2 Atomic structureAtoms are composed of protons (+), electrons (-) and neutrons. The nucleus contains the protons and neutrons and the electrons surround the nucleus.
3 Atomic structureThe outer layer of electrons in a metal is incomplete which allows them to pass from atom to atom
4 Atomic structureBecause electrons can pass from atom to atom. charge can pass through a conducting material such as a metal.Metals are conductors
5 These are called Insulators Atomic structureSome materials such as rubber and plastic have complete outer layers of electrons so they cannot pass from atom to atom. Charge cannot pass through these materials.These are called Insulators
6 Conventional current flows from positive to negative Because it is the electrons which move from atom to atom in reality negative charge flows from negative to positive.This has the same effect as positive charge moving from positive to negativeConventional current flows from positive to negative
7 The Ampere (Named after Andre Marie Ampere) The Ampere is a measure of how much electrical current is flowing and is measured in units of ampsQ I = tI = amps Q = charge (in coulombs)and t = time ( in seconds)
8 E V = ---- Q Potential Difference or Voltage Alessandro VoltaPotential difference, or voltage, is the electrical potential energy per coulomb of charge.E V = QV = voltage E = energy in Joules Q = charge (in coulombs)
9 I = current V = voltage R = resistance in ohms Georg OhmResistance is a measure of opposition to the flow of charge and is measured in ohms ()V I = RI = current V = voltage R = resistance in ohms
11 Power is the rate of using energy in joules per second Electrical PowerPower is the rate of using energy in joules per secondP = Etor E = Pxt
12 From previous slides we know that Electrical PowerFrom previous slides we know thatE V = QQ I = tand
13 Electrical Power Combine the two and cancel the Q from each E V = QQ I = tXLeaving E/t so electrical power is P = V x I
14 Electrical Power Equation variations P = V x IP = I2RP = V2/RThese were obtained by using Ohm’s law to substitute for V and I
15 Kirchoff’s Laws I2 I1 I3 I1 = I2 + I3 Kirchoff’s first Law The total current flowing into a junction is the total current flowing out of the circuitI2I1I3I1 = I2 + I3
16 Kirchoff’s Second Law 1. The sum of the potential differences around an electricalcircuit equals the supplyvoltage.
17 The total resistance is found by simply adding the resistance of each Resistors in seriesThe total resistance is found by simply adding the resistance of eachR1 + R2 +R3 etc
18 Resistors in seriesThe supply voltage (pd) is shared across the resistors. The voltage across each depends on the resistance of each
19 Resistors in seriesThe current in a series circuit is the same all the way round the circuit(as per Kirchoff’s first Law). Current flowing into the resistor is the same as the current flowing out of the resistor)
20 The total resistance is calculated as below Resistors in parallelThe total resistance is calculated as below
21 Resistors in parallelThe current in a parallel circuit is shared between each resistor. (The amount in each depends on the resistance)
22 Resistors in parallelThe supply voltage (pd) across each resistor is the same as the supply voltage
23 Combined resistorsTo calculate the total resistance of the circuit calculate the parallel set first and treat it as a single resistor in series with the other resistor
24 Example R1 = 50Ω, R2 = 100Ω and supply voltage = 12V. From the following diagram determine:a) Total resistance.b) Total (supply) current.c) Voltage across each resistor.d) Power loss in resistor R1.R1 = 50Ω, R2 = 100Ω and supply voltage = 12V.
25 Example R1 = 50Ω, R2 = 100Ω and supply voltage = 12V. Total resistance = R1 + R2 =150ΩTotal (supply) current V/I = 12/150 =0.08 amps.Voltage across R1 = 50 x 0.08 = 4 voltsVoltage across R2 = 100 x 0.08 = 8 voltsPower loss in R1 = V x I = 4 x 0.08 = 0.32 WattsR1 = 50Ω, R2 = 100Ω and supply voltage = 12V.
26 R1 100Ω, R2 = 1kΩand supply voltage = 12V. ExampleFrom the following diagram determine:a) Total resistance.b) Total (supply) current.c) Current through each resistor.R1 100Ω, R2 = 1kΩand supply voltage = 12V.
28 Example Total current = V/R = 12/90.9 = 0.132 amps Current through R1, V/R1 = 12/100 = 0.12 ampsCurrent through R2, V/R2 = 12/1000 = amps
29 Example From the diagram below, determine: a) The total resistance, and the supply current.b) The voltage across the R1 resistor.c) The current through R2 , and the power dissipated in it.R1 = 200Ω R2 and R3 are both 100Ω and the supply voltage is 12 volts
30 Example Resistance of the parallel resistors 1/total = 1/100 +1/100 =2/100Total resistance = 100/2= 50ΩTotal resistance in circuit = = 250ΩCurrent = V/R=12/250=0.048 amps
31 ExampleVoltage across R1I x R=0.048 x 200= 9.6 volts
32 Example Voltage across R1 & R2 V = I x R 0.048 x 50 = 2.4 volts Current through R2I = V/R=2.4/100=0.024amps
33 ExamplePower dissipatedP = V x I2.4 x 0.024= watts