4 | Fundamentals of Electrical Circuits 1

UNIT 4: Circuit Theorems

4.0 Intended Learning Outcomes

a. Explain the following laws: Linearity Property; Superposition; Source
Transformation ; Thevenin’s Theorem , and Norton’s Theorem;
b. Apply the laws above in the circuit.

4.1 Introduction

The growth in areas of application of electric circuits has led to an evolution from
simple to complex circuits. To handle the complexity, engineers over the years
have developed some theorems to simplify circuit analysis. Such theorems include
Thevenin’s and Norton’s theorems. Since these theorems are applicable to linear
circuits, we first discuss the concept of circuit linearity. In addition to circuit
theorems, we discuss the concepts of superposition, source transformation, and
maximum power transfer in this module. Are you ready guys? Let’s go.

4.2 Linearity Property

Linearity is the property of an element describing a linear relationship between

cause and effect. Although the property applies to many circuit elements, we shall
limit its applicability to resistors in this chapter. The property is a combination of
both the homogeneity (scaling) property and the additivity property.

A linear circuit is one whose output is linearly related (or directly proportional) to
its input.

Example 4-1 for the circuit in Fig. 4.1, find io when vs = 12 V and vs = 24 V.

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Figure 4.1 for Example 4-1.

Find:
(a) io when vs= 12V and vs=24 V

Solution:

Applying KVL to the two loops , we obtain,

(a)
(b)

But vx = 2i1. Substitute in Equation (b) it becomes

(c)

Adding Eqs. (a) and (c) it becomes

Substituting this in Eq. (a), we get

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SAQ 4-1 For the circuit in Fig. 4.2, find vo when is = 15 and is = 30 A.

Figure 4.2 for SAQ 4-1.

ASAQ 4-1 10 V, 20 V.

4.3 SUPERPOSITION

If a circuit has two or more independent sources, one way to determine the value
of a specific variable (voltage or current) is to use nodal or mesh analysis as in
Chapter 3. Another way is to determine the contribution of each independent
source to the variable and then add them up. The latter approach is known as the
superposition.

The superposition principle states that the voltage across (or current through)
an element in a linear circuit is the algebraic sum of the voltages across (or
currents through) that element due to each independent source acting alone.

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The principle of superposition helps us to analyze a linear circuit with more than
one independent source by calculating the contribution of each independent
source separately. However, to apply the superposition principle, we must keep
two things in mind:

1. We consider one independent source at a time while all other independent

sources are turned off. This implies that we replace every voltage source by 0 V (or
a short circuit), and every current source by 0 A (or an open circuit). This way we
obtain a simpler and more manageable circuit.

2. Dependent sources are left intact because they are controlled by circuit
variables.

With these in mind, we apply the superposition principle in three steps:

StepstoApplySuperpositionPrinciple:

1. Turn off all independent sources except one source. Find the output (voltage or
current) due to that active source using nodal or mesh analysis.
2. Repeat step 1 for each of the other independent sources.
3. Find the total contribution by adding algebraically all the contributions due to
the independent sources.

Analyzing a circuit using superposition has one major disadvantage: it may very
likely involve more work. If the circuit has three independent sources, we may
have to analyze three simpler circuits each providing the contribution due to the
respective individual source. However, superposition does help reduce a complex
circuit to simpler circuits through replacement of voltage sources by short circuits
and of current sources by open circuits.

Keep in mind that superposition is based on linearity. For this reason, it is not
applicable to the effect on power due to each source, because the power absorbed
by a resistor depends on the square of the voltage or current. If the power value is
needed, the current through (or voltage across) the element must be calculated

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first using superposition.

Example 4-2 Use the superposition theorem to find v in the circuit in Fig. 4.3

Figure 4.3 for Example 4-2

Find:
(a) v

Solution:

Since there are two sources, let

v = v1 + v2

where v1 and v2 are the contributions due to the 6-V voltage source and the 3-A
current source, respectively. To obtain v1, we set the current source to zero, as
shown in Fig. 4.4(a). Applying KVL to the loop in Fig. 4.4(a) gives

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Figure 4.4 for Example 4.2 (a) solving for v1 (b) solving for v2

To get v2, we set the voltage source to zero, as in Fig. 4.4(b). Using current
division,

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ASAQ 4-2 Using the superposition theorem, find vo 3 Ω 5 Ω in the circuit in Fig.
4.5.

