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Learn about electric circuits, current flow, voltage, resistance, and more. Explore Ohm's Law, parallel vs. series circuits, and factors affecting resistance. Conduct experiments and understand the relationship between current and voltage.
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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
Current • Flow of electrons
Current in a series circuit If the current here is 2 amps… The current here will be… The current here will be… And the current here will be… 2A 2A 2A In other words, the current in a series circuit is THE SAME at any point
Current in a parallel circuit 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… A PARALLEL circuit is one where the current has a “choice of routes”
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”
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 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 V Resistance = Voltage (in V) (in ) Current (in A) I R The resistance of a component can be calculated using Ohm’s Law:
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 = 5 + 5 + 5 = 15
VARIATION OF CURRENT (I) WITH P.D. (V) A Nichrome wire + 6 V - V
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 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 10º C 10ºC Ω Digital thermometer Wire wound on frame Water Glycerol Heat source
Graph and Precautions 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. R
Factors affecting Resistance of a conductor • Material, temperature, Area cross section and length R = L = Resistivity AUnit: ohm meter m
RESISTIVITY OF THE MATERIAL OF A WIRE Micrometer Nichrome wire Crocodile clips l Metre stick Bench clamp Stand
ρ 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
Resistors in series and Parallel I I1 I2 IT V R1 R2 R3 R1 V1 R2
Resistors in series and Parallel I I1 I2 IT V R1 R2 R3 R1 V1 R2
Resistors in series and Parallel I I1 I2 IT V R1 R2 R3 R1 V1 R2
Wheatstone Bridge r1 r3 r2 r4 B A C I D • Uses • Temperature control • Fail-Safe Device (switch circuit off) • Measure an unknown resistance • R1 = R3 (When it’s balanced) R2 R4 • Metre Bridge R1 = R2 (|AB|) |BC|
Effects of an Electric Current • Heat • Chemical • Magnetic
Current-voltage graphs I I V V 1. Active Electrodes 2. Inert Electrodes e.g. Platinum in Water e.g. Copper in Copper Sulphate
Resistance in Semiconductors Resistance Resistance Temperature Temperature 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
Fuse – Safety device Fuses are designed to melt when too largeacurrenttries to pass through them to protect devices. Prevent Fires Modern fuse boxes contain MCB (Miniature circuit breakers) 2A 5A
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 The casing touches the bare wire and it becomes live • A combination of fuse and Earth The fuse will melt to prevent electrocution and the electricity is carried to earth That Hurts! A.C. Supply
Wiring a plug 1. 4. 5. 2. 6. 3. Earth wire Live wire Fuse Neutral wire Cable grip Insulation
Uses of Capacitors • Storing charge for quick release – Camera Flash • Charging and discharging at fixed intervals – Hazard Lights • Smoothing rectified current – See Semiconductors
Parallel Plate Capacitors • 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 - + - + - +
Equations Permitivity in Fm-1 Capacitance In Farads Area In m2 Distance in meters For the parallel plate capacitor A C = d
Example 1 A C = d 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. 0 8.5x10-12Fm-1x 0.04m2 C = =3.4x10-11F. 0.01m
Equations Charge in Coulombs Capacitance In Farads Potential Difference in volts Capacitance on any conductor Q C = 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
Equations Capacitance In Farads Energy Stored Voltage Squared Energy stored on a capacitor C (V)2 Work Done ½ =
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
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 Step up transformer Step down transformer Homes Power station 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
Joules law A 10°C Lid Digital thermometer Calorimeter Water Heating coil Lagging
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 I2 is Joules law The power lost (Rate at which heat is produced) is proportional to the square of the current. I2
Coulomb's Law Force = f Q1.Q2 d2 • Force between two charged bodies Q1 d Q2 Put this as a sentence to get a law!
Coulomb Calculations Force =f Force = f Q1.Q2 = Q1.Q2 d2 4d2 • We replace the proportional with a equals and a constant to get an equation = permitivity as in capacitors
Coulomb's Law Calculations Force = f = Q1.Q2 4d2 • Force between these bodies 2C d=2m 4mC = 3.4 x 10-11
Coulomb's Law Calculations • Force between these bodies 2C d=2m 4mC Electric Field Strength = E = F/q Electric Field Strength = E = 7.49 x 10-15 N/2C = 3.75 x 10-15 N/C