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Transcript
Things to know (a)deduce from Faraday’s experiments on electromagnetic induction or other appropriate experiments: (i) that a changing magnetic field can induce an e.m.f. in a circuit (ii) that the direction of the induced e.m.f. opposes the change producing it (iii) the factors affecting the magnitude of the induced e.m.f. (b) describe a simple form of a.c. generator (rotating coil or rotating magnet) and the use of slip rings 1 (where needed) • (c) sketch a graph of voltage output against time for a simple a.c. generator • (d) describe the structure and principle of operation of a simple iron-cored transformer as used for voltage transformations • (e) recall and apply the equations VP / Vs = NP / Ns and VPIP = VsIs to new situations or to solve related problems (for an ideal transformer) • (f) describe the energy loss in cables and deduce the advantages of high voltage transmission 2 Electromagnetic Induction Definition: Electromagnetic induction is the production of electricity using magnetism. •Need to know: Describe an experiment which shows that a changing magnetic field can induce an e.m.f. in a circuit 3 A stationary magnet is near the coil Hollow paper or plastic tube N S -1 0 1 -2 2 Sensitive Galvanometer In this experiment, no battery is connected to the coil. Hence no e.m.f. is found in the coil. 4 motion Hollow paper or plastic tube N -1 0 1 -2 2 G S Induced I When the magnet is moving towards the coil, an electric current is induced simultaneously. 5 Faster motion Hollow paper or plastic tube N -1 0 1 -2 2 G S Induced I more When the magnet is moving faster, the induced current is more. 6 Not moving Hollow paper or plastic tube N -1 0 1 -2 2 S No current G When the magnet is not moving, no current is induced even though the magnetic flux is linked with the coil closely. 7 Faraday’s Law of Electromagnetic Induction: The magnitude (how strong) of the induced emf (or induced current) is directly proportional to the rate of change of the magnetic flux linked with the coil or the rate at which the magnetic flux and wire are cutting each other. This means that when the magnetic field is not moving in relation to the coil, there will be NO induced emf at all. 8 Self Test Question Not moving N S -1 0 1 -2 2 G There is plenty of magnetic flux linkage with the coil, but there is no motion. Is there any induced current in the coil now? Answer: _________ Please draw the needle of the galvanometer. 9 What law did you apply when you answer the question in the previous slide? 10 Self Test Question Moving constantly E D C B N AS -1 0 1 -2 2 G Deflection of G or emf induced E D C B A 11 Self Test Question: Sketch the graph as the magnet moves from A to E stant speed moving N E D S C B A B A -1 0 1 -2 2 G Deflection of G or emf induced Playing back of the graph E ` D C 12 By now, you have learned that the size or strength of the induced current (or induced e.m.f.) is determined by the speed of change of the magnetic flux linkage with the coil. There is still one more thing about electromagnetic induction you need to investigate. Look at the next slide. 13 When a current is induced in a coil, it has to flow in the certain direction. What factor determines the direction of the induced current? 14 Lenz’s Law of electromagnetic induction: The direction of the induced current is such that its own magnetic effect always opposes the change producing it. This law is actually related to the Law of Conservation of Energy. The coil needs to oppose something in order to obtain energy from it. The coil itself cannot CREATE energy! 15 Beware of a different way the coil can be wound: The paper tube can be taken away to test you 16 Can you spot the difference of winding? Note: The dotted parts are at the back. The solid lines are at the front. 