The Stages of Maturation in James Joyce’s Araby John...

When comparing the views of both James Joyce and John Updike on maturation from adolescence to adulthood it will be important to continually compare two of their similar works in Joyce’s â€Å"Araby† and Updike’s â€Å"AP†. James Joyce and John Updike follow similar views with the latter using Joyce as a foundation and following in similar footsteps; both authors follow a process of maturation based on the allure of love, while doing it at different stages of each of the protagonists’ lives resulting in similar views with different stages. First, both authors use the characters telling their own story in first person narration to express the protagonist’s inner thoughts and level of maturity. Second, again both Joyce and Updike use several literary metaphor’s to express the sexual and provocative intention of the main characters. Lastly, Joyce and Updike use different age, maturity and settings in their stories to define their rate of maturation and the results of the story. To begin, the first person narrative used in James Joyce’s â€Å"Araby† allows for Joyce to develop a younger character whose feelings, thoughts and maturity level is open to the understanding and criticism of the audience. The expansive level of personal description by Joyce’s narrator also indicates that the Joyce is trying to both develop the story itself but also the protagonist in terms of maturity level, general intelligence and knowledge. One of the main similarities between the two stories is the fact thatShow MoreRelatedANALIZ TEXT INTERPRETATION AND ANALYSIS28843 Words   |  116 Pagesarranged sequence of interrelated events that constitute the basic narrative structure of a novel or a short story. Events of any kind, of course, inevitably involve people, and for this reason it is virtually impossible to discuss plot in isolation from character. Character and plot are, in fact, intimately and reciprocally related, especially in modern fiction. A major function of plot can be said to be the representation of characters in action, though as we will see the action involved can be internal

What Is Ballfield Dirt Made From

Theres one out in the bottom of the first inning with a man on first base. The pitcher eyes the runner inching away from first. As he turns and throws a low curveball to the plate, the runner sprints for a second. The catcher is up with the ball and makes a strong throw, the runner slides under the infielders glove and is called safe in a cloud of dust. The crowd roars approval. The groundskeeper frowns. Thats too much dust. Runners and fielders sprint, brake, slide, and fall on the infield dirt through all nine innings. All of them rely on it for good footing. Fielders expect batted balls to bounce true on it. Each segment of the infield skin, as its called, has special problems and particular solutions. Maintaining it demands skilled hands and a geotechnical awareness. Ballfield Dirt Ingredients Ordinary soils contain organic matter and are too crumbly for sports. Ballfield dirt is a blend of water and three grades of sediment: sand, silt, and clay. Clay is mineral particles smaller than 2 micrometers, or 0.002 mm; it is plastic when wet and solid when dry. Clay furnishes strength and holds moisture. Sand (0.05 to 2 mm) and silt (0.002 to 0.05 mm) soften clays hardness and allow moisture in and out. Infield Skin The base layer of an infield skin is 10 to 15 centimeters thick and consists of 60 to 80 percent sand, 10 to 20 percent clay and the remainder silt. Given the right moisture content, this material delivers Traction—shoes dont slip or catchPlayability—balls bounce trueResilience—the ground gives when a players body strikes it A top layer of loose conditioning material, a centimeter or so thick, keeps cleats from sticking in the clay and allows players to fall safely and slide under control. It also shades the underlying soil and improves drainage in case of rain. A conditioner is made by calcining clay, roasting it at about 600 to 800Â °C to drive out the water chemically locked in the mineral. The clay expands into a lightweight, hard granular material. Also used is vitrified clay, roasted at a higher temperature and similar to the material in bricks and tiles. Finally, there is calcined diatomite, which is a nearly pure microscopic silica. The Pitchers Mound The mound and batting areas take a beating from players who dig in with their cleats, so these areas use a stronger mixture with a high clay fraction. Actual unfired bricks, 80 percent clay or more, are commonly used to build up these areas with a thin layer of infield mix on top. Watering Ballfield Dirt Daily water is the key to good ballfield dirt. If the field is too dry or too wet, it affects the quality of play and can even lead to injuries. The crew that sprays the infield before the game has already watered it several times that day. They will water it again when the game is over, or first thing the next morning. The ground can never dry out or the infield skin must be rebuilt. Watering has to take into account the regions climate, the weather that day, the presence of clouds or shadows, the wind, and even the style of play the team favors. Drainage is important for an infield skin, but not the way you might think. The clay content of infield mix does not let water percolate through it quickly; instead, the field is built with a slight slope, less than 1Â °, to direct rainwater off to the side. Maintaining Ballfield Dirt Before a game, the grounds crew loosens the upper part of the soil to soften it and prepare it for watering. They also rake and level the infield skin, then add top dressing as needed. They repeat this during the game to maintain consistent playability. If rain delays affect the game, the crew covers the infield with tarps to keep excess moisture from the skin. Afterward, they may need to remove puddles. A fine-grained calcined conditioner works for this purpose. A product made of ground corncobs is also used, but that is raked up before resuming play. The crew may also sometimes need to restore the mound or batting areas with fresh clay. Groundskeepers test their dirt every season, measuring its grain-size profile. They may have a soil lab do this work, though its basically a low-tech job involving screens, water, and beakers. But observing the soils behavior under different moisture conditions cannot be outsourced, and good groundskeepers are constantly in touch with the players and coaches as well as with the dirt itself. Umpires Mud Lets not forget the umpires. Before every game, they open a bag of regulation baseballs and rub the gloss off them using Major League Baseballs official rubbing mud, a brown, nearly pure silt from a New Jersey streambed. See the photos for my tests on this material. The true baseball fan can purchase dirt from Chicagos hallowed Wrigley Field, encapsulated in metal and accompanied by a handsome photo. Just the thing to hold as you root one more time for the Cubs.

