Lecture 07Dis week …J and EConsider a wireMicro-View “Resistivity”PowerThe figure below gives the electrical potential V(x) along a copper wire carrying a uniform current, from a point at higher potential (x=0m) to a point at a lower potential (x=3m). The wire has a radius of 2.45 mm. What is the current in the wire?We have all we need….Series CombinationsParallel Combination??What’s This???Moving on …..Power Source in a CircuitA REAL Power Source is NOT an ideal batteryA Physical BatteryBack to PotentialConsider a “circuit”.To rememberNEW LAWS PASSED BY THIS SESSION OF THE FLORIDUH LEGISLATURE.TWO resistors againA single “real” resistor can be modeled as follows:Take a trip around this circuit.Circuit ReductionMultiple BatteriesReductionAnother Reduction ExampleNOTICE ASSUMED DIRECTION OF TRAVELSlide 28The Unthinkable ….RC CircuitClose the SwitchReally Close the SwitchThis is a differential equation.Slide 34Time ConstantResult q=CE(1-e-t/RC)q=CE(1-e-t/RC) and i=(CE/RC) e-t/RCDischarging a CapacitorSlide 39If the resistance is doubled what is the power dissipated by the circuit?Lecture 07Current & CircuitsOctober 11, 2005Dis week …Monday – finish resistance and current and begin electric circuit.There is a new WA on board.Friday – Quiz on Monday-Wednesday’s materialNEXT FRIDAY – Examination #2Studying is a good idea!J and EEJand 0A 1letELVJLVALAVRAVAiJJAiConsider a wiredddneJAnetNecurrenttAnNMicro-View “Resistivity”meEameEad a depends on the material and is the mean time between collisionsease of motion – mobilityresistance to motion - scatteringmneEEEmnenevJd221PowerVi+-ELECTRONsBattery supplies energy to the resistor which, in turn, dissipates it in the form of heat.Work done on charge Q = Q x V RiiRiiVVtQtQVPPPOWERtimeWork2/REMEMBER: P=iV and P=i2RThe figure below gives the electrical potential V(x) along a copper wire carrying a uniform current, from a point at higher potential (x=0m) to a point at a lower potential (x=3m). The wire has a radius of 2.45 mm. What is the current in the wire?copper12 volts 0 voltsWhat does the graph tell us??*The length of the wire is 3 meters.*The potential difference across the wire is 12  volts.*The wire is uniform.Let’s get rid of the mm radius and convert it to area in square meters:A=r2 = 3.14159 x 2.452 x 10-6 m2orA=1.9 x 10-5 m 2Material is Copper so resistivity is (from table) = 1.69 x 10-8 ohm metersWe have all we need….ma 49.41067.21012RVi:Law sOhm' From67.2 109.1 0.3m-ohm 1069.13658ohmsvoltsmxmxALRSeries CombinationsR1 R2i iV1 V2ViiRseriesRgeneralRRRiRiRiRVVVandiRViRV)(:2121212211Parallel Combination??R1, I1R2, I2ViiRRgeneralRRRsoRVRVRViiiiRV11111..212121What’s This???In Fig. 28-39, find the equivalent resistance between points (a) F and H and [2.5] (b) F and G . [3.13]Moving on …..Fun and Frolic With Electric CircuitsPower Source in a CircuitVThe ideal battery does work on charges moving them (inside) from a lower potential to one that is V higher.A REAL Power Sourceis NOT an ideal batteryVE or Emf is an idealized device that does an amount of work E to move a unit charge from one side to another.By the way …. this is called a circuit!Internal ResistanceA Physical BatteryInternal ResistanceRrEmfiBack to PotentialRepresents a charge in spaceChange in potential as one circuitsthis complete circuit is ZERO!Consider a “circuit”.This trip around the circuit is the same as a path through space.THE CHANGE IN POTENTIAL FROM “a” AROUND THE CIRCUIT AND BACK TO “a” is ZERO!!To rememberIn a real circuit, we can neglect the resistance of the wires compared to the resistors.We can therefore consider a wire in a circuit to be an equipotential – the change in potential over its length is slight compared to that in a resistorA resistor allows current to flow from a high potential to a lower potential.The energy needed to do this is supplied by the battery. VqW NEW LAWS PASSED BY THIS SESSION OF THE FLORIDUH LEGISLATURE.LOOP EQUATIONThe sum of the voltage drops (or rises) as one completely travels through a circuit loop is zero.Sometimes known as Kirchoff’s loop equation.NODE EQUATIONThe sum of the currents entering (or leaving) a node in a circuit is ZEROTWO resistors againiR1 R2V1 V2Vjj2121RRResistors SERIESfor GeneralRRRoriRiRiRVA single “real” resistor can be modeledas follows:RabVpositionADD ENOUGH RESISTORS, MAKING THEM SMALLERAND YOU MODEL A CONTINUOUS VOLTAGE DROP.Take a trip around this circuit.Consider voltage DROPS:-E +ir +iR = 0orE=ir + iRriseCircuit Reductioni=E/ReqMultiple BatteriesReductionComputes iAnother Reduction ExamplePARALLEL12121600503012011RRNOTICE ASSUMED DIRECTION OF TRAVELVoltage Drops:-E1 –i1R1 + i2R2 + E2 +i1R1 = 0From “a”-i3R1 + E2 – E2 –i2R2 =0NODEI3 +i2 = i1In the figure, all the resistors have a resistance of 4.0  and all the (ideal) batteries have an emf of 4.0 V. What is the current through resistor R?The Unthinkable ….RC CircuitInitially, no current through the circuitClose switch at (a) and current begins to flow until the capacitor is fully charged.If capacitor is charged and switch is switched to (b) discharge will follow.Close the SwitchI need to use E for ENote RC = (Volts/Amp)(Coul/Volt) = Coul/(Coul/sec) = (1/sec)Really Close the SwitchI need to use E for ERERCqdtdqorECqdtdqRCqiREdtdqi since0Equation LoopNote RC = (Volts/Amp)(Coul/Volt) = Coul/(Coul/sec) = (1/sec)This is a differential equation.To solve we need what is called a particular solution as well as a general solution.We often do this by creative “guessing” and then matching the guess to reality.You may or may not have studied this topic … but you WILL!RCREaeCERERCqdtdqCEqRERCqdtdqKeqqatpatp1RCEEaE/R0CEa0for t/)e-CE(1)()e-CE(1q and -CEKKCE0solution from and 0q 0,When tand 0dq/dt charged,fully is device When the:solution particularat Look Solution Generalat-at-Time