African Americans in American Society 1920s - 2619 Words

African Americans in the 1920s * â€Å"Cast down your bucket where you are. Cast it down among the eight millions of Negroes†¦Ã¢â‚¬  – Booker T. Washington, 1895 Atlanta Compromise Throughout US history, there is an abundance of racism, segregation and discrimination towards the African American people. In 1619, the first African slaves were brought to Jamestown to produce tobacco, tea, cotton, coffee and other precious commodities. In this time period, 12 million Africans were forcibly transported to the Americas, where they worked as slaves until 1865, where the 13th Amendment abolished slavery. Although suppressed by whites and organisations such as the Ku Klux Klan, African Americans in the 1920s began to work towards social,†¦show more content†¦Despite the cultural developments in Harlem and the formation of the African American identity, white culture found it extremely difficult to accept their black neighbours, leading to racial tensions, and often as a result, lynchings. One such example of racial tension leading to horrific consequences was the Tulsa Race Riots. In 1921, Tulsa Oklahoma was experiencing an economic boom thanks to the di scovery of oil. Due to this African Americans also prospered, although confined to the Greenwood section of the city, also referred to as the Black Wall Street, due to a number of wealthy black entrepreneurs residing there. At this time, membership in the Ku Klux Klan was rising and there was an active chapter in Tulsa. On Memorial Day, a riot was triggered by a report in several white newspapers that a white, female elevator operator had been allegedly raped by black youths. In response to this, rumours circulated around the city that a mob was going to attempt to lynch the youths, then a group of armed African Americans bolted to the local police station in order to stop the lynching mob, that did not exist. AShow MoreRelatedDifferences Between 1920 And The 1920s1367 Words   |  6 Pagesprosperous and advantageous positions in society. With changing definitions representing changing social, political and economic policy, it is no surprise that freedom and prosperity exclude diff erent groups of people, such as immigrants and African Americans during both the 1920s and the 1950s. While both of these periods saw great changes to policies and increased prosperity under these times, the shadows of war and discrimination still left many behind. During the 1920s, freedom was defined by economicRead MoreSimilarities And Differences Between The 1920s And 1950s908 Words   |  4 PagesThe 1920’s were a very important era in America for better or worse. There were many issues in relation to race and how people of different ethnic groups were treated. African American had a cultural rejuvenation that being the Harlem Renaissance. The advent of the Ford Model T change the way how people traveled. Many may say an era like the 1950’s were highly comparable. Race related issues were on a decline as America as whole sought to be more accepting and the oppressed started to speak out onRead MoreEssay on Sula1337 Words   |  6 Pages they are also African Ame rican, which further exposes them to mistreatment and pre-determined societal roles. African Americans during the 1920’s were experiencing great social injustices and mistreatment, along with the likes of women who were also experiencing inequality to a lesser degree during this time as well. In her novel Sula, by addressing and shedding light on the many acts of racism and sexism that occurred during the 1920’s, Toni Morrison shows how African American women experiencedRead MoreRacial Discrimination1008 Words   |  5 PagesPrior to the 1920s, it was a time of racial hatred. Starting before the 1800s to the 1920s was the time of slavery. Many whites disliked black people. According to the article â€Å"Blacks Set Out in Search of a Better Life in 1920s American Society† published on VOA, â€Å"Many whites joined the Ku Klux Klan organization†¦ terrorized blacks. Klan members sometimes burned fiery crosses in front of the houses of black families. And t hey sometimes beat and murdered blacks†¦ hated blacks most of all† (VOA). RacialRead MoreThe Counter Culture of the 1920s Essay1493 Words   |  6 PagesThe counter culture of the 1920’s has affected the way the American lifestyle is today. Counter culture is a culture that primarily consists of younger people, with values and lifestyles opposing those of the original established culture. (Dictionary.com) A need for change. The 1920’s are also known as the â€Å"Jazz Age,† which was coined by F. Scott Fitzgerald, and the â€Å"Roaring Twenties.