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Transcript
Electricity N Bronks 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 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! Current • Flow of electrons Current Charge Passed a point time I Q t Current and Charge Since one Ampere flows when one coulomb of charge passes a given point in a circuit each second, Coulomb Amp = second Charge (C) Current (A) = time (s) or Q I= t Also Charge (C) = no. of electrons x charge of one electron Example 1: How many electrons are there in 20 Coulombs ? No. of electrons = total charge / charge of one electron No. of electrons = 20 / 1.6x10-19 No. of electrons = 1.25 x1020 electrons Example 2: The current in a circuit is 5A. What is the charge flowing in : a. 1 second ? 5 Coulomb b. 10 seconds ? 50 Coulomb Summary Question 1. a. 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 100s Ans: a. i. I = 0.35A t = 10s Q = It Q = 0.35 x 10 = 3.5 C ii. I = 0.35A Q = It t = 10 min = 600s Q = 0.35 x 600 = 210 C b. i. Q = 15C I = Q/t ii. Q = 15C I = Q/t t = 5s I = 15/5 = 3A t = 100s I = 15/100 = 0.15A 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. 5.0A Ans: a. I = 1.0µA t= 10min = 600s Q = It = 1x10-6 x 600 = 6x10-4 C No. of electrons = total charge / charge of one electron No. of electrons = 6x10-4 / 1.6x10-19 = 3.75x1015 electrons I = 5A t= 10min = 600s Q = It = 5 x 600 = 3000 C No. of electrons = total charge / charge of one electron No. of electrons = 3000 / 1.6x10-19 = 1.88x1022 electrons Summary Questions page 47 3. In an electron beam experiment, the beam current is 1.2mA. Calculate a. The charge flowing along the beam each minute b. The number of electrons that pass along the beam each minute Ans: a. I = 1.2x10-3 A t= 1min = 60s Q = It = 1.2x10-3 x60 = 0.072 C b. no. of electrons = total charge /charge of one electron no. of electrons = 0.072/1.6x10-19 = 4.5x1017 electrons 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: a. The total charge the battery can deliver before it needs to be recharged. b. The maximum time it could be used for without being recharged if the current through it was i. 0.5A, ii. 0.1A Ans: a. I = 0.2A t = 4000s Q = It = 0.2 x 4000 = 800C b. i. Q = 800C I = 0.5A t = Q/I = 800/0.5 = 1600s ii. Q = 800C I = 0.1A t = Q/I = 800/0.1 = 8000s Quiz 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= Q t Q=It Q = 0.4 x 3 x 60 = 72 Coulomb Current and charge quiz 2. Calculate the number of electrons flowing through a resistor when a current of 2.3 flows for 5 minutes Q I= t Q=Ixt Q = 2.3 x 5 x 60 = 690 Coulomb no. of electrons = 690 /1.6x10-19 = 4.31x1021 Current and charge quiz 3. What is the current in a circuit if 2.5x1020 electrons pass a given point every 8 seconds Charge (C) = no. of electrons x charge of one electrons Charge (C) = 2.5x1020 x 1.6x10-19 = 40 Coulombs I= Q t Current = 40/8 = 5 Amps Current and charge quiz 4. How long does it take for a current of 0.3A to supply a charge of 48C? Q I= t t = Q/I t = 48/0.3 = 160 seconds Current and charge quiz 5. How many electrons pass a point when a current of 0.4A flows for 900 seconds? Q I= t Q=Ixt = 0.4 x 900 = 360 Coulomb no. of electrons = total charge / charge of one electron no. of electrons = 360 / 1.6 x 10-19 = 2.25 x 1021 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=Ixt = 120x10-3 x 60 = 7.2 Coulomb 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=Ixt = 1 x 36 x 60 x 60 = 129600 Coulomb H/W • 2004 HL Q4 More basic ideas… Another battery means more current as there is a greater push on the electrons The extra resistance from the extra bulb means less current Current in a series circuit If the current here is 2 amps… The current here will be… 2A The current here will be… 2A And the current here will be… 2A In other words, the current in a series circuit is THE SAME at any point 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… Current in a parallel circuit If the current here is 6 amps And the current here will be… 6A The current here will be… 2A The current here will be… 2A The current here will be… 2A Voltage in a series circuit If the voltage across the battery is 6V… V Voltmeter always in parallel …and these bulbs are all identical… …what will the voltage across each bulb be? V V 2V Voltage in a series circuit If the voltage across the battery is 6V… …what will the voltage across two bulbs be? V V 4V Voltage in a parallel circuit If the voltage across the batteries is 4V… What is the voltage here? 4V V And here? V 4V Summary In a SERIES circuit: Current is THE SAME at any point Voltage SPLITS UP over each component In a PARALLEL circuit: Current SPLITS UP down each “strand” Voltage is THE SAME across each”strand” An example question: 6V A3 3A A1 V1 A2 V2 V3 Advantages of parallel circuits… There are two main reasons why parallel circuits