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Electric fire Cooker Kettle Washing machine etc.
4 Demo Heating Effect Of Electric Current Switch on … 10 minutes è Rise in temp. Conclusion = ?? The amount of heat depends on ?? Current Time Resistance Methods = ???
5 James Joule (1818 – 1889) W a I 2 W a R W = I 2 Rt W a t W ÷ t = P = I 2 R P = power = rate at which heat is produced
6 P = I 2 R
or P a I 2 The rate at which heat is produced P = power is directly proportional to I = current the current squared (R is constant) (R is constant) 7 P a I 2 è If current = 2I … heat is 4 times greater If current = 3I … heat is 9 times greater etc.
8 Problem 1 Find the heat produced in a 20 W resistor by a current of 3 A flowing for 40 s. W = I 2 R t = (3) 2 (20) (40) = 7,200 J
9 Problem 2 Find the rate at which heat is produced (i.e. the power dissipated) by a current of 80 mA in a 2 kW resistor. P = I 2 R = (80 x 10 ­3 ) 2 (2 x 10 3 ) = 12.8 W
10 Problem 3 When a current of 2 A flows in a wire, heat is produced at the rate of 60 W. What current would produce heat at the rate of 540 W? P = I 2 R è 60 = (2) 2 R è R = 15 W P = I 2 R è 540 = I 2 (15) è I = 6 A or P a I 2 540 W is 9 times 60 W è new I 2 is 9 times old I 2 è new I is 3 times old I = 6 A
11 Expt. To Verify Joule’s Law Apparatus Ø Calorimeter + water + lagging + stirrer Ø Heating coil Ø Thermometer Ø Battery + ammmeter + rheostat + switch
12 Expt. To Verify Joule’s Law Procedure 1. Switch on … constant 1 A … 10 mins. Temperature rises … 2. Measure … current (I) and rise in temp. (Dq) 3. I … from ammeter Dq = (final temp. ­ initial temp.) of water 4. Repeat, using 1.5 A, 2.0 A, 2.5 A, etc. … to graph (Same mass, time each run)
13 Expt. To Verify Joule’s Law Results Dq
Conclusion “SLTO” è Dq a I 2 I 2 or Table with … Dq ÷ I 2 = a constant è Dq a I 2
P a I 2 Table 14 Expt. To Verify Joule’s Law Precautions 1. Lagging & lid … 2. Constant I … 3. Fixed time … Fixed mass of … 4. etc.
15 Advantage Of Using High Voltage 25 kV 400 kV Power wasted (heat) = I 2 R Keep I low …
16 Advantage Of Using High Voltage Power transmitted = P = VI When V is high, I is low (Transformers …) Power wasted (as heat) = P / = I 2 R Since I is low – power wasted is low.
17 Demo Ø Switch on … Ø Cathode … increase in mass (gains copper) Ø Anode … decrease in mass (loses copper) Overall Copper transferred from anode to cathode (No change in the copper sulphate)
18 Demo Overall Copper transferred from anode to cathode (No change in the copper sulphate)
19 2. Copper Sulphate Dissociates Into Ions CuSO 4 à Cu ++ + SO 4 ­ ­ 3. Movement of ions … 1.
Cu ++ ® … to cathode (cation) SO 4 ­ ­ ¬ … to anode (anion) 4. Electrode Reactions Anode Cu ­ 2e ­ à Cu ++ Cathode Cu ++ + 2e ­ à Cu 5. Overall = ??? 20 § Electrolysis § Voltameter § Electrolyte § Active electrodes § Electrodes § Inactive electrodes § Anode § Ion § Cathode Inert
21 H 2 O à H 2 + O 2 2H 2 O à 2H 2 + O 2
22 Ø Electroplating (cathode) (spoon / silver plating / EPNS) (taps / chromium plated) Ø Extracting metals from their ores Ø Purifying metals Ø Hair removal Ø Electrolytic capacitors
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Apparatus Ø Ø Ø Ø Ø Copper voltameter Voltmeter Ammeter Potential divider Power supply
25 Procedure 1. Apply different pd’s to voltameter (potential divider) Take readings … 2. Measure … current (I) and pd (V) repeatedly (rheostat) 3. I … from ammeter V … from voltmeter 4. Repeat (V = 0.0, 0.2, 0.4, 0.6 volts … etc.) … to graph
26 Results I Conclusion “SLTO” è I a V V or Conclusion V ÷ I = a constant è I a V
I/A 0.2 0.4 0.6 0.8 1.0 1.2 1.4 V/V 1.0 2.0 3.0 4.0 5.0 6.0 7.0 V ÷ I 5.0 5.0 5.0 5.0 5.0 5.0 5.0 27 Precautions q High resistance voltmeter … q Limit the size of the current so that the copper sticks … q etc.
