Figure 4.1

← Back to Fundamentals of Electric Circuits Overview Throughout this book we consider only linear circuits. Note that since p = i 2 R = v 2 ∕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 Fig. 4.1. The linear circuit has no independent sources inside it. It is excited by a v oltage source vs, which serv es as the input. The circuit is terminated by a load R. We may tak e the current i through R as the output. Suppose v^s = 10 V gives i = 2 A. According to the linearity principle, v^s = 1 V will give i = 0.2 A. By the same token, i = 1 mA must be due to v^s = 5 mV.

For example, when current i1 flows through resistor R, the power is ^p1 = Ri¹ 2 , and when current i2 flows through R, the power is ^p2 = Ri² 2 . If current i¹ + ⁱ2 flows through R, the power absorbed is ^p3 = ^R (i¹ + ⁱ2) ² = Ri^{1 2} + Ri^{2 2} + 2Ri1i² ≠ ^p1 + ^p2. Thus, the power relation is nonlinear.

Figure 4.1

A linear circuit with input vs and output i.

^For the circuit in Fig. 4.2, find Io when v^{s =} 12 V and v^{s =} 24 V. Example 4.1

Solution:

Applying KVL to the two loops, we obtain

$$12i1 - 4i2 + vs = 0$$

\n(4.1.1)
\n

$$-4i1 + 16i2 - 3vx - vs = 0$$

\n(4.1.2)

But v^x = 2i1. Equation (4.1.2) becomes

$$-10i_1 + 16i_2 - v_s = 0 \tag{4.1.3}$$

Adding Eqs. (4.1.1) and (4.1.3) yields

$$2i_1 + 12i_2 = 0 \qquad \Rightarrow \qquad i_1 = -6i_2$$

Substituting this in Eq. (4.1.1), we get

$$-76i_2 + v_s = 0 \qquad \Rightarrow \qquad i_2 = \frac{v_s}{76}$$

When v^s = 12 V,

When v^s = 24 V,

$$I_o = i_2 = \frac{24}{76}$$

A

___ 12 76 A

Io = i² =

showing that when the source value is doubled, Io doubles.

For Practice Prob. 4.1.

Example 4.2

Assume Io = 1 A and use linearity to find the actual value of Io in the

Figure 4.4

circuit of Fig. 4.4.

Answer: 40 V, 60 V.

Solution:

Answer: 16 V.

If Io = 1 A, then V¹ = (3 + 5)Io = 8 V and I¹ = V1∕4 = 2 A. Applying KCL at node 1 gives

$$I_2 = I_1 + I_o = 3 \text{ A}$$ $$V_2 = V_1 + 2I_2 = 8 + 6 = 14 \text{ V}, \qquad I_3 = \frac{V_2}{7} = 2 \text{ A}$$

Applying KCL at node 2 gives

$$I_4 = I_3 + I_2 = 5 \text{ A}$$

Therefore, Is = 5 A. This shows that assuming Io = 1 gives Is = 5 A, the actual source current of 15 A will give Io = 3 A as the actual value.

Assume that Vo = 1 V and use linearity to calculate the actual value of Vo in the circuit of Fig. 4.5.

Practice Problem 4.2

For Practice Prob. 4.2.

4.3 Superposition

If a circuit has tw o or more independent sources, one w ay 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 v ariable and then add them up. The latter approach is known as the superposition principle.

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 element 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 contrib ution 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 e very voltage source by 0 V (or a short circuit), and e very 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 the y are controlled by circuit variables.

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

Steps to Apply Superposition Principle:

• 1. Turn off all independent sources e xcept 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.
• 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 ha ve to analyze three simpler circuits each pro viding the contribution due to the respective individual source. However, superposition does help reduce a comple x 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 ef fect on po wer due to each source, be cause the po wer 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.

Superposition is not limited to circuit analysis but is applicable in many fields where cause and effect bear a linear relationship to one another.

Other terms such as killed, made inactive, deadened, or set equal to zero are often used to convey the same idea.

Figure 4.6 For Example 4.3.

8 Ω

i 2 i 3

4 Ω

(b)

Example 4.3 Use the superposition theorem to find v in the circuit of Fig. 4.6.

