Solution of Fundamental of Electric Circuits Free Essays

Part 1, Problem 1 what number coulombs are spoken to by these measures of electrons: (a) 6. 482 ? 1017 (b) 1. 24 ? 1018 (c) 2. We will compose a custom exposition test on Arrangement of Fundamental of Electric Circuits or then again any comparable point just for you Request Now 46 ? 1019 (d) 1. 628 ? 10 20 Chapter 1, Solution 1 (a) q = 6. 482ãâ€"1017 x [-1. 602ãâ€"10-19 C] = - 0. 10384 C (b) q = 1. 24ãâ€"1018 x [-1. 602ãâ€"10-19 C] = - 0. 19865 (c) q = 2. 46ãâ€"1019 x [-1. 602ãâ€"10-19 C] = - 3. 941 C (d) q = 1. 628ãâ€"1020 x [-1. 602ãâ€"10-19 C] = - 26. 08 C Chapter 1, Problem 2. Decide the current coursing through a component if the charge stream is given by (a) q(t ) = (3t + 8) mC (b) q(t ) = ( 8t 2 + 4t-2) (c) q (t ) = 3e - t ? 5e ? 2 t nC (d) q(t ) = 10 sin 120? pC (e) q(t ) = 20e ? 4 t cos 50t ? C ( ) Chapter 1, Solution 2 (a) (b) (c) (d) (e) I = dq/dt = 3 mA I = dq/dt = (16t + 4) An I = dq/dt = (- 3e-t + 10e-2t) nA i=dq/dt = 1200? cos 120? t pA I =dq/dt = ? e ? 4t (80 cos 50 t + 1000 sin 50 t ) ? A PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights saved. No piece of this Manual might be shown, duplicated or disseminated in any structure or using any and all means, without the earlier composed authorization of the dist ributer, or utilized past the constrained dispersion to instructors and teachers allowed by McGraw-Hill for their individual course readiness. In the event that you are an understudy utilizing this Manual, you are utilizing it without authorization. Section 1, Problem 3. Discover the charge q(t) moving through a gadget if the current is: (an) I (t ) = 3A, q(0) = 1C (b) I ( t ) = ( 2t + 5) mA, q(0) = 0 (c) I ( t ) = 20 cos(10t + ? /6) ? A, q(0) = 2 ? C (d) I (t ) = 10e ? 30t sin 40tA, q(0) = 0 Chapter 1, Solution 3 (a) q(t) = ? i(t)dt + q(0) = (3t + 1) C (b) q(t) = ? (2t + s) dt + q(v) = (t 2 + 5t) mC q(t) = ? 10e - 30t sin 40t + q(0) = (c) q(t) = ? 20 cos (10t + ? /6 ) + q(0) = (2sin(10t + ? /6) + 1) ? C (d) 10e - 30t ( ? 0 sin 40 t †40 cos t) 900 + 1600 = ? e †30t (0. 16cos40 t + 0. 12 sin 40t) C Chapter 1, Problem 4. A current of 3. 2 A courses through a conductor. Ascertain how much charge goes through any cross-area of the conductor in 20 seconds. Part 1, Solution 4 q = it = 3. 2 x 20 = 64 C Chapter 1, Problem 5. Decide the all out charge moved over the time period ? t ? 10s when 1 I (t ) = t A. 2 Chapter 1, S olution 5 1 t 2 10 q = ? idt = ? tdt = 25 C 2 4 0 PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights saved. No piece of this Manual might be shown, imitated or circulated in any structure or using any and all means, without the earlier composed authorization of the distributer, or utilized past the constrained appropriation to instructors and teachers allowed by McGraw-Hill for their individual course arrangement. In the event that you are an understudy utilizing this Manual, you are utilizing it without consent. 10 Chapter 1, Problem 6. The charge entering a specific component is appeared in Fig. 1. 23. Locate the current at: (a) t = 1 ms (b) t = 6 ms (c) t = 10 ms Figure 1. 