2 Words: volts, amps, ohms, voltage, ammeter, voltmeter Basic ideas…Electric current is when electrons start to flow around a circuit. We use an _________ to measure it and it is measured in ____.Potential difference (also called _______) is how big the push on the electrons is. We use a ________ to measure it and it is measured in ______, a unit named after Volta.Resistance is anything that resists an electric current. It is measured in _____.Words: volts, amps, ohms, voltage, ammeter, voltmeter
3 Electrons are flowing from the negative to positive side of the battery through the wires Note current moves from positive to negative, however electrons are actually are moving in the opposite direction!
5 Current and ChargeSince one Ampere flows when one coulomb of charge passes a given point in a circuit each second,Amp =CoulombsecondCurrent (A) =Charge (C)time (s)I =QtorAlsoCharge (C) =no. of electrons x charge of one electron
6 Example 1:How many electrons are there in 20 Coulombs ?No. of electrons = total charge / charge of one electronNo. of electrons = 20 / 1.6x10-19No. of electrons = 1.25 x1020 electronsExample 2:The current in a circuit is 5A. What is the charge flowing in :1 second ?10 seconds ?5 Coulomb50 Coulomb
7 Summary Questiona. The current in a certain wire is 0.35A. Calculate the charge passing a point in the wire i. in 10 s, ii in 10 min.b. Calculate the average current in a wire through which a charge of 15C passes in i. 5s, ii 100sAns:i. I = 0.35A t = 10sQ = It Q = 0.35 x 10 = 3.5 Cii. I = 0.35A t = 10 min = 600sQ = It Q = 0.35 x 600 = 210 Cb. i. Q = 15C t = 5sI = Q/t I = 15/5 = 3Aii. Q = 15C t = 100sI = Q/t I = 15/100 = 0.15A
8 Summary Questions page 47 2. Calculate the number of electrons passing a point in the wire in 10 minutes when the current is a. 1.0µA b. 5.0AAns:I = 1.0µA t= 10min = 600sQ = It = 1x10-6 x 600 = 6x10-4 CNo. of electrons = total charge / charge of one electronNo. of electrons = 6x10-4 / 1.6x10-19 = 3.75x1015 electronsI = 5A t= 10min = 600sQ = It = 5 x 600 = 3000 CNo. of electrons = 3000 / 1.6x10-19 = 1.88x1022 electrons
9 Summary Questions page 47 3. In an electron beam experiment, the beam current is 1.2mA. CalculateThe charge flowing along the beam each minuteThe number of electrons that pass along the beam each minuteAns:I = 1.2x10-3 A t= 1min = 60sQ = It = 1.2x10-3 x60 = Cb. no. of electrons = total charge /charge of one electronno. of electrons = 0.072/1.6x10-19 = 4.5x1017 electrons
10 Summary Questions page 47 A certain type of rechargeable battery is capable of delivering a current of 0.2A for 4000s before its voltage drops and it needs to be recharged.Calculate:The total charge the battery can deliver before it needs to be recharged.The maximum time it could be used for without being recharged if the current through it was i. 0.5A, ii. 0.1AAns:I = 0.2A t = 4000sQ = It = 0.2 x 4000 = 800Ci. Q = 800C I = 0.5At = Q/I = 800/0.5 = 1600sii. Q = 800C I = 0.1At = Q/I = 800/0.1 = 8000s
12 Current and charge quiz 1. Calculate the charge passing through a lamp in three minutes when a steady current of 0.4 A is flowing.I =QtQ = I tQ = 0.4 x 3 x 60 = 72 Coulomb
13 Current and charge quiz 2. Calculate the number of electrons flowing through a resistor when a current of 2.3 flows for 5 minutesI =QtQ = I x tQ = 2.3 x 5 x 60 = 690 Coulombno. of electrons = 690 /1.6x10-19 = 4.31x1021
14 Current and charge quiz 3. What is the current in a circuit if 2.5x1020 electrons pass a given point every 8 secondsCharge (C) =no. of electrons x charge of one electronsCharge (C) =2.5x1020 x 1.6x = 40 CoulombsI =QtCurrent = 40/8 = 5 Amps
15 Current and charge quiz 4. How long does it take for a current of 0.3A to supply a charge of 48C?I =Qtt = Q/It = 48/0.3 = 160 seconds
16 Current and charge quiz 5. How many electrons pass a point when a current of 0.4A flows for 900 seconds?I =QtQ = I x t = 0.4 x = 360 Coulombno. of electrons = total charge / charge of one electronno. of electrons = 360 / 1.6 x = x 1021