Solution of Fundamental of Electric Circuits Free Essays

Chapter 1, Problem 1 How many coulombs are represented by these amounts of electrons: (a) 6. 482 ? 1017 (b) 1. 24 ? 1018 (c) 2. We will write a custom essay sample on Solution of Fundamental of Electric Circuits or any similar topic only for you Order 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 (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. Determine the current flowing through an element if the charge flow is given by (a) q(t ) = (3t + 8) mC (b) q(t ) = ( 8t 2 + 4t-2) C (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) A 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 reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Chapter 1, Problem 3. Find the charge q(t) flowing through a device if the current is: (a) 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 flows through a conductor. Calculate how much charge passes through any cross-section of the conductor in 20 seconds. Chapter 1, Solution 4 q = it = 3. 2 x 20 = 64 C Chapter 1, Problem 5. Determine the total charge transferred over the time interval of 0 ? t ? 10s when 1 i (t ) = t A. 2 Chapter 1, Solution 5 1 t 2 10 q = ? idt = ? tdt = = 25 C 2 4 0 0 PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 10 Chapter 1, Problem 6. The charge entering a certain element is shown in Fig. 1. 23. Find the current at: (a) t = 1 ms (b) t = 6 ms (c) t = 10 ms Figure 1. 23 Chapter 1, Solution 6 (a) At t = 1ms, i = (b) At t = 6ms, i = dq 80 = = 40 A dt 2 q = 0A dt dq 80 = = –20 A dt 4 (c) At t = 10ms, i = PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Chapter 1, Problem 7. The charge flowing in a wire is plotted in Fig. 1. 24. Sketch the corresponding current. Figure 1. 4 Chapter 1, Solution 7 ? 25A, dq ? i= = – 25A, dt ? ? 25A, ? 0 t I = inv(Z)*V I= 1. 6196 mA –1. 0202 mA –2. 461 mA 3 mA –2. 423 mA PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Chapter 3, Problem 54. Find the mesh currents i1, i2, and i3 in the circuit in Fig. 3. 99. Figure 3. 99 Chapter 3, Solution 54 Let the mesh currents be in mA. For mesh 1, ? 12 + 10 + 2 I 1 ? I 2 = 0 ? ? 2 = 2 I 1 ? I 2 For mesh 2, ? 10 + 3I 2 ? I 1 ? I 3 = 0 For mesh 3, ? 12 + 2 I 3 ? I 2 = 0 ? ? ? ? (1) 10 = ? I 1 + 3I 2 ? I 3 (2) 12 = ? I 2 + 2 I 3 (3) Putting (1) to (3) in matrix form leads to ? 2 ? 1 0 I 1 ? ? 2 ? ? ? ? ? ? ? 1 3 ? 1 I 2 ? = ? 10 ? ? 0 ? 1 2 I ? ?12 ? ? 3 ? ? ? Using MATLAB, ? ? AI = 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 PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Chapter 3, Problem 55. In the circuit of Fig. 3. 100, solve for i1, i2, and i3. Figure 3. 100 PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Chapter 3, Solution 55 10 V b I2 i1 I2 + c 1A 1A 4A 6? I1 d I3 2? i2 4A a 12 ? I4 i3 4? +– I3 I4 8V 0 It is evident that I1 = 4 For mesh 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 node c, I2 = I 3 + 1 Solving (1), (2), (3), and (4) yields, I1 = 4A, I2 = 3A, I3 = 2A, and I4 = 4A At node b, At node a, At node 0, i1 = I2 – I1 = -1A i2 = 4 – I4 = 0A i3 = I4 – I3 = 2A PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Chapter 3, Problem 56. Determine v1 and v2 in the circuit of Fig. 3. 101. Figure 3. 101 PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Chapter 3, Solution 56 + v1 – 2? 2? i2 2? 2? 2? + v2 12 V + – i1 i3 – For loop 1, 12 = 4i1 – 2i2 – 2i3 which leads to 6 = 2i1 – i2 – i3 For loop 2, 0 = 6i2 –2i1 – 2 i3 which leads to 0 = -i1 + 3i2 – i3 For loop 3, 0 = 6i3 – 2i1 – 2i2 which leads to 0 = -i1 – i2 + 3i3 In matrix form (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 , therefore 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 reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Chapter 3, Problem 57. In the circuit in Fig. 3. 102, find the values 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 3 iR = io , V1 = i R R ? 28 = (18) R 3+ R 3+ R This leads to R = 84/26 = 3. 23 k ? PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators pe rmitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Chapter 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 loop 1, 120 + 40i1 – 10i2 = 0, which leads to -12 = 4i1 – i2 For loop 2, 50i2 – 10i1 – 10i3 = 0, which leads to -i1 + 5i2 – i3 = 0 For loop 3, -120 – 10i2 + 40i3 = 0, which leads to 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 reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Chapter 3, Problem 59. Rework Prob. 3. 30 using mesh analysis. Chapter 3, Problem 30. Using nodal analysis, find vo and io in the circuit of Fig. 3. 79. Figure 3. 79 PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Chapter 3, Solution 59 40 ? –+ I0 10 ? 20 ? i2 120 V + 100V + i1 – 4v0 + – 2I0 i2 i3 v0 80 ? – i3 For loop 1, -100 + 30i1 – 20i2 + 4v0 = 0, where v0 = 80i3 or 5 = 1. 5i1 – i2 + 16i3 For the supermesh, 60i2 – 20i1 – 120 + 80i3 – 4 v0 = 0, where v0 = 80i3 or 6 = -i1 + 3i2 – 12i3 Also, 2I0 = i3 – i2 and I0 = i2, hence, 3i2 = i3 ? 3 ? 2 32 ? ? ? 1 3 ? 12? ? ? ?1 ? 3 ? 0 ? ? ? i1 ? ?10? ?i ? = ? 6 ? ? 2? ? ? ?i 3 ? ? 0 ? ? ? ? ? (1) (2) (3) From (1), (2), and (3), 3 ? 32 3 10 32 3 ? 2 10 ? = ? 1 3 ? 12 = 5, ? 2 = ? 1 6 ? 12 = ? 28, ? 3 = ? 1 3 6 = ? 84 0 3 ? 1 0 0 ? 1 0 3 0 I0 = i2 = ? 2/? = -28/5 = -5. 6 A v0 = 8i3 = (-84/5)80 = -1. 344 kvolts PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the pri or written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Chapter 3, Problem 60. Calculate the power dissipated in each resistor in the circuit in Fig. 3. 104. Figure 3. 104 Chapter 3, Solution 60 0. 5i0 4? 10 V 8? v1 1? 10 V + v2 2? – i0 At node 1, (v1/1) + (0. 5v1/1) = (10 – v1)/4, which leads to v1 = 10/7 At node 2, (0. 5v1/1) + ((10 – v2)/8) = v2/2 which leads to v2 = 22/7 P1? = (v1)2/1 = 2. 041 watts, P2? = (v2)2/2 = 4. 939 watts P4? = (10 – v1)2/4 = 18. 38 watts, P8? = (10 – v2)2/8 = 5. 88 watts PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Chapter 3, Problem 61. Calculate the current gain io/is in the circuit of Fig. 3. 105. Figure 3. 105 Chapter 3, Solution 61 v1 is 20 ? v2 10 ? i0 + v0 – 30 ? – + 5v0 40 ? At node 1, is = (v1/30) + ((v1 – v2)/20) which leads to 60is = 5v1 – 3v2 But v2 = -5v0 and v0 = v1 which leads to v2 = -5v1 Hence, 60is = 5v1 + 15v1 = 20v1 which leads to v1 = 3is, v2 = -15is i0 = v2/50 = -15is/50 which leads to i0/is = -15/50 = –0. 3 (1) PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Chapter 3, Problem 62. Find the mesh currents i1, i2, and i3 in the network of Fig. 3. 106. Figure 3. 106 Chapter 3, Solution 62 4 k? A 8 k? B 2 k? 100V + – i1 i2 i3 + – 40 V We have a supermesh. Let all R be in k? , i in mA, and v in volts. For the supermesh, -100 +4i1 + 8i2 + 2i3 + 40 = 0 or 30 = 2i1 + 4i2 + i3 At node A, At node B, i1 + 4 = i2 i2 = 2i1 + i3 (1) (2) (3) Solving (1), (2), and (3), we get i1 = 2 mA, i2 = 6 mA, and i3 = 2 mA. PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Chapter 3, Problem 63. Find vx, and ix in the circuit shown in Fig. 3. 107. Figure 3. 107 Chapter 3, Solution 63 10 ? A 5? 50 V + – i1 i2 + – 4ix For the supermesh, -50 + 10i1 + 5i2 + 4ix = 0, but ix = i1. Hence, 50 = 14i1 + 5i2 At node A, i1 + 3 + (vx/4) = i2, but vx = 2(i1 – i2), hence, i1 + 2 = i2 Solving (1) and (2) gives i1 = 2. 105 A and i2 = 4. 105 A vx = 2(i1 – i2) = –4 volts and ix = i2 – 2 = 2. 105 amp PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. (1) (2) Chapter 3, Problem 64. Find vo, and io in the circuit of Fig. 3. 108. Figure 3. 