Reduce each of the networks shown in Figure P2.1 to a single equivalent resistance by combining resistances in series and parallel. (a) 3 7 30 24 3 4 12 (b) 10 5 24 60 6 6 15 9 Figure P2.1
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Textbook Solutions for Electrical Engineering: Principles & Applications
Question
Find the Thvenin and Norton equivalent circuits for the circuit shown in Figure P2.85. Take care that you orient the polarity of the voltage source and the direction of the current source correctly relative to terminals a and b. What effect does the 9- resistor have on the equivalent circuits? Explain your answer. 6 12 V 1 A 9 30 + b a Figure P2.85
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Solved: Find the Thvenin and Norton equivalent circuits
Chapter 2 textbook questions
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
A 4- resistance is in series with the parallel combination of a 20- resistance and an unknown resistance Rx. The equivalent resistance for the network is 8 . Determine the value of Rx. *
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Find the equivalent resistance between terminals a and b in Figure P2.3. 20 30 40 4 60 b a Figure P2.3
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Find the equivalent resistance looking into terminals a and b in Figure P2.4.
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Suppose that we need a resistance of 1.5 k and you have a box of 1-k resistors. Devise a network of 1-k resistors so the equivalent resistance is 1.5 k. Repeat for an equivalent resistance of 2.2 k. P2.
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
a. Determine the resistance between terminals c and d for the network shown in Figure P2.6. b. Repeat after removing the short circuit between terminals a and b. 5 20 20 5 b a c d Figure P2.6
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Two resistances R1 and R2 are connected in series. We know that R1 = 60 and that the voltage across R2 is three times the value of the voltage across R1. Determine the value of R2.
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Find the equivalent resistance between terminals a and b for each of the networks shown in Figure P2.8.
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
What resistance in parallel with 70 results in an equivalent resistance of 20 ?
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Two resistances having values of 2R and 3R are in parallel. R and the equivalent resistance are both integers.What are the possible values for R?
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
A network connected between terminals a and b consists of two parallel combinations that are in series. The first parallel Denotes that answers are contained in the Student Solutions files. See Appendix E for more information about accessing the Student Solutions. 110 Chapter 2 Resistive Circuits (a) (b) 12 17 30 35 12 12 14 a b 24 48 14 8 10 9 15 5 6 20 a b (c) a b 12 6 30 30 30 Figure P2.8 combination is composed of a 10- resistor and a 15- resistor. The second parallel combination is composed of a 14- resistor and a 35- resistor. Draw the network and determine its equivalent resistance. P2
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
The heating element of an electric cook top has two resistive elements, R1 = 40 and R2 = 100 , which can be operated separately, in series, or in parallel from voltages of either 120 V or 240 V. For the lowest power, R1 is in series with R2, and the combination is operated from 120 V. What is the lowest power? For the highest power, how should the elements be operated? What power results? List three more modes of operation and the resulting power for each.
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Find the equivalent resistance for the infinite network shown in Figure P2.13(a). Because of its form, this network is called a semiinfinite ladder. (Hint: If another section is added to the ladder as shown in Figure P2.13(b), the equivalent resistance is the same. Thus, working from Figure P2.13(b), we can write an expression for Req in terms of Req. Then, we can solve for Req.) (a) (b) 8 8 5 8 8 8 5 8 Req 8 8 Req 5 Req Ladder network of (a) Figure P2.13
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
If we connect n 1000- resistances in parallel, what value is the equivalent resistance?
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
We are designing an electric space heater to operate from 120 V. Two heating elements with resistances R1 and R2 are to be used that can be operated in parallel, separately, or in series. The highest power is to be 960 watts, and the lowest power is to be 180 watts. What values are needed for R1 and R2?What intermediate power settings are available?
