Presentation on theme: "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."— Presentation 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!
Flow of electrons Current
Current and Charge Since one Ampere flows when one coulomb of charge passes a given point in a circuit each second, Also Current (A) = Charge (C) time (s) I = Q t or Charge (C) = no. of electrons x charge of one electron Amp = Coulomb second
Example 1: How many electrons are there in 20 Coulombs ? No. of electrons = total charge / charge of one electron No. of electrons = 20 / 1.6x No. of electrons = 1.25 x10 20 electrons Example 2: The current in a circuit is 5A. What is the charge flowing in : a.1 second ? b.10 seconds ? 5 Coulomb 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 t = 10 min = 600s Q = It Q = 0.35 x 600 = 210 C b. i. Q = 15C t = 5s I = Q/t I = 15/5 = 3A ii. Q = 15C t = 100s I = Q/t I = 15/100 = 0.15A
Summary Questions page 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 / 1.6x = 3.75x10 15 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 / 1.6x = 1.88x10 22 electrons
Summary Questions page 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 = C b. no. of electrons = total charge /charge of one electron no. of electrons = 0.072/1.6x = 4.5x10 17 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
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. Q = I t Q = 0.4 x 3 x 60 = 72 Coulomb I = Q t
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 x t Q = 2.3 x 5 x 60 = 690 Coulomb no. of electrons = 690 /1.6x = 4.31x10 21 I = Q t
Current and charge quiz 3. What is the current in a circuit if 2.5x10 20 electrons pass a given point every 8 seconds Charge (C) = no. of electrons x charge of one electrons Charge (C) = 2.5x10 20 x 1.6x = 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? t = Q/I t = 48/0.3 = 160 seconds I = Q t
Current and charge quiz 5. How many electrons pass a point when a current of 0.4A flows for 900 seconds? Q = I x t = 0.4 x 900 = 360 Coulomb I = Q t no. of electrons = total charge / charge of one electron no. of electrons = 360 / 1.6 x = 2.25 x 10 21
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 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 = I x t = 1 x 36 x 60 x 60 = 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… And the current here will be… In other words, the current in a series circuit is THE SAME at any point 2A
Current in a parallel circuit A PARALLEL circuit is one where the current has a “choice of routes” Here comes the current… And the rest will go down here… Half of the current will go down here (assuming the bulbs are the same)…
Current in a parallel circuit If the current here is 6 amps The current here will be… And the current here will be… 6A 2A
Voltage in a series circuit V VV If the voltage across the battery is 6V… …and these bulbs are all identical… …what will the voltage across each bulb be? 2V Voltmeter always in parallel
Voltage in a series circuit V V If the voltage across the battery is 6V… …what will the voltage across two bulbs be? 4V
Voltage in a parallel circuit If the voltage across the batteries is 4V… What is the voltage here? And here? VV 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: V1V1 V2V2 6V 3A A1A1 A2A2 V3V3 A3A3
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
Georg Simon Ohm Resistance Resistance is anything that will RESIST a current. It is measured in Ohms, a unit named after me. That makes me so happy The resistance of a component can be calculated using Ohm’s Law: Resistance = Voltage (in V) (in )Current (in A) V RI
An example question: V A What is the resistance across this bulb? As R = volts / current = 10/2 = 5 Assuming all the bulbs are the same what is the total resistance in this circuit? Total R = = 15 Voltmeter reads 10V Ammeter reads 2A
More examples… 12V 3A 6V 4V 2A 1A 2V What is the resistance of these bulbs?
Practice with Ohm’s Law OhmsVoltsAmps
VARIATION OF CURRENT (I) WITH P.D. (V) A V + 6 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 V I V I V 1. Resistor 3. Diode 2. Bulb Current increases in proportion to voltage As voltage increases the bulb gets hotter and resistance increases 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 Water Wire wound on frame Glycerol Heat source 10 º C Digital thermometer Ω 10 º C
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 Precautions - Heat the water slowly so temperature does not rise at end of experiment - 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. R
Length Resistance of a uniform conductor is directly proportional to its length. i.e. R L Factors affecting Resistance of a conductor Cross-sectional area Resistance of a uniform conductor is inversely proportional to its cross- sectional area. i.e.R 1 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 = Resistivity A Unit: ohm meter m
RESISTIVITY OF THE MATERIAL OF A WIRE Micrometer Metre stick l Bench clamp Stand Nichrome wire Crocodile clips
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.
1.Calculate the resistivity where 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 ρ
H/W 2004 HL Q4
Resistors in series and Parallel V1V1 II1I1 V I2I2 ITIT R1R1 R1R1 R2R2 R3R3 R2R2
Wheatstone Bridge Uses –Temperature control –Fail-Safe Device (switch circuit off) –Measure an unknown resistance –R 1 = R 3 (When it’s balanced R 2 R 4 Galvanometer reads zero) Metre Bridge R 1 = R 2 (|AB|) |BC| I r 1 r 2 r 4 r 3 ACBD
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 H 2 SO 4 Active electrodes are electrodes that take part in the chemical reaction e.g. copper in CuSO 4
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 V I V 1. Active Electrodes 2. Inert Electrodes e.g. Copper in Copper Sulphate e.g. Platinum in Water
Current Carriers MediumCarrier Solid (Metal)Electrons Liquid (Electrolyte)Ions GasElectrons and Ions
Resistance in Semiconductors 2) Thermistor – resistance DECREASES when temperature INCREASES – Due to more charge carriers being liberated by heat 1) Normal conductor like metal resistance increases as vibrating atoms slow the flow of electrons Resistance Temperature Resistance 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 2) Residual Current Circuit Breaker In some parts of Europe they have no earth wire just two layer of insulating material the sign is 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 A combination of fuse and Earth A.C. Supply That Hurts! The fuse will melt to prevent electrocution and the electricity is carried to earth The casing touches the bare wire and it becomes live Fuse on live wire !!
