Electricity N Bronks.

Electricity N Bronks

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

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 and Charge Since one Ampere flows when one coulomb of charge passes a given point in a circuit each second, Amp = Coulomb second Current (A) = Charge (C) time (s) I = Q t or 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 : 1 second ? 10 seconds ? 5 Coulomb 50 Coulomb

Summary Question 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: 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 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: 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 = 3000 / 1.6x10-19 = 1.88x1022 electrons

3. In an electron beam experiment, the beam current is 1.2mA. Calculate The charge flowing along the beam each minute The number of electrons that pass along the beam each minute Ans: 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 = 0.072/1.6x10-19 = 4.5x1017 electrons

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. 0.1A Ans: I = 0.2A t = 4000s Q = It = 0.2 x 4000 = 800C 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. I = Q t Q = I t Q = 0.4 x 3 x 60 = 72 Coulomb

2. Calculate the number of electrons flowing through a resistor when a current of 2.3 flows for 5 minutes I = Q t Q = I x t Q = 2.3 x 5 x 60 = 690 Coulomb no. of electrons = 690 /1.6x10-19 = 4.31x1021

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 1.6x = 40 Coulombs I = Q t Current = 40/8 = 5 Amps

4. How long does it take for a current of 0.3A to supply a charge of 48C? I = Q t t = Q/I t = 48/0.3 = 160 seconds

5. How many electrons pass a point when a current of 0.4A flows for 900 seconds? I = Q t Q = I x t = 0.4 x = 360 Coulomb no. of electrons = total charge / charge of one electron no. of electrons = 360 / 1.6 x = x 1021

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

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

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… And the current here will be… 2A 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… The current here will be… 2A 2A

Voltage in a series circuit
Voltmeter always in parallel V If the voltage across the battery is 6V… …and these bulbs are all identical… V V …what will the voltage across each bulb be? 2V

Voltage in a series circuit
If the voltage across the battery is 6V… V …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? V 4V And here? V 4V

Summary Current is THE SAME at any point
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: Extra appliances (like bulbs) can be added without affecting the output of the others 2) When one breaks they don’t all fail

Resistance V R I Resistance = Voltage (in V) (in ) Current (in A)
Georg Simon Ohm 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: V R I Resistance = Voltage (in V) (in ) Current (in A)

An example question: Ammeter reads 2A A V Voltmeter reads 10V
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

What is the resistance of these bulbs?
More examples… 3A 6V 12V 4V 2A 1A 2V 3A What is the resistance of these bulbs?

Practice with Ohm’s Law
Volts Amps 4 100 25 15 150 10 2 30 9 45 5 6 48 8

VARIATION OF CURRENT (I) WITH P.D. (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

(a) A METALLIC CONDUCTOR
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 3. Diode 1. Resistor A diode only lets current go in one direction Current increases in proportion to voltage 2. Bulb As voltage increases the bulb gets hotter and resistance increases

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 10ºC Digital thermometer Ω 10º 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  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.

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 cross-sectional area. i.e. R  1 A

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 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.

ρ Precautions Ensure wire is straight and has no kinks like ....
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

H/W 2004 HL Q4

Resistors in series and Parallel
IT V R1 R2 R3 I2 R1 V1 I1 R2

H/W 2005 HL Q9

Wheatstone Bridge Uses Temperature control
Fail-Safe Device (switch circuit off) Measure an unknown resistance R1 = R3 (When it’s balanced R2 R4 Galvanometer reads zero) Metre Bridge R1 = R2 (|AB|) |BC| I r1 r2 r4 r3 A C B D

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 V I V 1. Active Electrodes 2. Inert Electrodes e.g. Platinum in Water e.g. Copper in Copper Sulphate

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 2) Thermistor – resistance DECREASES when temperature INCREASES – Due to more charge carriers being liberated by heat 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 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 A combination of fuse and Earth That Hurts!
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. 2. 3. 4. 5. 6. Earth wire Live wire Fuse Neutral wire
Cable grip Insulation

Capacitors A device for storing charge.
A pair of metal plates are separated by a narrow gap electrons - + - - - -

capacitor charge ‘right click’ on the switch for action

charged capacitor ‘right click’ on the switch for action (or lack of it in this case)

capacitor discharge electrons ‘right click’ on the switch for action

Charge & Discharge ‘right click’ on the switch for action

Capacitor Construction
Two metal plates Separated by insulating material ‘Sandwich’ construction ‘Swiss roll’ structure Capacitance set by... Dielectric material is good at holding onto the e-filed between the plates

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
The size of the capacitor depends on The Distance the plates are apart d - + - + - + d

2 /.The area of overlap A - + A - + - +

3/.The material between () - + + - - + - + High  material Called a DIELECTRIC - + - + - +

Finding Capacitance

For the parallel plate capacitor
Equations For the parallel plate capacitor Permitivity in Fm-1 Capacitance In Farads  A C = Area In m2 d Distance in meters

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. C d A = 0 8.5x10-12Fm-1x C = 0.04m2 =3.4x10-11F. 0.01m

Capacitance experiment on the internet

Capacitance on any conductor
Equations Capacitance on any conductor Charge in Coulombs Capacitance In Farads Q C = V Potential Difference in volts

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

Energy stored on a capacitor
Equations Energy stored on a capacitor Capacitance In Farads Energy Stored C (V)2 Work Done Voltage Squared =

Energy stored = ½ C V2 = ½ x 2.025x10-9x (150)2
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 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 V DC stands for “Direct Current” – the current only flows in one direction: Time AC stands for “Alternating Current” – the current changes direction 50 times every second (frequency = 50Hz) Find Root Mean Square of voltage by Vrms= Vpeak/ √2 1/50th s 240V T V

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 Transmitted is = P = V.I
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 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 = I2R = (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 = I2R = (1)2.12 = 12watts 10000 times less!

Joules law A Lid Digital thermometer Calorimeter Water 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 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 law The power lost (Rate at which heat is produced) is proportional to the square of the current. I2

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 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 d Q2 Force = f
 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 4d2  = permitivity as in capacitors

Coulomb's Law Calculations
Force between these bodies 2C d=2m 4mC Force = f = Q1.Q2 4d2  = 3.4 x 10-11

Force between these bodies 2C d=2m 4mC Force = f = x 0.004 4 x3.4 x 10-11x 22

Force between these bodies 2C d=2m 4mC Force = f = 7.49 x N

Force between these bodies 2C d=2m 4mC 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

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 = 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

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- Velocity Force e- Velocity Force e- Velocity Force e- e-

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- Velocity Force

See particles motion

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

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… The magnet is moved in the opposite direction The other pole is inserted first The size of the induced current can be increased by: Increasing the speed of movement Increasing the magnet strength 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:
Increasing the speed of movement Increasing the magnetic field strength Increasing the number of turns on the coil Increasing the area of the coil

Electric motor

Faraday’s Law Basically More turns (N) more EMF
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 preventing more induction Aluminum Ring

Mutual induction Main use in a transformer
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 D.C. Source 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 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

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