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
The equivalent resistance between terminals a and b in Figure P2.16 is Rab = 40 . Determine the value of R. 70 20 30 45 a R b Figure P2.16
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Sometimes, we can use symmetry considerations to find the resistance of a circuit that cannot be reduced by series or parallel combinations. A classic problem of this type is illustrated in Figure P2.17. Twelve 1- resistors are arranged on the edges of a cube, and terminals a and b are connected to diagonally opposite corners of the cube. The problem is to find the resistance between the terminals. Approach the problem this way: Assume that 1 A of current enters terminal a and exits through terminal b. Then, the voltage between terminals a and b is equal to the unknown resistance. By symmetry considerations, we can find the current in each resistor. Then, using KVL, we can find the voltage between a and b. b a Figure P2.17
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
a. Three conductances G1, G2, and G3 are in series. Write an expression for the equivalent conductance Geq = 1/Req in terms of G1, G2, and G3. b. Repeat part (a) with the conductances in parallel.
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Most sources of electrical power behave as (approximately) ideal voltage sources. In this case, if we have several loads that we want to operate independently, we place the loads in parallel with a switch in series with each load. Thereupon, we can switch each load on or off without affecting the power delivered to the other loads. How would we connect the loads and switches if the source were an ideal independent current source? Draw the diagram of the current source and three loads with onoff switches such that each load can be switched on or off without affecting the power supplied to the other loads. To turn a load off, should the corresponding switch be opened or closed? Explain.
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Often, we encounter delta-connected loads, such as that illustrated in Figure P2.20, in three-phase power distribution systems (which are treated in Section 5.7). If we only have access to the three terminals, a method for determining the resistances is to repeatedly short two terminals together and measure the resistance between the shorted terminals and the third terminal. Then, the resistances can be calculated from the three measurements. Suppose that the measurements are Ras = 24 , Rbs = 30 , and Rcs = 40 , where Ras is the resistance between terminal a and the short between b and c, etc. Determine the values of Ra, Rb, and Rc. (Hint: You may find the equations easier to deal with if you work in terms of conductances rather than resistances. Once the conductances are known, you can easily invert their values to find the resistances.) Ra Rc Rb b c a Figure P2.20 P
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
The resistance between terminals a and b with c open circuited for the network shown in Figure P2.21 is Rab = 30 . Similarly, the resistance between terminals b and c with a open is Rbc =50, and between c and a with b open, the resistance is Rca = 40 . Now, suppose that a short circuit is connected from terminal b to terminal c, and determine the resistance between terminal a and the shorted terminals bc. Ra Rc Rb b a c Figure P2.21
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
From memory,list the steps in solving a circuit by network reduction (series/parallel combinations). Does this method always provide the solution? Explain.
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Find the values of i1 and i2 in Figure P2.23. + 10 V i1 i2 6 vx 8 8 + Figure P2.23
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Find the values of i1 and i2 in Figure P2.24. Find the power for each element in the circuit, and state whether each is absorbing or delivering energy. Verify that the total power absorbed equals the total power delivered. i1 2 A i2 4 A vx + 5 15 Figure P2.24
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Find the values of v and i in Figure P2.25. 5 25 10 20 20 10 20 20 v + 8 A i Figure P2.25
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Find the voltages v1 and v2 for the circuit shown in Figure P2.26 by combining resistances in series and parallel. 12 6 10 5 25 v1 30 + v2 + + vs = 12 V Figure P2.26
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Consider the circuit shown in Figure P2.26. Suppose that the value of vs is adjusted until v1 = 10 V. Determine the new values for v2 and vs. (Hint: Start at the location of v2 and compute currents and voltages, moving to the right and left.)