17 Current and charge quiz 6. A torch bulb passes a current of 120 mA.How many coulombs of charge flow through the lamp in 1 minute?Q = I x t = 120x10-3 x = 7.2 Coulomb
18 Current and charge quiz 7. A car battery is rated as 36 A h.In principle this means it could pass a current of 1 A for 36 h before it runs down. How much charge passes through the battery if it is completely run down?Q = I x t = 1 x 36 x 60 x = Coulomb
20 More basic ideas…Another battery means more current as there is a greater push on the electronsThe extra resistance from the extra bulb means less current
21 Current in a series circuit If the current here is 2 amps…The current here will be…2AThe current here will be…And the current here will be…2A2AIn other words, the current in a series circuit is THE SAME at any point
22 Current in a parallel circuit A PARALLEL circuit is one where the current has a “choice of routes”Here comes the current…Half of the current will go down here (assuming the bulbs are the same)…And the rest will go down here…
23 Current in a parallel circuit If the current here is 6 ampsAnd the current here will be…6AThe current here will be…2AThe current here will be…The current here will be…2A2A
24 Voltage in a series circuit Voltmeter always in parallelVIf the voltage across the battery is 6V……and these bulbs are all identical…VV…what will the voltage across each bulb be?2V
25 Voltage in a series circuit If the voltage across the battery is 6V…V…what will the voltage across two bulbs be?4V
26 Voltage in a parallel circuit If the voltage across the batteries is 4V…What is the voltage here?V4VAnd here?V4V
27 Summary Current is THE SAME at any point In a SERIES circuit:Current is THE SAME at any pointVoltage SPLITS UP over each componentIn a PARALLEL circuit:Current SPLITS UP down each “strand”Voltage is THE SAME across each”strand”
29 Advantages of parallel circuits… There are two main reasons why parallel circuits are used more commonly than series circuits:Extra appliances (like bulbs) can be added without affecting the output of the others2) When one breaks they don’t all fail
30 Resistance V R I Resistance = Voltage (in V) (in ) Current (in A) Georg Simon OhmResistance is anything that will RESIST a current. It is measured in Ohms, a unit named after me.That makes me so happyThe resistance of a component can be calculated using Ohm’s Law:VRIResistance = Voltage (in V)(in ) Current (in A)
31 An example question: Ammeter reads 2A A V Voltmeter reads 10V What is the resistance across this bulb?As R = volts / current = 10/2 = 5Assuming all the bulbs are the same what is the total resistance in this circuit?Total R = = 15 Voltmeter reads 10V
32 What is the resistance of these bulbs? More examples…3A6V12V4V2A1A2V3AWhat is the resistance of these bulbs?
33 Practice with Ohm’s Law VoltsAmps410025151501023094556488
35 VARIATION OF CURRENT (I) WITH P.D. (V) +6 V-Nichrome wire
36 Method1. Set up the circuit as shown and set the voltage supply at 6 V d.c.2.Adjust the by moving the slider of the potential divider to obtain different values for the voltage V and hence for the current I.3.Obtain at least six values for V and I using the voltmeter and the ammeter.4.Plot a graph of V against I
37 (a) A METALLIC CONDUCTOR Variations(a) A METALLIC CONDUCTORWith a wire(b) A FILAMENT BULB(c) COPPER SULFATE SOLUTION WITH COPPER ELECTRODES(d) SEMICONDUCTOR DIODEDone both ways with a milli-Ammeter and the a micro Ammeter
38 Current-voltage graphs IVIVIV3. Diode1. ResistorA diode only lets current go in one directionCurrent increases in proportion to voltage2. BulbAs voltage increases the bulb gets hotter and resistance increases
39 Factors affecting Resistance of a conductor Resistance depends onTemperatureMaterial of conductorLengthCross-sectional areaTemperatureThe resistance of a metallic conductor increases as the temperature increases e.g. copperThe resistance of a semiconductor/insulator decreases as the temperature increases e.g. thermistor.
