MASSACHUSETTS INSTITUTE OF TECHNOLOGYDepartment of Electrical Engineering and Computer Science6.002 – Electronic CircuitsFall 2002Quiz 1 SolutionsName:Recitation Section:Recitation Instructor: Teaching Assistant:Enter all your work and your answers directly in the spaces provided on theprinted pages. Make sure that your name is on all sheets. Use the backs of theprinted pages as scratch paper, but we will only grade the work that you neatly transferto the spaces on the printed pages. Answers must be derived or explained, not just simplywritten down. The quiz is closed book, but calculators are allowed.This quiz contains 9 pages including the cover sheet. Make sure that your quiz contains all9 pages and that you hand in all 9 pages.Problem Points Grade Grader1 302 403 30Total 100Name: Solutions 2Problem 1: (30 points) For parts (A)-(C), use the associated branch variables as defined inFigure 1.Figure 1: Circuit for Problem 1(A), 1(B) and 1(C)(A) Write element laws for the resistor R1, the current source IA, and the voltage source VB.v1= R1i1i7= −IAv6= −VB(B) Write a complete set of independent KVL equations expressed only in terms of the branchvoltages v1, v2, ..., and v8.There are several possible answers, one set of independent KVL equations is:v1+ v2+ v6+ v5=0v3+ v7+ v2+ v6=0v3+ v4+ v8=0(C) Write a complete set of independent KCL equations expressed only in terms of the branchcurrents i1, i2, ..., and i8.Any five of the following is a complete set of independent KCL equations:i1− i5=0i6− i2=0i2− i1− i7=0i4+ i7− i3=0i8− i4=0i5− i6+ i3− i8=0Name: Solutions 3(D) The same circuit is shown in Figure 2, labeled for node analysis. Write out the node equationsnecessary to solve for the three unknown node voltages e1, e2and e3. DO NOT SOLVEthese equations.Figure 2: Circuit for Problem 1(D)Thenodeequationsare:e1− VAR1+e1− VBR2− IA=0e2R3+e2− e3R4+ IA=0e3− e2R4− IB=0In matrix form, we have:1R1+1R20001R3+1R4−1R40 −1R41R4e1e2e3VAR1+VBR2+ IA−IAIBName: Solutions 4(E) The current i1in Figure 3 can be written in the form:i1= aVA+ bVB+ cIA+ dIBDetermine the coefficients a, b, c and d in terms of the resistor variables in the circuit.Figure 3: Circuit for Problem 1(E)Here, superposition is used to determine each constant. The four figures below depict thecircuit when only one source is active.+−Ai11RR2i11RIAR2R3i11RR2R3i11R+−VBR2R4IBWhen VAis the only source activei1= −VAR1+ R2Name: Solutions 5When only VBis activei1=VBR1+ R2.When only IAis activei1=R2R1+ R2IA.From the figures above, IBdoes not contribute to i1.a = −1R1+ R2b =1R1+ R2c =R2R1+ R2d =0Name: Solutions 6Problem 2: (40 po ints) Network 1, shown in Figure 4, is described by its v-i relationshipmeasured at the terminals. Network 2, shown in Figure 5, is described by a schematic diagram ofits components.(A) Find the Th´evenin and Norton equivalent circuits that have the same v-i relationship as Net-work 1.Figure 4: Network for Problem 2(A)vOC=1.6ViSC= −8mARTH= −vOCiSC=0.2kΩNortonThevenin8m AΩ0.2k-+v0.2kΩ-+v1.6Vi i-+Name: Solutions 7(B) Find the Th´evenin and Norton equivalent circuits that have the same v-i relationship as Net-work 2.Figure 5: Network for Problem 2(B)Suppressing all the sources and looking at the resistance into the port gives RTH:RTH=2kΩ||3kΩ=65kΩ=1.2kΩUsing superposition, vOCand iSCcan be determined:vOC= −25V +2 · 32+3· 2V =2ViSC= −2mA +13mA = −53mAmA ΩNortonThevenin-+v-+v531.2k1.2kΩ2Vi i-+Name: Solutions 8(C) Suppose that the two networks are connected together through a resistor as shown in Figure 6.Find the current i1and the voltage v1.Figure 6: Network for Problem 2(C)First, redraw the circuit using the Th´evenin equivalents of each network:110.2kΩv2V0.6kΩ+-Ω1.2k1.6V-++-ii1=2V − 1.6V2kΩ= −0.2mAUsing superposition and voltage dividers:v1=1.821.6+0.222=1.44 + 0.2=1.64Vi1= −0.2mAv1=1.64VName: Solutions 9(D) Suppose that the two networks are connected together through a resistor as shown in Figure 7.Find the current i2and the voltage v2.Figure 7: Network for Problem 2(D)First, redraw the circuit using the Norton equivalents of each network:2-1.2kΩv53mA1.2kΩi2+Ω0.2k8m Av2=53mA +8mA(0.2kΩ||1.2kΩ||1.2kΩ)=293mA(0.15kΩ) = 1.45Vi2=1.45V1.2kΩ=2924mA ≈ 1.208mAi2=1.208mAv2=1.45VName: Solutions 10Problem 3: (30 points) Determine all node potentials in the network shown in Figure 8 in termsof the conductances of the resistors (GA, GB, GC, GD, GE,andGF), the current sources (ICandIF), and the voltage sources (VA, VB, VDand VE).Figure 8: Circuit for Problem 3The potentials e1and e2can be determined immediately from the circuit.e1= VAe2= VA+ VBNoting that e3, e4,ande5are related in the following way,e5= e3− VD− VEe4= e3− VDe5= e4− VE,the circuit can be simplified as shown below.Name: Solutions 11SupernodeEDV + VV + VABe25e3eCICGIFGF+-+-Now, we write a node equation for the supernode encompassing nodes 3, 4, and 5:(e3− VA− VB)GC+ e5GF= IC− IFUsing e5= e3− VD− VEin the equation above gives:e3GC+ e3GF= GC(VA+ VB)+GF(VD+ VE)+IC− IFe3=GC(VA+ VB)+GF(VD+ VE)+IC− IFGC+ GFUsing the relationship e4= e3− VDgives an expression for e4:e4=GC(VA+ VB)+GF(VD+ VE)+IC− IFGC+ GF− VDe4=GC(VA+ VB− VD)+GFVE+ IC− IFGC+ GFUsing the relationship e5= e4− VEgives an expression for e5:e5=GC(VA+ VB− VD)+GFVE+ IC− IFGC+ GF− VEe5=GC(VA+ VB− VD− VE)+IC− IFGC+
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