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### HW1%20solution0

Course: IE 370, Spring 2011
School: Purdue
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Purdue - IE - 370
Purdue - IE - 370
Purdue - IE - 370
Purdue - IE - 370
ECE 201 Spring 2010Homework 5 SolutionsProblem 26Since V2 = 60 V , the current through 60 branch is 1 A. The resistors 90 and 180 are in parallel. Their equivalent resistance is90 18090 + 180= 60Req =Now 60 and 60 are in series. Thus the voltage
Purdue - IE - 370
Purdue - IE - 370
Purdue - IE - 370
Purdue - IE - 370
ECE 201 Spring 2010Homework 9 SolutionsProblem 37(a)The following loop equation can be written,(I1 0.75)200 300I1 (I1 + 0.1)500 = 0 I1 = 0.1 AP = VIPs1 = 200(0.75 0.1)0.75= 97.5 WPs2 = 500(0.1 + 0.1)0.1= 10 W(b)Again, writing the loop equatio
Purdue - IE - 370
=9t.lV PtM&quot;'V'wr6)-.,C; I (VO tot1. \,.o'IOtA'&quot;: I.-/\)/Vio\ lvu-rcfw_f):c)' i V, +t [10J:I. +I,) ;:, JA( lot)J - Y (.;2) :6.) (VOkJ/)0_ .JI5&quot;, J,G,t)+; )&quot;'10:((lb-t.&quot;r :,':.k1 &quot; &quot; :&lt; cfw_V/&quot;/ &lt;#A .214
Purdue - IE - 370
Purdue - IE - 370
Purdue - IE - 370
ECE 201 Spring 2010Homework 13 SolutionsProblem 9(a)2R and 6R in series gives 8R. 8R and 8R in parallel gives 4R. Thus 12R is inseries with Vs in the simplied circuit. Thus the Thevenin voltage is givenbyVs1 6R12R 2Vs=4= 30 VVoc =To nd the
Purdue - IE - 370
Purdue - IE - 370
1086iout(t) (mA)420-2-4-6-8-1000.0050.010.015t(s)1009080W L(t) (nJ)70605040302010000.0050.01t(s)0.015
Purdue - IE - 370
Purdue - ECE - 201
ECE 201 Spring 2010Homework 17 SolutionsProblem 27(a)Let V be the common voltage across C1 and C2 . ThusdVdtdVC2dt(C 1 + C 2 )= is (t)= iC 2 (t) iC 2 (t) =C2is (t)C1 + C2(b)Using KVL across the second loop and relation from part (a), we
Purdue - IE - 370
Purdue - IE - 370
20181614Vc (t)(V)12108642002468t(s)108t(s)1012141610987Vc (t)(V)654321002461214160.20.180.160.14iL(t)(A)0.120.10.080.060.040.02000.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009t(s)0.010.00
Purdue - IE - 370
III8-20 (d) Vc(t)8-20 (d) Vc(t)8-20 (d) Vc(t)151010II5Vc (t) (V)15Vc (t) (V)2010I2015Vc (t) (V)2055000-5-5-500.10.20.30.400.1t (s)0.20.30.40I0.20.30.4t (s)II8-20 (d) c (t)I8-20 (d) c (t)I8-20 (d) c (t)I0
Purdue - IE - 370
ECE 201 Spring 2010Homework 21 SolutionsProblem 31(a)Using voltage division,20RV050R= 0.4V0vC (0) == vC (0+)(b)The Thevenin equivalent is given byR3R3 + (R1 |R2 )= 0.8V0Voc = V0Rth = R1 |R2 |R3= 4R(c)vC () = 0.8V0vC (0+) = 0.4V0 vC (
Purdue - IE - 370
Purdue - IE - 370
21.5Vc (t) (V)10.50-0.5-101234t(s)567810987Vc (t)(V)654321000.20.40.60.81t(s)1.21t(s)1.21.41.61.820-0.05-0.1iL(t)(A)-0.15-0.2-0.25-0.3-0.35-0.400.20.40.60.81.41.61.821210Vc (t)(V)8642
Purdue - IE - 370
Purdue - IE - 370
ECE 201 Spring 2010Homework 25 SolutionsProblem 43(a)To nd the initial conditions at t=0-, we can write the following KVL equations,20 + 40iL (0) 60(0.1 iL (0) = 0 iL (0) = iL (0+) = iC (0+) = 0.14 A vC (0+) = vC (0) = 20 0.14 40= 14.4 VAfter t=0
Purdue - IE - 370
Purdue - IE - 370
Purdue - IE - 370
Purdue - IE - 370