† It was a decade of change. (Hakim, 41) The counterculture of the 1920’s resulted from the Age of Jazz, FlappersRead MoreThe Aftermath of World War I927 Words   |  4 Pages The aftermath of World War I left a lasting impression on the 1920s because America entered the Great War later than the big European countries. This gave them an upper hand in their economic position since they did no t spend as much money as France, Germany, and Britain did. This ignited their unprecedented affluence which had a domino effect in America’s society in terms of government’s relationship to business. Another effect of the First World War is the Red Scare and America’s prejudiceRead MoreThe Critical Impacts Of The Harlem Renaissance1066 Words   |  5 Pagesunderlines the trouble of ethnic issue knowledgeable by African Americans all through the twentieth century. There were numerous critical impacts, for instance, artistic the growth. The Harlem Renaissance was an energetic affiliation amongst the 1920s where African Americans started composed and transported artistry and writing one of a caring to their race, motivating a countless many darks kin to complete in a white overwhelming society. While the American war was a disaster for the confederate south leavingRead MoreThe 1920s transformations greatly affected the American society. There was a dramatic social change700 Words   |  3 PagesThe 1920s transformations greatly affected the American society. There was a dramatic social change and great economic growth in the 1920s that was made possible by the technological revolution. Productivity rose by more than sixty percent and the mass culture’s influence contributed to the progress and advancement of technology and goods. The 1920s was a time of culture wars and an age of incredible affluence and expansion of human rights. Although there were many aspects of this culture that brokeRead MoreCultural Confrontations of the 1920’s: KKK, Scopes Trial Essay1388 Words   |  6 PagesCultural Confrontations of the 1920’s The 1920s were a time of change for the United States. Following the First World War there was a rush of new cultural, social, and artistic dynamism, partly fuelled by the Progressivism movement that was cut short when American entered the Great War. This decade was defined by a change from more rural farm life to industrialism in big cities. The shift from the frugality and traditional family values or previous generations to the happy-go-lucky consumerismRead MoreHarlem Renaissance : The Cultural And Artistic Explosion745 Words   |  3 Pagesthe end of World War 1. The time of the 1920’s was a time of change for everyone. During the 1920’s, the Harlem Renaissance was the most influential movement where African Americans came together and created multiple things that was unique to their race such as; music, literature, poetry, and much more which really impacted the way African Americans stood up together in a country controlled by whites. After the end of slavery, many African Americans from the south migrated to industrial northern

Juvenile Probation Officers Work With High Risk Teens

As described by Kelly Peterson juvenile probation officers work with high risk teens along with their peers, family, work, school and involved activities. People like Kelly do there best to try and keep juveniles out of the adult system and further criminal systems. Over the course of Kelly’s visit she talked about many things some main points being; Her caseload and how she manages it, the main kinds of cases she deals with, and the court experience of people in the juvenile system. Kelly stated that at one time she could have up to thirtyfive cases. She would usually have cases of kids who were the ages fifteen or above, but sometimes she would even have a twelve year old on her caseload. Kelly said it is somewhat easy for her to keep her cases organized because she is assigned certain schools or areas. For example if a kid committed a crime near the town of mapleton she would than take on that case because it fell in her designated area. Crimes that were committed somewhere beside her area would be handed to another juvenile probation officer. Another thing that Kelly talked about was the kinds of case she gets and what crimes she comes across most often. Kelly mostly comes across crimes that involved chemical dependence. For example either on alcohol, THC or synthetic drugs. Another big one that is a problem is truancy. She explained that truancy case usually come from families that don t take thing seriously or from families that are not stable or connected. KellyShow MoreRelatedJuvenile Justice And Its Effects On Society1722 Words   |  7 PagesJuvenile justice is compared in chapter thirteen. In the nineteenth century, there was an increased number of children at risk and chronic poverty. This overall was a general concern because there was an increase of people in the â€Å"dangerous classes†. There was a child saving movement, in which the poor children represented a threat to the moral fabric of society. The nineteenth century was a ti me where they had a house of refuge. In this house of refuge, they had a society for the prevention of pauperismRead MoreThe Effects Of Juvenile Delinquency On Teens864 Words   |  4 PagesTheses can also affect teens that increase-doing crimes in which it called juvenile delinquency. There are two terms