108 PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Chapter 3, Solution 64 i1 50 ? A i2 10 ? + ? i0 10 ? i2 i1 100V + + – 4i0 i3 40 ? – 0. 2V0 2A B i1 i3 For mesh 2, 20i2 – 10i1 + 4i0 = 0 (1) (2) But at node A, io = i1 – i2 so that (1) becomes i1 = (16/6)i2 For the supermesh, -100 + 50i1 + 10(i1 – i2) – 4i0 + 40i3 = 0 or At node B, But, 50 = 28i1 – 3i2 + 20i3 i3 + 0. 2v0 = 2 + i1 v0 = 10i2 so that (4) becomes i3 = 2 + (2/3)i2 (3) (4) (5) Solving (1) to (5), i2 = 0. 11764, v0 = 10i2 = 1. 1764 volts, i0 = i1 – i2 = (5/3)i2 = 196. 07 mA PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Chapter 3, Problem 65. Use MATLAB to solve for the mesh currents in the circuit of Fig. 3. 109. Figure 3. 109 Chapter 3, Solution 65 For mesh 1, –12 + 12I1 – 6I2 – I4 = 0 or 12 = 12I 1 ? 6 I 2 ? I 4 For mesh 2, For mesh 3, For mesh 4, For mesh 5, –6I1 + 16I2 – 8I3 – I4 – I5 = 0 –8I2 + 15I3 – I5 – 9 = 0 or 9 = –8I2 + 15I3 – I5 –I1 – I2 + 7I4 – 2I5 – 6 = 0 or 6 = –I1 – I2 + 7I4 – 2I5 –I2 – I3 – 2I4 + 8I5 – 10 = 0 or 10 = ? I 2 ? I 3 ? 2 I 4 + 8I 5 (2) (3) (4) (5) (1) PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Casting (1) to (5) in matrix form gives 1 0 I1 ? ?12 ? ? 12 ? 6 0 ? ? ? ? ? ? 6 16 ? 8 ? 1 ? 1 I 2 ? ? 0 ? ? 0 ? 8 15 0 ? 1 I ? = ? 9 ? 3 ? ? ? ? 7 ? 2 I 4 ? ? 6 ? ? ? 1 ? 1 0 ? 0 ? 1 ? 1 ? 2 8 I ? ?10 ? 5 ? ? ? ? ? ? AI = B Using MATLAB we input: Z=[12,-6,0,-1,0;-6,16,-8,-1,-1;0,-8,15,0,-1;-1,-1,0,7,-2;0,-1,-1,-2,8] and V=[12;0;9;6;10] This leads to Z=[12,-6,0,-1,0;-6,16,-8,-1,-1;0,-8,15,0,-1;-1,-1,0,7,-2;0,-1,-1,-2,8] Z= 12 -6 0 -1 0 -6 0 -1 16 -8 -1 -8 15 0 -1 0 7 -1 -1 -2 0 -1 -1 -2 8 gt; V=[12;0;9;6;10] V= 12 0 9 6 10 I=inv(Z)*V I= 2. 1701 1. 9912 1. 8119 2. 0942 2. 2489 Thus, I = [2. 17, 1. 9912, 1. 8119, 2. 094, 2. 249] A. PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribut ion to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Chapter 3, Problem 66. Write a set of mesh equations for the circuit in Fig. 3. 110. Use MATLAB to determine the mesh currents. 10 ? 10 ? 8? 12 V + _ 6? 4? I1 + _ 2? I2 24 V + _ 6? 8? 40 V 2? I4 4? 8? 30 V + _ I3 8? 4? I5 + _ 32 V Figure 3. 110 For Prob. 3. 66. Chapter 3, Solution 66 The mesh equations are obtained as follows. ?12 + 24 + 30I1 ? 4I2 ? 6I3 ? 2I4 = 0 or 30I1 – 4I2 – 6I3 – 2I4 = –12 ? 24 + 40 ? 4I1 + 30I2 ? 2I4 ? 6I5 = 0 or –4I1 + 30I2 – 2I4 – 6I5 = –16 –6I1 + 18I3 – 4I4 = 30 –2I1 – 2I2 – 4I3 + 12I4 –4I5 = 0 –6I2 – 4I4 + 18I5 = –32 (1) 2) (3) (4) (5) PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitt ed by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Putting (1) to (5) in matrix form ? 30 ? 4 ? 6 ? 2 0 ? ? ? 12 ? 30 0 ? 2 ? 6? ? ? 16 ? ? ? ? ? ? ? 6 0 18 ? 4 0 ? I = ? 30 ? ? ? ? ? 2 ? 2 ? 4 12 ? 4? ? 0 ? ? 0 ? 6 0 ? 4 18 ? 32? ? ? ? ? ZI = V Using MATLAB, Z = [30,-4,-6,-2,0; -4,30,0,-2,-6; -6,0,18,-4,0; -2,-2,-4,12,-4; 0,-6,0,-4,18] Z= 30 -4 -6 -2 0 -4 30 0 -2 -6 -6 0 18 -4 0 -2 0 -2 -6 -4 0 12 -4 -4 18 V = [-12,-16,30,0,-32]’ V= -12 -16 30 0 -32 ;; I = inv(Z)*V I= -0. 2779 A -1. 0488 A 1. 4682 A -0. 4761 A -2. 2332 A PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Chapter 3, Problem 67. Obtain the node-voltage equations for the circuit in Fig. 3. 111 by inspection. Then solve for Vo. 2A 4? 2? + Vo _ 3 Vo 10 ? 5? 4A Figure 3. 111 For Prob. 3. 67. Chapter 3, Solution 67 Consider the circuit below. A V1 4? V2 2? + Vo – V3 3 Vo 10 ? 5? 4A 0 ? ? 0. 35 ? 0. 25 2 + 3Vo ? 0. 25 0. 95 ? 0. 5? V = ? ? 0 ? ? ? ? ? 0 ? ? ? 0. 5 0. 5 ? 6 ? ? ? ? PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond th e limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Since we actually have four unknowns and only three equations, we need a constraint equation. Vo = V2 – V3 Substituting this back into the matrix equation, the first equation becomes, 0. 35V1 – 3. 25V2 + 3V3 = –2 This now results in the following matrix equation, 3 ? ? 0. 35 ? 3. 25 ? ? 2? 0. 25 0. 95 ? 0. 5? V = ? 0 ? ? ? ? ? ? 0 ? 6? ? 0. 5 0. 5 ? ? ? ? ? Now we can use MATLAB to solve for V. ;; Y=[0. 35,-3. 25,3;-0. 25,0. 95,-0. 5;0,-0. 5,0. 5] Y= 0. 3500 -3. 2500 3. 0000 -0. 2500 0. 9500 -0. 5000 0 -0. 5000 0. 5000 ;; I=[-2,0,6]’ I= -2 0 6 V=inv(Y)*I V= -164. 105 -77. 8947 -65. 8947 Vo = V2 – V3 = –77. 89 + 65. 89 = –12 V. Let us now do a quick check at node 1. –3(–12) + 0. 1(–164. 21) + 0. 25(–164. 21+77. 89) + 2 = +36 – 16. 421 – 21. 58 + 2 = –0. 001; answer checks! PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Chapter 3, Problem 68. Find the voltage Vo in the circuit of Fig. 3. 112. 3A 10 ? + 4A 40 ? Vo _ 25 ? 20 ? + _ 24 V Figure 3. 112 For Prob. 3. 68. Chapter 3, Solution 68 Consider the circuit below. There are two non-reference nodes. 3A V1 10 ? + Vo 25 ? 4A 40 ? Vo _ 20 ? + _ 24 V PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. ? +4+3 ? ? 7 ? ?0. 125 ? 0. 1? ? ? 0. 1 0. 19 ? V = 3 + 24 / 25? = 2. 04? ? ? ? ? ? ? Using MATLAB, we get, Y=[0. 125,-0. 1;-0. 1,0. 19] Y= 0. 1250 -0. 1000 -0. 1000 0. 1900 I=[7,-2. 04]’ I= 7. 0000 -2. 400 ;; V=inv(Y)*I V= 81. 8909 32. 3636 Thus, Vo = 32. 36 V. We can perform a simple check at node Vo, 3 + 0. 1(32. 36–81. 89) + 0. 05(32. 36) + 0. 04(32. 36–24) = 3 – 4. 953 + 1. 618 + 0. 3344 = – 0. 0004; answer checks! PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any m eans, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Chapter 3, Problem 69. For the circuit in Fig. 3. 113, write the node voltage equations by inspection. Figure 3. 113 Chapter 3, Solution 69 Assume that all conductances are in mS, all currents are in mA, and all voltages are in volts. G11 = (1/2) + (1/4) + (1/1) = 1. 75, G22 = (1/4) + (1/4) + (1/2) = 1, G33 = (1/1) + (1/4) = 1. 25, G12 = -1/4 = -0. 25, G13 = -1/1 = -1, G21 = -0. 25, G23 = -1/4 = -0. 25, G31 = -1, G32 = -0. 25 i1 = 20, i2 = 5, and i3 = 10 – 5 = 5 The node-voltage equations are: 1 ? ? v 1 ? ?20? ? 1. 75 ? 0. 25 ? ? 0. 25 1 ? 0. 25? ? v 2 ? = ? 5 ? ? ? ? ? ? 0. 25 1. 25 ? ? v 3 ? ? 5 ? ? ? ?1 ? ? ? PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Chapter 3, Problem 70. Write the node-voltage equations by inspection and then determine values of V1 and V2 in the circuit in Fig. 3. 114. V1 ix 4A 1S 2S 4ix V2 5S 2A Figure 3. 114 For Prob. 3. 70. Chapter 3, Solution 70 ? 4I x + 4 ? ?3 0? ?0 5 ? V = ? ? 4 I ? 2 ? x ? ? ? ? With two equations and three unknowns, we need a constraint equation, Ix = 2V1, thus the matrix equation becomes, ? ? 5 0? ?4? V=? ? ? 8 5? ? ? ? ? 2? This results in V1 = 4/(–5) = –0. 8V and V2 = [–8(–0. 8) – 2]/5 = [6. 4 – 2]/5 = 0. 88 V. PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Chapter 3, Problem 71. Write the mesh-current equations for the circuit in Fig. 3. 115. Next, determine the values of I1, I2, and I3. 5? I1 I3 3? 10 V + _ 1? 2? 4? I2 + _ 5V Figure 3. 15 For Prob. 3. 71. PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you a re using it without permission. Chapter 3, Solution 71 ? 9 ? 4 ? 5? ? 10 ? 4 7 ? 1? I = 5? ? ? ? ? 5 ? 1 9 ? ? 0 ? ? ? ? ? We can now use MATLAB solve for our currents. ;; R=[9,-4,-5;-4,7,-1;-5,-1,9] R= 9 -4 -5 -4 7 -1 -5 -1 9 ;; V=[10,-5,0]’ V= 10 -5 0 I=inv(R)*V I= 2. 085 A 653. 3 mA 1. 2312 A PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Chapter 3, Problem 72. By inspection, write the mesh-current equations for the circuit in Fig. 3. 116. Figure 3. 116 Chapter 3, Solution 72 R11 = 5 + 2 = 7, R22 = 2 + 4 = 6, R33 = 1 + 4 = 5, R44 = 1 + 4 = 5, R12 = -2, R13 = 0 = R14, R21 = -2, R23 = -4, R24 = 0, R31 = 0, R32 = -4, R34 = -1, R41 = 0 = R42, R43 = -1, we note that Rij = Rji for all i not equal to j. v1 = 8, v2 = 4, v3 = -10, and v4 = -4 Hence the mesh-current equations are: 0 ? i1 ? ? 