40 VARIATION OF THE RESISTANCE OF A METALLIC CONDUCTOR WITH TEMPERATURE WaterWire wound on frameGlycerolHeat source10ºCDigitalthermometerΩ10º C
41 Method1. Set up as shown.2. Use the thermometer to note the temperature of the glycerol, which is also the temperature of the coil.3. Record the resistance of the coil of wire using the ohmmeter.4. Heat the beaker.5. For each 10 C rise in temperature record the resistance and temperature using the ohmmeter and the thermometer.6. Plot a graph of resistance against temperature.
42 Graph and Precautions R Precautions - Heat the water slowly so temperature does not rise at end of experiment-Wait until glycerol is the same temperature as water before taking a reading.
43 Factors affecting Resistance of a conductor LengthResistance of a uniform conductor is directly proportional to its length.i.e. R LCross-sectional areaResistance of a uniform conductor is inversely proportional to its cross-sectional area.i.e. R 1A
44 Factors affecting Resistance of a conductor MaterialThe material also affects the resistance of a conductor by a fixed amount for different materials. This is known as resistivity ().R = L = ResistivityA Unit: ohm meter m
45 RESISTIVITY OF THE MATERIAL OF A WIRE MicrometerNichrome wireCrocodile clipslMetre stickBench clampStand
46 Method1. Note the resistance of the leads when the crocodile clips are connected together. Could also be precaution.2. Stretch the wire enough to remove any kinks or ‘slack’ in the wire.3.Read the resistance of the leads plus the resistance of wire between the crocodile clips from the ohmmeter. Subtract the resistance of the leads to get R.4.Measure the length l of the wire between the crocodile clips, with the metre stick.5.Increase the distance between the crocodile clips. Measure the new values of R and l and tabulate the results.6.Make a note of the zero error on the micrometer. Find the average value of the diameter d.
47 ρ Precautions Ensure wire is straight and has no kinks like .... 1. Calculate the resistivitywhere A =2. Calculate the average value.Precautions Ensure wire is straight and has no kinks like ....Take the diameter of the wire at different angles
53 Wheatstone Bridge Uses Temperature control Fail-Safe Device (switch circuit off)Measure an unknown resistanceR1 = R3 (When it’s balancedR2 R4 Galvanometer reads zero)Metre BridgeR1 = R2 (|AB|)|BC|Ir1r2r4r3ACBD
54 Effects of an Electric Current HeatChemicalMagnetic
55 Chemical Effects of an Electric Current Electrolysis is the chemical effect of an electric currentVoltameter consists of electrodes, an electrolyte and a containerInactive electrodes are electrodes that don’t take part in the chemical reaction e.g. platinum in H2SO4Active electrodes are electrodes that take part in the chemical reaction e.g. copper in CuSO4
56 Chemical EffectsIon is an atom or molecule that has lost or gained 1 or more electronsCharge Carriers in an electrolyte are + and – ionsUsesElectroplating to make metal look better, prevent corrosionPurifying metalsMaking electrolytic capacitors
57 Current-voltage graphs IVIV1. Active Electrodes2. Inert Electrodese.g. Platinum in Watere.g. Copper in Copper Sulphate
58 Current Carriers Medium Carrier Solid (Metal) Electrons Liquid (Electrolyte)IonsGasElectrons and Ions
59 Resistance in Semiconductors 1) Normal conductor like metal resistance increases as vibrating atoms slow the flow of electrons2) Thermistor – resistance DECREASES when temperature INCREASES – Due to more charge carriers being liberated by heatResistanceTemperatureResistanceTemperature
60 Fuse – Safety deviceFuses are designed to melt when too large a current tries to pass through them to protect devices.Prevent FiresModern fuse boxes contain MCB (Miniature circuit breakers) that trip when too much current flows to protect the circuit2A5A
61 Which FuseA i-pod charge uses 200W and is plugged into the mains at 230v. What fuse is in the plug?P=I.V200=I.230I = 200/230 = 0.87A is current usedSo the most the fuse should be is a 1A
62 Other safety devices…1) Insulation and double insulationIn some parts of Europe they have no earth wire just two layer of insulating material the sign is2) Residual Current Circuit BreakerAn RCCB (RCB) detects any difference in current between the live and neutral connectors and the earth it switches off the current when needed. They can also be easily reset.
