Slide 1EE100 Summer 2008 Bharathwaj MuthuswamyEE100 Su08 Lecture #3 (June 27th2008)• Administrivia– Videos for lectures 1 and 2 are up (WMV format). Quality is pretty good ☺.• For today:– Questions?– Wrap up chapter 2.– Start chapter 3. Refer to pdf slides for week 2.– An outline of Labs #1 and #2.Slide 2EE100 Summer 2008 Bharathwaj MuthuswamyGeneralization of KCL• The sum of currents entering/leaving a closed surface is zero. Circuit branches can be inside this surface, i.e. the surface can enclose more than one node!This could be a big chunk of a circuit, e.g. a “black box”i1i2i3i4Slide 3EE100 Summer 2008 Bharathwaj MuthuswamyGeneralized KCL Examples5µA2µAi50 mAiSlide 4EE100 Summer 2008 Bharathwaj Muthuswamy•Use reference polarities to determine whether a voltage is dropped • No concern about actual voltage polaritiesUsing Kirchhoff’s Voltage Law (KVL)Consider a branch which forms part of a loop. One possibility for sign convention:+v1_loopMoving from + to -We add V1–v2+loopMoving from - to +We subtract V2Slide 5EE100 Summer 2008 Bharathwaj MuthuswamyFormulations of Kirchhoff’s Voltage LawFormulation 1: Sum of voltage drops around loop = sum of voltage rises around loopFormulation 2:Algebraic sum of voltage drops around loop = 0• Voltage rises are included with a minus sign.Formulation 3:Algebraic sum of voltage rises around loop = 0• Voltage drops are included with a minus sign.(Conservation of energy)(Handy trick: Look at the first sign you encounter on each element when tracing the loop.)Slide 6EE100 Summer 2008 Bharathwaj MuthuswamyA Major Implication of KVL• KVL tells us that any set of elements which are connected at both ends carry the same voltage.• We say these elements are connected in parallel.Applying KVL in the clockwise direction, starting at the top:vb– va= 0 vb= va+va_+vb_Slide 7EE100 Summer 2008 Bharathwaj MuthuswamyPath 1:Path 2:Path 3:vcva+−+−321+ −vbv3v2+ −+-Three closed paths:abcKVL ExampleSlide 8EE100 Summer 2008 Bharathwaj MuthuswamyI-V Characteristic of ElementsFind the I-V characteristic.+_vsii+Vab_vabRSlide 9EE100 Summer 2008 Bharathwaj MuthuswamyMore Examples• Are these interconnections permissible?Slide 10EE100 Summer 2008 Bharathwaj MuthuswamySlide 11EE100 Summer 2008 Bharathwaj MuthuswamySummary• An electrical system can be modeled by an electric circuit(combination of paths, each containing 1 or more circuit elements) – Lumped model•The Current versus voltage characteristics (I-V plot) is a universal means of describing a circuit element.• Kirchhoff’s current law (KCL) states that the algebraic sum of all currents at any node in a circuit equals zero.– Comes from conservation of charge• Kirchhoff’s voltage law (KVL) states that the algebraic sum of all voltages around any closed path in a circuit equals zero.– Comes from conservation of potential energySlide 12EE100 Summer 2008 Bharathwaj MuthuswamyChapters 3 and 4• Outline– Resistors in Series – Voltage Divider– Conductances in Parallel – Current Divider– Node-Voltage Analysis– Mesh-Current Analysis– Superposition– Thévenin equivalent circuits– Norton equivalent circuits– Maximum Power TransferSlide 13EE100 Summer 2008 Bharathwaj MuthuswamyConsider a circuit with multiple resistors connected in series.Find their “equivalent resistance”.