9Electric Fields in Circuits Point away from positive terminal, towards negativeChanneled by conductor (wire)Electrons flow opposite field lines (neg. charge)Eelectrons & direction of motionEElectric field directionEE
10Batteries produce a voltage Typical Alkaline cells produce 1.5 VoltsAAA cellsAA cellsD cellsPutting batteries in series, voltages addPutting batteries in parallel, same voltageas a single cell, but can drawmore current, lasts longer (more water in reservoir)
13Relationship between Voltage, Current and Resistance: Ohm’s Law There is a simple relationship between voltage, current and resistance:V is in Volts (V)I is in Amperes, or amps (A)R is in Ohms ()V = I ROhm’s LawVIR
14ExamplesWhat is the ratio of the currents that flow in these 2 circuits?4 V Ohms V Ohms
15Class Problem - Ohm’s Law (V = I·R) How much voltage is being supplied to a circuit that contains a 1 Ohm resistance, if the current that flows is 1.5 Amperes?If a 12 Volt car battery is powering headlights that draw 0.5 Amps of current, what is the total resistance in the circuit?
16Class Problem(How much voltage is being supplied to a circuit that contains a 1 Ohm resistance, if the current that flows is 1.5 Amperes?)Use the relationship between Voltage, Current and Resistance, V = IR.Total resistance is 1 OhmCurrent is 1.5 AmpsSo V = IR = (1.5 Amps)(1 Ohms) = 1.5 Volts
17Class ProblemIf a 12 Volt car battery is powering headlights that draw 0.5 Amps of current, what is the total resistance in the circuit?Again need V = IRKnow I, V, need RRearrange equation: R = V divided by I= (12 Volts)/(0.5 Amps)= 24 Ohms
18How about multiple resistances? Resistances in series simply addVoltage across each one is DV = IRR1=10 R2=20 V = 3.0 VoltsTotal resistance is 10 + 20 = 30 So current that flows must be I = V/R = 3.0 V / 30 = 0.1 AWhat are the Voltages across R1 and R2?
19Voltage is potential, bulb presents resistance Battery is like reservoir of elevated waterThe higher, the bigger the potential, or voltageImagine bulbs as small tubes that let water drainResistance is represented by length of tubeshorter: less resistance: more current flowsVoltagelonger: more resistance: less current flows
20Parallel Resistances are a little trickier.... Rule for resistances in parallel:1/Rtot = 1/R1 + 1/R210 Ohms Ohms OhmsCan arrive at this by applying Ohm’s Law to find equal currentin each leg. To get twice the current of a single10 , could use 5 .
22Power Dissipation Physical model – electrons bumping into things! Kinetic Energy is turned into thermal energy (heat)Power = Voltage CurrentP = V IA device with a voltage drop of 1 Volt that passes a current of 1 Amp uses 1 Watt of power.
23Multi-bulb circuitsRank the expected brightness of the bulbs in the circuits shown, e.g. A>B, C=B, etc. WHY?!__++ABC
24Answer:Bulbs B and C have the same brightness, since the same current is flowing through them both.Bulb A is brighter than B and C are, since there is less total resistance in the single-bulb loop, soA > B=C.
25Adding BulbsWhere should we add bulb C in order to get A to shine more brightly?_+CAB
26AnswerThe only way to get bulb A to shine more brightly is to increase the current flowing through A.The only way to increase the current flowing through A is to decrease the total resistance in the circuit loopSince bulbs in parallel produce more paths for the current to take, the best (and only) choice for C is to put it in parallel with B, as illustrated on next page
29ExercisesIf you double the voltage across a light bulb, while keeping the current the same, by what factor does the power consumption increase?If you double the current through a resistor, by what factor does the consumption change?
30AnswersIf you double the voltage while keeping the current fixed, the power consumption doublesP = IVIf you double the current though a resistor, the power used goes up by a factor of 4! This is because both the current and the voltage doubleP = I V = I (IR) = I2 R
31Flashlights “A holder for dead batteries” How does a flashlight work? Light source?Power source?Control device?
32Incandescent Bulb Tungsten Filament Sealed Bulb Electrical contacts 120 W bulb at 120 V must be conducting 1 Amp (P = VI)Bulb resistance is then about 120 Ohms (V = IR)
33What limits bulb’s lifetime? Heated tungsten filament drives off tungsten atomsTradeoff between filament temperature and lifetimeEventually the filament burns out, and current no longer flows – no more light!How “efficient” do you think incandescent bulbs are?
34EfficiencyRatio between energy doing what you want vs. energy suppliedEfficiency = (energy emitted as visible light)/(total supplied)For incandescent bulbs, efficiency is at most 10% percentWhere does the rest of the energy go?
35Decorative LightsStrings of lights used to decorate contain many bulbsIn some light sets, a single bulb going out can shut off the entire setHow do you think sets like this are wired up?How might you design a light set that still works, even if a bulb goes out?Series combo: one goesno light
36Fault-tolerant light sets Wire up the bulbs in parallel, then if one goes out it still works
37Lights in your CarThe car has a battery as part of its electrical system(as well as a generator, voltage regulator, etc..)Lights in a car include:Interior light, turn on when the door opensTurn signalsBrake lightsHeadlights (high and low beam)The illustrations that follow are by no means the only way to accomplish these tasks!
38Brake lightsPedalSwitchPlus red filter to get desired color
39Interior car lights Manual Switch Door switch Switches wired in parallel: either one will do!(Example of OR logic circuit)
40Class ProblemThe simple series circuit consists of three identical lamps powered by battery. When a wire is connected between points a and b, a) What happens to the brightness of lamp 3? b) Does current in the circuit increase, decrease or remain the same? c) What happens to the brightness of lamps 1 and 2? d) Does the voltage drop across lamps 1 and 2 increase, decrease, or remain the same? e) Is the power dissipated by the circuit increased, decreased, or does it remain the same?
41Class Problema) Lamp 3 is short-circuited. It no longer glows because no current passes through it. b) The current in the circuit increases. Why? Because the circuit resistance is reduced. Whereas charge was made to flow through three lamps before, now it flows through only two lamps. So more energy is now given to each lamp. c) Lamps 1 and 2 glow brighter because of the increased current through them. d) The voltage drop across lamps 1 and 2 is greater. Whereas voltage supplied by the battery was previously divided between three lamps, it is now divided only between two lamps. So more energy is now given to each lamp. e) The power output of the two-lamp circuit is greater because of the greater current. This means more light will be emitted by the two lamps in series than from the three lamps in series. Three lamps connected in parallel, however, put out more light. Lamps are most often connected in parallel.