63 Electrical Safety A combination of fuse and Earth That Hurts! Fuse on live wire !!A combination of fuse and EarthThe casing touches the bare wire and it becomes liveThatHurts!A.C. SupplyThe fuse will melt to prevent electrocution and the electricity is carried to earth
64 Wiring a plug 1. 2. 3. 4. 5. 6. Earth wire Live wire Fuse Neutral wire Cable gripInsulation
65 Capacitors A device for storing charge. A pair of metal plates are separated by a narrow gapelectrons-+----
66 capacitor charge‘right click’ on the switch for action
67 charged capacitor‘right click’ on the switch for action (or lack of it in this case)
68 capacitor dischargeelectrons‘right click’ on the switch for action
69 Charge & Discharge‘right click’ on the switch for action
70 Capacitor Construction Two metal platesSeparated by insulating material‘Sandwich’ construction‘Swiss roll’ structureCapacitance set by...Dielectric material is good at holding onto the e-filed between the plates
71 Uses of Capacitors Storing charge for quick release – Camera Flash Charging and discharging at fixed intervals – Hazard LightsSmoothing rectified current – See Semiconductors
79 For the parallel plate capacitor EquationsFor the parallel plate capacitorPermitivity inFm-1CapacitanceIn FaradsAC=AreaIn m2dDistance inmeters
80 Example 1 C d A = 0 C = 8.5x10-12Fm-1x 0.04m2 =3.4x10-11F. 0.01m The common area of the plates of an air capacitor is 400cm2 if the distance between the plates is 1cm and ε0=8.5x10-12Fm-1.CdA=08.5x10-12Fm-1xC=0.04m2=3.4x10-11F.0.01m
82 Capacitance on any conductor EquationsCapacitance on any conductorCharge inCoulombsCapacitanceIn FaradsQC=VPotentialDifferencein volts
83 Placing a charge of 35μC on a conductor raises it's potential by 100 V Placing a charge of 35μC on a conductor raises it's potential by 100 V. Calculate the capacitance of the conductor.Info Q = 35μC and V = 100V find C=?Using Q=VC or C = Q/V= 35 x 10-6/100= 35 x 10-8 Farads
84 Energy stored on a capacitor EquationsEnergy stored on a capacitorCapacitanceIn FaradsEnergyStoredC(V)2Work DoneVoltageSquared=
85 Energy stored = ½ C V2 = ½ x 2.025x10-9x (150)2 Example 3Find the capacitance and energy stored of a parallel plate capacitor with 2mm between the plates and 150cm2 overlap area and a dielectric of relative Permittivity of 3. The potential across the plates is 150V.A = 150cm2=0.015m2, d = 2x10-3m,ε = 3xε0 = 27x10-12Fm-1As C = ε0A/d = 27x10-12 x 0.015/0.002 = 2.025x10-9 FEnergy stored = ½ C V2 = ½ x 2.025x10-9x (150)2= 2.28x10-5 Joules
86 Types of Batteries Type of Battery Contains Uses Wet cell rechargeable Lead and acidCars, industryDry cell rechargeableNickel, cadmium, lithiumMobile phones, power toolsDry cell non-rechargeableZinc, carbon, manganese, lithiumTorches, clocks, hearing aidsWhy use rechargeable batteries?Long long-term expenseCan be used many timesLess energy to produceWhy use standard batteries?No need for chargerLess expensiveRechargeables contain carcinogens