Wiring a plug Earth wire Neutral wire Insulation Live wire Fuse Cable grip
Capacitors A device for storing charge. A pair of metal plates are separated by a narrow gap electrons
capacitor discharge electrons
Charge & Discharge
Capacitor Construction Two metal plates Separated by insulating material ‘Sandwich’ construction ‘Swiss roll’ structure Capacitance set by...
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
Parallel Plate Capacitors The size of the capacitor depends on 1.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
Equations C d A = For the parallel plate capacitor Distance in meters Area In m 2 Permitivity in Fm -1 Capacitance In Farads
Example 1 00 C 0.01m 0.04m 2 = The common area of the plates of an air capacitor is 400cm 2 if the distance between the plates is 1cm and ε 0 =8.5x Fm -1. C d A = 8.5x Fm -1 x =3.4x F.
Capacitance experiment on the internet Capacitance experiment on the internet
Equations C V Q = Capacitance on any conductor Potential Difference in volts Charge in Coulombs Capacitance In Farads
Placing a charge of 35 μ C on a conductor raises it's potential by 100 V. Calculate the capacitance of the conductor. Info Q = 35 μ C and V = 100V find C=? Using Q=VC or C = Q/V = 35 x /100 = 35 x Farads
Equations C ½Work Done (V) 2 = Energy stored on a capacitor Voltage Squared Capacitance In Farads Energy Stored
Example 3 Find the capacitance and energy stored of a parallel plate capacitor with 2mm between the plates and 150cm 2 overlap area and a dielectric of relative Permittivity of 3. The potential across the plates is 150V. A = 150cm 2 =0.015m 2,d = 2x10 -3 m, ε = 3x ε 0 = 27x Fm -1 As C = ε 0 A/d = 27x x 0.015/0.002 = 2.025x10 -9 F Energy stored = ½ C V 2 = ½ x 2.025x10 -9 x (150) 2 = 2.28x10 -5 Joules
Types of Batteries Type of BatteryContainsUses Wet cell rechargeable Lead and acidCars, industry Dry cell rechargeable Nickel, cadmium, lithium Mobile phones, power tools Dry cell non- rechargeable Zinc, carbon, manganese, lithium Torches, clocks, hearing aids Why use rechargeable batteries? Long long-term expense Can be used many times Less energy to produce Why use standard batteries? No need for charger Less expensive 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 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) Find Root Mean Square of voltage by V rms = V peak / √2 1/50 th s 240V V V Time T
The National Grid 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 Power station Step up transformer Step down transformer Homes
The National Grid Power Transmitted is = P = V.I JOULES LAW gives us the power turned into heat Power Lost = I 2 R So if we have a high voltage we only need a small current. We loss much less energy Power station Step up transformer Step down transformer Homes
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 current Power transmitted = V. I = V. 100 So V=1000Volts But the loss is from Joules law = I 2 R = (100) 2.12 = watts
Power loss in Transmission lines If we want the same power but use only 1A as the current Power transmitted = V. I = V. 1 So V=100000Volts But the loss is from Joules law = I 2 R = (1) 2.12 = 12watts times less!
Joules law Heating coil Lagging Calorimeter Water A Lid Digital thermometer 10°C
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 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 I 2 is Joules law The power lost (Rate at which heat is produced) is proportional to the square of the current. ∆∆ I2I2
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 Force = f Q1.Q2 d2d2 Q1Q2 d Put this as a sentence to get a law!
Coulomb Calculations We replace the proportional with a equals and a constant to get an equation Force =f Q1.Q2 d2d2 Force = f= Q1.Q2 4 d 2 = permitivity as in capacitors
Coulomb's Law Calculations Force between these bodies Force = f = Q1.Q2 4 d 2 2C4mC d=2m = 3.4 x
Coulomb's Law Calculations Force between these bodies Force = f = 2 x x3.4 x x 2 2 2C4mC d=2m
Coulomb's Law Calculations Force between these bodies Force = f = 7.49 x N 2C4mC d=2m
Coulomb's Law Calculations Force between these bodies 2C4mC d=2m Electric Field Strength = E = F/q Electric Field Strength = E = 7.49 x N /2C = 3.75 x 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
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
Current in a Magnetic Field N S Force Current Magnetic Field A conductor carrying a current in a magnetic field will always feel a force THE MOTOR EFFECT 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 F=2x10 -7 N/m 1m The current flowing is 1A when the force between two infinitely long conductors 1m apart in a vacuum is 2x10 -7 N 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-e- Velocity Force e-e- Velocity Force e-e- Velocity Force e-e- e-e-
Moving Charge A positive will move the other way e-e- Velocity Force + 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 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 Using Force = F = B.q.v = 10x.1x80 = 80N
Demo CRT and magnet
Induction is where changes in the current flow in a circuit are caused by changes in an external field. N Moving Magnet 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/faradayhttp://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 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. Aluminum Ring The ring moves away as the induced current is preventing 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 D.C. Source Current Back EMF 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 WeberphiSIunitWeber Magnetic field (B) is the flux through a square meter (the unit of magnetic field is the Weber per square meter, or Tesla.)esla As the flux expands the density through any square meter decreases