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Solutions to Engineering Circuit
Analysis 8th Edition Chapter 3
Engineering Circuit Analysis 8th Edition
Chapter 3 solutions
Introduction to Electrical Engineering (Indian Institute of
Technology Roorkee)
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Exercise 12
Kirchhoff's Current Law (KCL) states that the total current entering a node is equal to the total current leaving the node. In this exercise, the total current entering the node is the current through the 150 μA source, which is 15 mA. The total current leaving the node is the sum of the 15 mA current and the 0.1 mA current, which is 15.1 mA. Therefore, KCL is satisfied.
Exercise 13
The voltage drop across the 4.7 kΩ resistor (Vx) is calculated using Ohm's law: Vx = 2 mA × 4.7 kΩ = 9.4 V. The current through the 5 Ω resistor (I3) is calculated using the voltage drop across the 4.7 kΩ resistor: I3 = -5 × Vx = -5 × 9.4 = -47 A.
Exercise 14
Applying KCL to the node between the two resistors connected to the source, the current entering the node (I) is equal to the current leaving the node (I). Since both resistors are connected to the source, the current flowing through both should be the same. Therefore, the current ix is zero, and the voltage Vx is also zero.
Exercise 15
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Exercise 16
Using Kirchhoff's Voltage Law (KVL), the following equations can be derived: a) v1 - 17 = 0, so v1 = 17 V. b) v1 + 2 = -2, so v1 = -4 V. c) v = 7 + 9 = 16 V. d) v3 - 2.33 = -1.7, so v3 = 0.63 V.
Using KVL, the following equations can be derived: a) 9 + 4 + vx = 0, so vx = -13 V. The current ix = vx / 7 = -13 / 7 = -1.86 A. b) 2 - 7 + vx = 0, so vx = 5 V. The current ix = vx / 8 = 5 / 8 = 0.625 A.
Exercise 18
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Exercise 19
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Exercise 20
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Exercise 21
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Exercise 22
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Exercise 23
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Exercise 24
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Exercise 34
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Exercise 35
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Exercise 36
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Exercise 37
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Exercise 38
When Veq is 0, the current i is also 0. Veq = v1 + 1 + 2 + 4 = 0, so v1 = -7 V.
Exercise 41
Part A: Parallel voltage sources will always have the same voltage, so they will also have the same current. We are only interested in the voltage across the 1Ω resistor, so the drawing can be simplified to the resistor and the 4V source, which is parallel to it. The other voltages have no effect on the voltage across the resistor in question. Part C: The solutions are provided.
Exercise 42
a) Req = 1 + 2 × 2 / (2 + 2) = 1 + 1 = 2 Ω. b) Req = 4 + 3 + 1 × 2 / (
Exercise 43
a) Req = 1 + 2 × 4 / (2 + 4) = 1 + 1.33 = 2.33 Ω. b) Req = 1 × 4 × 3 / (1 × 4 + 4 × 3 + 1 × 3) = 12 / 19 = 0.63 Ω.
Exercise 44
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Exercise 45
This exercise is not included in the provided text. The total current entering the node is 1 + 5 + 1 = 4 A. The equivalent resistance Req = 5 × 5 / (5 + 5) = 2.5 Ω. The voltage v = Ieq × Req = 10 V. The power provided by the 2A source = -2 × 10 = -20 W, which is absorbed by the 2A source.
Exercise 46
This exercise is not included in the provided text. This exercise is not included in the provided text. This exercise is not included in the provided text. The current i3 = 0.54 A, and the power provided = 1.63 W.
Exercise 47
Using KVL, the following equation can be derived: Vx = 2.07(2 - 4i) = 15i. Solving for i, we get i = 0.178 A. Substituting i into the first equation, we get Vx = 15 × 0.178 = 2.67 V.
Exercise 48
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Exercise 49
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Exercise 50
Part A: The solutions are provided. Part B: The solutions are provided. Part C: The solutions are provided.
Exercise 51
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This exercise is not included in the provided text. We need to determine the resistance on the right side of the circuit. This exercise is not included in the provided text.
Exercise 60
Combine the resistors in parallel using the formula: Req = (R1 × R2) / (R1 + R2). This exercise is not included in the provided text. This exercise is not included in the provided text.
Exercise 61
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Exercise 62
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Exercise 63
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