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Find the values of vs, v1, and i2 in Figure P2.28. i1 i2 25 3 A 6 20 5 + v1 + v2 + vs Figure P2.28
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Determine the values of i1 and i2 in Figure P2.29. i2 i1 i3 v2 + 2 4 7 18 30 V + Figure P2.29
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Find the voltage v and the currents i1 and i2 for the circuit shown in Figure P2.30. 6 10 40 18 9 v + i1 i2 5 A Figure P2.30
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Solve for the values of i1, i2, and the powers for the sources in Figure P2.31. Is the current source absorbing energy or delivering it? Is the voltage source absorbing energy or delivering it? Check to see that power is conserved in the circuit. i1 i2 + 4 12 V 2 A Figure P2.31
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Consider the circuit shown in Figure P2.32. With the switch open, we have v2 = 10 V. On the other hand, with the switch closed, we have v2 = 8 V. Determine the values of R2 and RL. + v2 + R2 6 16 V RL Figure P2.32
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Consider the circuit shown in Figure P2.33. Find the values of v1, v2, vab, vbc, and vca. 30 V 7 3 A 8 3 4 v1 + 2 v2 a b + + i1 c Figure P2.33
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
We know that the 10-V source in Figure P2.34 is delivering 4 W of power. All four resistors have the same value R. Find the value of R. i + R R 10 V R R Figure P2.34
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Find the values of i1 and i2 in Figure P2.35. 6 12 12 6 i1 i2 + 20 V Figure P2.35
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Use the voltage-division principle to calculate v1, v2, and v3 in Figure P2.36.
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Use the current-division principle to calculate i1 and i2 in Figure P2.37. i1 i2 3 A R2 = 5 R1 = 10 Figure P2.37
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Use the voltage-division principle to calculate v in Figure P2.38. + v + R2 = 20 R1 = 20 10 V R3 = 20 Figure P2.38
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Use the current-division principle to calculate the value of i3 in Figure P2.39. R2 = 25 R1 = 100 30 mA R3 = 50 i3 i2 Figure P2.39
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
We want to design a voltage-divider circuit to provide an output voltage vo = 5V from a 9-V battery as shown in Figure P2.40. The current taken from the 9-V source with no load connected is to be 10 mA. a. Find the values of R1and R2. b. Now suppose that a load resistance of 1 k is connected across the output terminals (i.e., in parallel with R2). Find the loaded value of vo. c. How could we change the design so the voltage remains closer to 5V when the load is connected? How would this affect the life of the battery? R1 R2 + 9 V + vo Figure P2.40
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
A series-connected circuit has a 240-V voltage source, a 10- resistance, a 5- resistance, and an unknown resistance Rx. The voltage across the 5- resistance is 30 V. Determine the value of the unknown resistance. P
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
A parallel circuit (i.e., all elements are in parallel with one another) has a 60- resistance, a 20- resistance, an unknown resistance Rx, and 30 mA current source. The current through the unknown resistance is 10 mA. Determine the value of Rx.
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
The circuit of Figure P2.43 is similar to networks used in some digital-to-analog converters. For this problem, assume that the circuit continues indefinitely to the right. Find the values of i1, i2, i3, and i4. How is in+2 related to in? What is the value of i18? (Hint: See Problem P2.13 for hints on how to handle semi-infinite networks.) R2 4 k i2 i3 i1 R4 4 k R6 4 k R8 4 k R1 = 2 k R3 = 2 k R5 = 2 k R7 = 2 k + 16 V i5 i7 i4 i6 Figure P2.43
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
A worker is standing on a wet concrete floor, holding an electric drill having a metallic case. The metallic case is connected through the ground wire of a three-terminal power outlet to power-system ground. The resistance of the ground wire is Rg. The resistance of the workers body is Rw = 500 . Due to faulty insulation in the drill, a current of 2 A flows into its metallic case. The circuit diagram for this situation is shown in Figure P2.44. Find the maximum value of Rg so that the current through the worker does not exceed 0.1 mA. Rw = 500 Rg + v i = 2 A Concrete floor Power-system ground Metallic case Figure P2.44
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
We have a 15-V source and a load that absorbs power and requires a current varying between 0 and 100 mA.The voltage across the load must remain between 4.7 and 5.0 V for all values of load current. Design a voltagedivider network to supply the load. You may assume that resistors of any value desired are available. Also, give the minimum power rating for each resistor.
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
A load resistance of 150 needs to be supplied with 5 V. A 12.6-V voltage source and resistors of any value needed are available. Draw a suitable circuit consisting of the voltage source, the load, and one additional resistor. Specify the value of the resistor.