Solution:

Since there are two sources, let

$$v = v_1 + v_2$$

where v1 and v2 are the contributions due to the 6-V v oltage source and the 3-A current source, respecti vely. To obtain v1, we set the current source to zero, as sho wn in Fig. 4.7(a). Applying KVL to the loop in Fig. 4.7(a) gives

$$12i_1 - 6 = 0 \qquad \Rightarrow \qquad i_1 = 0.5 \text{ A}$$

Thus,

$$v_1 = 4i_1 = 2 \text{ V}$$

We may also use voltage division to get v1 by writing

$$v_1 = \frac{4}{4+8}(6) = 2 \text{ V}$$

To get v2, we set the voltage source to zero, as in Fig. 4.7(b). Using current division,

$$i_3 = \frac{8}{4+8}(3) = 2 \text{ A}$$

Hence,

3 A

v2

• –

v² = 4i³ = 8 V

And we find

Answer: 16 V.

$$v = v_1 + v_2 = 2 + 8 = 10 \text{ V}$$

Figure 4.7 For Example 4.3: (a) calculating v1, (b) calculating v2.

Figure 4.8 For Practice Prob. 4.3.

Find io in the circuit of Fig. 4.9 using superposition.

Solution:

The circuit in Fig. 4.9 in volves a dependent source, which must be left intact. We let

$$i_o = i'_o + i''_o \tag{4.4.1}$$

where i′ ^o and i o ″ are due to the 4-A current source and 20-V voltage source respectively. To obtain i′ ^o, we turn off the 20-V source so that we have the circuit in Fig. 4.10(a). We apply mesh analysis in order to obtain i′ ^o. For loop 1,

$$i_1 = 4 \text{ A} \tag{4.4.2}$$

For loop 2,

$$l_1 = 4 \,\mathrm{A}$$ $$-3i_1 + 6i_2 - 1i_3 - 5i'_o = 0 \tag{4.4.3}$$

Figure 4.10

For Example 4.4: Applying superposition to (a) obtain i′ ^o, (b) obtain i ″ o.

For loop 3,

−5i1 − 1i2 + 10i3 + 5i′ ^o = 0 (4.4.4) But at node 0, i3 = i1 − i′ ^o = 4 − i′ ^o (4.4.5)

Substituting Eqs. (4.4.2) and (4.4.5) into Eqs. (4.4.3) and (4.4.4) gi ves two simultaneous equations

$$3i_2 - 2i'_o = 8 \tag{4.4.6}$$ $$i_2 + 5i'_o = 20 \tag{4.4.7}$$

which can be solved to get

$$i'_{o} = \frac{52}{17} A$$

(4.4.8)

To obtain i″ ^o, we turn of f the 4-A current source so that the circuit becomes that shown in Fig. 4.10(b). For loop 4, KVL gives

$$6i_4 - i_5 - 5i''_o = 0 \tag{4.4.9}$$

and for loop 5,

$$-i_4 + 10i_5 - 20 + 5i''_o = 0 \tag{4.4.10}$$

But i5 = −i″ ^o. Substituting this in Eqs. (4.4.9) and (4.4.10) gives

$$6i_4 - 4i''_o = 0 \tag{4.4.11}$$ $$i_4 + 5i''_o = -20 \tag{4.4.12}$$

which we solve to get

$$i_o'' = -\frac{60}{17} \,\mathrm{A} \tag{4.4.13}$$

Now substituting Eqs. (4.4.8) and (4.4.13) into Eq. (4.4.1) gives

$$i_o = -\frac{8}{17} = -0.4706 \text{ A}$$

Example 4.5

For the circuit in Fig. 4.12, use the superposition principle to find i.

