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CIRCUIT THEOREMS

Module by: Dinh Sy Hien

Summary: 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. The concepts we develop are applied in the last section to source modeling and resistance measurement.

INTRODUCTION

A major advantage of analyzing circuits using Kirchhoff’s law as we did in Chapter 3 is that we can analyze a circuit without tampering with its original configuration. A major disadvantage of this approach is that, for a large, complex circuit, tedious computation is involved.
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 chapter. The concepts we develop are applied in the last section to source modeling and resistance measurement.

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 additive property.
The homogeneity property requires that if the input (also called the excitation) is multiplied by a constant, then the output (also called response) is multiplied by the same constant. For a resistor, for example, Ohm’s law relates the input I to the output v,
v = iR v = iR size 12{v= ital "iR"} {} (1)
If the current is increased by constant k, then the voltage increases correspondingly by k, that is,
kiR = kv kiR = kv size 12{ ital "kiR"= ital "kv"} {} (2)
The additive property requires that the response to a sum of inputs is the sum of the responses to each input applied separately. Using the voltage current relationship of a resistor, if
v 1 = i 1 R v 1 = i 1 R size 12{v rSub { size 8{1} } =i rSub { size 8{1} } R} {} (3)
and
v 2 = i 2 R v 2 = i 2 R size 12{v rSub { size 8{2} } =i rSub { size 8{2} } R} {} (4)
then applying (i1+i2)(i1+i2) size 12{ \( i rSub { size 8{1} } +i rSub { size 8{2} } \) } {} gives
v = ( i 1 + i 2 ) R = i 1 R + i 2 R = v 1 + v 2 v = ( i 1 + i 2 ) R = i 1 R + i 2 R = v 1 + v 2 size 12{v= \( i rSub { size 8{1} } +i rSub { size 8{2} } \) R=i rSub { size 8{1} } R+i rSub { size 8{2} } R=v rSub { size 8{1} } +v rSub { size 8{2} } } {} (5)
We say that a resistor is a linear element because the voltage-current relationship satisfies both the homogeneity and the additive properties.
In general, a circuit is linear if it is both additive and homogeneous. A linear circuit consists of only linear elements, linear dependent sources, and independent sources.
A linear circuit is one whose output is linearly related (or directly proportional) to its input.
Note that since p=i2R=v2/Rp=i2R=v2/R size 12{p=i rSup { size 8{2} } R= {v rSup { size 8{2} } } slash {R} } {} (making it a quadratic function rather than a linear one), the relationship between power and voltage (or current) is nonlinear. Therefore, the theorems covered in this chapter are not applicable to power.
To illustrate the linearity principle, consider the linear circuit shown in Figure 1. The linear circuit has no independent sources inside it. It is excited by a voltage source vs, which serves as the input. The circuit is terminated by a load R. We may take the current i through R as the output. Suppose vS=10VvS=10V size 12{v rSub { size 8{S} } ="10"V} {} gives i = 2 A. According to the linearity principle, vS=1VvS=1V size 12{v rSub { size 8{S} } =1V} {} will give i = 0.2 A. By the same token, i = 1 mA must be due to vS=5mVvS=5mV size 12{v rSub { size 8{S} } =5 ital "mV"} {}.
Figure 1: A linear circuit with input Vs and output i.

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 idea of superposition rests on the linearity property.
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 elements due to each independent source acting alone.
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 this in mind, we apply the superposition principle in three steps:
Steps to apply superposition principle:
  1. Turn off all independent sources except one source. Find the output (voltage or current) due to that active source using the techniques covered in Chapters 2 and 3.
  2. Repeat step 1 for each of the other independent sources.
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 respectively 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 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 first using superposition.