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Suppose that we wish to supply 500 mW to a 200- load resistance RL. A 100-mA current source and resistors of any value needed are available. Draw a suitable circuit consisting of the current source, the load, and one additional resistor. Specify the value of the resistor.
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
On your own, using analytical thinking and memory, list the steps to follow in analyzing a general circuit with the node-voltage technique.
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Write equations and solve for the node voltages shown in Figure P2.49. Then, find the value of i1.
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Solve for the node voltages shown in Figure P2.50. What are the new values of the node voltages after the direction of the current source is reversed? How are the values related?
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Given R1 = 4 , R2 = 5 , R3 = 8 , R4 = 6 , R5 = 8 , and Is = 4 A, solve for the node voltages shown in Figure P2.51. v1 v2 v3 R4 R2 R3 R5 R1 Is Figure P2.51 P2.
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Given R1 = 15 , R2 = 5 , R3 = 20 , R4 = 10 , R5 = 8 , R6 = 4 , and Is = 2 A, solve for the node voltages shown in Figure P2.52. *P2.
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Solve for the node voltages shown in Figure P2.53. Then, find the value of is.
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Determine the value of i1 in Figure P2.54 using node voltages to solve the circuit. Select the location of the reference node to minimize the number of unknown node voltages. What effect does the 17- resistance have on the answer? Explain.
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
In solving a network, what rule must you observe when writing KCL equations? Why?
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Use the symbolic features of MATLAB to find an expression for the equivalent resistance for the network shown in Figure P2.56. (Hint: First, connect a 1-A current source across terminals a and b. Then, solve the network by the node-voltage technique. The voltage across the current source is equal in value to the equivalent resistance.) Finally, use the subs command to evaluate for R1 = 15 , R2 = 15 , R3 = 15 , R4 = 10 , and R5 = 10 .
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Solve for the values of the node voltages shown in Figure P2.57. Then, find the value of ix. 5 10 20 ix 1 A v1 v2 0.5ix Figure P2.57
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Solve for the node voltages shown in Figure P2.58. 5 5 10 15 10 1 A v2 v1 2vx 2 A + + vx Figure P2.58
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Solve for the node voltages shown in Figure P2.59.
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Solve for the power delivered to the 8- resistance and solve for the node voltages shown in Figure P2.60.
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Find the equivalent resistance looking into terminals ab for the network shown in 6 A 5 10 2 A v2 v1 aix 10 ix + 4 A a = 5 Figure P2.59 10 8 20 20 ix 5 A v1 v2 3 A + aix a = 5 Figure P2.60 Figure P2.61. (Hint: First, connect a 1-A current source across terminals a and b. Then, solve the network by the node-voltage technique. The voltage across the current source is equal in value to the equivalent resistance.) a b 36 26 5 + vx avx a = 0
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Find the equivalent resistance looking into terminals ab for the network shown in Figure P2.62. (Hint: First, connect a 1-A current source across terminals a and b. Then, solve the network by the node-voltage technique. The voltage across the current source is equal in value to the equivalent resistance.)
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
We have a cube with 1- resistances along each edge as illustrated in Figure P2.63, in which we are looking into the front face which has corners at nodes 1, 2, 7, and the reference node. Nodes 3, 4, 5, and 6 are the corners on the rear face of the cube. (Alternatively, you can consider it to be a planar network.) We want to find the resistance between adjacent nodes, such as node 1 and the reference node. We do this by connecting a 1-A current source as shown and solving for v1, which by symmetry is equal in value to the resistance between any two adjacent nodes. a. Use MATLAB to solve the matrix equation GV = I for the node voltages and determine the resistance. b. Modify your work to determine the resistance between nodes at the ends of a diagonal across a face, such as node 2 and the reference node. c. Finally, find the resistance between opposite corners of the cube. (Comment: Part (c) is the same as Problem 2.17 in which we suggested using symmetry to solve for the resistance. Parts (a) and (b) can also be solved by use of symmetry and the fact that nodes having the same value of voltage can be connected by short circuits without changing v1 1 A v2 v7 v5 v6 v3 v4 Figure P2.63 Each resistance is 1 . the currents and voltages. With the shorts in place, the resistances can be combined in series and parallel to obtain the answers. Of course, if the resistors have arbitrary values, the MATLAB approach will still work, but considerations of symmetry will not.)
