Electricity - 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 ______. Resistance is anything that resists an electric current. It is measured in _____.” (Words: volts, amps, ohms, voltage, ammeter, voltmeter)

Charge (Q) As we said, electricity is when electrons move around a circuit and carry energy with them. Each electron has a negative CHARGE. Charge is measured in Coulombs (C). We can work out how much charge flows in a circuit using the equation: Q TI Charge = current x time (in C) (in A) (in s) Charge on electron = 1.6x C One coulomb = I amp per second

Example questions Charge (C)Current (A)Time (s) )A circuit is switched on for 30s with a current of 3A. How much charge flowed? 2)During electrolysis 6A was passed through some copper chloride and a charge of 1200C flowed. How long was the experiment on for? 3)A bed lamp is switched on for 10 minutes. It works on a current of 0.5A. How much charge flowed?

More basic ideas… If a battery is added the current will ________ because there is a greater _____ on the electrons If a bulb is added the current will _______ because there is greater ________ in the circuit

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

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…

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

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

Georg Simon Ohm Resistance Resistance is anything that will RESIST a current. It is measured in Ohms, a unit named after me. 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 1)What is the resistance across this bulb? 2)Assuming all the bulbs are the same what is the total resistance in this circuit? Voltmeter reads 10V Ammeter reads 2A

Current-voltage graphs I V I V I V 1. Resistor 3. Diode 2. Bulb Explain the shape of each graph

Voltage - Current Graphs: Diode Only lets current flow in one direction. Forward biased – forward bias: high resistance (initially gives small current) at ≈ 0.6V, resistance decreases rapidly (current increases) reverse bias: high resistance (gives ~ zero or slightly negative current. At breakdown, resistance ~ zero (and very large current) I V

Voltage - Current Graphs: Filament Lamp As Voltage increases current increases. Current heats the filament Therefore resistance increases Same applies in negative direction I V

Current/Voltage Graphs: Ohmic conductor Straight line through origin – constant gradient V and I increasing proportionally I V

Resistors in Series Suppose all the resistors could be replaced with one resistor of equivalent resistance R T V T = IR T and V T =V 1 +V 2 +V 3 Eqn1 V 1 =IR 1 V 2 =IR 2 V 3 =IR 3 Therefore substituting into 1 I R T = IR 1 + IR 2 + IR 3 ( I cancels ) Therfore in series: R T = R 1 + R 2 + R 3

Resistors in Parallel I = I 1 +I 2 +I 3 Equation 1 Suppose all the resistors could be replaced with one resistor of equivalent resistance R T. I=V/R T I 1 = V/R 1 I 2 = V/R 2 I 3 = V/R 3 Substituting into Equation 1 V/R T = V/R 1 + V/R 2 + V/R 3 Cancelling V Therefore 1/R T = 1/R 1 + 1/R 2 + 1/R 3

Questions 1.Find the combined resistance of two resistors of 4Ω and 6Ω connected in a) seriesb) parallel. 2.Find the combined resistance of three resistors of 4Ω and 5Ω and 20Ω connected in a) seriesb) parallel. 3.What is the combined resistance of 5 resistors, each of value 10Ω, connected in a) series and b) parallel. 4.What resistance must be connected in parallel with a resistor of 16Ω to give a combined resistance of 3.2Ω.

Resistivity The resistance of a piece of conductor depends on length, area, temp and material. Assuming temp is constant R = Resistance (Ω) L = Length (m) A = Area (m 2) Ρ is a constant of Proportionality Resistivity units are Ωm.

Table of Resistivities MaterialResistivity(Ωm) Silver1.6x10 -8 Copper1.7x10 -8 Aluminium2.7x10 -8 Silicon2300 Glass Polystyrene10 15

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) 1/50 th s 240V V V Time T

Power and fuses Power is “the rate of doing work”. The amount of power being used in an electrical circuit is given by: P IV Power = voltage x current in W in V in A Using this equation we can work out the fuse rating for any appliance. For example, a 3kW (3000W) fire plugged into a 240V supply would need a current of _______ A, so a _______ amp fuse would be used (fuse values are usually 3, 5 or 13A).

Power and fuses Copy and complete the following table: AppliancePower rating (W) Voltage (V)Current needed (A) Fuse needed (3, 5 or 13A) Toaster Fire Hairdryer Hoover Computer Stereo80240

Energy and charge The amount of energy that flows in a circuit will depend on the amount of charge carried by the electrons and the voltage pushing the charge around: W QV Energy transferred = charge x voltage (in J) (in C) (in V)

Electromotive force and Internal Resistance Let an amount of energy E be supplied when a charge Q passes through the source. We call the energy supplied per unit charge the e.m.f. of the source Є. Є = E/Q Where E is measures in Joules, J Q is measured in coulombs, C

Electromotive Force and Internal Resistance For the dry cell battery, as the current is increased, the reading of the voltmeter across the battery decreased. Missing Voltage is proportional to Current E.M.F. of battery ( ε – V) proportional current –( ε – V) = Ir ( r is the internal resistance of the cell. Chemicals inside the battery offer some resistance to current flow) –ε = V + Ir = IR + Ir –therefore ε = I(R+r)

Example questions 1)In a radio circuit a voltage of 6V is applied and a charge of 100C flows. How much energy has been transferred? 2)In this circuit the radio drew a current of 0.5A. How long was it on for? 3)A motor operates at 6V and draws a current of 3A. The motor is used for 5 minutes. Calculate: a) The motor’s resistance, b) the charge flowing through it, c) the energy supplied to it 4)A lamp is attached to a 12V circuit and a charge of 1200C flows through it. If the lamp is on for 10 minutes calculate a) the current, b) the resistance, c) the energy supplied to the bulb.