Lecture 12: TUE 02 FEBHow to Solve Multi-Loop CircuitsStep I: Simplify “Compile” CircuitsNote: Skip Compile and Go Straight to Loop and Junction Rules if Number of Batteries is LargePowerPoint PresentationSlide 6Slide 7ExampleSlide 9Slide 10Non-Ideal BatteriesMore Light BulbsRC Circuits: Charging a CapacitorRC Circuits: Discharging a CapacitorSlide 15Slide 16Slide 17Lecture 12: TUE 02 FEBLecture 12: TUE 02 FEBDC circuitsDC circuitsCh27.4-9Ch27.4-9Physics 2102Jonathan DowlingQuickTime™ and a decompressorare needed to see this picture.QuickTime™ and a decompressorare needed to see this picture.QuickTime™ and a decompressorare needed to see this picture.How to Solve Multi-Loop CircuitsHow to Solve Multi-Loop CircuitsStep I: Simplify “Compile” Step I: Simplify “Compile” CircuitsCircuits Resistors CapacitorsKey formula: V=iR Q=CVIn series: same current dQ/dt same charge Q Req=∑Rj 1/Ceq= ∑1/Cj In parallel: same voltage same voltage 1/Req= ∑1/Rj Ceq=∑Cj P = iV = i2R = V2/RU = QV/2 = Q2/2C = CV2Note: Skip Compile and Go Note: Skip Compile and Go Straight to Loop and Junction Straight to Loop and Junction Rules if Number of Batteries is Rules if Number of Batteries is LargeLargeOne Battery: Compile FirstThree Batteries: Straight to Loop & JunctionAround every loop add +E if you cross a battery from minus to plus, –E if plus to minus, and –iR for each resistor. Then sum to Zero: +E1 –E2 – iR1 – iR2 = 0.Step II: Apply Loop RuleStep II: Apply Loop RuleR1R2E1E2+–+–Conservation of ENERGY!Step II: Apply Junction RuleStep II: Apply Junction RuleAt every junction sum the ingoing currents and outgoing currents and set them equal. i1 = i2 + i3 i1i2i3Conservation of CHARGE!Step III: Equations to UnknownsStep III: Equations to UnknownsContinue Steps I–III until you have as many equations as unknowns! +E1 –E2 – i1R1 – i2 R2 = 0andi = i1 + i2Solve for i2 , i3 Given: E1 , E2 , i , R1 , R2ExampleExampleFind the equivalent resistance between points (a) F and H and (b) F and G. (Hint: For each pair of points, imagine that a battery is connected across the pair.) Compile R’s in SeriesCompile equivalent R’s in ParallelFHFHFHSlide RuleSeriesParallelExampleExampleAssume the batteries are ideal, and have emf E1=8V, E2=5V, E3=4V, and R1=140R2=75andR3=2. What is the current in each branch?What is the power delivered by each battery?Which point is at a higher potential, a or b?Apply loop rule three times and junction rule twice.ExampleExample•What’s the current through resistor R1?•What’s the current through resistor R2? •What’s the current through each battery?Apply loop rule three times and junction rule twice.Non-Ideal BatteriesNon-Ideal Batteries•You have two ideal identical batteries, and a resistor. Do you connect the batteries in series or in parallel to get maximum current through R?•Does the answer change if you have non-ideal (but still identical) batteries?Apply loop and junction rulesuntil you have current in R.More Light BulbsMore Light Bulbs•If all batteries are ideal, and all batteries and light bulbs are identical, in which arrangements will the light bulbs as bright as the one in circuit X?•Does the answer change if batteries are not ideal?Calculate i and V across each bulb.P=iV= “brightness”orCalculate each i with R’s the same:P=i2Ri(t)E/RRC Circuits: RC Circuits: ChargingCharging a a CapacitorCapacitorIn these circuits, current will change for a while, and then stay constant.We want to solve for current as a function of time i(t)=dq/dt. The charge on the capacitor will also be a function of time: q(t). The voltage across the resistor and the capacitor also change with time.To charge the capacitor, close the switch on a. Time constant: =RCA differential equation for q(t)! The solution is: € E +VR(t) + VC(t) = 0E − i(t)R − q(t) /C = 0E − dq /dt( )R − q(t) /C = 0 € E +VR(t) + VC(t) = 0E − i(t)R − q(t) /C = 0E − dq /dt( )R − q(t) /C = 0CEVC=Q/CVR=iR € q(t) = CE 1− e−t / RC( )→ i(t) ≡ dq /dt = E /R( )e−t / RC € q(t) = CE 1− e−t / RC( )→ i(t) ≡ dq /dt = E /R( )e−t / RCQuickTime™ and a decompressorare needed to see this picture.i(t)E/RRC Circuits: RC Circuits: DisDischarging a charging a CapacitorCapacitorAssume the switch has been closed on a for a long time: the capacitor will be charged with Q=CE. Assume the switch has been closed on a for a long time: the capacitor will be charged with Q=CE. Then, close the switch on b: charges find their way across the circuit, establishing a current.C++++€ VR+ VC= 0−i(t)R + q(t) /C = 0 → dq /dt( )R + q(t) /C = 0€ VR+ VC= 0−i(t)R + q(t) /C = 0 → dq /dt( )R + q(t) /C = 0Solution: € q(t) = q(0)e−t / RC= CEe−t / RCi(t) = dq /dt = q(0) /RC( )e−t / RC= E /R( )e−t / RC € q(t) = q(0)e−t / RC= CEe−t / RCi(t) = dq /dt = q(0) /RC( )e−t / RC= E /R( )e−t / RCQuickTime™ and aSorenson Video 3 decompressorare needed to see this picture.ExampleExampleThe three circuits below are connected to the same ideal battery with emf E. All resistors have resistance R, and all capacitors have capacitance C. •Which capacitor takes the longest in getting charged?•Which capacitor ends up with the largest charge? • What’s the final current delivered by each battery?•What happens when we disconnect the battery?Compile R’s into into Req. Then apply charging formula with ReqC = ExampleExampleIn the figure, E= 1 kV, C = 10 µF, R1 = R2 = R3 = 1 M. With C completely uncharged, switch S is suddenly closed (at t = 0).• What’s the current through each resistor at t=0?• What’s the current through each resistor after a long time?• How long is a long time?Compile R1, R2, and R3 into Req. Then apply discharging formula with ReqC =
View Full Document
Unlocking...