Figure 4.5 for SAQ 4-2

ASAQ 4-2 Answer: 12 V.

4.4 SOURCE TRANSFORMATION

We have noticed that series-parallel combination and wye-delta transformation help

simplify circuits. Source transformation is another tool for simplifying circuits. Basic to
these tools is the concept of equivalence.

We recall that an equivalent circuit is one whose v-i characteristics are identical with
the original circuit. In our previous discussion, we saw that node-voltage (or mesh-
current) equations can be obtained by mere inspection of a circuit when the sources
are all independent current (or all independent voltage) sources. It is therefore
expedient in circuit analysis to be able to substitute a voltage source in series with a
resistor for a current source in parallel with a resistor, or vice versa, as shown in Fig.
4.6. Either substitution is known as a source transformation.

Figure 4.6 Transformation of independent sources.

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A source transformation is the process of replacing a voltage source vs in series

with a resistor R by a current source is in parallel with a resistor R, or vice versa.

We should keep the following points in mind when dealing with source
transformation.

1. Note from Fig. 4.6 that the arrow of the current source is directed toward the
positive terminal of the voltage source.
2. Note from Equation,

That source transformation is not possible when R=0, which is the case with an
ideal voltage source. However, for a practical, non-ideal voltage source, R ≠ 0.
Similarly, an ideal current source with R =∞ cannot be replaced by a finite voltage
source.

Example 4.3 Use source transformation to find vo in the circuit in Fig. 4.7

Figure 4.7 for example 4.3

Find:
(a) vo

Solution:

We first transform the current and voltage sources to obtain the circuit in Fig. 4.8(a).
Combining the 4 ohm and 2 ohm resistors in series and transforming the 12-voltage

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source gives us Fig. 4.8(b). We now combine the 3 ohm and 6 ohm resistors in
parallel to get 2 ohm. We also combine the 2-A and 4-A current sources to get a 2-A
source. Thus, by repeatedly applying source transformations, we obtain the circuit in
Fig. 4.8(c).

Figure 4.8 for example 4.3

We use current division in Fig. 4.8(c) to get

And

Alternatively, since the 8 ohm and 2 ohm resistors in Fig. 4.8(c) are in parallel, they
have the same voltage vo across them. Hence,

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SAQ 4-3 Find io in the circuit of Fig. 4.9 using source transformation.

Figure 4.9 for SAQ 4-3

ASAQ 4-3 Answer: 1.78 A.

4.5 THEVENIN’S THEOREM

It often occurs in practice that a particular element in a circuit is variable (usually

called the load) while other elements are fixed. As a typical example, a household
outlet terminal may be connected to different appliances constituting a variable
load. Each time the variable element is changed, the entire circuit has to be
analyzed all over again. To avoid this problem, Thevenin’s theorem provides a
technique by which the fixed part of the circuit is replaced by an equivalent circuit.

According to Thevenin’s theorem, the linear circuit in Fig. 4.10(a) can be replaced
by that in Fig. 4.10(b). (The load in Fig. 4.10 may be a single resistor or another
circuit.) The circuit to the left of the terminals a-b in Fig. 4.10(b) is known as the
Thevenin equivalent circuit; it was developed in 1883 by M. Leon Thevenin (1857–
1926), a French telegraph engineer.

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Figure 4.10 Replacing a linear two-terminal circuit by its Thevenin equivalent: (a)
original circuit, (b) the Thevenin equivalent circuit.

Thevenin’s theorem states that a linear two-terminal circuit can be replaced by an

equivalent circuit consisting of a voltage source VTh in series with a resistor RTh,
where VTh is the open-circuit voltage at the terminals and RTh is the input or
equivalent resistance at the terminals when the independent sources are turned off.