87 There are 2 types of currents: Direct Current (DC) – Where electrons flow in the same direction in a wire.
88 There are 2 types of currents: Alternating Current (AC) – electrons flow in different directions in a wire
89 DC and ACVDC stands for “Direct Current” – the current only flows in one direction:TimeAC stands for “Alternating Current” – the current changes direction 50 times every second (frequency = 50Hz)Find Root Mean Square of voltage byVrms= Vpeak/ √21/50th s240VTV
91 The National GridPower stationStep up transformerStep down transformerHomesIf electricity companies transmitted electricity at 240 volts through overhead power lines there would be too much energy lost by the time electricity reached our homes.This is explained by JOULES LAW
92 The National Grid Power Transmitted is = P = V.I Power stationStep up transformerStep down transformerHomesPower Transmitted is = P = V.IJOULES LAW gives us the power turned into heatPower Lost = I2RSo if we have a high voltage we only need a small current. We loss much less energy
93 Power loss in Transmission lines A power company wants to send w of power by a line with a resistance of 12 ohms. If it uses 100A as the currentPower transmitted = V . I= V . 100So V=1000VoltsBut the loss is from Joules law = I2R= (100)2.12 = watts
94 Power loss in Transmission lines If we want the same power but use only 1A as the currentPower transmitted = V . I= V . 1So V=100000VoltsBut the loss is from Joules law = I2R= (1)2.12 = 12watts10000 times less!
95 Joules law A Lid Digital thermometer Calorimeter Water Heating coil LaggingCalorimeterWaterALidDigitalthermometer10°C
96 Method1. Put sufficient water in a calorimeter to cover the heating coil. Set up the circuit as shown.2. Note the temperature.3. Switch on the power and simultaneously start the stopwatch. Allow a current of 0.5 A to flow for five minutes. Make sure the current stays constant throughout; adjust the rheostat if necessary.4. Note the current, using the ammeter.5. Note the time for which the current flowed.6. Stir and note the highest temperature. Calculate the change in temperature ∆.
97 Calculation and Graph Plot a graph of ∆(Y-axis) against I 2 (X-axis). Repeat the above procedure for increasing values of current I, taking care not to exceed the current rating marked on the rheostat or the power supply. Take at least six readings.Plot a graph of ∆(Y-axis) against I 2 (X-axis).A straight-line graph through the origin verifies that ∆ I 2 i.e. Joule’s law.Electrical Power lost as Heat P I2 is Joules lawThe power lost (Rate at which heat is produced) is proportional to the square of the current.I2
99 Experiment to Show shape of Electric Field The electrodes connected to high voltage source is placed in the shallow glass dish containing a mixture of semolina and castor oil. The semolina aligns itself along the lines of the electric field.
100 The Electroscope The electroscope detects charge +-The electroscope detects chargeThe Gold leaf and post repel each other++++
102 Electric and Magnetic Fields Electric Field- region of space where a charged particle feels a electrostatic force.Magnetic field – region where a magnet feels a force other than gravity.Field lines are the path a positive charge or north pole would travel