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 concept of equivalence. We recall that an equivalent circuit is one whose v-i characteristics are identical with the original circuit.
In Section 3.6, 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 Figure 15. Either substitution is known as source transformation.
Figure 2: Transformation of independent sources.
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.
The two circuits in Figure 2 are equivalent – provided they have the same voltage–current relation at terminals a-b. It is easy to show that they are indeed equivalent. If the sources are turned off, the equivalent resistance at terminals a-b in both circuits is R. Also, when terminals a-b are short circuited, the short circuit current flowing from a to b is iSG=vS/RiSG=vS/R size 12{i rSub { size 8{ ital "SG"} } = {v rSub { size 8{S} } } slash {R} } {} in the circuit in the left hand side and iSG=iSiSG=iS size 12{i rSub { size 8{ ital "SG"} } =i rSub { size 8{S} } } {} for the circuit on right hand side. Thus, vS/R=iSvS/R=iS size 12{ {v rSub { size 8{S} } } slash {R} =i rSub { size 8{S} } } {} in order for the two circuits to be equivalent. Hence, source transformation requires that
vS=iSRvS=iSR size 12{v rSub { size 8{S} } =i rSub { size 8{S} } R} {}(6)
or
iS=vSRiS=vSR size 12{i rSub { size 8{S} } = { {v rSub { size 8{S} } } over {R} } } {}
Source transformation also applies to dependent sources, provided we carefully handle the dependent variable. As shown in Figure 3, a dependent voltage source in series with a resistor can be transformed to a dependent current source in parallel with the resistor or vice versa where we make sure that Equation 6 is satisfied.
Figure 3: Transformation of dependent sources.
Like wye-delta transformation we studied in Chapter 2, a source transformation does not affect the remaining part of the circuit. When applicable, source transformation is a powerful tool that allows circuit manipulations to ease circuit analysis. However, we should keep the following points in mind when dealing with source transformation.
1. Note from Figure 2 (or Figure 3) that the arrow of the current source is directed toward the positive terminal of the voltage source.
2. Note from Equation 6 that source transformation is not possible when R = 0, which is the case with an ideal voltage source. However, for a practical, nonideal voltage source, R≠0R≠0 size 12{R <> 0} {}. Similarly, an ideal current source with R=∞R=∞ size 12{R= infinity } {} cannot be replaced by a finite voltage source.