• KCL tells us that the same current (I) flows through every resistor• KVL tells usEquivalent resistance of resistors in series is the sumR2R1VSSIR3R4−+Resistors in SeriesSlide 14EE100 Summer 2008 Bharathwaj MuthuswamyI = VSS/ (R1+ R2+ R3+ R4)Voltage Divider+–V1+–V3R2R1VSSIR3R4−+Slide 15EE100 Summer 2008 Bharathwaj MuthuswamySS432122VRRRRRV⋅+++=Correct, if nothing elseis connected to nodesWhy? What is V2?SS432122VRRRRRV⋅+++≠When can the Voltage Divider Formula be Used?+–V2R2R1VSSIR3R4−+R2R1VSSIR3R4−+R5+–V2Slide 16EE100 Summer 2008 Bharathwaj MuthuswamySlide 17EE100 Summer 2008 Bharathwaj Muthuswamy• KVL tells us that thesame voltage is droppedacross each resistorVx= I1R1= I2R2• KCL tells usR2R1ISSI2I1xResistors in ParallelConsider a circuit with two resistors connected in parallel.Find their “equivalent resistance”.Slide 18EE100 Summer 2008 Bharathwaj MuthuswamyWhat single resistance Reqis equivalent to three resistors in parallel?+−VIV+−IR3R2R1Reqeq≡General Formula for Parallel ResistorsEquivalent conductance of resistors in parallel is the sumSlide 19EE100 Summer 2008 Bharathwaj MuthuswamySlide 20EE100 Summer 2008 Bharathwaj MuthuswamyVx= I1R1= ISSReqCurrent DividerR2R1ISSI2I1xSlide 21EE100 Summer 2008 Bharathwaj MuthuswamyR2R1 II2I1I3R3+−V⎟⎟⎠⎞⎜⎜⎝⎛+⎟⎠⎞⎜⎝⎛+⎟⎠⎞⎜⎝⎛=321R1R1R1IV⎥⎦⎤⎢⎣⎡++==3213331/R1/R1/R1/RIRVIGeneralized Current Divider FormulaConsider a current divider circuit with >2 resistors in parallel:Slide 22EE100 Summer 2008 Bharathwaj MuthuswamyTo measure the voltage drop across an element in a real circuit, insert a voltmeter (digital multimeter in voltage mode) in parallel with the element. Voltmeters are characterized by their “voltmeter input resistance” (Rin). Ideally, this should be very high (typical value 10 MΩ)Ideal VoltmeterRinMeasuring VoltageSlide 23EE100 Summer 2008 Bharathwaj Muthuswamy⎥⎦⎤⎢⎣⎡+=212SS2RRRVV⎥⎦⎤⎢⎣⎡+=′1in2in2SS2RR||RR||RVVExample: V1VK900R,K100R,V10V212SS=⇒===VSSR1R2210 , ?inRMV′==Effect of Voltmeterundisturbed circuitcircuit with voltmeter inserted_++–V2VSSR1R2Rin_++–V2′Compare to R2Slide 24EE100 Summer 2008 Bharathwaj MuthuswamySlide 25EE100 Summer 2008 Bharathwaj MuthuswamyTo measure the current flowing through an element in a real circuit, insert an ammeter (digital multimeter in current mode) in series with the element. Ammeters are characterized by their “ammeter input resistance” (Rin). Ideally, this should be very low (typical value 1Ω).Ideal AmmeterRinMeasuring CurrentSlide 26EE100 Summer 2008 Bharathwaj MuthuswamyRinV1ImeasR1R2ammetercircuit with ammeter inserted_+V1IR1R2undisturbed circuitExample: V1= 1 V, R1= R2= 500 Ω, Rin= 1Ω211RRVI+=in211measRRRVI++=11, ?500 500measVImAI===Ω+ΩEffect of AmmeterMeasurement error due to non-zero input resistance:_+Compare to R2+ R2Slide 27EE100 Summer 2008 Bharathwaj MuthuswamySimplify a circuit before applying KCL and/or KVL:−+7 VUsing Equivalent ResistancesR1= R2= 3 kΩR3= 6 kΩR4= R5= 5 kΩR6= 10
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