that define juvenile delinquency, 1: conduct by a juvenile characterized by antisocial behavior that is a beyond parental control and therefore subject to legal action; 2: a violation of law committed by a juvenile and not punishable by death or live imprisonment. (Cite). In this topic, I will explain more about family factors in which it has major influence on teens such as the way parentsRead MoreA Historical Look At The Concept Of Juvenile Justice1394 Words   |  6 Pages1. Chapter 13 is a historical look at the concept of juvenile justice. What did you learn from reading this chapter? At the beginning of the nineteenth century delinquent, neglected, and runaway children in the United States were left with very little guidance and help from the government and the court system. Unlike today there was no voice for those kids who were neglected and left to find there own ways of survival which most likely ment that they would have to turn to crime to survive. If thatRead MoreThe Juvenile Justice System Police2399 Words   |  10 Pagescontact with a police officer often is a young person s introduction to the juvenile justice system police account for most referrals to juvenile court. Law enforcement s role with boys and girls under the age of eighteen is challenging because there are laws that federally protect youth that commit serious crimes and attempts to aid them in a road to recovery to return to their communities. Police officers generally summon young offenders to the police department s juvenile division to questionRead MoreJuvenile Justice and Rehabilitati on2500 Words   |  10 PagesJuvenile Justice and Rehabilitation When discussing rehabilitation or punishment for juvenile delinquents, I believe there should be rehabilitation over punishment. Granted there are numerous cases that completely warrant punishment, but punishment isn’t always the answer. Adults are usually given second, third and fourth chances to change their lives. And sometimes rehabilitation isn’t involved. I believe since adolescents still have plenty of time to get counseling or learn about themselvesRead MoreJuvenile Justice System Is The Fundamental System1840 Words   |  8 PagesThe Juvenile Correction System is the fundamental system used to address and deal with youth who are caught and convicted of crimes, such as murder, robbery, and aggravated assault. The juvenile justice system gets involved in delinquent behavior through police, court, and correctional commitment. Throughout history, many individuals have tried to change the policies and process of the juvenile correction system. Some agree with the s tructure of incarceration and the treatment these teens â€Å"deserve†Read MoreAnalysis Of The Giddings State School Capital Offenders Programs1623 Words   |  7 Pages Rehabilitation for at risk teens has been an ongoing issue that runs deep in certain communities. When kids at young ages are exposed to stress and have to cope early on with dysfunction they are denied the opportunity to mature and conditioned to commit thinking errors that perpetuate a young offender into an adult offender. To find ways to break this cycle John Hubner accounts his time on the Giddings State School Capital Offenders Program and how a group of counselors are able to combine manyRead MoreJustice System Position Paper1845 Words   |  8 PagesPhoenix Introduction to Juvenile Justice Timothy Cariker March 5, 2011 Justice system position paper The purpose of this paper is to state my belief that juveniles should and can be rehabilitated. The goals and efforts of rehabilitation are to keep the juvenile offenders out of the correctional facility. Peers impact delinquencyRead MoreFalling Back By Jamie Fader1585 Words   |  7 Pagesinvolved in drug offenses and violence within their suburban communities and were now in the process of behavioral change in order to help them reflect and be able to make better decisions which would lead them to a better life. Fader observed these juveniles as they transitioned back to urban Philadelphia where they would resume their daily lives and also struggle to adopt adult masculine roles: â€Å"After these young men demonstrated sufficient change to earn their release or, more typically staff membersRead MoreJuvenile Delinquents and Drug Abuse Essay examples1501 Words   |  7 PagesDoes only the juvenile drinking or drugging up suffer, or do others get involved? The answer is, not only do the users suffer, but so do their family, friends, and the community. However, due to the rise of juveniles becoming involved in substance abuse, the juvenile justice system has resulted in an increased burden. Over the past fifteen years, the fad of drug use among kids has steadily been increasing. Persistent substance abuse among youth is often accompanied by an array of problems, including

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

Perceptual color space represe... free essay sample