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Figure P2.64 shows an unusual voltagedivider circuit. Use node-voltage analysis and the symbolic math commands in MATLAB to solve for the voltage-division ratio Vout/Vin in terms of the resistances. Notice that the node voltage variables are V1, V2, and Vout. + + R1 R1 R1 2R1 R1 V2 V1 R2 Vin Vout Figure P2.64
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Solve for the node voltages in the circuit of Figure P2.65. (Disregard the mesh currents, i1, i2, i3, and i4 when working with the node voltages.) + 4 k 20 V 1k i1 i2 2 k 5 k 3 k 2 k i3 i4 5 mA v2 v4 v1 v3 Figure P2.65
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
List the steps in analyzing a planar network using the mesh-current method. Attempt this as a closed-book exam problem.
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Solve for the power delivered to the 15- resistor and for the mesh currents shown in Figure P2.67. + + 20 V 5 10 i1 15 i2 10 V Figure P2.67
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Determine the value of v2 and the power delivered by the source in the circuit of Figure P2.26 by using mesh-current analysis.
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Use mesh-current analysis to find the value of i1 in the circuit of Figure P2.49.
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Solve for the power delivered by the voltage source in Figure P2.70, using the meshcurrent method. + i1 i2 i3 1 7 5 11 3 62 V Figure P2.70
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Use mesh-current analysis to find the value of v in the circuit of Figure P2.38.
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Use mesh-current analysis to find the value of i3 in the circuit of Figure P2.39. (Choose your mesh-current variables as iAand iB to avoid confusion with the current labels on the circuit diagram.)
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Use mesh-current analysis to find the values of i1 and i2 in Figure P2.30. Select i1 clockwise around the left-hand mesh, i2 clockwise around the right-hand mesh, and i3 clockwise around the center mesh.
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Find the power delivered by the source and the values of i1 and i2 in the circuit of Figure P2.23, using mesh-current analysis.
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Use mesh-current analysis to find the values of i1 and i2 in Figure P2.29. First, select iA clockwise around the left-hand mesh and iB clockwise around the right-hand mesh. After solving for the mesh currents, iA and iB, determine the values of i1 and i2.
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Use mesh-current analysis to find the values of i1 and i2 in Figure P2.28. First, select iA clockwise around the left-hand mesh and iB clockwise around the right-hand mesh. After solving for the mesh currents, iA and iB, determine the values of i1 and i2.
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
The circuit shown in Figure P2.77 is the dc equivalent of a simple residential power distribution system. Each of the resistances labeled R1 and R2 represents various parallelconnected loads, such as lights or devices plugged into outlets that nominally operate at 120 V, while R3 represents a load, such as the heating element in an oven that nominally operates at 240 V. The resistances labeled Rw represent the resistances of wires. Rn represents the neutral wire. a. Use mesh-current analysis to determine the voltage magnitude for each load and the current in the neutral wire. b. Now, suppose that due to a fault in the wiring at the distribution panel, the neutral wire becomes an open circuit. Again, compute the voltages across the loads and comment on the probable outcome for a sensitive device such as a computer or plasma television that is part of the 20- load. Rw = 0.1 R1 = 20 Rn = 0.1 Rw = 0.1 120 V R2 = 10 R3 = 16 120 V + + Figure P2.77
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Use MATLAB and mesh-current analysis to determine the value of v2 in the circuit of Figure P2.51. The component values are R1 = 4 , R2 = 5 , R3 = 8 , R4 = 6 , R5 = 8 , and Is = 4 A. P2.
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Connect a 1-V voltage source across terminals a and b of the network shown in Figure P2.56. Then, solve the network by the meshcurrent technique to find the current through the source. Finally, divide the source voltage by the current to determine the equivalent resistance looking into terminals a and b. The resistance values are R1 = 15 , R2 = 15 , R3 = 15 , R4 = 10 , and R5 = 10 . P2.