ECE 201 Spring 2010Homework 29 SolutionsProblem 40(a)The following dierential equation can be written using KCL at the invertingterminal of the op amp and using the virtual ground concept,vs (t) vout (t)dvout (t)++C=0R1R2dtvs (t) = 100u(t)=
Purdue - IE - 370
Purdue - IE - 370
Purdue - IE - 370
Purdue - IE - 370
ECE 201 Spring 2010Homework 33 SolutionsProblem 30(a)11+R j Lj 0.25= 0.05 1Zin (j ) =Yin (j )j 20=5 + jYin (j ) =(b)Iin Zin (j )j L10(i + j )= 10 j 20 j /4= 5 2eIL = iL (t) = 10 cos(5t /4) mAProblem 40Let the impedances of R1 ,
Purdue - IE - 370
Purdue - IE - 370
Purdue - IE - 370
11-2(a) Instantaneous Power p(t) and Average Power pave11-2(a) Instantaneous Current i(t)0.090.0060.080.0040.070.0020.06power (W)0.10.008current (A)0.010-0.0020.050.04-0.0040.03-0.0060.02-0.008Instantaneous PowerAverage Power0.01
Purdue - IE - 370
ECE 201 Spring 2010Homework 37 SolutionsProblem 5(a)For Figure (a),1 T2i (t) dtT0319 dt +=9025=3= 5/ 32Ief f = Ief f616 dt3For Figure (b),119 dt +40= 25/42Ief f =316 dt2 Ief f = 2.5(b)Current through RL is given by i(t)
Purdue - IE - 370
Purdue - IE - 370
Purdue - IE - 370
ECE 201 Spring 2010Homework 40 SolutionsProblem 30(a)Zth = (R1 |R2 ) j/ CVoc= (20 j 10) R2= VsR1 + R2= 20 VrmsFor maximum power transfer, ZL = Zth = (20 + j 10) .(b)2Voc4RL= 5WPavg =Problem 37(a)Here we cannot apply the maximum power t
Concordia AB - CHEM - 11
Acids / Base EquilibriumUnit IVI. Arrhenius acids and bases (1884)According to Arrhenius acids are substances that dissociate in water to produce Hydrogen ions (H+), andbases are substances that dissociate in water to produce hydroxide ions (OH-)Acid
Concordia AB - CHEM - 11
Bufferschemicals that resist changes in pHExample: The pH of blood is 7.4.Many buffers are present to keep pH stable.H1+ + HCO31H2CO3H2O + CO2hyperventilating: CO2 leaves blood too quickly[ CO2 ]shift right[ H1+ ]pH(more basic)alkalosis: bloo
Concordia AB - CHEM - 11
Name: _Acids and BasesIts a chemical world!Acids can be identified by theirchemical formulas The chemical formula for an acid willstart with hydrogen HNO3 is nitric acid; H 2SO4 is sulfuricacid; HCl is hydr ochloric acid; H3PO4 isphosphoric acid
Concordia AB - CHEM - 11
Acids and BasesAcids: The term acid, in fact, comes from the Latin termacere, which means sour. Acids taste sour, are corrosiveto metals, change litmus (a dye extracted from lichens)red, and become less acidic when mixed with bases.Bases: Bases (also
Concordia AB - CHEM - 11
Chemistry 112 Laboratory: Chemistry of Acids &amp; BasesPage 43Chemistry of Acids and BasesThe chemistry of acids and bases is an area of fundamental importance inchemistry. In this experiment you will study acid-base equilibria by determining the pH of
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Chemistry 11Lab: Types of Chemical Reactions(April 5, 2008)Lab #5C:Types of Chemical ReactionsOBJECTIVES To observe a variety of chemical reactions To interpret and explain observations with balanced chemical equations To classify each reaction as
