What we are going to achieve today <ul><li>Creating power from lemons </li></ul><ul><li>Find out about internal resistance of batteries </li></ul><ul><li>Know what is the electromotive force is </li></ul><ul><li>Use an oscilloscope to measure the frequency and voltage from an signal generator </li></ul>
Using lemons, limes and potatoes to power an LED zinc copper
Making Batteries There is nothing special about batteries – but these have a high internal resistance. R l What do you think the best way to minimise the internal resistance of your battery? Think about resistivity (lemon is a poor conductor)
Batteries What is the main energy transfer in a battery? Electrical energy This is powered by two 1.5 V AA cells in series - what is the supply voltage? What is the emf of the supply? After a while the battery needs to be replaced; why? What determines how quickly it runs down? What determines how much current is drawn from the supply?
EMF Electromotive Force <ul><li>EMF is the external work expended per unit of charge to produce an electric potential difference across two open-circuited terminals </li></ul><ul><li>This is the same definition as voltage but on an open circuit (no current flow) </li></ul><ul><li>Why is this definition important? </li></ul>
Batteries have internal resistance The circuit now has two resistors in The internal resistance of the battery, r is very small. R is much larger The total resistance of this circuit is R total = R +r I = E / (R+r)
Batteries have internal resistance As charge goes around the circuit the sum of emfs must equal the sum of voltage drops leading to EMF = I R + I r The terminal voltage is equal to I R so this can be rearranged to give: V = E – I r and interpreted as terminal voltage = emf – ‘lost volts’
R -small R total = r + R -small The current will now be larger as the total resistance of the circuit is much lower The voltage lost across r V = I (large) r The voltage lost will now be a problem This case is of a small load resistance connected to a battery is seen in a starter motor on a car http://www.youtube.com/watch?v=al6Yz3Nv7dY http://www.youtube.com/watch?v=Ut7yBdIehYY&NR=1
Starter motor on a conventional car The headlamps are connected in parallel across a twelve-volt battery. The starter motor is also in parallel controlled by the ignition switch. Since the starter motor has a very low resistance it demands a very high current (say 60 A). The battery itself has a low internal resistance (say 0.01 Ω). The headlamps themselves draw a much lower current (they have a higher resistance) Lamp R Starter motor Ignition switch 12V What will happen to the lights?
Quick Questions <ul><li>1. A 9.0 V battery has an internal resistance of 12.0 . </li></ul><ul><li>(a) What is the potential difference across its terminals when it is supplying a current of 50.0 mA? </li></ul><ul><li>What is the maximum current this battery could supply? </li></ul><ul><li>Draw a sketch graph to show how the terminal potential difference varies with the current supplied if the internal resistance remains constant. How could the internal resistance be obtained from the graph? </li></ul><ul><li>2. A cell in a deaf aid supplies a current of 25.0 mA through a resistance of 400 . When the wearer turns up the volume, the resistance is changed to 100 and the current rises to 60 mA. What is the emf and internal resistance of the cell? </li></ul>V = E – I r V = IR E = I(R +r) You need to set up a simultaneous equation
Answers 1. (a) pd = E – I r = 9 – (50 x 10 -3 x 12) = 8.4 V (b) Max current = E/r = 9 / 12 = 0.75 A 2. E = I(R +r) E = 25 x 10 -3 (400 + r) and E = 60 x 10 -3 (100 + r) So 25 x 10 -3 (400 + r) = 60 x 10 -3 (100 + r) so r = 114.3 E = 10 + (25 x 10 -3 x 114.3) = 12.86 V I V
Oscilloscope Use the signal generator to create A/C (alternating current) on the Oscilloscope Draw 2 signals of 2 different frequencies Work out the frequency of the Oscilloscope Add the magnitude of the wave height
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