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Connect a 1-V voltage source across the terminals of the network shown in Figure P2.1(a). Then, solve the network by the meshcurrent technique to find the current through the source. Finally, divide the source voltage by the current to determine the equivalent resistance looking into the terminals. Check your answer by combining resistances in series and parallel.
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Use MATLAB to solve for the mesh currents in Figure P2.65.
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
List the steps in determining the Thvenin and Norton equivalent circuits for a general two-terminal circuit. Try this as if it were a closed-book exam question.
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Find the Thvenin and Norton equivalent circuits for the two-terminal circuit shown in Figure P2.83. 1 A 10 5 + 10 V Figure P2.83
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
We can model a certain battery as a voltage source in series with a resistance. The opencircuit voltage of the battery is 9 V. When a 100- resistor is placed across the terminals of the battery, the voltage drops to 6 V. Determine the internal resistance (Thvenin resistance) of the battery.
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Find the Thvenin and Norton equivalent circuits for the circuit shown in Figure P2.85. Take care that you orient the polarity of the voltage source and the direction of the current source correctly relative to terminals a and b. What effect does the 9- resistor have on the equivalent circuits? Explain your answer. 6 12 V 1 A 9 30 + b a Figure P2.85
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Find the Thvenin and Norton equivalent circuits for the circuit shown in Figure P2.86. 3 A 9 36 9 a b Figure P2.86
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Find the Thvenin and Norton equivalent circuits for the two-terminal circuit shown in Figure P2.87. 64 V 10 30 30 10 30 + Figure P2.87
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
A somewhat discharged automotive battery has an open-circuit voltage of 12.5 V and supplies 50 A when a 0.1- resistance is connected across the battery terminals. Draw the Thvenin and Norton equivalent circuits, including values for the circuit parameters. What current can this battery deliver to a short circuit? Considering that the energy stored in the battery remains constant under open-circuit conditions, which of these equivalent circuits seems more realistic? Explain.
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
A certain two-terminal circuit has an opencircuit voltage of 9 V. When a 200- load is attached, the voltage across the load is 7 V. Determine the Thvenin resistance for the circuit.
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
If we measure the voltage at the terminals of a two-terminal network with two known (and different) resistive loads attached, we can determine the Thvenin and Norton equivalent circuits. When a 1-k load is attached to a two-terminal circuit, the load voltage is 8 V. When the load is increased to 2 k, the load voltage becomes 10V. Find theThvenin voltage and resistance for the circuit.
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Find the Thvenin and Norton equivalent circuits for the circuit shown in Figure P2.91. a b 10 5 10 25 V 0.5ix ix + Figure P2.91
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Find the maximum power that can be delivered to a resistive load by the circuit shown in Figure P2.83. For what value of load resistance is the power maximum?
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Find the maximum power that can be delivered to a resistive load by the circuit shown in Figure P2.86. For what value of load resistance is the power maximum?
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
A battery can be modeled by a voltage source Vt in series with a resistance Rt. Assuming that the load resistance is selected to maximize the power delivered, what percentage of the power taken from the voltage source Vt is actually delivered to the load? Suppose that RL = 9Rt; what percentage of the power taken from Vt is delivered to the load? Usually, we want to design battery-operated systems so that nearly all of the energy stored in the battery is delivered to the load. Should we design for maximum power transfer?
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Figure P2.95 shows a resistive load RL connected to a Thvenin equivalent circuit. For what value of Thvenin resistance is the power delivered to the load maximized? Find the maximum power delivered to the load. (Hint: Be careful; this is a tricky question if you dont stop to think about it.) + Rt Vt = 20 V RL = 5 Figure P2.95
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Starting from the Norton equivalent circuit with a resistive load RL attached, find an expression for the power delivered to the load in terms of In, Rt, and RL. Assuming that In and Rt are fixed values and that RL is a variable, show that maximum power is delivered for RL = Rt. Find an expression for maximum power delivered to the load in terms of In and Rt.