Example 4. 4 Find the Thevenin equivalent circuit of the circuit shown in Fig. 4.11,
to the left of the terminals a-b. Then find the current through RL = 6, 16, and 36Ω .

Figure 4.11 for Example 4.4

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Find:
(a) Thevennin’s equivalent circuit

Solution:

We find RTh by turning off the 32-V voltage source (replacing it with a short circuit)
and the 2-A current source (replacing it with an open circuit). The circuit becomes
what is shown in Fig. 4.12(a). Thus,

Figure 4.12 For Example 4.4: (a) finding RTh, (b) finding VTh.

To find VTh, consider the circuit in Fig. 4.12(b). Applying mesh analysis to the two
loops, we obtain

Solving for i1, we get i1 = 0.5 A. Thus,

Alternatively, it is even easier to use nodal analysis. We ignore the 1 Ω resistor since
no current flows through it. At the top node, KCL gives

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Or

as obtained before. We could also use source transformation to find VTh. The
Thevenin equivalent circuit is shown in Fig. 4.13. The current through RL is

The equivalent circuit is,

Figure 4.13 The Thevenin equivalent circuit for Example 4.4

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SAQ 4-4 Using Thevenin’s theorem, find the equivalent circuit to the left of the
terminals in the circuit in Fig. 4.14. Then find i.

Figure 4.14 for SAQ 4-4

ASAQ 4-4

4.6 NORTON’S THEOREM

In 1926, about 43 years after Thevenin published his theorem, E. L. Norton, an

American engineer at Bell Telephone Laboratories, proposed a similar theorem.

Norton’s theorem states that a linear two-terminal circuit can be replaced by an

equivalent circuit consisting of a current source IN in parallel with a resistor RN,
where IN is the short-circuit current through the terminals and RN is the input or
equivalent resistance at the terminals when the independent sources are turned off.

Thus, the circuit in Fig. 4.15(a) can be replaced by the one in Fig. 4.15(b).

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Figure 4.15 (a) Original circuit, (b) Norton equivalent circuit.

For now, we are mainly concerned with how to get RN and IN. We find RN in the
same way we find RTh. In fact, from what we know about source transformation, the
Thevenin and Norton resistances are equal; that is,

To find the Norton current IN, we determine the short-circuit current flowing from
terminal a to b in both circuits in Fig. 4.15. It is evident that the short-circuit current
in Fig. 4.15(b) is IN. This must be the same short-circuit current from terminal a to b
in Fig. 4.15(a), since the two circuits are equivalent. Thus,

shown in Fig. 4.16. Dependent and independent sources are treated the same way as
in Thevenin’s theorem.

Figure 4.16 Finding Norton current IN.

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As shown in Fig. 4.16. Dependent and independent sources are treated the same way
as in Thevenin’s theorem.

Observe the close relationship between Norton’s and Thevenin’s theorems: RN = RTH
And,

Since VTh, IN, and RTh are related, to determine the Thevenin or Norton equivalent
circuit requires that we find:

We can calculate any two of the three using the method that takes the least effort
and use them to get the third using Ohm’s law. We can have

Example 4.5 Find the Norton equivalent circuit of the circuit in Fig. 4.17

Figure 4.17 for example 4.5

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Find:
(a) Norton’s equivalent circuit

Solution:

We find RN in the same way we find RTh in the Thevenin equivalent circuit.
Set the independent sources equal to zero. This leads to the circuit in Fig. 4.18(a),
from which we find RN. Thus,

To find IN, we short-circuit terminals a and b, as shown in Fig. 4.18(b). We ignore the
5Ω resistor because it has been short-circuited. Applying mesh analysis, we obtain

Alternatively, we may determine IN from VTh/RTh. We obtain VTh as the open-circuit

voltage across terminals a and b in Fig. 4.18(c). Using mesh analysis, we obtain

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Figure 4.18 for example 4.5 ; finding: (a) RN, (b) IN = Isc, (c) VTh = Voc.

Since,

RTh = Voc/Isc = 4/1 = 4 Ω. Thus, the Norton equivalent circuit is as shown in figure
4.19.