are used more commonly than series circuits: 1) Extra appliances (like bulbs) can be added without affecting the output of the others 2) When one breaks they don’t all fail Resistance Resistance is anything that will RESIST a current. It is measured in Ohms, a unit named after me. That makes me so happy Georg Simon Ohm 1789-1854 The resistance of a component can be calculated using Ohm’s Law: Resistance (in ) = V Voltage (in V) Current (in A) I R An example question: Ammeter reads 2A A What is the resistance across this bulb? V As R = volts / current = 10/2 = 5 Assuming all the bulbs are the same what is the total resistance in this circuit? Voltmeter reads 10V Total R = 5 + 5 + 5 = 15 More examples… 3A 6V 12V 3A 2A 4V 2V 1A What is the resistance of these bulbs? Practice with Ohm’s Law Ohms 4 15 2 9 6 Volts 100 150 30 45 48 Amps 25 10 15 5 8 VARIATION OF CURRENT (I) WITH P.D. (V) A + 6V - V Nichrome wire Method 1. 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 Variations (a) A METALLIC CONDUCTOR With a wire (b) A FILAMENT BULB (c) COPPER SULFATE SOLUTION WITH COPPER ELECTRODES (d) SEMICONDUCTOR DIODE Done both ways with a milli-Ammeter and the a micro Ammeter Current-voltage graphs I I I V V V 1. Resistor Current increases in proportion to voltage 2. Bulb As voltage increases the bulb gets hotter and resistance increases 3. Diode A diode only lets current go in one direction Factors affecting Resistance of a conductor • Resistance depends on – Temperature – Material of conductor – Length – Cross-sectional area Temperature The resistance of a metallic conductor increases as the temperature increases e.g. copper The resistance of a semiconductor/insulator decreases as the temperature increases e.g. thermistor. VARIATION OF THE RESISTANCE OF A METALLIC CONDUCTOR WITH TEMPERATURE 10º C 10ºC Ω Digital thermometer Wire wound on frame Water Glycerol Heat source Method 1. 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. 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. Factors affecting Resistance of a conductor Length Resistance of a uniform conductor is directly proportional to its length. i.e. R L Cross-sectional area Resistance of a uniform conductor is inversely proportional to its crosssectional area. i.e. R1 A Factors affecting Resistance of a conductor • Material The material also affects the resistance of a conductor by a fixed amount for different materials. This is known as resistivity (). R = L A = Resistivity Unit: ohm meter m RESISTIVITY OF THE MATERIAL OF A WIRE Micrometer Nichrome wire Crocodile clips l Metre stick Bench clamp Stand Method 1. 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. R 1. Calculate the resistivity ρñ A, 2 d l where A = 4 2. Calculate the average value. Precautions Ensure wire is straight and has no kinks like .... Take the diameter of the wire at different angles Resistors in series and Parallel I IT R1 V1 R2 R3 VT V1 V2 V3 V I2 R1 I1 R2 IT I1 I 2 I 3 Resistors in series and Parallel I IT R1 V1 R2 R3 VT V1 V2 V3 IRT IR1 IR2 IR3 RT R1 R2 R3 V I2 R1 I1 R2 Resistors in series and Parallel I IT R1 V1 R2 R3 IT I1 I 2 I 3 V V V V R T R1 R 2 R 3 1 1 1 1 RT R1 R2 R3 V I2 R1 I1 R2 H/W • 2005 HL Q9 Wheatstone Bridge Uses – Temperature control – Fail-Safe Device (switch circuit off) – Measure an unknown resistance B A I C D – R1 = R3 (When it’s balanced R2 R4 Galvanometer reads zero) Metre Bridge R1 = R2 (|AB|) |BC| Effects of an Electric Current • Heat • Chemical • Magnetic Chemical Effects of an Electric Current • Electrolysis is the chemical effect of an electric current • Voltameter consists of electrodes, an electrolyte and a container • Inactive electrodes are electrodes that don’t take part in the chemical reaction e.g. platinum in H2SO4 • Active electrodes are electrodes that take part in the chemical reaction e.g. copper in CuSO4 Chemical Effects • Ion is an atom or molecule that has lost or gained 1 or more electrons • Charge Carriers in an electrolyte are + and – ions Uses Electroplating to make metal look better, prevent corrosion Purifying metals Making electrolytic capacitors Current-voltage graphs I I V V 1. Active Electrodes 2. Inert Electrodes e.g. Copper in Copper Sulphate e.g. Platinum in Water Current Carriers Medium Carrier Solid (Metal) Electrons Liquid (Electrolyte) Ions Gas Electrons and Ions Resistance in Semiconductors 1) Normal conductor like metal resistance increases as vibrating atoms slow the flow of electrons Resistance 2) Thermistor – resistance DECREASES when temperature INCREASES – Due to more charge carriers being liberated by heat Resistance Temperature Temperature Fuse – Safety device Fuses are designed to melt when too large a current tries to pass through them to protect devices. Prevent Fires Modern fuse boxes contain MCB (Miniature circuit breakers) that trip