THEVENIN’S THEOREM

It often occurs in practice that a particular element in the circuit is variable (usually called the load) while other elements are fixed. As a typical example, a household outer 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 Figure 4(a) can be replaced by that in Figure 4(b). The load in Figure 4 may be a single resistor or another circuit. The circuit to the left of the terminals a-b in Figure 3(b) is known as the Thevenin equivalent circuit; it was developed in 1883 by M. Leon Thevenin (1857-1926), a French telegraph engineer.
Figure 4: 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 VThVTh size 12{V rSub { size 8{ ital "Th"} } } {} in series with a resistor RThRTh size 12{R rSub { size 8{ ital "Th"} } } {}, where VThVTh size 12{V rSub { size 8{ ital "Th"} } } {} is the open-circuit voltage at the terminals and RThRTh size 12{R rSub { size 8{ ital "Th"} } } {} is the input or equivalent resistance at the terminals when the independent sources are turned off.
The proof of the theorem will be given later, in section 7. Our major concern right now is how to find the Thevenin equivalent voltage VThVTh size 12{V rSub { size 8{ ital "Th"} } } {} and resistance RThRTh size 12{R rSub { size 8{ ital "Th"} } } {}. To do so, suppose the two circuits in Figure 5 are equivalent. Two circuits are said to be equivalent if they have the same voltage-current relation at their terminals. Let us find out what will make open-circuits in Figure 5 equivalent. If the terminals a-b are made open-circuit voltage (by removing the load), no current flows, so that the open-circuit voltage across the terminals a-b in Figure 5(a) must be equal to the voltage source VThVTh size 12{V rSub { size 8{ ital "Th"} } } {} in Figure 5(b), since the two circuits are equivalent. Thus VThVTh size 12{V rSub { size 8{ ital "Th"} } } {} is open-circuit voltage across the terminals as shown in Figure 6(a); that is,
V Th = V ∞ V Th = V ∞ size 12{V rSub { size 8{ ital "Th"} } =V rSub { size 8{ infinity } } } {} (7)
Figure 5: Finding Vth and Rth.
Again, with the load disconnected and terminals a-b open-circuited, we turn off all independent sources. The input resistance (or equivalent resistance) of the dead circuit at the terminals a-b in Figure 3(a) must be equal to RThRTh size 12{R rSub { size 8{ ital "Th"} } } {} in Figure 3(b) because the two circuits are equivalent. Thus, RThRTh size 12{R rSub { size 8{ ital "Th"} } } {} is the input resistance at the terminals when the independent sources are turned off, as shown in Figure 5(b), that is,
R Th = R in R Th = R in size 12{R rSub { size 8{ ital "Th"} } =R rSub { size 8{ ital "in"} } } {} (8)
To apply this idea in finding the Thevenin resistance RThRTh size 12{R rSub { size 8{ ital "Th"} } } {}, we need to consider two cases,
CASE 1. If network has no dependent sources, we turn off all independent sources, RThRTh size 12{R rSub { size 8{ ital "Th"} } } {} is the input resistance of the network looking between terminals a and b, as shown in Figure 5(b).
CASE 2. If the network has dependent sources, we turn off all independent sources. As with superposition, dependent sources are not to be turned off because they are controlled by circuit variables. We apply a voltage sources v0 at terminals a and b and determine the resulting current i0. Then RTh=v0/i0RTh=v0/i0 size 12{R rSub { size 8{ ital "Th"} } = {v rSub { size 8{0} } } slash {i rSub { size 8{0} } } } {}, as shown in Figure 6(a). Alternatively, we may insert a current source i0 and at terminals a-b as shown in Figure 6(b) and find the terminal voltage v0v0 size 12{v rSub { size 8{0} } } {}. Again RTh=v0/i0RTh=v0/i0 size 12{R rSub { size 8{ ital "Th"} } = {v rSub { size 8{0} } } slash {i rSub { size 8{0} } } } {}. Either of the two approaches will give the same result. In either approach we may assume any value or v0v0 size 12{v rSub { size 8{0} } } {} and i0i0 size 12{i rSub { size 8{0} } } {}. For example, we may use v0=1Vv0=1V size 12{v rSub { size 8{0} } =1V} {} or i0=1Ai0=1A size 12{i rSub { size 8{0} } =1A} {}, or even use unspecified values of v0v0 size 12{v rSub { size 8{0} } } {} or i0i0 size 12{i rSub { size 8{0} } } {}.
Figure 6: Finding Rth circuit has dependent sources.
It often occurs that RThRTh size 12{R rSub { size 8{ ital "Th"} } } {} takes a negative value. In this case, the negative resistance (v = - iR) implies that the current is supplying power. This is possible in a circuit with dependent sources.
Thevenin’s theorem is very important in circuit analysis. It helps simplify a circuit. A large circuit may be replaced by a simple independent voltage source and a single resistor. This replacement technique is powerful tool in circuit design.
As mentioned earlier, a linear circuit with a variable load can be replaced by the Thevenin equivalent, exclusive of the load. The equivalent network behaves the same way externally as the original circuit. Consider a linear circuit terminated by a load RLRL size 12{R rSub { size 8{L} } } {}, as shown in Figure 6(a). The current ILIL size 12{I rSub { size 8{L} } } {}through the load and the voltage VLVL size 12{V rSub { size 8{L} } } {}across the load are easily determined once the Thevenin equivalent of the current at the load’s terminals is obtained, as shown in Figure 7(b). From Figure 7(b), we obtain
I L = V Th R Th + R L I L = V Th R Th + R L size 12{I rSub { size 8{L} } = { {V rSub { size 8{ ital "Th"} } } over {R rSub { size 8{ ital "Th"} } +R rSub { size 8{L} } } } } {} (9)
V L = R L I L = V Th R Th + R L V Th V L = R L I L = V Th R Th + R L V Th size 12{V rSub { size 8{L} } =R rSub { size 8{L} } I rSub { size 8{L} } = { {V rSub { size 8{ ital "Th"} } } over {R rSub { size 8{ ital "Th"} } +R rSub { size 8{L} } } } V rSub { size 8{ ital "Th"} } } {} (10)
Note from Figure 7(b) that the Thevenin equivalent is a simple voltage divider, yielding VLVL size 12{V rSub { size 8{L} } } {} by mere inspection.
Figure 7: A circuit with a load: a) original circuit, b) Thevenin equivalent.