23 Chapter 1, Solution 6 (an) At t = 1ms, I = (b) At t = 6ms, I = dq 80 = 40 A dt 2 q = 0A dt dq 80 = â€20 A dt 4 (c) At t = 10ms, I = PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights held. No piece of this Manual might be shown, recreated or conveyed in any structure or using any and all means, without the earlier composed consent of the distributer, or utilized past the restricted disper sion to instructors and teachers allowed by McGraw-Hill for their individual course planning. On the off chance that you are an understudy utilizing this Manual, you are utilizing it without consent. Part 1, Problem 7. The charge streaming in a wire is plotted in Fig. 1. 24. Sketch the relating current. Figure 1. 4 Chapter 1, Solution 7 ? 25A, dq ? i= = †25A, dt ? ? 25A, ? 0 t I = inv(Z)*V I= 1. 6196 mA â€1. 0202 mA â€2. 461 mA 3 mA â€2. 423 mA PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights saved. No piece of this Manual might be shown, replicated or disseminated in any structure or using any and all means, without the earlier composed authorization of the distributer, or utilized past the constrained appropriation to instructors and teachers allowed by McGraw-Hill for their individual course arrangement. In the event that you are an understudy utilizing this Manual, you are utilizing it without consent. Section 3, Problem 54. Discover the work flows i1, i2, and i3 in the circuit in Fig. 3. 99. Figure 3. 99 Chapter 3, Solution 54 Let the work flows be in mA. For work 1, ? 12 + 10 + 2 I 1 ? I 2 = 0 ? ? 2 = 2 I 1 ? I 2 For work 2, ? 10 + 3I 2 ? I 1 ? I 3 = 0 For work 3, ? 12 + 2 I 3 ? I 2 = 0 ? ? ? ? (1) 10 = ? I 1 + 3I 2 ? I 3 (2) 12 = ? I 2 + 2 I (3) Putting (1) to (3) in framework structure prompts ? 2 ? 1 0 I 1 ? ? 2 ? ? ? ? ? ? ? 1 3 ? 1 I 2 ? = ? 10 ? ? 0 ? 1 2 I ? ?12 ? ? 3 ? ? ? Utilizing MATLAB, ? ? Computer based intelligence = B ? 5. 25 ? I = A B = ? 8. 5 ? ? ? ?10. 25? ? ? ?1 ? ? I 1 = 5. 25 mA, I 2 = 8. 5 mA, I 3 = 10. 25 mA Restrictive MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights saved. No piece of this Manual might be shown, repeated or dispersed in any structure or using any and all means, without the earlier composed consent of the distributer, or utilized past the restricted dissemination to instructors and teachers allowed by McGraw-Hill for their individual course planning. On the off chance that you are an understudy utilizing this Manual, you are utilizing it without consent. Section 3, Problem 55. In the circuit of Fig. 3. 100, tackle for i1, i2, and i3. Figure 3. 100 PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights held. No piece of this Manual might be shown, duplicated or dispersed in any structure or using any and all means, without the earlier composed consent of the distributer, or utilized past the constrained circulation to instructors and teachers allowed by McGraw-Hill for their individual course arrangement. On the off chance that you are an understudy utilizing this Manual, you are utilizing it without authorization. Section 3, Solution 55 10 V b I2 i1 I2 + c 1A 4A 6? I1 d I3 2? i2 4A a 12 ? I4 i3 4? +†I3 I4 8V 0 It is obvious that I1 = 4 For work 4, 12(I4 †I1) + 4(I4 †I3) †8 = 0 6(I2 †I1) + 10 + 2I3 + 4(I3 †I4) = 0 or - 3I1 + 3I2 + 3I3 †2I4 = - 5 (1) (2) (3) (4) For the supermesh At hub c, I2 = I 3 + 1 Solving (1), (2), (3), and (4) yields, I1 = 4A, I2 = 3A, I3 = 2A, and I4 = 4A At hub b, At hub an, At hub 0, i1 = I2 †I1 = - 1A i2 = 4 †I4 = 0A i3 = I4 †I3 = 2A PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights held. No piece of this Manual might be shown, imitated or appropriated in any structure or using any and all means, without the earlier composed consent of the distributer, or utilized past the restricted dispersion to instructors and teachers allowed by McGraw-Hill for their individual course readiness. On the off chance that you are an understudy utilizing this Manual, you are utilizing it without authorization. Section 3, Problem 56. Decide v1 and v2 in the circuit of Fig. 3. 