IX. Electricity· Electrical circuits: basic elements.· Current.· Potential difference.· Resistance.· Ohms law.· Effects of electrical resistors in series and in parallel.· Effects of electricity: chemical, thermal, magneticand mechanical.· Joule Effect.· Electrical power.
Charge and currentThe charge of electricity is measured in the unit coulomb (C).There is a link between the charge and current:If the charge flows at this rate: Then the current is:1 coulomb per second 1 ampere2 coulombs per second 2 amperesetc.The equation is therefore the following:Charge = current x time (C) = (A) x (s)
An electrical cell makes electrons (e-) move. Anexample of an electrical cell is a battery.CurrentA flow of charge is called a current. A charge isthe flow of electrons through matter.The SI unit to measure the strength of a currentis ampere (A)More or less 6.000.000.000.000.000.000 e- flowround a circuit per second to give a strength of 1A.For smaller currents milliamperes (mA) are usedas a unit. 1000mA = 1A
Current directionMost circuit diagrams, like the one below, have an arrowindicating the direction of the current flow. The flow alwaysgoes from positive to negative round the circuit.As electrons are negatively charged they are repelled by thenegatively charged side of the cell and go therefore in theopposite direction as the current flow.
The potential differece (PD) or voltage The scientific name for voltage is potential difference (PD) The higher the voltage the more energy a cell gives to the electrons it pushes out. The PD is measured in volt (V). If the PD accross a cell is 1 volt, then 1 joule of potential energy are given to each coulomb of charge. So 1 volt = 1 joule per coulomb 1V=(j/C) Cells in parallel and in series To produce a higher PD, cells can be put in series .
When batteries are connected in series youhave to sum up their voltage.The voltage tells you how much energy isgiven per Coulomb.So if you connect batteries in series, thevoltage given is the sum of all.
ResistanceThe resistance of a material indicates how well it is able to prevent the electrons from flowing through it. The resistance is calculated with the following equation: PD accros the conductor (V) Resistance ( Ω ) = Current through conductor (A) So the lower the resistance, the less PD is needed to give the same current For example if a PD of 6 V is needed to make a current of 3A flow through a wire the resistance would be: 6V/3A= 2Ω 1 kilohm (kΩ) = 1000 Ω 1 megaohm (MΩ) = 1.000.000 Ω
Ohm’s law1. Calculate the resistance of a conductor through which a current of 2 A flows with a PD of 12V.2. Calculate the PD between the extremes of aconductor with a resistance of 10 Ω and a current of7.5 A.3. Calculate the intensity of a current which circulates through a conductor with a resistance of 10 Ω, if the PD between its extremes 0.02 mV.
Series and Parallel circuitsBulbs in series·The bulbs share the PD (V)from the battery, bulbs glowdimly·If one bulb would be removed,the would go out because thecircuit is broken.Bulbs in parallel·Each gets the full PD from the battery because they are bothconnected directly to it.·If one bulb would be removed teh other would keep on workingbecasue the circuit isnt broken
Circuit in seriesIntensity/Current (A)I = I = I = I etc T 1 2 3PD (V)V = V + V + V etc T 1 2 3Resistance (Ω)R = R + R + R etc T 1 2 3Circuits in parallelIntensity/Current (A)I = I + I + I etcT 1 2 3 For examplePD (V)V = V = V V etc T 1 2= 3Resistance (Ω) 1 = 1 + 1 + 1 etc R R T R 1 2 R3
9)a) The combined resistance.b) The current which flows through the circuit.c) The voltage in each resistance.10. Given the following circuit, calculate:a) The combined resistance of the circuit.b) The current that circulated through the circuit.
PowerPower = energy transformed time taken W= Joules/SecondsFor electrical energy we use the unit kwh (Kilowattper hour)1 kw= 1000 watts1kwh= 1Kw per hor kwh= kw h
Electrical PowerPower = PD x currentWatt = Volt x AmpereorP= VISo for a battery:Power = 12V x 2A = 24WFor bulb A: 8V x 2A = 16WFor bulb B: 4V x 2A = 8W
4.- Calculate the energy consumed by an iron of1000 W if it is switched on during 30 minutes. Sol.0.5 kwh5.- We have a light bulb of 75 W and 220 V.Calculate a) the current that passes through the bulb; b)the resistance of the bulb; c) the energy consumedduring 30 days if it will be switched on for 4 hours a day.Sol. a)
1. Calculate the energy consumed by an iron of 1000 W if it is plugged to a 220 V socket during 30 minutes.2. We have a light bulb of 75 W and 220 V. Calculate a) the current that passes through the bulb; b) the resistance of the bulb; c) the energy consumed during 30 days if it will be switched on for 4 hours a day.3. Calculate how much energy it will cost to dry yourhair every day if you use a hair dryer of 1100 W during 20minutes each. The Price per KWh is €0.15.4. Calculate the amount of energy consumed and its price per month if you use a 350 W television during 3 hours each day. The Price per KWh is €0.15.