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Example 4.11 Find the Norton equivalent circuit of the circuit in Fig. 4.39 at terminals a-b. 8Ω ww Solution: We find Ry 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. 4.40(a), from which we find RN. Thus, 2 A 5Ω 12 V 20 X 5 5 || (8 + 4 + 8) = 5 || 20 = = 4 2 25 8Ω RN = Figure 4.39 For Example 4.11. To find Iy, we short-circuit terminals a and b, as shown in Fig. 4.40(b). We ignore the 5.0 resistor because it has been short.circuited

Example 4.11 Find the Norton equivalent circuit of the circuit in Fig. 4.39 at terminals a-b. 8Ω ww Solution: We find Ry 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. 4.40(a), from which we find RN. Thus, 2 A 5Ω 12 V 20 X 5 5 || (8 + 4 + 8) = 5 || 20 = = 4 2 25 8Ω RN = Figure 4.39 For Example 4.11. To find Iy, we short-circuit terminals a and b, as shown in Fig. 4.40(b). We ignore the 5.0 resistor because it has been short.circuited

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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Determine the Thevenin equivalent of the circuit in Fig. 4.35(a) at Example 4.10 terminals a-b. Solution: 1. Define. The problem is clearly defined; we are to determine the Thevenin equivalent of the circuit shown in Fig. 4.35(a). 2. Present. The circuit contains a 2-2 resistor in parallel with a 4-N resistor. These are, in turn, in parallel with a dependent current source. It is important to note that there are no independent sources. 3. Alternative. The first thing to consider is that, since we have no independent sources in this circuit, we must excite the circuit externally. In addition, when you have no independent sources you will not have a value for VTh; you will only have to find RTh- -o a The simplest approach is to excite the circuit with either a 1-V voltage source or a 1-A current source. Since we will end up with an equivalent resistance (either positive or negative), I prefer to use the current source and nodal analysis which will yield a voltage at the output terminals equal to the resistance (with 1 A flowing in, v, is equal to 1 times the equivalent resistance). As an alternative, the circuit could also be excited by a 1-V voltage source and mesh analysis could be used to find the equivalent resistance. 4. Attempt. We start by writing the nodal equation at a in Fig. 4.35(b) assuming i, = 1 A. 2i, 4Ω (a) a 2i, 2 2 2i, + (v, – 0)/4 + (v. 0)/2 + (-1) = 0 (4.10.1) b (b) Since we have two unknowns and only one equation, we will need a constraint equation. 4Ω 9Ω a ix = (0 – vo)/2 = -vo/2 (4.10.2) i, 8i, C 22 i2 Substituting Eq. (4.10.2) into Eq. (4.10.1) yields 10 V 2(-v./2) + (v. 0)/4 + (v, – 0)/2 + (-1) = 0 = (-1+ + )v, - 1 v, = -4 V or b Since v. = 1 × RTh, then RTh = Vo/1 = -4 N. The negative value of the resistance tells us that, according to the passive sign convention, the circuit in Fig. 4.35(a) is supplying power. Of course, the resistors in Fig. 4.35(a) cannot supply power (they absorb power); it is the dependent source that supplies the power. This is an example of how a dependent source and resistors could be used to simulate negative resistance. 5. Evaluate. First of all, we note that the answer has a negative value. We know this is not possible in a passive circuit, but in this circuit we do have an active device (the dependent current (c) -4 2 9Ω a 10 V (d) Figure 4.35 For Example 4.10. source). Thus, the equivalent circuit is essentially an active ww ww
Transcribed Image Text:Determine the Thevenin equivalent of the circuit in Fig. 4.35(a) at Example 4.10 terminals a-b. Solution: 1. Define. The problem is clearly defined; we are to determine the Thevenin equivalent of the circuit shown in Fig. 4.35(a). 2. Present. The circuit contains a 2-2 resistor in parallel with a 4-N resistor. These are, in turn, in parallel with a dependent current source. It is important to note that there are no independent sources. 3. Alternative. The first thing to consider is that, since we have no independent sources in this circuit, we must excite the circuit externally. In addition, when you have no independent sources you will not have a value for VTh; you will only have to find RTh- -o a The simplest approach is to excite the circuit with either a 1-V voltage source or a 1-A current source. Since we will end up with an equivalent resistance (either positive or negative), I prefer to use the current source and nodal analysis which will yield a voltage at the output terminals equal to the resistance (with 1 A flowing in, v, is equal to 1 times the equivalent resistance). As an alternative, the circuit could also be excited by a 1-V voltage source and mesh analysis could be used to find the equivalent resistance. 4. Attempt. We start by writing the nodal equation at a in Fig. 4.35(b) assuming i, = 1 A. 2i, 4Ω (a) a 2i, 2 2 2i, + (v, – 0)/4 + (v. 0)/2 + (-1) = 0 (4.10.1) b (b) Since we have two unknowns and only one equation, we will need a constraint equation. 4Ω 9Ω a ix = (0 – vo)/2 = -vo/2 (4.10.2) i, 8i, C 22 i2 Substituting Eq. (4.10.2) into Eq. (4.10.1) yields 10 V 2(-v./2) + (v. 0)/4 + (v, – 0)/2 + (-1) = 0 = (-1+ + )v, - 1 v, = -4 V or b Since v. = 1 × RTh, then RTh = Vo/1 = -4 N. The negative value of the resistance tells us that, according to the passive sign convention, the circuit in Fig. 4.35(a) is supplying power. Of course, the resistors in Fig. 4.35(a) cannot supply power (they absorb power); it is the dependent source that supplies the power. This is an example of how a dependent source and resistors could be used to simulate negative resistance. 5. Evaluate. First of all, we note that the answer has a negative value. We know this is not possible in a passive circuit, but in this circuit we do have an active device (the dependent current (c) -4 2 9Ω a 10 V (d) Figure 4.35 For Example 4.10. source). Thus, the equivalent circuit is essentially an active ww ww
Example 4.11 Find the Norton equivalent circuit of the circuit in Fig. 4.39 at terminals a-b. 8Ω O a Solution: We find Ry 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. 4.40(a), from which we find RN. Thus, 2 A 5Ω 12 V ww 20 X 5 RN = 5 || (8 + 4 + 8) = 5 || 20 = 25 = 4 0 Figure 4.39 For Example 4.11. To find IN, we short-circuit terminals a and b, as shown in Fig. 4.40(b). We ignore the 5-0 resistor because it has been short-circuited
Transcribed Image Text:Example 4.11 Find the Norton equivalent circuit of the circuit in Fig. 4.39 at terminals a-b. 8Ω O a Solution: We find Ry 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. 4.40(a), from which we find RN. Thus, 2 A 5Ω 12 V ww 20 X 5 RN = 5 || (8 + 4 + 8) = 5 || 20 = 25 = 4 0 Figure 4.39 For Example 4.11. To find IN, we short-circuit terminals a and b, as shown in Fig. 4.40(b). We ignore the 5-0 resistor because it has been short-circuited
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