4. 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)
5. 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?
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7. 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)
8. 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’
9. 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
10. 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?
13. 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
14. 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 How do you use an oscilloscope as a dc and ac voltmeter? How do you use it to measure frequency?
15. Voltage what the average value of an ac voltage or current is over a whole number of cycles? power is calculated by P = IV , both I and V change sign together, so power is always positive
16. Positive power calling the maximum value of I p V p and a minimum of zero 0V P = I p V p Average power = ½ I p V p P = 1/2 I p V p = 1/2 I p 2 R = 1/2 V p 2 / R (using V = I R)
17. RMS voltage (root mean squared) RMS links the dc equivalent values to ac peak values. The point is that a sinusoidal ac supply of peak value V p delivers the same average power as steady dc of value V p / √2 The same power would be delivered by a dc with values I and V if: P = I 2 R= 1/2 I p 2 R and P = V 2 / R = 1/2 V p 2 / R I 2 R= 1/2 I p 2 R (divide both sides by R) V 2 / R = 1/2 V p 2 / R (multiply by R) These lead to the equations: I 2 = I p 2 / 2 V 2 = V p 2 / 2 (square root) I = I p / √2 V = V p / √2
18. RMS Graph listed voltages for power outlets, e.g. 120 V (USA) or 230 V (Europe), are almost always quoted in RMS values, and not peak values
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20. Answers and Worked Solutions 1. (a) 110 x √ 2 = 155.6 V (b) 2 x 155.6 = 311 V (c) 1 / 50 Hz = 0.02s (d) P = V 2 /R R = 12100/100 = 121 2. 12 V peak ac is equivalent to 12/√ 2 V = 8.5 V dc equivalent. So 12 V peak to peak is dimmer. 3. 230 x √ 2 V = 325V, so graph varies between + 325 V and -325 V. One time period = 1 / 50 Hz = 0.02s 4. P = V 2 /R so the ratio of powers is ratio of voltages squared. 20 V peak ac is equivalent to 20/ √ 2 V dc i.e. 14.14 V so dc power / ac power = 20 2 /14.14 2 = 2 or dc V 2 / (ac peak / √ 2) 2 = (√ 2) 2 =2
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