.Calculating VTh(Open-Circuit):
Applying KVL at left loop:
Applying KVL at right loop:
Solving, 2.
Calculating (Short-Circuit):Applying KVL for the outer right loop (with no resistor, no voltage drop)
3.Calculating :
.
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The Deactivation Method (for independent Sources Only)
The technique for determining RTh that
we discussed and illustrated earlier is not
always the easiest method available. Two other
methods are generally simpler to use.
The first is useful if the network contains only
independent sources. To calculate RThfor such
a network, we first deactivate all independent
sources and then calculate the resistance seen
looking into the network at the terminal pair. A
voltage source is deactivated by replacing it
with a short circuit. A current source is
deactivated by replacing it with a short circuit. For example, consider the circuit
shown in Fig. 4.52. Deactivating the independent sources simplifies the circuit to the
one shown in Fig. 4.53. The resistance seen looking into the terminals a, b is denoted
Rab, which consists of the 4 resistor in series with the parallel combinations of the
5 and 20 resistors. Thus,
4.11 The Test Method (for Independent and Dependent Sources)
If the circuit or network contains dependent and independent sources, an
alternativeprocedure for finding the Thevenin resistance RThis as follows. We first
deactivate all independent sources, and we then apply either a test voltage source or a
test current source to the Thevenin terminals a, b. The Thevenin resistance equals the
ratio of the voltage across the test sources to the current delivered by the test source.
.
Figure 4.52A circuit used to illustrate
a Thvenin equivalent
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Finding the Thevenin Equivalent Using a Test Source
Example 4.11
Find the Thevenin resistance RTh for the circuit in Fig. 4.49, using the alternative
method described.
Solution
We first deactivate the independent voltage source from the circuit and then excite
the circuit from the terminals a, b with either a test voltage source or a test current
source. If we apply a test voltage source, we will know the voltage of the dependent
voltage source and hence the controlling current i. Therefore we opt for the rest
voltage source.
The externally applied test voltage source is denoted , and the current that itdelivers to the circuit is labeled . To find the Thevenin resistance, we simply solvethe circuit for the ratio of the voltage to the current at the test source; that is,
.
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In general, these computations are easier than those involved in computing the
short-circuit current. Moreover, in a network containing only resistors and
independent sources, you must use the alternative method because the ratio of the
Thevenin voltage to the short-circuit current is indeterminate. That is, it is the
ratio 0/0.
(Thevenin Method)
Method Indep. Only Dep. Only Indep + Dep.
Basic Method ---
Source Transformation --- ---
Deactivation --- ---
Test (alternative)
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Assessment Problem 4.16
Solution
To find RTh, replace the 72 V source with a short circuit:
The 5 and 20 resistors are in parallel, with
an equivalent resistance of 5 || 20 = 4 .
The equivalent 4 resistance is in series with
the 8 resistor for an equivalent resistance of
4 + 8 = 12 . The 12 equivalent resistance is
in parallel with the 12 resistor, so
Using node voltage analysis to find :The node voltage equations are:
Solving,
and
.
Find the Thvenin equivalent circuit with respect
to the terminals a, b for the circuit shown.
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Assessment Problem 4.17
Find the Norton equivalent circuit with respect
to the terminals a, b for the circuit shown.
Solution
We perform a source transformation, turning the parallel combination of the 15 A
source and 8 resistor into a series combination of a 120 V source and an 8
resistor. Next, combine the 2 , 8 and 10 resistors in series to give an equivalent20 resistance. Then transform the series combination of the 120 V source and the
20 equivalent resistance into a parallel combination of a 6 A source and a 20
resistor.
Finally, combine the 20 v and 12 parallel resistors to give RN= 20||12 = 7.5 .
Thus, the Norton equivalent circuit is the parallel combination of a 6 A source and a7.5 resistor.
.
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Assessment Problem 4.18
A voltmeter with an internal resistance of 100 k
is used to measure the voltage in the circuitshown. What is the voltmeter reading?
Solution
Using source transformations, convert the series
combination of the -36 V source and 12 k
resistor into a parallel combination of a -3 mA
source and 12 k resistor.
