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Electrical Engineering: Principles and Applications, 6e Allan R. Hambley Copyright ©2014 by Pearson Education, Inc. All rights reserved. CHAPTER 2 RESISTIVE CIRCUITS

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Page 1: Electrical and Computer Engineering Department - CHAPTER ...ece.uprm.edu/~mtoledo/web/4075/S2014/ch02.pdfTitle ch02.key Created Date 1/27/2014 1:34:58 PM

Electrical Engineering: Principles and Applications, 6e Allan R. Hambley

Copyright ©2014 by Pearson Education, Inc. All rights reserved.

CHAPTER 2RESISTIVE CIRCUITS

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Electrical Engineering: Principles and Applications, 6e Allan R. Hambley

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Figure 2.1 Series resistances can be combined into an equivalent resistance.

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Figure 2.2 Parallel resistances can be combined into an equivalent resistance.

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Figure 2.3 Resistive network for Example 2.1.

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Figure 2.4 Resistive networks for Exercise 2.1.

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Figure 2.5 A circuit and its simplified versions. See Example 2.2.

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Figure 2.6 After reducing the circuit to a source and an equivalent resistance, we solve the simplified circuit. Then, we transfer results back to the original circuit. Notice that the logical flow in solving for currents and voltages starts from the simplified circuit in (c).

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Figure 2.7 Circuits for Exercise 2.2.

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Figure 2.8 Circuit used to derive the voltage-division principle.

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Figure 2.9 Circuit for Example 2.3.

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Figure 2.10 Circuit used to derive the current-division principle.

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Figure 2.11 Circuit for Example 2.4.

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Figure 2.12 Circuit for Example 2.5.

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Figure 2.13 The voltage-division principle forms the basis for some position sensors. This figure shows a transducer that produces an output voltage vo proportional to the rudder angle θ.

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Figure 2.14 Circuits for Exercise 2.3.

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Figure 2.15 Circuits for Exercise 2.4.

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Figure 2.16 The first step in node analysis is to select a reference node and label the voltages at each of the other nodes.

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Figure 2.17 Assuming that we can determine the node voltages v1 , v2 , and v3 , we can use KVL to determine vx , vy , and vz . Then using Ohm’s law, we can find the current in each of the resistances. Thus, the key problem is in determining the node voltages.

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Figure 2.18 Circuit for Example 2.6.

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Figure 2.19 Circuit for Exercise 2.6.

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Figure 2.20 Circuit for Example 2.7.

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Figure 2.21 Circuit for Example 2.8.

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Figure 2.22 Circuit of Example 2.8 with a different choice for the reference node. See Exercise 2.9.

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Figure 2.23 Circuit for Example 2.9.

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Figure 2.24 Circuit for Exercise 2.11.

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Figure 2.25 A supernode is formed by drawing a dashed line enclosing several nodes and any elements connected between them.

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Figure 2.26 Node voltages v1 and v2 and the 10-V source form a closed loop to which KVL can be applied. (This is the same circuit as that of Figure 2.25.)

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Figure 2.27 Circuit for Exercise 2.13.

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Figure 2.28 Circuit containing a current-controlled current source. See Example 2.10.

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Figure 2.29 Circuit containing a voltage-controlled voltage source. See Example 2.11.

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Figure 2.30 Circuits for Exercise 2.14.

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Figure 2.31 Circuits for Exercise 2.15.

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Figure 2.32 Circuit for illustrating the mesh-current method of circuit analysis.

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Figure 2.33 Two circuits and their mesh-current variables.

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Figure 2.34 Circuit of Example 2.13.

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Figure 2.35 Circuit of Exercise 2.19.

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Figure 2.36 Circuit of Example 2.14.

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Figure 2.37 In this circuit, we have i1 = 2 A.

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Figure 2.38 A circuit with a current source common to two meshes.

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Figure 2.39 The circuit for Exercise 2.22.

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Figure 2.40 The circuit for Exercise 2.23.

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Figure 2.41 A circuit with a voltage-controlled current source. See Example 2.15.

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Figure 2.42 A two-terminal circuit consisting of resistances and sources can be replaced by a Thévenin equivalent circuit.

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Figure 2.43 Thévenin equivalent circuit with open-circuited terminals. The open-circuit voltage voc is equal to the Thévenin voltage Vt.

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Figure 2.44 Thévenin equivalent circuit with short-circuited terminals. The short-circuit current is isc = Vt/Rt.

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Figure 2.45 Circuit for Example 2.16.

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Figure 2.46 Circuit for Exercise 2.26.

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Figure 2.47 When the source is zeroed, the resistance seen from the circuit terminals is equal to the Thévenin resistance.

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Figure 2.48 Circuit for Example 2.17.

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Figure 2.49 Circuits for Exercise 2.28.

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Figure 2.50 Circuit for Example 2.18.

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Figure 2.51 The Norton equivalent circuit consists of an independent current source In in parallel with the Thévenin resistance Rt.

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Figure 2.52 The Norton equivalent circuit with a short circuit across its terminals.

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Figure 2.53 Circuit of Example 2.19.

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Figure 2.54 Circuits for Exercise 2.29.

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Figure 2.55 A voltage source in series with a resistance is externally equivalent to a current source in parallel with the resistance, provided that In = Vt /Rt .

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Figure 2.56 Circuit for Example 2.20.

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Figure 2.57 Circuit for Exercise 2.30.

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Figure 2.58 Circuits for analysis of maximum power transfer.

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Figure 2.59 Circuit for Example 2.21.

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Figure 2.60 Circuit used to illustrate the superposition principle.

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Figure 2.61 A resistance that obeys Ohm’s law is linear.

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Figure 2.62 Circuit for Example 2.22 and Exercise 2.31.

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Figure 2.63 Circuit for Exercise 2.32.

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Figure 2.64 The Wheatstone bridge. When the Wheatstone bridge is balanced, ig=0 and vab=0.

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Figure 2.65a Strain measurements using the Wheatstone bridge.

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Figure 2.65b Strain measurements using the Wheatstone bridge.

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Figure P2.1

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Figure P2.3

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Figure P2.4

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Figure P2.6

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Figure P2.8a and b

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Figure P2.8c

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Figure P2.13

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Figure P2.16

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Figure P2.17

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Figure P2.20

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Figure P2.23

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Figure P2.24

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Figure P2.25

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Figure P2.26

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Figure P2.28

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Figure P2.29

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Figure P2.30

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Figure P2.31

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Figure P2.32

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Figure P2.33

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Figure P2.39

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Figure P2.50

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Figure P2.58

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Figure P2.59

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Figure P2.60

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Figure P2.63

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Figure P2.65

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Figure P2.83

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Figure P2.95

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Figure P2.97

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Figure P2.103

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Figure P2.104

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Table T2.1

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Table T2.1 (continued)

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Figure T2.2

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Figure T2.3

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Figure T2.4

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Figure T2.5

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Figure T2.6