103 Coulomb's Law Force between two charged bodies Q1 d Q2 Force = f Q1.Q2d2Put this as a sentence to get a law!
104 Coulomb CalculationsForce =f Q1.Q2d2We replace the proportional with a equals and a constant to get an equationForce = f= Q1.Q24d2 = permitivity as in capacitors
105 Coulomb's Law Calculations Force between these bodies2Cd=2m4mCForce = f= Q1.Q24d2 = 3.4 x 10-11
106 Coulomb's Law Calculations Force between these bodies2Cd=2m4mCForce = f= x 0.0044 x3.4 x 10-11x 22
107 Coulomb's Law Calculations Force between these bodies2Cd=2m4mCForce = f= 7.49 x N
108 Coulomb's Law Calculations Force between these bodies2Cd=2m4mCElectric Field Strength = E = F/qElectric Field Strength =E = 7.49 x N /2C= 3.75 x N /C
109 PrecipitatorCarbon and ash - can be removed from waste gases with the use of electrostatic precipitators
110 PrecipitatorDirt particles are charged then made to stick to oppositely charged plates
112 Potential Difference (V) Potential difference is the work done per unit charge to transfer a charge from one point to another (also Voltage)i.e V = WQ
113 Potential Difference (V) V = WQUnit Volt V or J C-1Volt is the p.d. between two points if one joule of work is done bringing one coulomb from one point to the otherPotential at a point is the p.d. between a point and the Earth, where the Earth is at zero potential
115 Current in a Magnetic Field A conductor carrying a current in a magnetic field will always feel a forceCurrentN SMagnetic FieldForceThe force is perpendicular to the current and the field. – This is THE MOTOR EFFECT
116 Fleming’s Left Hand Rule I used my left hand to show the direction the wire would move
117 The Size of the Force Force = F = B.I.l = 4.5x3x0.8 = 10.8N Where B = Magnetic Field Density in Tesla (T)I= Current in Amps (A)……………………………L = length if the conductor in metres…Example What is the force acting on a conductor of length 80cm carrying a current of 3A in a 4.5T magnetic field?Using Force = F = B.I.l= 4.5x3x0.8= 10.8N
118 Two Parallel WiresWires also produce magnetic fields when a current flowsAttraction
119 Two Parallel Wires The fields act like magnets when the current flows Repulsion
120 The Ampere Basic unit of electricity 1m F=2x10-7N/m The current flowing is 1A when the force between two infinitely long conductors 1m apart in a vacuum is 2x10-7N Per metre of length.
123 Moving ChargeWhen any charged particle moves it is like a small current of electricityIt feels the same forceThe crosses show a magnetic field into the screene-VelocityForcee-VelocityForcee-VelocityForcee-e-
124 Moving Charge A positive will move the other way All charged particles moving in magnetic fields always have a force at right angles to their velocity so follow a circular path due to FLH Rule+e-VelocityForce
126 Force 0n a Particle Force = F = B.q.v = 10x.1x80 = 80N Where B = Magnetic Field Density in Tesla (T)q=charge on the particle (C)v=velocity of the particle…Example What is the force acting on a particle travelling at 80m/s carrying a charge of 0.1C in a 10T magnetic field?Using Force = F = B.q.v= 10x.1x80= 80N
129 Inductionis where changes in the current flow in a circuit are caused by changes in an external field.NMoving MagnetCircuit turning off and on
130 Electromagnetic induction The direction of the induced current is reversed if…The magnet is moved in the opposite directionThe other pole is inserted firstThe size of the induced current can be increased by:Increasing the speed of movementIncreasing the magnet strengthIncreasing the number of turns on the coil
131 Demo Coils and spot galvo Internethttp://phet.colorado.edu/en/simulation/faraday
132 Generators (dynamos) Induced current can be increased in 4 ways: Increasing the speed of movementIncreasing the magnetic field strengthIncreasing the number of turns on the coilIncreasing the area of the coil
134 Faraday’s Law Basically More turns (N) more EMF Faster movement more EMFRate of change of FLUX DENSITY is proportional to induced EMFInduced EMF = E = - Nd ( =B.A)dt
135 Lenz’s Law The induced EMF always opposes the current/Motion You get ought for noughtA version of Newton III and of energy conversionThe induction always tries to stop the motion or change in the field.The ring moves away as the induced current is preventing more inductionAluminum Ring
136 Mutual induction Main use in a transformer Induction in a second circuit caused by changes in a first circuitMain use in a transformerAs the current changes the field changes giving a EMF in the second circuit.
137 Transformers = This how A.C. changes voltage up or down V In V Out Turns 2Turns 1=
138 Self Inductionproperty whereby an electromotive force (EMF) is induced in a circuit by a variation of current in the circuit its selfCurrentD.C. SourceBack EMFAnother example on LENZ’S LAW
139 Flux DensityMagnetic flux, represented by the Greek letter Φ (phi), total magnetism produced by an object. The SI unit of magnetic flux is the WeberMagnetic field (B) is the flux through a square meter (the unit of magnetic field is the Weber per square meter, or Tesla.)As the flux expands the density through any square meter decreases
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