when too much current flows to protect the circuit 2A 5A Which Fuse • A i-pod charge uses 200W and is plugged into the mains at 230v. What fuse is in the plug? • P=I.V • 200=I.230 • I = 200/230 = 0.87A is current used • So the most the fuse should be is a 1A Other safety devices… 1) Insulation and double insulation In some parts of Europe they have no earth wire just two layer of insulating material the sign is 2) Residual Current Circuit Breaker An 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. Electrical Safety Fuse on live wire !! • A combination of fuse and Earth The casing touches the bare wire and it becomes live That Hurts! A.C. Supply The fuse will melt to prevent electrocution and the electricity is carried to earth Wiring a plug 1. Earth wire 4. Live wire 5. Fuse 2. Neutral wire 3. Insulation 6. Cable grip Capacitors • A device for storing charge. • A pair of metal plates are separated by a narrow gap - + + + + + + electrons - - - - capacitor charge charged capacitor capacitor discharge electrons Charge & Discharge Capacitor Construction • Two metal plates • Separated by insulating material • ‘Sandwich’ construction • ‘Swiss roll’ structure • Capacitance set by... A C d Uses of Capacitors • Storing charge for quick release – Camera Flash • Charging and discharging at fixed intervals – Hazard Lights • Smoothing rectified current – See Semiconductors Smoothing • Add capacitor variable capacitor smoothing capacitors Parallel Plate Capacitors • 1. The size of the capacitor depends on The Distance the plates are apart d - + - + - + d Parallel Plate Capacitors 2 /.The area of overlap A - + + + A Parallel Plate Capacitors • 3/.The material between () - - - + + + + + + + High material Called a DIELECTRIC Finding Capacitance Vs capacitanc e VA Equations For the parallel plate capacitor Capacitance In Farads Permitivity in Fm-1 C = A d Distance in meters Area In m2 Example 1 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. C = 0 A d C= 8.5x10-12Fm-1x 0.04m2 =3.4x10-11F. 0.01m Capacitance experiment on the internet Equations Capacitance on any conductor Capacitance In Farads C = Q V Potential Difference in volts Charge in Coulombs 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 Equations Energy stored on a capacitor Energy Stored Capacitance In Farads 2 C (V) Work Done = ½ Voltage Squared Example 3 Find 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-1 As C = ε0A/d = 27x10-12 x 0.015/0.002 = 2.025x10-9 F Energy stored = ½ C V2 = ½ x 2.025x10-9x (150)2 = 2.28x10-5 Joules Types of Batteries Type of Battery Contains Uses Wet cell rechargeable Lead and acid Cars, industry Dry cell rechargeable Nickel, cadmium, lithium Mobile phones, power tools Dry cell nonrechargeable Zinc, carbon, manganese, lithium Torches, clocks, hearing aids Why use rechargeable batteries? Why use standard batteries? • Long long-term expense • No need for charger • Can be used many times • Less expensive • Less energy to produce • Rechargeables contain carcinogens There are 2 types of currents: • Direct Current (DC) – Where electrons flow in the same direction in a wire. There are 2 types of currents: • Alternating Current (AC) – electrons flow in different directions in a wire DC and AC V DC stands for “Direct Current” – the current only flows in one direction: AC stands for “Alternating Current” – the current changes direction 50 times every second (frequency = 50Hz) Time 1/50th s 240V Find Root Mean Square of voltage by T Vrms= Vpeak/ √2 V Used to compared power output of D.C. with A.C. The National Grid Power station Step up transformer Step down transformer Homes If 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 The National Grid Power station Step up transformer Step down transformer Homes Power Transmitted is = P = V.I JOULES LAW gives us the power turned into heat Power Lost = I2R So if we have a high voltage we only need a small current. We loss much less energy Power loss in Transmission lines A power company wants to send 100000w of power by a line with a resistance of 12 ohms. If it uses 100A as the current Power transmitted = V . I 100000 = V . 100 So V=1000Volts But the loss is from Joules law = I2R = (100)2.12 = 120000watts Power loss in Transmission lines If we want the same power but use only 1A as the current Power transmitted = V . I 100000 = V . 1 So V=100000Volts But the loss is from Joules law = I2R = (1)2.12 = 12watts 10000 times less! Joules law 10°C A Digital thermometer Calorimeter Heating coil Lid Water Lagging Method 1. 