Combine the two parallel current sources and the
two parallel resistors to give a -3 + 18 = 15 mA
source in parallel with a 12 k || 60 k = 10 k
resistor. Transform the 15 mA source in parallel
with the 10 k resistor into a 150 V source in
series with a 10 k resistor, and combine this
10 k resistor in series with the 15 k resistor.
The Thevenin equivalent is thus a 150 V source in series with a 25 k resistor
Using voltage division:
.
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Assessment Problem 4.19
Find the Thvenin equivalent circuit with respect
to the terminals a, b for the circuit shown.
Solution
Calculating the open circuit voltage, which is also vTh,
Solving,
Using the test source method to calculate RTh,replace
the voltage source with a short circuit, the current
source with an open circuit:
Applying KCL equation at the middle node:
Solving,
The Thevenin equivalent is an 8 V source in series with a 1 resistor.
.
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Problem 4.63
Find the Thvenin equivalent with respect to the
terminals a, b for the circuit in Fig. P4.63.
Solution
Using deactivating method, we can calculate RTh
as follows:
Using voltage divider:
The final circuit of Thevenin is
.
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Problem 4.64
Find the Thvenin equivalent with respect to the
terminals a, b for the circuit in Fig. P4.64.
Solution
Applying node-voltage to get v1:
Using deactivation method to get :
The final circuit of Thevenin is
.
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Problem 4.65
Find the Thvenin equivalent with respect to the
terminals a, b for the circuit in Fig. P4.65.
Solution
Using source transformation, the circuit becomes:
Applying node-voltage method:
Solving,
and
Using deactivation method to get RTh:
The final Thevenin circuit is:
.
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Problem 4.66
Find the Norton equivalent with respect to the
terminals a, b in the circuit in Fig. P4.66.
Solution
The 8 mA current source and the 20 k resistor
will have no effect on the behavior of the circuit
with respect to the terminals a, b. This is because
they are in parallel with an ideal voltage source.
Which can be transformed to:
Which can be simplified to Norton equivalent:
.
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Problem 4.67
A voltmeter with a resistance of 100 is used tomeasure the voltage in the circuit in Fig. P4.67.
a)What is the voltmeter reading?
b)What is the percentage of error in the
voltmeter reading if the percentage of error is
defined as [(measured-actual)/actual] 100?
Solution:
a)
Using source transformation method, we can get through the following steps:
1.
Converting 30 V along with series 10 kresistor to 3 mA current source parallel to
10 kresistor, then combining 10 k|| 40 kto 8 k.
2.Converting 3 mA current source along with
parallel 8 k resistor to 24 V voltage source in
series along with 8 k resistor.
3.Converting 24 V voltage source in series
along with 12 k resistor to 2 mA current
source in parallel along with 12 k resistor,
and adding current sources together.
.
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Continue Problem 4.67
4.Combining (10 mA with 12 k) to (120 Vwith 12 k, adding (12 k with 3 k) to15 k, transferring (120 V with 15 k) to
(8 mA with 15 k), combining resulting
(15 k || 10 k) to 6 k, finally transferring
(8 mA and 6 k) to (48 V and 6 k).
b)
( )
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a)
Find the Thvenin equivalent with respect to
the terminals a, b for the circuit in Fig. P4.68
by finding the open-circuit voltage and the
short-circuit current.
b)
Solve for the Thvenin resistance by
removing the independent sources. Compare
your result to the Thvenin resistance found in (a)
Solution
a)
Open circuit:
Short circuit:
b)
.
8.4
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Problem 4.71
Determine the Thvenin equivalent with
respect to the terminals a, b the circuit shown
in Fig. P4.71.
Solution
Open circuit:
Applying Ohms law for right loop:
Applying KVL for middle loop:
Applying KCL at middle node:
Solving,
.Short circuit:
The final Thevenin circuit is:
.
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Problem 4.72
Find the Thvenin equivalent with respect to
the terminals a, b for the circuit seen in
Fig.P4.72
Solution
Open circuit:
Short circuit:
Applying mesh method to get isc
Solving,
.
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4.73 When a voltmeter is used to measure the
voltage in Fig. P.4.73, it reads 7.5V.a. What is the resistance of the voltmeter?
b.