8 ? ? 7 ? 2 0 ? ? 2 6 ? 4 0 ? ?i ? ? 4 ? ? ? 2 ? = ? ? 0 ? 4 5 ? 1? ?i 3 ? ? ? 10? ? ? ? ? 0 ? 1 5 ? ?i 4 ? ? ? 4 ? ? 0 PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Chapter 3, Problem 73. Write the mesh-current equations for the circuit in Fig. 3. 117. Figure 3. 117 Chapter 3, Solution 73 R11 = 2 + 3 +4 = 9, R22 = 3 + 5 = 8, R33 = 1+1 + 4 = 6, R44 = 1 + 1 = 2, R12 = -3, R13 = -4, R14 = 0, R23 = 0, R24 = 0, R34 = -1 v1 = 6, v2 = 4, v3 = 2, and v4 = -3 Hence, ? 9 ? 3 ? 4 0 ? ? i1 ? ? 6 ? 3 8 0 0 ? ?i 2 ? ? 4 ? ? = ? ? ? 4 0 6 ? 1? ?i3 ? ? 2 ? ? ? ? ? 0 ? 1 2 ? ?i 4 ? 3? ?0 PROPRIETARY MATERIAL. 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Chapter 3, Problem 74. By inspection, obtain the mesh-current e quations for the circuit in Fig. 3. 11. Figure 3. 118 Chapter 3, Solution 74 R11 = R1 + R4 + R6, R22 = R2 + R4 + R5, R33 = R6 + R7 + R8, R44 = R3 + R5 + R8, R12 = -R4, R13 = -R6, R14 = 0, R23 = 0, R24 = -R5, R34 = -R8, again, we note that Rij = Rji for all i not equal to j. ? V1 ? V ? 2? The input voltage vector is = ? ? V3 ? ? ? ? ? V4 ? ?R 1 + R 4 + R 6 ? ? R4 ? ? R6 ? ? 0 ? ? R4 R2 + R4 + R5 0 ? R5 ? R6 0 R6 + R7 + R8 ? R8 0 ? ? i 1 ? ? V1 ? ? ? i ? ? ? V ? ? R5 2? 2 ? = ? ? R8 ? ?i 3 ? ? V3 ? ? ? ? R 3 + R 5 + R 8 ? ?i 4 ? ? ? V4 ? PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Chapter 3, Problem 75. Use PSpice to solve Prob. 3. 58. Chapter 3, Problem 58 Find i1, i2, and i3 the circuit in Fig. 3. 103. Figure 3. 103 PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Chapter 3, Solution 75 * Schematics Netlist * R_R4 R_R2 R_R1 R_R3 R_R5 V_V4 v_V3 v_V2 v_V1 $N_0002 $N_0001 30 $N_0001 $N_0003 10 $N_0005 $N_0004 30 $N_0003 $N_0004 10 $N_0006 $N_0004 30 $N_0003 0 120V $N_0005 $N_0001 0 0 $N_0006 0 0 $N_0002 0 3 i1 i2 Clearly, i1 = –3 amps, i2 = 0 amps, and i3 = 3 amps, which agrees with the answers in Problem 3. 44. Chapter 3, Problem 76. PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the l imited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Use PSpice to solve Prob. 3. 27. Chapter 3, Problem 27 Use nodal analysis to determine voltages v1, v2, and v3 in the circuit in Fig. 3. 76. Figure 3. 76 Chapter 3, Solution 76 PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. * Schematics Netlist * I_I2 R_R1 R_R3 R_R2 F_F1 VF_F1 R_R4 R_R6 I_I1 R_R5 0 $N_0001 DC 4A $N_0002 $N_0001 0. 25 $N_0003 $N_0001 1 $N_0002 $N_0003 1 $N_0002 $N_0001 VF_F1 3 $N_0003 $N_0004 0V 0 $N_0002 0. 5 0 $N_0001 0. 5 0 $N_0002 DC 2A 0 $N_0004 0. 25 Clearly, v1 = 625 mVolts, v2 = 375 mVolts, and v3 = 1. 625 volts, which agrees with the solution obtained in Problem 3. 27. Chapter 3, Problem 77. Solve for V1 and V2 in the circuit of Fig. 3. 119 using PSpice. PROPRIETARY MATERIAL. 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2 ix V1 5? V2 5A 2? ix 1? 2A Figure 3. 119 For Prob. 3. 77. Chapter 3, Solution 77 PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. As a check we can write the nodal equations, ? 1. 7 ? 0. 2? ?5? V=? ? 1. 2 1. 2 ? ? ? ? ? 2? Solving this leads to V1 = 3. 111 V and V2 = 1. 4444 V. The answer checks! Chapter 3, Problem 78. PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Solve Prob. 3. 20 using PSpice. Chapter 3, Problem 20 For the circuit in Fig. 3. 9, find V1, V2, and V3 using nodal analysis. Figure 3. 69 Chapter 3, Solution 78 PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. The schematic is shown below. When the circuit is saved and simulated the node voltages are displaced on the pseudocomponents as shown. Thus, V1 = ? 3V, V2 = 4. 5V, V3 = ? 15V, . Chapter 3, Problem 79. Rework Prob. 3. 28 using PSpice. PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Chapter 3, Problem 28 Use MATLAB to find the voltages at nodes a, b, c, and d in the circuit of Fig. 3. 77. Figure 3. 77 Chapter 3, Solution 79 The schematic is shown below. When the circuit is saved and simulated, we obtain the node voltages as displaced. Thus, Va = ? 5. 278 V, Vb = 10. 28 V, Vc = 0. 6944 V, Vd = ? 26. 88 V Chapter 3, Problem 80. PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Find the nodal voltage v1 through v4 in the circuit in Fig. 3. 120 using PSpice. Figure 3. 120 Chapter 3, Solution 80 PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. * Schematics Netlist * H_H1 VH_H1 I_I1 V_V1 R_R4 R_R1 R_R2 R_R5 R_R3 $N_0002 $N_0003 VH_H1 6 0 $N_0001 0V $N_0004 $N_0005 DC 8A $N_0002 0 20V 0 $N_0003 4 $N_0005 $N_0003 10 $N_0003 $N_0002 12 0 $N_0004 1 $N_0004 $N_0001 2 Clearly, v1 = 84 volts, v2 = 4 volts, v3 = 20 volts, and v4 = -5. 333 volts Chapter 3, Problem 81. Use PSpice to solve the problem in Example 3. 4 PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Example 3. 4 Find the node voltages in the circuit of Fig. 3. 12. Figure 3. 12 Chapter 3, Solution 81 PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Clearly, v1 = 26. 67 volts, v2 = 6. 667 volts, v3 = 173. 33 volts, and v4 = -46. 67 volts which agrees with the results of Example 3. 4. This is the netlist for this circuit. * Schematics Netlist * R_R1 R_R2 R_R3 R_R4 R_R5 I_I1 V_V1 E_E1 0 $N_0001 2 $N_0003 $N_0002 6 0 $N_0002 4 0 $N_0004 1 $N_0001 $N_0004 3 0 $N_0003 DC 10A $N_0001 $N_0003 20V $N_0002 $N_0004 $N_0001 $N_0004 3 Chapter 3, Problem 82. If the Schematics Netlist for a network is as follows, draw the network. PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. R_R1 R_R2 R_R3 R_R4 R_R5 V_VS I_IS F_F1 VF_F1 E_E1 1 2 2 3 1 4 0 1 5 3 2 0 0 4 3 0 1 3 0 2 2K 4K 8K 6K 3K DC DC VF_F1 0V 1 100 4 2 3 3 Chapter 3, Solution 82 2i0 + v0 – 3 k? 1 4A 2 k? 2 + 3v0 3 6 k? 4 4 k? 8 k? 100V + – 0 This network corresponds to the Netlist. Chapter 3, Problem 83. PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. The following program is the Schematics Netlist of a particular circuit. Draw the circuit and determine the voltage at node 2. R_R1 R_R2 R_R3 R_R4 V_VS I_IS 1 2 2 3 1 2 2 0 3 0 0 0 20 50 70 30 20V DC 2A Chapter 3, Solution 83 The circuit is shown below. 1 20 ? 2 70 ? 3 20 V + – 50 ? 2A 30 ? 0 When the circuit is saved and simulated, we obtain v2 = –12. 5 volts Chapter 3, Problem 84. Calculate vo and io in the circuit of Fig. 3. 121. Figure 3. 121 Chapter 3, Solution 84 From the output loop, v0 = 50i0x20x103 = 106i0 (1) From the input loop, 3Ãâ€"10-3 + 4000i0 – v0/100 = 0 (2) From (1) and (2) we get, i0 = 0. 5? A and v0 = 0. 5 volt. Chapter 3, Problem 85. PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. An audio amplifier with resistance 9? supplies power to a speaker. In order that maximum power is delivered, what should be the resistance of the speaker? Chapter 3, Solution 85 The amplifier acts as a source. Rs + Vs RL For maximum power transfer, R L = R s = 9? Chapter 3, Problem 86. For the simplified transistor circuit of Fig. 3. 122, calculate the voltage vo. Figure 3. 122 Chapter 3, Solution 86 Let v1 be the potential across the 2 k-ohm resistor with plus being on top. Then, [(0. 03 – v1)/1k] + 400i = v1/2k (1) Assume that i is in mA. But, i = (0. 03 – v1)/1 Combining (1) and (2) yields, v1 = 29. 963 mVolts and i = 37. nA, therefore, v0 = -5000x400x37. 4Ãâ€"10-9 = -74. 8 mvolts (2) Chapter 3, Problem 87. For the circuit in Fig. 3. 123, find the gain vo/vs. PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution t o teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Figure 3. 