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 theorem states that a linear two-terminal circuit can be replaced by an equivalent circuit consisting of a current source ININ size 12{I rSub { size 8{N} } } {} in parallel with a resistor RNRN size 12{R rSub { size 8{N} } } {}, where ININ size 12{I rSub { size 8{N} } } {} is the short-circuit current through the terminals and RNRN size 12{R rSub { size 8{N} } } {} is the input or equivalent resistance at the terminals when the independent sources are turned off.
Thus, the circuit in Figure 8(a) can be replaced by the one in Figure 8(b).
Figure 8: a) Original circuit, b) Norton equivalnent circuit.
The proof of Noton’s theorem will be given in the next section. For now, we are mainly concerned with how to get RNRN size 12{R rSub { size 8{ ital "N"} } } {} and ININ size 12{R rSub { size 8{ ital "N"} } } {}. We find RNRN size 12{R rSub { size 8{ ital "N"} } } {}. in the same way we find RThRTh size 12{R rSub { size 8{ ital "Th"} } } {}. In fact, from what we know about source tranformation, the Thevenin and Norton resistances are equal; that is,
R N = R Th R N = R Th size 12{R rSub { size 8{N} } =R rSub { size 8{ ital "Th"} } } {} (11)
To find the Norton current IN, we determine the short-circuit current flowing from terminal a to b in both circuits in Figure 8. It is evident that the circuit the short-circuit current in Figure 8(b) is IN. this must be the same short-circuit current from terminal a and b in Figure 8(a), since the two circuits are equivalent. Thus,
I N = i SG I N = i SG size 12{I rSub { size 8{N} } =i rSub { size 8{ ital "SG"} } } {} (12)
shown in Figure 9. Dependent and independent sources are treated the same way as in Thevenin’s theorem.
Figure 9: Finding Norton current In.
Observe the close relationship between Norton’s and Thevenin’s theorems: RN=RThRN=RTh size 12{R rSub { size 8{N} } =R rSub { size 8{ ital "Th"} } } {} as in Equation 11, and
I N = V Th R Th I N = V Th R Th size 12{I rSub { size 8{N} } = { {V rSub { size 8{ ital "Th"} } } over {R rSub { size 8{ ital "Th"} } } } } {} (13)
This is essentially source transformation. For this reason, source transformation is often called Thevenin-Norton transformation.
Since VThVTh size 12{V rSub { size 8{ ital "Th"} } } {}, ININ size 12{I rSub { size 8{N} } } {} and RThRTh size 12{R rSub { size 8{ ital "Th"} } } {} are related according to Equation 13, to determine the Thevenin or Norton equivalent circuit requires that we find:
  • The open-circuit voltage vocvoc size 12{v rSub { size 8{ ital "oc"} } } {} across terminal a and b.
  • The short-circuit current iSGiSG size 12{i rSub { size 8{ ital "SG"} } } {} at terminals a and b.
  • The equivalent or input resistance RinRin size 12{R rSub { size 8{ ital "in"} } } {} at terminals a and b when all independent sources are turned off.
  • 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. Also, since
V Th = v oc V Th = v oc size 12{V rSub { size 8{ ital "Th"} } =v rSub { size 8{ ital "oc"} } } {} (14)
I N = i sc I N = i sc size 12{I rSub { size 8{N} } =i rSub { size 8{ ital "sc"} } } {} (15)
R Th = v oc i sc = R N R Th = v oc i sc = R N size 12{R rSub { size 8{ ital "Th"} } = { {v rSub { size 8{ ital "oc"} } } over {i rSub { size 8{ ital "sc"} } } } =R rSub { size 8{N} } } {}
the open-circuit and short circuit tests are sufficient to find any Thevenin or Norton equivalent.

DERIVATIONS OF THEVENIN’S AND NORTON’S THEOREMS

In this section, we will prove Thevenin’s and Norton’s theorem using the superposition principle.
Consider the linear circuit in Figure 10(a). It is assumed that the circuit contains resistors, and dependent and independent sources. We have access to the circuit via terminals a and b, through which current from an external source is applied. Our objective is to ensure that the voltage-current relation at terminals a and b is identical to that of the Thevenin equivalent in Figure 10(b). For the sake of simplicity, suppose the linear circuit in Figure 10(a) contains two independent voltage sources vs1 and vs2 and two independent current sources iS1iS1 size 12{i rSub { size 8{S1} } } {} and iS1iS1 size 12{i rSub { size 8{S1} } } {}. We may obtain any circuit variable, such as the terminal voltage v, by applying superposition. That is, we consider the contribution due to each independent source including the external source i. by superposition, the terminal voltage v is
v = A 0 i + A 1 v s1 + A 2 v s2 + A 3 i s1 + A 4 i s2 v = A 0 i + A 1 v s1 + A 2 v s2 + A 3 i s1 + A 4 i s2 size 12{v=A rSub { size 8{0} } i+A rSub { size 8{1} } v rSub { size 8{s1} } +A rSub { size 8{2} } v rSub { size 8{s2} } +A rSub { size 8{3} } i rSub { size 8{s1} } +A rSub { size 8{4} } i rSub { size 8{s2} } } {} (16)
Where A0A0 size 12{A rSub { size 8{0} } } {}, A1A1 size 12{A rSub { size 8{1} } } {}, A2A2 size 12{A rSub { size 8{2} } } {}, A3A3 size 12{A rSub { size 8{3} } } {}, and A4A4 size 12{A rSub { size 8{4} } } {} are constants. Each term on the right hand side of Equation 16 is the contribution of the related independent source; that is, A0iA0i size 12{A rSub { size 8{0} } i} {} is contribution to v due to the voltage source vS1vS1 size 12{v rSub { size 8{S1} } } {}, and so on. We may collect terms for the internal independent sources together as B0B0<