What is the percentage of error in thevoltage measurement?
Solution
a)
Use source transformationsto simplify the left side of the circuit, as follows:
1. Transfer (16 V in series with 4 k) to (4 mA in series with 4 k).
3. Transfer (4 mA in parallel with 2.4 k) to (9.6 V in series with 2.4 k)
4. Add (9.6 V to 0.4 V) and (2.4 k to 0.1 k)to get the following shape:
Applying KVL for left loop:
b)
( )
.
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Problem 4.77
Find the Thvenin equivalent with respect to the
terminals a, b in the circuit in Fig. P4.77.
Solution
VTh = 0, since there are no independent sources,
we must apply the test method:
Solving,
The equivalent Thevenin circuit is:
Note: is zero since there are no independent sources.
.
1A
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Problem 4.78
Find the Thvenin equivalent with respect to the
terminals a, b for the circuit seen in Fig. P.4.78.
Solution:
= 0 since there are no independent resourcesin the circuit. Therefore, we must apply the test
methodas there are no independent resources.
Applying node-voltage method
Substituting in :
Substituting in :
1
2
3
3 2
4 1
4
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4.12 Maximum Power Transfer
Maximum power transfer can best be
described with the aid of the circuit shown in
Fig. 4.58. We assume a resistive network
containing independent and dependent
sources and a designated pair of terminals, a,
b to which a load, RLis to be connected.
The problem is to determine the value of RL
that permits maximum power delivery to RL.
The first step in this process is to recognize
that a resistive network can always be
replaced by its Thevenin equivalent.
Therefore, we redraw the circuit shown in
Fig. 4.58 as the one shown in Fig. 4.59.
Replacing the original network by its
Thevenin equivalent greatly simplifies the
task of finding RL.
Derivation of RL requires expressing the
power dissipated in RL as a function of the
three circuit parameters VTh,
RTh, and RL.
Thus,
( )
.
Fig. 4.59
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Next, we recognize that for a given circuit, VThand RThwill be fixed.
Therefore the power dissipated is a function of the single variable RL.
To find the value of RLthat maximizes the power, we use the elementary calculus.
We begin by writing an equation for the derivative of p with respect to RL:
[ ]
The derivative is zero and p is maximized when
Solving this equation yields
.
.
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Example 4.12: Calculating The Condition for Maximum Power Transfer
a)For the circuit shown, find the value of RL that
results in maximum power being transferred to RL.
b)Calculate the maximum power that can be
delivered to RL.
c)When RLis adjusted for maximum power transfer, what percentage of the power
delivered by the 360 V source reaches RL?
Solution
a)
Applying the node-voltage equation
Using Deactivation method:
c) When RLequals , the voltage is
.
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Assessment Problem 4.20
Find the Thvenin equivalent circuit with respect to
the terminals a, b for the circuit shown. (Hint: Define
the voltage at the leftmost node as
, and write two
nodal equations with as the right node voltage.)
Solution
Using the node voltage method
Solving with:
Using test source method to calculate the testcurrent and thus . Replace the current sourcewith a short circuit and apply the test source.
Applying KCL equation at the rightmost node:
Solving with:
Thus, the Thevenin equivalent is a 30 V source in series with a 10 resistor.
.
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Assessment Problem 4.21
a)Find the value or R that enables the
circuit shown to deliver maximum power
to the terminals a, b.
b)Find the maximum power delivered to R.
Solution
Applying node-voltage to get VTh:
Solving,
Creating a short circuit between nodes a and b and use the
mesh current method to get
Solving,
.
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Continue Assessment Problem 4.21
Thus,
a)
For maximum power transfer, b)
The Thevenin voltage,
, is divided equally between the Thevenin
resistance and the load resistance, so
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Assessment Problem 4.22
Assume that the circuit in Assessment problem 4.21
is delivering maximum power to the load resistor R.
a)How much power is the 100 V source
delivering to the network?
b)Repeat (a) for the dependent voltage source.
c)
What percentage of the total power generated
by these two sources is delivered to the load
resistor R?
Solution
According to Assessment 4.21, R = 3 :
Using mesh current method:
Solving,
a) b)
.