123 Chapter 3, Solution 87 v1 = 500(vs)/(500 + 2000) = vs/5 v0 = -400(60v1)/(400 + 2000) = -40v1 = -40(vs/5) = -8vs, Therefore, v0/vs = –8 Chapter 3, Problem 88. Determine the gain vo/vs of the transistor amplifier circuit in Fig. 3. 124. Figure 3. 124 Chapter 3, Solution 88 Let v1 be the potential at the top end of the 100-ohm resistor. (vs – v1)/200 = v1/100 + (v1 – 10-3v0)/2000 For the right loop, v0 = -40i0(10,000) = -40(v1 – 10-3)10,000/2000, or, v0 = -200v1 + 0. 2v0 = -4Ãâ€"10-3v0 (2) (1) Substituting (2) into (1) gives, (vs + 0. 004v1)/2 = -0. 004v0 + (-0. 04v1 – 0. 001v0)/20 This leads to 0. 125v0 = 10vs or (v0/vs) = 10/0. 125 = -80 Chapter 3, Problem 89. PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to t eachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. For the transistor circuit shown in Fig. . 125, find IB and VCE. Let ? = 100 and VBE = 0. 7V. _ 3V + _ 1 k? 0. 7 V + 100 k? | | 15 V Figure 3. 125 For Prob. 3. 89. Chapter 3, Solution 89 Consider the circuit below. _ 0. 7 V C + 100 k? + IC VCE _ 3V + _ E For the left loop, applying KVL gives VBE = 0. 7 ? 3 ? 0. 7 + 100 x103 IB + VBE = 0 IB = 30 ? A For the right loop, ? VCE + 15 ? Ic(1Ãâ€"10 3 ) = 0 But IC = ? IB = 100 x30 ? A= 3 mA | | 15 V 1 k? VCE = 15 ? 3 x10 ? 3 x103 = 12 V Chapter 3, Problem 90. PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Calculate vs for the transistor in Fig. 3. 126, given that vo = 4 V, ? = 150, VBE = 0. 7V. Figure 3. 126 Chapter 3, Solution 90 1 k? 10 k? IB + VBE + VCE – i2 + vs + – – i1 18V 500 ? + V0 – IE – For loop 1, -vs + 10k(IB) + VBE + IE (500) = 0 = -vs + 0. + 10,000IB + 500(1 + ? )IB which leads to vs + 0. 7 = 10,000IB + 500(151)IB = 85,500IB But, v0 = 500IE = 500x151IB = 4 which leads to IB = 5. 298Ãâ€"10-5 Therefore, vs = 0. 7 + 85,500IB = 5. 23 volts Chapter 3, Problem 91. PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. For the transistor circuit of Fig. 3. 127, find IB, VCE, and vo. Take ? = 200, VBE = 0. 7V. Figure 3. 127 Chapter 3, Solution 91 We first determine the Thevenin equivalent for the input circuit. RTh = 6||2 = 6Ãâ€"2/8 = 1. 5 k? and VTh = 2(3)/(2+6) = 0. 75 volts 5 k? IC 1. 5 k? IB + VBE + VCE – i2 + + 0. 75 V – – i1 9V 400 ? + V0 – IE B – For loop 1, -0. 75 + 1. 5kIB + VBE + 400IE = 0 = -0. 75 + 0. 7 + 1500IB + 400(1 + ? )IB B B IB = 0. 05/81,900 = 0. 61 ? A B v0 = 400IE = 400(1 + ? IB = 49 mV B For loop 2, -400IE – VCE – 5kIC + 9 = 0, but, IC = ? IB and IE = (1 + ? )IB B B VCE = 9 – 5k? IB – 400(1 + ? )IB = 9 – 0. 659 = 8. 641 volts B B Chapter 3, Problem 92. PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, w ithout the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Find IB and VC for the circuit in Fig. 3. 128. Let ? = 100, VBE = 0. 7V. Figure 3. 128 Chapter 3, Solution 92 10 k? I1 5 k? VC IC IB + + VBE 4 k? VCE – – 12V + + V0 – IE – I1 = IB + IC = (1 + ? )IB and IE = IB + IC = I1 Applying KVL around the outer loop, 4kIE + VBE + 10kIB + 5kI1 = 12 12 – 0. 7 = 5k(1 + ? )IB + 10kIB + 4k(1 + ? )IB = 919kIB IB = 11. 3/919k = 12. 296 ? A Also, 12 = 5kI1 + VC which leads to VC = 12 – 5k(101)IB = 5. 791 volts Chapter 3, Problem 93 Rework Example 3. 1 with hand calculation. PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. I f you are a student using this Manual, you are using it without permission. In the circuit in Fig. 3. 34, determine the currents i1, i2, and i3. Figure 3. 34 Chapter 3, Solution 93 ? 4? v1 i1 2? i 2? 3v0 v2 i3 + 8? 2? 3v0 i2 4? i + v0 + + v2 24V + + v1 – – – – (a) (b) From (b), -v1 + 2i – 3v0 + v2 = 0 which leads to i = (v1 + 3v0 – v2)/2 At node 1 in (a), ((24 – v1)/4) = (v1/2) + ((v1 +3v0 – v2)/2) + ((v1 – v2)/1), where v0 = v2 or 24 = 9v1 which leads to v1 = 2. 667 volts At node 2, ((v1 – v2)/1) + ((v1 + 3v0 – v2)/2) = (v2/8) + v2/4, v0 = v2 v2 = 4v1 = 10. 66 volts Now we can solve for the currents, i1 = v1/2 = 1. 333 A, i2 = 1. 333 A, and i3 = 2. 6667 A. PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Chapter 4, Problem 1. Calculate the current io in the circuit of Fig. 4. 69. What does this current become when the input voltage is raised to 10 V? Figure 4. 69 Chapter 4, Solution 1. + ? 8 (5 + 3) = 4? , i = io = 1 1 = 1+ 4 5 1 i= = 0. 1A 2 10 Since the resistance remains the same we get i = 10/5 = 2A which leads to io = (1/2)i = (1/2)2 = 1A. PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Chapter 4, Problem 2. Find vo in the circuit of Fig. 4. 70. If the source current is reduced to 1 ? A, what is vo? Figure 4. 70 Chapter 4, Solution 2. 6 (4 + 2) = 3? , i1 = i 2 = io = 1 A 2 1 1 i1 = , v o = 2i o = 0. 5V 2 4 If is = 1? A, then vo = 0. 5? V PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Chapter 4, Problem 3. (a) In the circuit in Fig. 4. 71, calculate vo and Io when vs = 1 V. (b) Find vo and io when vs = 10 V. (c) What are vo and Io when each of the 1-? resistors is replaced by a 10-? resistor and vs = 10 V? Figure 4. 71 Chapter 4, Solution 3. + ? + vo + ? (a) We transform the Y sub-circuit to the equivalent ? . 3R 2 3 3 3 3 = R, R + R = R R 3R = 4R 4 4 4 2 vs vo = independent of R 2 io = vo/(R) When vs = 1V, vo = 0. 5V, io = 0. 5A (b) When vs = 10V, vo = 5V, io = 5A (c) When vs = 10V and R = 10? vo = 5V, io = 10/(10) = 500mA PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Chapter 4, Problem 4. Use linearity to determine io in the circuit in Fig. 4. 72. Figure 4. 72 Chapter 4, Solution 4. If Io = 1, the voltage across the 6? resistor is 6V so that the current through the 3? resistor is 2A. + v1 3 6 = 2? , vo = 3(4) = 12V, i1 = Hence Is = 3 + 3 = 6A If Is = 6A Is = 9A Io = 1 Io = 9/6 = 1. 5A vo = 3A. 4 PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Chapter 4, Problem 5. For the circuit in Fig. 4. 73, assume vo = 1 V, and use linearity to find the actual value of vo. Figure 4. 73 Chapter 4, Solution 5. + ? If vo = 1V, ?1? V1 = ? ? + 1 = 2V ? 3? 10 ? 2? Vs = 2? ? + v1 = 3 ? 3? If vs = 10 3 vo = 1 vo = 3 x15 = 4. 5V 10 Then vs = 15 PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Chapter 4, Problem 6. For the linear circuit shown in Fig. 4. 74, use linearity to complete the following table. Experiment 1 2 3 4 Vs 12 V -1V -Vo 4V 16 V –2V + Vs + _ Linear Circuit Vo – Figure 4. 74 For Prob. 4. 6. Chapter 4, Solution 6. Due to linearity, from the first experiment, 1 Vo = Vs 3 Applying this to other experiments, we obtain: Experiment 2 3 4 Vs 48 1V -6 V Vo 16 V 0. 333 V -2V PROPRIETARY MATERIAL.  © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Chapter 4, Problem 7. Use linearity and the assumption that Vo = 1V to find the actual value of Vo in Fig. 4. 75. . 1? 4? + 4V + _ 3? 2? Vo _ Figure 4. 75 For Prob. 4. 7. Chapter 4, Solution 7. If Vo = 1V, then the current through the 2-? and 4-? resistors is ? = 0. 5. The voltage across the 3-? resistor is ? (4 + 2) = 3 V. The total current through the 1-? resistor is 0. 5 +3/3 = 1. 5 A. Hence the How to cite Solution of Fundamental of Electric Circuits, Essay examples

Case Study for Project Plan of Perth Stadium †Free Sample