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 ∆. Calculation and Graph ∆ I2 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 law The power lost (Rate at which heat is produced) is proportional to the square of the current. H/W • 2006 HL Q 4 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. The Electroscope - The electroscope detects charge The Gold leaf and post repel each other - + + + + H/W • 2006 HL Q9 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 Coulomb's Law • Force between two charged bodies Q1 Force = f d Q2 Q1.Q2 d2 Put this as a sentence to get a law! Coulomb Calculations Force =f Q1.Q2 d2 • We replace the proportional with a equals and a constant to get an equation Force = f = Q1.Q2 4d2 = permitivity as in capacitors Coulomb's Law Calculations • Force between these bodies d=2m 2C Force = f 4mC = Q1.Q2 4d2 = 3.4 x 10-11 Coulomb's Law Calculations • Force between these bodies d=2m 2C Force = f 4mC = 2 x 0.004 4 x3.4 x 10-11x 22 Coulomb's Law Calculations • Force between these bodies d=2m 2C Force = f 4mC = 7.49 x 10-15 N Coulomb's Law Calculations • Force between these bodies 2C d=2m 4mC Electric Field Strength Electric Field Strength = E = F/q = E = 7.49 x 10-15 N /2C = 3.75 x 10-15 N /C Precipitator • Carbon and ash - can be removed from waste gases with the use of electrostatic precipitators Precipitator • Dirt particles are charged then made to stick to oppositely charged plates Photocopier • • • • • • Charging: Exposure: Developing: Transfer: Fusing: Cleaning: 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=W Q Potential Difference (V) V=W Q Unit Volt V or J C-1 Volt is the p.d. between two points if one joule of work is done bringing one coulomb from one point to the other Potential at a point is the p.d. between a point and the Earth, where the Earth is at zero potential Current in a Magnetic Field N S N S Current in a Magnetic Field A conductor carrying a current in a magnetic field will always feel a force Current N S Magnetic Field Force The force is perpendicular to the current and the field. – This is THE MOTOR EFFECT Fleming’s Left Hand Rule I used my left hand to show the direction the wire would move The Size of the Force Force = F = B.I.l 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 Two Parallel Wires • Wires also produce magnetic fields when a current flows Attraction Two Parallel Wires • The fields act like magnets when the current flows Repulsion 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. Demo • OHP and coils and compass Moving Charge • When any charged particle moves it is like a small current of electricity • It feels the same force • The crosses show a magnetic field into the screen e- Force Velocity e- e- Moving Charge • A positive will move the other way + e- Force Velocity 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 See particles motion Force 0n a Particle Force = F = B.q.v 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 Use to define the tesla Defining Units • Find a formula • Say 1 thingys is when all other thingys are 1 • F = B.q.v • 1 Tesla in when 1N of force is on a charge of 1C moving at 1m/s Demo • CRT and magnet Induction is where changes in the current flow in a circuit are caused by changes in an external field. Moving Magnet N Circuit turning off and on Electromagnetic induction The direction of the induced current is reversed if… 1) The magnet is moved in the opposite direction 2) The other pole is inserted first The size of the induced current can be increased by: 1) Increasing the speed of movement 2) Increasing the magnet strength 3) Increasing the number of turns on the coil Demo • Coils and spot galvo • Internethttp://phet.colorado.edu/en/ simulation/faraday Generators (dynamos) Induced current can be increased in 4 ways: 1) Increasing the speed of movement 2) Increasing the magnetic field strength 3) Increasing the number of turns on the coil 4) Increasing the area of the coil Electric motor Faraday’s Law Basically 1. More turns (N) more EMF 2. Faster movement more EMF Rate of change of FLUX DENSITY is proportional to induced EMF Induced EMF = E = - Nd ( =B.A) dt Lenz’s Law The induced EMF always opposes the current/Motion You get ought for nought A version of Newton III and of energy conversion The induction always tries to stop the motion or change in the field. The ring moves away as the induced current is Aluminum preventing Ring more induction Mutual induction • Induction in a second circuit caused by changes in a first circuit • Main use in a transformer • As the current changes the field changes giving a EMF in the second circuit. Transformers This how A.C. changes voltage up or down V In V Out = Turns 2 Turns 1 Self Induction • property whereby an electromotive force (EMF) is induced in a circuit by a variation of current in the circuit its self Current Back EMF D.C. Source Another example on LENZ’S LAW Flux Density • Magnetic flux, represented by the Greek letter Φ (phi), total magnetism produced by an object. The SI unit of magnetic flux is the Weber • Magnetic 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