101. Figure 3. 101 PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights held. No piece of this Manual might be shown, duplicated or dispersed in any structure or using any and all means, without the earlier composed authorization of the distributer, or utilized past the restricted circulation to instructors and teachers allowed by McGraw-Hill for their individual course planning. On the off chance that you are an understudy utilizing this Manual, you are utilizing it without consent. Part 3, Solution 56 + v1 †2? 2? i2 2? 2? 2? + v2 12 V + †i1 i3 †For circle 1, 12 = 4i1 †2i2 †2i3 which prompts 6 = 2i1 †i2 †i3 For circle 2, 0 = 6i2 â€2i1 †2 i3 which prompts 0 = - i1 + 3i2 †i3 For circle 3, 0 = 6i3 †2i1 †2i2 which prompts 0 = - i1 †i2 + 3i3 In lattice structure (1), (2), and (3) become, ? 2 ? 1 ? 1? ? i1 ? ?6? ? ? 1 3 ? 1? ?I ? = ? 0? ? 2 ? ? ? ? ? 1 ? 1 3 ? ?I 3 ? ?0? ? ? ? ? (1) (2) (3) 2 ? 1 ? 1 2 6 ? 1 ? = ? 1 3 ? 1 = 8, ? 2 = ? 1 3 ? 1 = 24 ? 1 ? 1 3 ? 1 0 3 2 ? 1 6 ? 3 = ? 1 3 0 = 24 , hence i2 = i3 = 24/8 = 3A, ? 1 ? 1 0 v1 = 2i2 = 6 volts, v = 2i3 = 6 volts PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights held. No piece of this Manual might be shown, repeated or appropriated in any structure or using any and all means, without the earlier composed authorization of the distributer, or utilized past the restricted dissemination to instructors and teachers allowed by McGraw-Hill for their individual course planning. On the off chance that you are an understudy utilizing this Manual, you are utilizing it without authorization. Section 3, Problem 57. In the circuit in Fig. 3. 102, discover the estimations of R, V1, and V2 given that io = 18 mA. Figure 3. 102 Chapter 3, Solution 57 Assume R is in kilo-ohms. V2 = 4k? x18mA = 72V , V1 = 100 ? V2 = 100 ? 72 = 28V Current through R is 3 iR = io , V1 = I R ? 28 = (18) R 3+ R 3+ R This prompts R = 84/26 = 3. 23 k ? Exclusive MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights held. No piece of this Manual might be shown, repeated or appropriated in any structure or using any and all means, without the earlier composed authorization of the di stributer, or utilized past the constrained dissemination to instructors and teachers allowed by McGraw-Hill for their individual course arrangement. On the off chance that you are an understudy utilizing this Manual, you are utilizing it without consent. Part 3, Problem 58. Find i1, i2, and i3 the circuit in Fig. 3. 103. Figure 3. 103 Chapter 3, Solution 58 30 ? i2 30 ? 10 ? 10 ? 30 ? i1 + i3 20 V †For circle 1, 120 + 40i1 †10i2 = 0, which prompts - 12 = 4i1 †i2 For circle 2, 50i2 †10i1 †10i3 = 0, which prompts - i1 + 5i2 †i3 = 0 For circle 3, - 120 †10i2 + 40i3 = 0, which prompts 12 = - i2 + 4i3 Solving (1), (2), and (3), we get, i1 = - 3A, i2 = 0, and i3 = 3A (1) (2) (3) PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights saved. No piece of this Manual might be shown, repeated