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Problem 4.79
The variable resistor (Ro) in the circuit in
Fig.P4.79 is adjusted until the power dissipated in
the resistor is 1.5 W. Find the values of Ro that
satisfy this condition.
Solution
We need to find the Thevenin equivalent with respect to .1)Transfer ( 100 V in series with 50 ) to ( 2 A in parallel with 50 ).
2)
50 || 200 = 40 .
3)Transfer ( 2A in parallel with 40 ) to ( 80 V in series with 40 ).
4)40 + 60 = 100 .
Applying KVL for the left loop:
Using the test-source methodto find the Thevenin
resistance gives
Using the node voltage method:
.
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Continue Problem 4.79
Solving,
Solving for :
( ) .
.
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Problem 4.80
The variable resistor (RL) in the circuit in
Fig.P4.80 is adjusted for maximum power
transfer to RL.
a)Find the numerical value of RL.
b)Find the maximum power transferred to RL.
Solution
a)
Finding the Thevenin equivalent:
Open circuit voltage:
Using mesh current method:
The dependent source constraint equation is:
Solving,
.
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Continue Problem 4.80:
Short-circuit current:
Using mesh current method:
The dependent source constraint equation is:
Solving,
b)
.
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Problem 4.81
The variable resistor in the circuit in Fig. P4.81 is
adjusted for maximum power transfer to Ro.
a)
Find the value of Ro.b)Find the maximum power that can be delivered to Ro.
Solution
a)
Finding Thevenin equivalent:
Open Circuit Voltage:
1)
Convert ( 9 mA in parallel with 2 k )
to ( 18 V in series with 2 k )
2)2 k + 4 k = 6 k
Applying node voltage method:
Solving,
Using deactivation method to calculate :
= b)
.
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Problem 4.86
The variable resistor (Ro) in the circuit in
Fig.P4.86 is adjusted for maximum power
transfer to Ro.
a)Find the value of Ro.
b)Find the maximum power that can be delivered to Ro.
Solution
a)
Finding the Thevenin equivalent:( Open Circuit Voltage ):
Using mesh current method:
Solving,
( Short Circuit Current ):
Using mesh current method:
.
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Continue Problem 4.86
Solving,
For maximum power transfer,
b)
.
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4.13 Superposition
Whenever we have more than one independent source, we can study the effect
of one-by-one source and add these effects together to result in the same result of the
overall system.
We demonstrate the superposition principle by
using it to find the branch currents in the circuit
shown in Fig. 4.62. We begin by finding the
branch currents resulting from the 120 V
voltage source.
To find the component of the branch currents
resulting from the current source, we deactivate
the ideal voltage source and solve the circuit
shown in Fig. 4.64.
.
Figure 4.63The circuit shown in Fig. 4.62
with the current source deactivated.
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The currents
and
in Fig. 4.62 are:
For circuits containing both independent and dependent sources, you must recognize
that the dependent sources are never deactivated.
.
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Example 4.13
Use the principle of superposition to find in thecircuit shown in Fig. 4.66
Solution
We begin by finding the component of resultingfrom the 10 V source. Fig. 4.67 shows the circuit.
With the 5 A source deactivated, must equal
. Hence,
must b zero, the branch
containing the two dependent sources is open, and
When the 10 V source is deactivated, the circuit
reduces to the one shown in Fig. 4.68. And applying
the node-voltage equations yield:
Solving with:
The value of is the sum of and , or 24 V
.
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Problem 4.91
a)Use the principle of superposition to find the
voltage in the circuit of Fig. P4.91.b)
Find the power dissipated in the 20 resistor.
Solution
a)
75 V source acting alone:
6 A source acting alone:
Our circuit reduces to
b)
.
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Problem 4.93
Use the principle of superposition to find the current io
in the circuit in Fig. P4.93.
Solution
45 V source acting alone:
10 V source acting alone:
8 A current source acting alone:
Transferring to resistors:
.
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Problem 4.94
Use the principle of superposition to find in thecircuit in Fig. P4.94.
Solution:
Voltage source acting alone:
Using node voltage method:
(
)
Current source acting alone:
Using node voltage method:
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