Slide 1Single Loop CircuitsSingle Loop Circuits – with a Current SourceSingle Loop Circuits – with a Voltage SourceWith Multiple Voltage SourcesVoltage DivisionVoltage Divider: A Practical ExampleVoltage Divider: An ExampleSlide 9Slide 10Slide 11Example: Two Resistors in ParallelSlide 13Equivalent Resistance of Parallel ResistorsWhat are I1 and I2 ?Circuits with More Than One SourceWhat if More Than One Source?Class ExamplesSuperposition Method – A More General Approach to Multiple SourcesHow to Apply SuperpositionSuperposition of Summing CircuitSlide 22Superposition ProcedureSlide 24Single Loop Circuits* with a current source* with a voltage source* with multiple sources* voltage divider circuits * Equivalent resistanceSuperposition method * Principle* Procedures* How to applyLecture 3. Single Loop Circuits & Superposition Method12Single Loop Circuits•The same current flows through each element of the circuit—the elements are in series.+–VSR1R2RnIV1V2ISR1R2RnV1IV2V3V3With an independent voltage source With an independent current source3•What is I?Single Loop Circuits – with a Current SourcesII •In terms of I, what is the voltage across each resistor? 111RIIRVs…ISR1R2RnV1IV2V3222RIIRVsnsnnRIIRV 4•To solve for I, apply KVL around the loop.+–VSR1R2RnI + –I R2+–I R1I Rn+–IR1 + IR2 + … + IRn – VS = 0Single Loop Circuits – with a Voltage SourceisnsRVRRRVI...21•In terms of I, what is the voltage across each resistor? 11IRV 22IRV …nnIRV 5With Multiple Voltage Sources•The current i(t) is:•Resistors in seriessresistanceofsumsourcesvoltageofsumRVtijSi)(jNequivalentRRRRR 216Voltage Division• Consider two resistors in series with a voltage v(t) across them:R1R2– v1(t)++–v2(t)+–v(t)2111)()(RRRtvtv2122)()(RRRtvtv• If n resistors in series:jiSRRRtvtvki)()(7Voltage Divider: A Practical ExampleElectrochemical Fabrication ofQuantum Point Contactor Atomic-scale wireMolecular JunctionAnode: Etching delocalized, but Cathode: Deposition localized at sharpest point, due to: • Self-focusing – directional growthDecreasing Gap!++++++--++++++++++--EVoltage Divider: An Example++++++VgapARextVextV00VRRRVextgapgapgap• Initially, Rgap >> Rext, Vgap ~ V0 full speed deposition.• Finally, Rgap << Rext, Vgap ~ 0 deposition terminates.• The gap resistance is determined by Rext.Voltage Divider: An Example2• Growth starts after applying 1.5 V1• Two electrodes with10 m initial separation3• Self-terminates after forming a tunneling gapVoltage Divider: An ExampleOhmic behaviorTime (sec.)G (2e2/h)Stepwise increase in ConductanceVoltage Divider: An Example12Example: Two Resistors in ParallelHow do you find I1 and I2?IR1R2V+–I1I21321212111RRVRVRVIIIIR1R2V+–I1I2212121111RRRRIRRIV• Apply KCL with Ohm’s LawExample: Two Resistors in ParallelEquivalent Resistance of Parallel Resistors• Two parallel resistors is often equivalent to a single resistor with resistance value of:2121RRRRReqinparRRRRR1111121• n-Resistors in parallel:152121212111RRRIRRRRRIRVIWhat are I1 and I2 ?•This is the current divider formula•It tells us how to divide the current through parallel resistors2112212122RRRIRRRRRIRVI16Circuits with More Than One SourceHow do we find I1 or I2?Is1Is2VR1R2+–I1I2172121212111RRVRVRVIIIIss 212121RRRRIIVssIs1Is2VR1R2+–I1I2• Apply KCL at the Top NodeWhat if More Than One Source?18Class Examples•Example: P1-33 (page 43).•Drill Problem P1-34 (page 43).19Superposition Method – A More General Approach to Multiple Sources“In any linear circuit containing multiple independent sources, the current or voltage at any point in the circuit may be calculated as the algebraic sum of the individual contributions of each source acting alone.”How to Apply Superposition•To find the contribution due to an individual independent source, zero out the other independent sources in the circuit–Voltage source  short circuit–Current source  open circuit•Solve the resulting circuit using your favorite technique(s)21Superposition of Summing Circuit+–Vout1k1k1kV1V2+–+–+–V’out1k1k1kV1+–V’’out1k1k1kV2++–+–22V’out = V1/3V’’out = V2/3Vout = V’out + V’’out = V1/3 + V2/3+–V’out1k1k1kV1+–V’’out1k1k1kV2++–+–Superposition of Summing Circuit (cont’d)23Superposition Procedure1. For each independent voltage and current source (repeat the following): a) Replace the other independent voltage sources with a short circuit (i.e., V = 0). b) Replace the other independent current sources with an open circuit (i.e., I = 0). Note: Dependent sources are not changed! c) Calculate the contribution of this particular voltage or current source to the desired output parameter. 2. Algebraically sum the individual contributions (current and/or voltage) from each independent source.24Class Examples•Example 2-9 (page 70).•Drill Problem