Question: The Project Charter should explain the background of the organisation where this project will be carried out, the objectives of the project and the importance of this project to the organisation. The project team should also include a Business Case, Milestone Schedule, as well as any major problems or anticipated Risks with this project if they are known at this stage. The textbook provides detailed examples of what needs to be included in the Project Charter. These examples will assist the project team in the development of the Project Charter for My Project.The project that your team uses could be from one of the following categories depending on the type of business a student is involved in: an organisational project that a student is involved in as the organisations main line of business to manage projects for your organisation or on behalf of others an activity that a student is convinced would benefit from being handled as a project an activity in which a student was involved in the past that was not carried out as a project, but one which you believe would have been managed better as a project. Answer: 1. Introduction: Currently in Western Australia, Brookfield Multiplex is reconstructing famous Perth Stadium. Westadium Consortium (WC) is handling the whole project and Brookfield Multiplex is handling construction works. The group of Lakhwinder and Navdeep is also a part of Westadium Consortium and the group is actively involved into completion of the new stadium. The current assignment deals with planning and management of Perth stadium reconstruction project. 2. Project Charter for New Perth stadium development: Charter Element Description Scope Stadium development, transportation infrastructure development, refreshment centers Business Case behind the project part of locality development Milestone Schedule Construction starts in 2015 Seat installation in 2016 Surface development in 2017 Analysis of risk Change in international sports authoritys guidelines Estimated Budget $1.1 billion Stakeholders of the project Westadium Consortium The government of Western Australia Local people, players, vistors etc. Lessons Importance of government support in large constructions Operating principle of teams Different departments operate in collaboration Table1: Project charter for New Perth Stadium Development project 2.1 Scope analysis: Current project for new Perth Stadium primarily aims at construction of a five-tiered multipurpose stadium, which will be able to different sports including Cricket, soccer and AFL Football, rugby and other entertainment events. Analysis of project scope is important to develop detailed plan (Kloppenborg, 2012). Currently the stadium will have capacity of 60000 seats (afl.com.au, 2015). However, seating capacity can be increased up to 70000 in future (afl.com.au, 2015). In the newly built stadium, comfort of supporters is prioritized the most and covered seats will account 85% of total stadium capacity (Get The Bigger Picture, 2015). The project also includes installation of two 240 sqm video screens (Perthstadium.com.au, 2015). The design is focused o the aesthetic look of stadium. The project also deals with constructions of a bronze-entrance, which will display geological features of Western Australia. Secondary Constructions included in the project are pedestrian paths, community halls, and places for BBQ, playgrounds and three recreational spaces. In this stadium, LED lights will be installed which will be changed according to the journey color of home teams at night. Apart from this, there will be 4G Wi-Fi connections and provision of future installation of more 1000 screens. WC is not only undertaking construction activities, it is also responsible for proper maintenance of the stadium for next 25 years (Brookfieldmultiplex.com, 2015). 2.2 Business Case analysis: According to Eric Kirkland, (2014) business case analysis enables to understand real purposes of conducting the project. The new Perth stadium construction is an integral part of Burswood Peninsula redevelopment activities. However, the new stadium can be used as a center for different sports and entertainment activities. As the new stadium has the infrastructure of hosting different sports and entertainment activities, many international and national events will be organized here. This new stadium will require large number of staffs when it will become operational. So the current project is expected to improve the employment statistics of the locality. It is also expected that the new stadium will enhance popularity of Perth to tourists and it will contribute significantly on the local economy. 2.3 Milestone schedule and success criteria: Milestones Date of Completion Stakeholders Success Criteria Current Condition: Pre construction site works November 2013 Ertech Keller Joint Venture(EKJV) 50% completion of the work RFP (Request For Proposal) Release and receiving proposals December ,2013 Western Australia Authority Receiving RFPs Assessing and selecting proposals Mid-2014 Western Australia Authority Announcing award of contract to Westadium Consortium Commencement of detailed design development Mid-2014 Westadium Consortium Detailed Design development Starting of construction process 2015 Westadium Consortium Brookfield Multiplex Completion of the stadium within stipulated time Future: Placing 8 tower cranes 2015 Westadium Consortium Brookfield Multiplex Starting Seat installation 2016 Westadium Consortium Brookfield Multiplex Finishing seat installation on time Starting installation of bronze faced 2016 Westadium Consortium Brookfield Multiplex Finising faced installation successfully Hiring 5700 workers By end of 2016 Westadium Consortium Brookfield Multiplex Enhancing speed of construction Development of playing surface By mid 2017 Westadium Consortium Brookfield Multiplex Cpmpletion development of grass surface Stadium becoming fully operational March,2018 Westadium Consortium Brookfield Multiplex Western Australia Authority Successful organization of events Maintenance of the stadium For Next 25 years Westadium Consortium Preventing the stadium from being damaged and keeping it operational Table 2: Milestone schedules and Success Criteria for stadium development project (Source: Perthstadium.com.au, 2015) Our current project is segmented to different subtasks. Setting realistic milestones is one of the most important tasks for smooth execution of the project (Kendrick, 2011).The current project commenced in July 2013 though inviting various organizations to undertake it. The tendering process took place during the period August 2013 to December; 2013.Finally in mid 2014, Westadium Consortium (WC) received the responsibility to undertake the project. Before WC undertook the project, some pre construction works took place at the site by EKJV and half of these works were completed within November,2013.Designing team of the organization started developing the detailed design of new stadium from mid 2014.Finally the construction process commenced since late 2014. Currently the new stadium is under construction. The whole construction is expected to be completed by December 2017 (PerthNow, 2015).Future target for this project is making the stadium fully operational from March 2018.Although the construction work has just been started, the first target is placing 8 tower cranes in the stadium area by 2015. The next target in the timeline is starting seat installation from early 2016.According to the set timeline, installation of the bronze faced, will also start from early 2016.By the end of 2016, large number of workers will be hired in the project to enhance speed. The playing surface will be finished by mid 2017. 