28 A Metallic Concuctor “SLTO” è Obeys Ohm’s Law Resistance does not change … Charge carriers = ?? Negative electrons
V Slope = V/I = R I 29 A Filament Bulb Ohm’s law not obeyed since temp. not constant R increases with temp. /current Charge carriers = ?? Negative electrons
30 Semiconductor (Thermistor) Ohm’s law not obeyed R decreases with temp. /current Charge carriers = ?? Free positive holes & Free negative electrons Greater temp. èMore free charge carriers èLower R
31 Electrolytes Ohm’s law obeyed Active Electrodes Ohm’s law not obeyed Inactive / Inert Electrodes
Charge carriers = ?? Positive & Negative Ions 32 Gas Ohm’s law not obeyed Charge carriers = ?? Ions (+ & ­) Electrons (­)
Street lamps 33 Vacuum Thermionic Emission
Ohm’s law not obeyed Charge carriers = ?? Electrons (­) TV 34 (i) Metallic Conductor (ii) (ii) Filament Bulb Apparatus or Coil of wire (metallic conductor) 35 (i) Metallic Conductor (ii) (ii) Filament Bulb Procedure 1. Apply different pd’s … potential divider … take readings 2. Measure … pd (V) & current (I) repeatedly 3. V … voltmeter I … ammeter 4. Repeat … other pd’s … to graph
36 (i) Metallic Conductor (ii) (ii) Filament Bulb Results Precautions 1. Use a high resistance voltmeter. 2. etc.
37 230 V a.c. (+325 V to – 325 V) Series or Parallel ???
38 Parallel Switch in live Fuse in live
39 Large current Its own separate fuse­circuit Cooker / Electric shower / Immersion heater
40 2 paths (½ I) Thinner cables Draw ­ NN
41 Protection against too large a current e
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§ Fuse = resistance wire § It gets hot when I < 13A passes through it. § When I > 13A flows … fuse overheats / melts / breaks circuit etc. 42 Miniature Circuit Breaker Protection against too large a current Similar to fuse
§ Electromagnet § Large current – electromagnet separates two contacts § Disconnects the live 43 t
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Residual Current Device Normally Live = neutral (current) Fault Live ¹ neutral Electromagnet switches off … Very sensitive & Very quick …faster 44 Residual Current Device
45 • All metals fixtures … • e.g. baths, sinks, pipes, water tanks … • are connected together and then to earth (bonded to earth) Protection If live … Safe … as fuse blows and … at zero potential (not 230 V)
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Large current to earth Blows the fuse Disconnects from mains
48 Problem 4 A 100 W lamp operates on 230 V mains. What size fuse, 3 A or 13 A, should be put in the plug? P = VI 100 = 230 x I 3 A fuse
I = 0.43 A 49 Problem 5 An electric cooker has four 500 W plates, a 2 kW grill and a 3 kW oven. It operates on 230 V. Is a 40 A fuse suitable for this cooker? Total power = 4(500) + 2000 + 3000 = 7000 W P = VI 7000 = 230 x I 40 A fuse is suitable
I = 30.4 A 50 ESB unit of energy The amount of energy “used” by a 1000 W appliance in 1 hour No. of kW­h = No. of kW x No. of hours
51 Problem 6 1000 W How many joules in a kilowatt­hour? = 1000 joules per second = 1000 x (60 x 60) joules per hour = 3.6 x 10 6 J = 3.6 MJ
52 Problem 7 What is the total cost of using the following: a 100 W lamp for 4 hours, a 2 kW fire for 30 minutes, a 500 W motor for 1 hour? The cost of a unit is 12 p. No. of units = (0.100 x 4) + (2 x 0.5) + (0.500 x 1) = 1.9 units Cost = 12 x 1.9 = 22.8 p
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