Perceptual color space represent color which is closely related with human vision. The main idea is to arrange all the colors by perceptual difference of color. In these color spaces-luminance (L) and chrominance is represented in XYZ coordinates. However, skin detection is not a physical property of an object in an image. It can be related as a perceptual concept of human vision. So, it is quite efficient to take a color space for skin detection that has the sensitivity of human vision. CIELAB and CIELUV are two examples of perceptually uniform color spaces. Perceptual uniformity defines that a small distortion to a component value can be equally observable across the range of values. Moreover, RGB and other well-known color spaces are not perceptually uniform. Non –linear transformation of perceptual color spaces corrects this shortcoming. But it needs complex transformation function for conversion. When an image is converted to CIE-LAB it may appear exactly as the previous color space. The reason is CIE-LAB and CIE-LUV contains all possible colors and since the translation of color is not related, so no deterioration in image quality. It may be little tricky since in usual color spaces (e.g., RGB, HSV) contains logical colors. But in LAB one of the channel contains no color and other two channels have dual color combinations without any contrast. The L channel is for lightness and there is no color value associated with it. It usually depicts the difference between presence of darkness and light. In addition to that, A is the color balance between green and Magenta and B is a balance between blue and yellow. For aforementioned advantages Zarit et al. [47], Yang and Ahuja [25], Schumeyer and Barner [48] used this color space. However, I also consider this color space for clustering skin pixels from an image.Color space comparisonsColor spaces obviously affect the performance of skin detectors. Several authors experimented different results for the influence of color space choice on skin segmentation algorithms performance. In [49], author claimed that TSL is the best color space for skin detection While combined with Gaussian and Mixture of Gaussian models. Comparing the performance of variety of color spaces on single Gaussian model has been yielded to YCbCr for the best performance [50]. On the other hand, Montenegro et al. compared RGB, HSV, YCbCr, CIE-Lab and CIE-Luv implementing SFA dataset. Matthews correlation coefficient (MCC) was used as a performance metric and Gaussian classifier as the method. CIE-Lab outperforms all others and normalized RGB performs the least [51]. Furthermore, performance of RGB based techniques suffers in decimating skin-pixels while objects increases [52]. A comparison study is done using Gaussian and histogram approaches for a dataset of 805 color images [53]. This works claimed that the choice of color space significantly changes the performance, and HSI color space accompanying with histogram model outperform others. In comparison of three different color spaces (HIS, RGB, CIE-Lab and YCbCr), in terms of classification error is obtained in HIS and YCbCr models [7]. A recent study [54] claims YPbPr outperforms other models. Besides Gaussian approaches, Gonzà ¡lez et al. [55] compared the performance of 10 common color spaces based on k-means clustering algorithm and conclude that HSV, YCgCr, and YDbDr are the best color spaces. Nalepa et al. [56] showed that statistically combination of RGB and HSV outperforms other color spaces. In other hand, artificial neural network has been subjected for skin segmentation in variety of color spaces yielding to the conclusion that YIQ is the best choice [57]. The idea to adapt an optimal chrominance color space rather than a segmentation model has been suggest in [58]. Here a non-liner transformation between YUV and new TSL* color space is used to boost the segmentation. In summary, performance of color space based skin detection is highly depend on various factors including used methods. To make a fair comparison, all influencing factors should be considered. As observed, different authors take account of different training, testing and validation data. As a result, optimum color space for skin segmentation has changed to adapting optimal skin detection models. Besides, in neural networks, the performance of classifier is directly depend on the number of neurons, as well as the initial guess of the weights. 2.2 Skin Detection MethodsSkin detection processes may be categorized into diverse categories which are not mutually distinct. Statistical strategies are based totally on records extracted from histogram of training skin and non-skin pixels. Non-parametric or parametric models are developed with the aim of acquiring the probability that a pixel belongs to skin or non-skin class. Artificial neural networks (ANN) are very beneficial gear in both estimation of skin distribution or direct classification of the pixels. Adaptive techniques often reach better accuracies with the cost of computation. SVM based techniques also are used for classification of skin pixels in applications. 