2.4 Risk analysis and Constraint identification: As we are part of a large project, analysis of risks is important for timely completion. As stated by Rybka and Bondar-Nowakowska, (2013) risk analysis is also important to limit the budget. Change in Guidelines of Sports authorities: The stadium design indicates that it will be used to organize different sports including cricket, football, rugby etc. Current design of the stadium is developed according to the guidelines of these sports authorities. As it is a large project, it will take time to be completed. If the sports authorities such as FIFA or ICC change their rules and standards regarding stadium, exiting design may need to be modified. It will cause adverse impacts on by both time and cost for successful completion of the project. Damage of property caused by Supporters: WC is not only responsible for developing the stadium; it is also responsible for maintaining it when it will become operational. Often it is observed that the agitated supporters cause loss or damage of properties in a stadium. It will also affect on the operating cost of new Perth Stadium. Apart from this, it should be ensured that the stadium is being developed in a sustainable manner and it is not affecting environment. On other hand, the Australian government also emphasizes on providing comfort to spectators. Any change in environmental standards may require change in construction design and material. Time constraint for the project: New Stadium development in Perth is a large project and it includes completion of several constructions such as multi-storied stadiums, road networks and recreation centers. The construction work has just started in 2014 and according to the timeline; WC will complete it by December, 2017.Due to the high complexity of designs, construction may not be finished within time. 2.5 Estimated Budget: According to the current plan, the stadium will require more than $1.1 billion to be built. However, about half of this fund will be collected from taxes paid by citizens. Rest amount is collected by a financial plan of WC. 2.6 Stakeholders of the project: Stakeholders of the current projects are constructors-Brookfield Multiplex, project managers -Westadium Consortium, the Government of Western Australia. Members of WC except Brookfield Multilex are BrookField Financial, Brookfield Johnson controls and John Liang. Ponnappa, (2014) opined that success of a project depends on contribution of stakeholders in it. John Liang is responsible for asset management and equity investment. Brookfield financial is responsible for fund allocation and Brookfield Johnson looks after facility management activities. All these members are stakeholders of new stadium development project. Apart from these local people of Perth can also be considered as stakeholders as the project may influence their lives also. Stakeholder Identification and Prioritization matrix Brookfield Multiplex Brookfield Financial Brookfield Johnson controls John Liang Western Australia Government Local people Importance of project to stakeholder Successful completion of construction works Maintaining cash flow and providing financial advises Managing core project activities and makes it profitable to WC Fund development Successful completion of the project Making the new stadium operational Successful completion may lead to financial growth and also can increase job opportunities Power High: Taking decisions regarding construction of stadium and surrounding area High: Making cash flow and budget allocation strategies High: Responsible for undertaking main tasks related to stadium development High: Develops strategies regarding stock market activities High: Made all policies regarding stadium development Low: Cannot take part in project works directly Interest High: Completing all primary and secondary construction works within time and budget High: Completing the project without hampering quality and making profit for WC High: Ensuring quality of tasks while maintaining profitability High: Development of fund High: Successful completion of project will lead to financial growth in local area. Medium: Successful completion of stadium can lead to prosperity indirectly Influence High Influence is high as construction is an important part of stadium development project High: Interrupted cash flow and budget allocation can interrupt progress of whole project High: Ensures that the tasks related to stadium development are executed efficiently High: Ensures uninterrupted process of stadium development High: Can change policies anytime Medium: Cannot take part directly in policy making process Impact High Quality of the project depends on construction work High: Timely completion depends on efficiency of financial allocations High: Controls efficiency and thus improves quality of new stadium High: Efficient funding ensures timely completion of stadium High: Change in government policies can cause alteration in design and time limit Low: Cannot alter project works Urgency High Delay in construction can cause delay in total project High Financial allocation is important to avoid unwanted delays High As it is a large construction project, quality is important High: Quality of the project depends on funding High: Main client of the project Low: Indirectly related to project Legitimacy It is a member of Westadium Consortium (WC)(Project manager) It is a member of Westadium Consortium (WC)(Project manager) Facilities manager and member of WC Equity Investor and member of WC Project owner Stadium construction will impact on lives of Perth Residents indirectly Total High High High High High Low Priority Key Key Key Key Key Other (secondary) Table 3: Stakeholder Identification and Prioritization matrix for stadium development 2.7 Lessons: The Western Australia is undertaking the project of stadium development in PPP model and it is part of Burswood re-development process. Our team got the opportunity to take part in this project; we learned valuable lessons while working for WC. These are as follows, Importance of Government support for large structure development: We, Lakhwinder and Navdeep, realized the importance of government support for large constructions while working in this project. In case of our project, new stadium development in Perth, huge fund is required for its successful execution. In this case, funding has not become of problem as the government collected almost 60% of total amount from taxpayers (PerthNow, 2015). Although there are environmental constraints, the governments focus on attracting visitors helped to designers to reduce complexity caused by strict environmental standards. Importance of provision of change: As the proposed stadium is a multipurpose one, its design needs to be changed due to modification in any of the sports guidelines. Apart from this, it may require to expand in future also. To avoid future problems the design is made flexible for changes. 2.8 Operating principle of teams: The current project is being executed through the collaboration of different departments. Regular meetings are held among teams o these to discuss status of work and new target development. However, each team sets their own course of action to complete their tasks. 