Figure 4.19 for example 4.5 Norton equivalent circuits for figure 4.17

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SAQ 4-5 Find the Norton equivalent circuit for the circuit in Fig. 4.20.

Figure 4.20 for SAQ 4-5

ASAQ 4-5

4.7 Maximum Power Transfer

In many practical situations, a circuit is designed to provide power to a load. While

for electric utilities, minimizing power losses in the process of transmission and
distribution is critical for efficiency and economic reasons, there are other
applications in areas such as communications where it is desirable to maximize the
power delivered to a load. We now address the problem of delivering the maximum
power to a load when given a system with known internal losses. It should be noted
that this will result in significant internal losses greater than or equal to the power
delivered to the load.

The Thevenin equivalent is useful in finding the maximum power a linear circuit can
deliver to a load. We assume that we can adjust the load resistance RL. If the entire
circuit is replaced by its Thevenin equivalent except for the load, as shown in Fig.
4.21, the power delivered to the load is,

(a)

For a given circuit, VTh and RTh are fixed. By varying the load resistance RL, the

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power delivered to the load varies as sketched in Fig. 4.22. We notice from Fig. 4.22
that the power is small for small or large values of RL but maximum for some value
of RL between 0 and ∞. We now want to show that this maximum power occurs
when RL is equal to RTh. This is known as the maximum power theorem.

Figure 4.21 The circuit used for maximum power transfer

Figure 4.22 Power delivered to the load as a function of RL.

Maximum power is transferred to the load when the load resistance equals the
Thevenin resistance as seen from the load (RL = RTh).

To prove the maximum power transfer theorem, we differentiate p in Eq. (a) with
respect to RL and set the result equal to zero. We obtain

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(b)

(c)

showing that the maximum power transfer takes place when the load resistance RL
equals the Thevenin resistance RTh. We can readily confirm that Eq. (c) gives the
maximum power by showing that d2p /dR2L < 0.

The maximum power transferred is obtained by substituting Eq. (c) into Eq. (a), for

(d)

Equation (d) applies only when RL = RTh. When RL ≠ RTh, we compute the power
delivered to the load using Eq. (a).

Example 4.6 Find the value of RL for maximum power transfer in the circuit of
Fig. 4.23. Find the maximum power.

Figure 4.23 for example 4.6

Find:
(a) maximum power

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Solution:

We need to find the Thevenin resistance RTh and the Thevenin voltage VTh across the
terminals a-b. To get RTh, we use the circuit in Fig. 4.24(a) and obtain

Figure 4.24 for example 4.6: (a) finding RTh, (b) finding VTh.

To get VTh, we consider the circuit in Fig. 4.24(b). Applying mesh analysis,

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SAQ 4-6 Determine the value of RL that will draw the maximum power from the
rest of the circuit in Fig. 4.25. Calculate the maximum power.

Figure 4.25 for ASAQ 4-6

ASAQ 4-6 Answer: 4.22Ω, 2.901 W.

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SUMMARY

Assignment No. 4

Name: _____________________________ Course, Year & Section: ________

Subject: EE 1 Instructor: Francisco Dequito,Jr.

Directions: Solve the following comprehensively. Write your answers on a short

bond paper.

1. Find vo in the circuit of Fig. 4.26 . If the source current is reduced to 1 μA, what
is vo?

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Figure 4.26

2. Given the circuit in Fig. 4.27, calculate ix and the power dissipated by the 10
Ω resistor using superposition.

Figure 4.27

3. Determine RTh and VTh at terminals 1-2 of each of the circuits in Fig. 4.28

Figure 4.28

4. Find the Thevenin equivalent at terminals a-b of the circuit in Fig. 4.29.

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Figure 4.29

5. Use source transformation to find ix in the circuit of Fig. 4.30.

Figure 4.30

4.8 Reference:

1. Fundamentals of Electric circuits. Charles Alexander, Matheu Sadiku.

McGraw Hill Book of Company.

4.9 Acknowledgment

Further revealed that the images, tables, figures and information contained in this
module were taken from the references cited above.

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