Solution:

In this case, we have three sources. Let

$$i = i_1 + i_2 + i_3$$

where i1, i2, and i3 are due to the 12-V , 24-V, and 3-A sources respec tively. To get i1, consider the circuit in Fig. 4.13(a). Combining 4 Ω (on the right-hand side) in series with 8 Ω gives 12 Ω. The 12 Ω in parallel with 4 Ω gives 12 × 4∕16 = 3 Ω. Thus,

$$i_1 = \frac{12}{6} = 2 \text{ A}$$

To get i2, consider the circuit in Fig. 4.13(b). Applying mesh analysis gives

$$16i_a - 4i_b + 24 = 0 \qquad \Rightarrow \qquad 4i_a - i_b = -6 \tag{4.5.1}$$ $$\mathcal{T}_b - 4i_a = 0 \qquad \Rightarrow \qquad i_a = \frac{7}{4} i_b \tag{4.5.2}$$

Substituting Eq. (4.5.2) into Eq. (4.5.1) gives

$$i_2 = i_b = -1$$

To get i3, consider the circuit in Fig. 4.13(c). Using nodal analysis gives

$$3 = \frac{v_2}{8} + \frac{v_2 - v_1}{4} \qquad \Rightarrow \qquad 24 = 3v_2 - 2v_1 \tag{4.5.3}$$ $$\frac{v_2 - v_1}{4} = \frac{v_1}{4} + \frac{v_1}{3} \qquad \Rightarrow \qquad v_2 = \frac{10}{3}v_1 \tag{4.5.4}$$

Substituting Eq. (4.5.4) into Eq. (4.5.3) leads to v¹ = 3 and

$$i_3 = \frac{v_1}{3} = 1 \text{ A}$$

Thus,

$$i = i_1 + i_2 + i_3 = 2 - 1 + 1 = 2 \text{ A}$$

Find I in the circuit of Fig. 4.14 using the superposition principle. Practice Problem 4.5

2 Ω

2 A

– 6 V ⁺

8 Ω

I

–

8 V

Figure 4.14 For Practice Prob. 4.5.

4.4 Source Transformation

6 Ω

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 Section 3.6, we sa w that node-v oltage (or mesh-current) equa tions can be obtained by mere inspection of a circuit when the sources are all independent current (or all independent v oltage) sources. It is therefore expedient in circuit analysis to be able to substitute a v oltage source in series with a resistor for a current source in parallel with a

resistor, or vice versa, as shown in Fig. 4.15. Either substitution is known as a source transformation.

Figure 4.15 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 tw o circuits in Fig. 4.15 are equi valent—provided the y ha ve the same voltage-current relation at terminals a-b. It is easy to sho w that they are indeed equi valent. 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 isc = vs∕R in the circuit on the left-hand side and isc = is for the circuit on the right-hand side. Thus, vs∕R = is in order for the two circuits to be equivalent. Hence, source transformation requires that

$$v_s = i_s R \qquad \text{or} \qquad i_s = \frac{v_s}{R} \tag{4.5}$$

Source transformation also applies to dependent sources, pro vided we carefully handle the dependent v ariable. As shown in Fig. 4.16, 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 Eq. (4.5) is satisfied.

Figure 4.16 Transformation of dependent sources.

Like the wye-delta transformation we studied in Chapter 2, a source transformation does not af fect the remaining part of the circuit. When applicable, source transformation is a po werful tool that allo ws circuit manipulations to ease circuit analysis. However, we should keep the following points in mind when dealing with source transformation.

• 1. Note from Fig. 4.15 (or Fig. 4.16) that the arrow of the current source is directed toward the positive terminal of the voltage source.
• 2. Note from Eq. (4.5) 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 ≠ 0. Similarly, an ideal current source with R = ∞ cannot be replaced by a finite voltage source. More will be said on ideal and nonideal sources in Section 4.10.1.

Use source transformation to find vo in the circuit of Fig. 4.17.

Solution:

We first transform the current and voltage sources to obtain the cir cuit in Fig. 4.18(a). Combining the 4-Ω and 2-Ω resistors in series and transforming the 12-V v oltage source gi ves us Fig. 4.18(b). We now combine the 3-Ω and 6-Ω resistors in parallel to get 2-Ω. 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. 4.18(c).

Example 4.6

We use current division in Fig. 4.18(c) to get

$$i = \frac{2}{2+8}(2) = 0.4 \text{ A}$$

and

$$v_o = 8i = 8(0.4) = 3.2$$

V

Alternatively, since the 8-Ω and 2-Ω resistors in Fig. 4.18(c) are in parallel, they have the same voltage vo across them. Hence,

$$v_o = (8 \parallel 2)(2 \text{ A}) = \frac{8 \times 2}{10}(2) = 3.2 \text{ V}$$

Find io in the circuit of Fig. 4.19 using source transformation. Practice Problem 4.6

Answer: 1.78 A.