2.2.1 Explicitly Defined Boundary ModelsThe main difference in skin color is in its intensity (brightness) rather than the chrominance (color) [59, 55, 57]. Those techniques have a pixel based processing scheme in which for any given pixel, methods are investigated to determine the class of that pixel. Theyre very prominent particularly due to their simple and quick training, low cost implementation and fast processing. However, several parameters are concerned in degrading the performance of classifiers, consisting of their static nature, excessive dependence on training images, effectiveness of rules and inability to cope with maximum skin detection challenges. Kovac et al. [60] proposed a way of explicitly defined boundary model using RGB color space in two of daylight and flashlight situations which has been reutilized in [61,62,63,64]. Orthogonal color areas are regularly utilized in case of explicitly described boundary models. Sagheer et al. [65] exploited cases of regular lights and distinctive lighting fixtures. In order to detect faces, in [66, 67], rules are dynamically reconfigured based on pixels value. Zahir et al. [68] proposed an easy boundary version using HSV color space for indoor and outdoor conditions. Combination of color spaces has been also effective to reinforce the performance of explicitly defined boundary methods. Thakur et al. [69] employed RGB, CbCr and HSV color method. Rules are set independently for triple color spaces and outcomes are clearly fused for taking final selection. Furthermore, fusion of the result of different color space to reduce false positives has been a common procedure; some examples are: RGB and YCbCr [70], YCbCr and YUV [8], HSV and YUV [71], HSV and YCbCr [72,73], RGB and YUV [74] and HSV and YCgCr [75].2.2. 2 Statistical ModelsSkin detection is a probabilistic problem and lots of techniques based on general distribution of skin and skin color, every in a selected color model, were developed. Based totally on an extensive training set, these strategies estimate the probability that a located pixel is associated with skin. 2.2.2.1 Non-parametric (histogram based) modelsThere is no any specific definition of probability density characteristic in non-parametric techniques. Single histogram primarily based Look-UP-Table (LUT) model is a common approach in modeling skin color cluster. In this technique, by using a set of training skin pixels, distribution of skin pixels in a selected color space is received. The training technique is simple but populating the histogram calls for a large skin dataset. But, Bayesian classifier considers two histograms of skin and non-skin pixels. In [76, 77], a Bayesian classifier based on YCbCr color space is employed. Erdem et al. [61] employed Bayesian based post filtering technique to reduce false positives of Viola-Johns face detector. Zarit et al. [78] performed a comparison study of overall performance of histogram method in Fleck HS, HSV, RGB, CIE Lab and YCbCr. Phung et al. [7] examined the tradeoff among the wide variety of bins in line with channel and detection rate.There are numerous gain and drawbacks associated with non-parametric techniques. The benefit includes- training and implementation, rapid training, independency at the shape of cluster [5] are the primary blessings. However, one main drawback of these strategies is their dependency at the training set which requires gathering of a massive range of skin pixels. Further, large memories are required to implement histogram based models especially whilst fine resolutions are required. 2.2.2.2 Parametric modelsParametric method which include single Gaussian method (SGMs), Gaussian mixture models (GMMs), cluster of Gaussian models (CGMs), Elliptical models (EMs), etc, are evolved to compensate LUT shortcomings consisting of excessive storage requirement. Similarly, they generalize thoroughly with a tremendously smaller amount of training set. In SGM, there should be a smooth Gaussian distribution around the mean vector. GMM is advanced to version more complex distributions as a normalized weighted sum of Gaussian PDFs [25]. It compensates the inability of SGM in managing out of control conditions in a general skin segmentation problem, similarly to the truth that SGM is not capable of approximating the actual distribution due to the asymmetry of distribution in its peak [79]. The most appealing features of GMM models are their easy evaluation procedure and occasional memory cost. The training technique is but, longer than the former methods. A comparative look at the overall performance of single and combination of Gaussian distributions in [80] confirmed that mixture method improve the performance only in a relevant operating area.Several statistical skin detection strategies have been elucidated in this subsection; all classified in non-parametric and parametric processes. Non-parametric methods are primarily based on histogram of skin and non-skin pixels in a predefined training set. But, in compare with parametric