3. Communication Plan: The communication plan for the project is presented here through a communication matrix. Communication plan is important to execute operational activities and it is helpful for strategic planning (Kloppenborg, 2012). SI Purpose Structure Time Method 1. Stake holders: Both members of team(Lakhwinder and Navdeep) Purpose: Planning to complete the assignment, resource identification, budget allocation Face to face communication Before starting construction Pull: Meeting 2. Stake holders : Navdeep and Lakhwinder Purpose: Resource allocation Face to face communication and documentation After design development Pull: meeting 3. Target Audience: Lakhwinder and Navdeep Purpose: Progress in work File sharing Once in a week Push: E-mail 4. Both members of team(Lakhwinder and Navdeep) Purpose: Issues causing delay and finding solutions Face to face communication and documentation After finishing milestones like placing tower cranes Pull: Meeting Table 3: Communication plan We, Lakhwinder and Navdeep implemented a meeting management process to finish our tasks. Figure1: Meeting management process used by us First meeting: Before starting the task, we arranged for a discovery meeting. In this meeting, we discussed on the sub-tasks, which we need to complete. In this meeting, we also decided the milestones, which will be completed to finish the whole project on time. Both of us contributed equally to set the milestones. However, we both took responsibility of some tasks separately and we developed our own plans to finish these tasks separately. There were some tasks, which we decided to complete as a group. Planning for these tasks was done in collaboration with each other. Identification of resources was another important part of our first meeting. We both used our knowledge to identify which resources are needed by us. As it was our first meeting, we also discussed on the budget of our project and developed plans for funding it. However such meetings are arranged whenever we needed to discuss on the current status of projects. Second meeting: our second meeting was arranged to allocate resources for the project. As I am the manager of our group, resource allocation was my responsibility. I took three days to collect all resources. Second meeting was held after three days of first meeting. In this meeting, I provided resources to another member of our group, Navdeep .Although the meeting was arranged by me, we both decided how to use these resources efficiently. Third meeting: After second meeting, we both started to complete our tasks. It was important to know progress of tasks. Navdeep communicated once in each week for informing me regarding the progress. Most of the time, we used emails for such communication. Although Navdeep initiated such communications, I also informed him about the progress in my part. Fourth meeting: Both of us faced different issues while completing our tasks. Once in every week we met to discuss on the issues, which are causing delay to finish tasks. In these meetings, we also developed strategies for eliminating problems. These meetings are also used to identify future issues and modification of our current strategies. 4. WBS for new Stadium development in Perth: Figure1: WBS for new stadium development in Perth WBS development helps to identify subtasks for completion of whole project (Randolph, 2014).WC is executing the project in three levels. Deliverable of the first level is finished stadium. However, completion of this level is dependent on activities of level two. Level two activities include inviting RFPs, selecting the project managers, detailed planning of the project and conducting operational activities. The Government of Western Australia started the process in 2013. The organizations, which are willing to undertake the project, got sufficient time to develop the proposal. However, after receiving these RFPs, the authority analysed all proposals and selected WC as the project manager. After completion of tendering process, WC started conducting project management activities for the new stadium. Developing design of the stadium was one of the most complex tasks among level 3 activities as the stadium should be able to host both sports and entertainment events. Seating capacity of the new stadium is also large and it is ensured that the capacity will be increased in future. As the Western Australia authority focused on y providing spectators high quality services, WC required to design large number of restaurants and refreshment centers. As the stadium capacity allows large number of visitors every day, transportation is one of the most important factors for its smooth operation. WC is also in charge of the transportation infrastructure development at the surrounding area. According to the proposal, the transportation infrastructure of new stadium will include both train and bus services. Apart from these, pedestrian bridge on Swan River will also be used to vacate the stadium after an event. Thus, designing process conducted by WC also included designing of a six-platform railway station in stadium area, bus terminuses and pedestrian bridge (Perthstadium.com.au, 2015). The stadium also required to be developed according to the guidelines of different sports events and it increased difficulty of the whole process. On completion of designing process, WC authority started other level 3 activities including resource allocation, setting milestones and hiring staffs. Before WC commenced the construction process, the local authority started conducting pre-construction activities in stadium ground. About 50% of these works finished before the main construction was started. Other activities of this level such as installation of seats, decoration of stadium and developing grass surface are yet to be done. References afl.com.au, (2015). Design plans for new Perth stadium unveiled - AFL.com.au. [online] Available at: https://www.afl.com.au/news/2014-07-17/new-perth-stadium-unveiled [Accessed 24 Feb. 2015]. Brookfieldmultiplex.com, (2015). Brookfield Multiplex :: Westadium Design For New Perth Stadium Revealed. [online] Available at: https://www.brookfieldmultiplex.com/newsfeed/view/westadium_design_for_new_perth_stadium_revealed_2014_07_17 [Accessed 24 Feb. 2015]. Eric Kirkland, C. (2014). Project Management: A Problem-Based Approach. Project Management Journal, 45(1) Get The Bigger Picture, (2015). New Perth Stadium. [online] Available at: https://getthebiggerpicture.wa.gov.au/new-perth-stadium/ [Accessed 24 Feb. 2015]. Kendrick, T. (2011). 101 project management problems and how to solve them. New York: AMACOM, American Management Association. Kloppenborg, T. (2012). Contemporary project management. Mason, Ohio: South-Western Cengage Learning. PerthNow, (2015). First look at new Perth Stadium train station at Burswood. [online] Available at: https://www.perthnow.com.au/news/western-australia/first-look-at-new-perth-stadium-train-station-at-burswood/story-fnhocxo3-1227222675576 [Accessed 24 Feb. 2015]. Perthstadium.com.au, (2015). Project timeline. [online] Available at: https://www.perthstadium.com.au/project-design/project-timeline [Accessed 24 Feb. 2015]. Perthstadium.com.au, (2015). Stadium Station design revealed. [online] Available at: https://www.perthstadium.com.au/news-and-information/latest-news/news-article/2015/02/17/stadium-station-design-revealed [Accessed 24 Feb. 2015]. Ponnappa, G. (2014). Project Stakeholder Management. Project Management Journal, 45(2), pp.e3-e3. Randolph, S. (2014). Maximizing Project Value: A Project Manager's Guide. Project Management Journal, 45(2), pp.e2-e2. Rybka, I. and Bondar-Nowakowska, E. (2013). Planning of the Risk Handling Methods Related to Alterations to Project Documentation. Procedia Engineering, 57, pp.952-957.