Example 4.7

Figure 4.20 For Example 4.7. Find vx in Fig. 4.20 using source transformation.

Solution:

The circuit in Fig. 4.20 in volves a v oltage-controlled dependent cur rent source. We transform this dependent current source as well as the 6-V independent v oltage source as sho wn in Fig. 4.21(a). The 18-V voltage source is not transformed because it is not connected in series with any resistor. The two 2-Ω resistors in parallel combine to gi ve a 1-Ω resistor, which is in parallel with the 3-A current source. The current source is transformed to a v oltage source as shown in Fig. 4.21(b). Notice that the terminals for vx are intact. Applying KVL around the loop in Fig. 4.21(b) gives

Figure 4.21 For Example 4.7: Applying source transformation to the circuit in Fig. 4.20.

Applying KVL to the loop containing only the 3-V v oltage source, the 1-Ω resistor, and vx yields

$$-3 + 1i + v_x = 0 \qquad \Rightarrow \qquad v_x = 3 - i \tag{4.7.2}$$

Substituting this into Eq. (4.7.1), we obtain

$$15 + 5i + 3 - i = 0 \qquad \Rightarrow \qquad i = -4.5 \text{ A}$$

Alternatively, we may apply KVL to the loop containing vx, the 4-Ω resistor, the v oltage-controlled dependent v oltage source, and the 18-V voltage source in Fig. 4.21(b). We obtain

$$-v_x + 4i + v_x + 18 = 0 \qquad \Rightarrow \qquad i = -4.5 \text{ A}$$

Thus, v^x = 3 − i = 7.5 V.

For Practice Prob. 4.7.

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 dif ferent appliances constituting a v ariable load. Each time the v ariable element is changed, the entire circuit has to be analyzed all over again. To avoid this problem, Thevenin’s theorem pro vides 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. 4.23(a) can be replaced by that in Fig. 4.23(b). (The load in Fig. 4.23 may be a single resistor or another circuit.) The circuit to the left of the terminals a-b in Fig. 4.23(b) is kno wn as the Thevenin equivalent cir cuit; it w as developed in 1883 by M. Leon Thevenin (1857–1926), a French telegraph engineer.

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.

The proof of the theorem will be gi ven later, in Section 4.7. Our major concern right now is how to find the Thevenin equivalent voltage VTh and resistance RTh. To do so, suppose the tw o circuits in Fig. 4.23 are equivalent. Two circuits are said to be equivalent if the y have the same voltage-current relation at their terminals. Let us find out what will make the tw o circuits in Fig. 4.23 equi valent. If the terminals a-b are made open-circuited (by remo ving the load), no current flows, so that the open-circuit voltage across the terminals a-b in Fig. 4.23(a) must be equal to the voltage source VTh in Fig. 4.23(b), since the two circuits are equivalent. Thus, VTh is the open-circuit v oltage across the terminals as shown in Fig. 4.24(a); that is,

Finding VTh and RTh.

Again, with the load disconnected and terminals a-b opencir cuited, we turn of f all independent sources. The input resistance (or equi valent resistance) of the dead circuit at the terminals a-b in Fig. 4.23(a) must be equal to RTh in Fig. 4.23(b) because the two circuits are equivalent. Thus, RTh is the input resistance at the terminals when the independent sources are turned off, as shown in Fig. 4.24(b); that is,

$$R_{\rm Th} = R_{\rm in} \tag{4.7}$$

Figure 4.23 Replacing a linear two-terminal circuit

by its Thevenin equivalent: (a) original circuit, (b) the Thevenin equivalent circuit.

Figure 4.25

Finding RTh when circuit has dependent sources.

Later we will see that an alternative way of finding RTh is ^RTh = ^voc ∕isc.

Figure 4.26

A circuit with a load: (a) original circuit, (b) Thevenin equivalent.

To apply this idea in finding the Thevenin resistance RTh, we need to consider two cases.

■ CASE 1 If the network has no dependent sources, we turn of f all independent sources. RTh is the input resistance of the netw ork looking between terminals a and b, as shown in Fig. 4.24(b).