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Chemistry 11Lab: Solution Chemistry(April 14, 2008)Lab #10D:Solubility Trends and Precipitate FormationOBJECTIVES To mix several pairs of solutions together and then note whether any precipitates form To deduce, from the experimental results, which
Concordia AB - CHEM - 11
Chemistry 12Name:_, Blk:_Experiment 18A: Factors Affecting Reaction RatePurpose:_ To observe and record the effects of concentration, surface area, andtemperature on reaction ratesMaterials: As per pg. 192 of Heath Laboratory Experiments.Procedure:
Concordia AB - CHEM - 11
UMass Lowell Freshman Chemistry LabPage 1 of 4Experiment14Double Displacement ReactionsOverviewIn this experiment you ultimately want to try to identify two &quot;unknown&quot; solutions of ionic substances, bycomparing the reactions of the &quot;unknown&quot; substan
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Exp 20A: Acid RainExperiment 20A: Formal Lab Write-upInclude ALL Questions and Follow-up QuestionsNote: you will need the following information to complete the follow-up questionsSolution A = NaOHSolution B = HClSolution C = HNO3Solution D = Ca(OH)
Concordia AB - CHEM - 11
Qualitative AnalysisPage 139 156Pre-lab, pages 143-144, AND 153-154Post Lab, page 156, all questionsIntroduction to Qualitative Analysis Qualitative analysis is used to separate and detectcations and anions in a sample substance. Qualitative analys
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Chemistry 12 Unit 4Introduction: (Post Lab Discussion)Operational Definitions: Describe what acids or bases doeg:Bases turn red litmus blueeg:Acids react with magnesiumConceptual Definitions: Define acids or bases in terms of their molecularstru
Concordia AB - CHEM - 11
Unit 11: Equilibrium / Acids and Basesreversible reaction:RPandPRAcid dissociation is a reversible reaction.2 H1+ + SO41H2SO4Rate at whichRate at which=RPPRequilibrium:- looks like nothing is happening, however- system is dynamic, NOT s
Concordia AB - CHEM - 11
Common AcidsStrong Acidshydrochloric acid:HClH1+ + Cl1- stomach acid; pickling: cleaning metals w/conc. HCl2 H1+ + SO42H2SO4sulfuric acid:- #1 chemical; (auto) battery acidnitric acid:H1+ + NO31HNO3- explosives; fertilizerWeak Acidsacetic a
Concordia AB - CHEM - 11
Various Definitions of Acids and BasesArrhenius acid:yields H1+ in solne.g., HNO3Arrhenius base:H1+ + NO31yields OH1 in solne.g., Ba(OH)2Ba2+ + 2 OH1Lewis acid:e pair acceptorLewis base:e pair donorTOPIC FORFUTURE CHEM.COURSESBronsted-Lowr
Concordia AB - CHEM - 11
pH CalculationsRecall that the hydronium ion (H3O1+) is the speciesformed when hydrogen ion (H1+) attaches to water (H2O).OH1 is the hydroxide ion.For this class, in any aqueous soln,[ H3O1+ ] [ OH1 ] = 1 x 1014( or [ H1+ ] [ OH1 ] = 1 x 1014 )If h
Concordia AB - CHEM - 11
Acid-Dissociation Constant, KaFor the generic reaction in soln:A+B[ PRODUCTS ][ REACTANTS ]KaKaFor strong acids, e.g., HClHClH1++ Cl[ H1 ] [ Cl1- ][ HCl ]Ka1&quot; BIG.&quot;C+D[C][D][ A ][B ]Assume 100%dissociation;Ka notapplicable forstron
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