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Use superposition to find the current i in Figure P2.97. First, zero the current source and find the value iv caused by the voltage source alone. Then, zero the voltage source and find the value ic caused by the current source alone. Finally, add the results algebraically. 30 V 10 5 i 3 A + Figure P2.97
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Solve for is in Figure P2.53 by using superposition.
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Solve the circuit shown in Figure P2.49 by using superposition. First, zero the 1-A source and find the value of i1 with only the 2-A source activated. Then, zero the 2-A source and find the value of i1 with only the 1-A source activated. Finally, find the total value of i1, with both sources activated, by algebraically adding the previous results.
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Solve for i1 in Figure P2.24 by using superposition
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Another method of solving the circuit of Figure P2.26 is to start by assuming that v2 = 1 V. Accordingly, we work backward toward the source, using Ohms law, KCL, and KVL to find the value of vs. Since we know that v2 is proportional to the value of vs, and since we have found the value of vs that produces v2 = 1 V, we can calculate the value of v2 that results when vs = 12 V. Solve for v2 by using this method.
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Use the method of Problem P2.101 for the circuit of Figure P2.23, starting with the assumption that i2 = 1 A.
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Solve for the actual value of i6 for the circuit of Figure P2.103 with Vs = 10 V, starting with the assumption that i6 = 1 A.Work back through the circuit to find the value ofVs that results in i6 = 1 A. Then, use proportionality to determine the value of i6 that results for Vs = 10 V. 12 6 8 a = 24 ix aix i6 + + Vs Figure P2.103
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Device A shown in Figure P2.104 has v = 2i 3 . a. Solve for v with the 2-A source active and the 1-A source zeroed. b. Solve for v with the 1-A source active and the 2-A source zeroed. c. Solve for v with both sources active. Why doesnt superposition apply? 2 A i A v 1 A + Figure P2.104
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
a. The Wheatstone bridge shown in Figure 2.64 on page 105 is balanced with R1 = 1 k, R3 = 3419 , and R2 = 1 k. Find Rx. b. Repeat if R2 is 100 k and the other values are unchanged. *P
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
The Wheatstone bridge shown in Figure 2.64 on page 105 has vs = 10 V, R1 = 10 k, R2 = 10 k, and Rx = 5932 . The detector can be modeled as a 5-k resistance. a. What value of R3 is required to balance the bridge? b. Suppose that R3 is 1 higher than the value found in part (a). Find the current through the detector. (Hint: Find the Thvenin equivalent for the circuit with the detector removed. Then, place the detector across the Thvenin equivalent and solve for the current.) Comment. P2.
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
In theory, any values can be used for R1 and R3 in theWheatstone bridge of Figure 2.64 on page 105. For the bridge to balance, it is only the ratio R3/R1 that is important. What practical problems might occur if the values are very small? What practical problems might occur if the values are very large?
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Derive expressions for the Thvenin voltage and resistance seen by the detector in the Wheatstone bridge in Figure 2.64 on page 105. (In other words, remove the detector from the circuit and determine the Thvenin resistance for the remaining two-terminal circuit.) What is the value of the Thvenin voltage when the bridge is balanced?
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Derive Equation 2.93 for the bridge circuit of Figure 2.65 on page 107.
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Consider a strain gauge in the form of a long thin wire having a length L and a cross-sectional area A before strain is applied. After the strain is applied, the length increases slightly to L + L and the area is reduced so the volume occupied by the wire is constant. Assume that L/L << 1 and that the resistivity of the wire material is constant. Determine the gauge factor G = R/R0 L/L (Hint: Make use of Equation 1.10.)
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Chapter 2: Problem 0 Electrical Engineering: Principles & Applications 6
Explain what would happen if, in wiring the bridge circuit of Figure 2.65 on page 107, the gauges in tension (i.e., those labeled R+ R) were both placed on the top of the bridge circuit diagram, shown in part (b) of the figure, and those in compression were both placed at the bottom of the bridge circuit diagram.
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