17 Please mark one arrow on the left end of the coil and one arrow through the bulb to show how the induced current should flow: Motion N S Induced CURRENT 18 Please mark + or – signs at the points X and Y to show the induced e.m.f. : Motion N S X + Y 19 Please mark + or – signs at the points X and Y to show the presence of induced e.m.f. : Motion N S X + Y Induced emf This induced emf is still there as long as the magnet is moving, even though the circuit is broken and the induced current cannot flow. 20 Coil is stationary N S G 21 Coil is in motion, approaching the magnet Induced current Induced current N N S G 22 Coil is in motion, approaching the magnet Induced current N N S 23 G Coil is approaching the magnet N Existing flux Induced current N S Can you see the Right Hand Rule ? G This is also called the Dynamo Rule 24 Coil is stationary again N Existing flux S No more induced current here 25 G Fleming’s Right Hand Rule is also called the Dynamo Rule. thuMb -- the Motion of the wire First finger -- magnetic Flux (Field) seCond finger -- induced Current This is actually the result of Lenz’s Law So, sometimes you use the right hand rule instead of Lenz’s Law. 26 Straight wire The straight moving vertically wire stops tomoving. magnetic field. Current is is No current induced inducedaway at all. from you This straight wire is This straight wire moving along theis not moving magnetic field No current is induced Wire moving Wire stops moving vertically to the flux Current is induced No current is towardsatyou induced all 27 28 29 Weak induced current Strong induced current wire cutting flux vertically wire cutting flux obliquely No current is induced in the wire wire moving alongside flux No current is induced in wire wire moving alongside flux 30 N Magnetic flux S 31 N S 32 B C Motion N S Motion A D 33 B Motion N Magnetic Flux C S A Motion D 34 B motion N S A C motion No induced current D 35 B N S A C D 36 Motion C B N S D A 37 C N Motion Magnetic Flux B S D A 38 C motion Magnetic flux N S D B No induced current A 39 C N S D B A 40 B C Motion N S Motion A D 41 B C N S A D Red arrows represents magnetic flux Induced current Time A D 42 B N C A S Red arrows represents magnetic flux D Induced current Time 43 B motion N A motion C S D Induced current Time 44 motion C N D B Flux S A Induced current Time 45 D A C motion N D B S A Induced current Time 46 B C N S A D Induced current Time A D 47 Induced current / emf Time 48 Induced emf 2.5V Time -2.5V 49 How would you demonstrate electromagnetic induction here? G The iron core 50 Input emf Np turns Ns turns output emf AC Source The iron core Insulated copper wire Insulated copper wire Primary windings Secondary windings 51 Primary emf Ns turns Np turns Secondary emf AC Source The iron core emf secondary emf primary = Ns Np 52 Primary emf Np turns Ns turns Secondary emf AC Source emf Time 53 emf Primary Alternating emf Peak 12V, 50 Hz Secondary Alternating emf Peak 18V 50 Hz Time 54 Primary emf Ns turns Np turns Secondary emf AC Source The iron core emf secondary emf primary = Ns Np Step-up Transformer 1 55 Primary emf Ns turns Np turns Secondary emf AC Source The iron core emf secondary emf primary = Ns Np Step-down Transformer 1 56 Input emf 150V AC Source Np 200 turns Calculate (i) the output emf (iii) the primary current Ns 600 turns output emf 60 (ii) the induced current 57 Np 200 turns Input emf 150V AC Source Ns 600 turns output emf Vs 600 60 Calculate (i) the output emf Vs Vp = Ns Np Vs = 450 V peak 150 = 200 58 Np 200 turns Input emf 150V AC Source Calculate (i) the output emf output emf Ns 600 turns 60 Vs = 450 V peak (ii) the induced current I= V R I= 450 60 = 7.5 A Peak 59 emf , current 450V Secondary Alternating emf Peak 450 V, freq 50 Hz 7.5A Time 60 Np 200 turns Input emf 150V AC Source Calculate (i) the output emf (iii) the primary current Power= VI, Ns 600 turns Vs = 450 V peak output emf 60 (ii) Is = 7.5 A Output power = Input power (assuming that the transformer is 100% effecient) Vsx Is = Vpx Ip , 450x 7.5 = 150x Ip 61 Np 200 turns Input emf 150V AC Source Calculate (i) the output emf Ns 600 turns output emf 60 Vs = 450 V peak (iii) the primary current Is Ip Np = Ns provided that the transformer has an efficiency of 100% 62 emf , current 22.5A 450V Secondary Alternating emf Peak 450V, freq 50 Hz 7.5A 150V Time 63 AC Source Primary emf Np turns The iron core Ns turns Secondary emf eddy current Power lost in a transformer, Textbook Page 350 Power losses are : (i) due to the electrical resistance in both the windings Heat is generated in the wires unnecessarily (ii) due to the production of the eddy currents in the iron core. 64 AC Source Primary emf Np turns Ns turns Secondary emf The iron core Such power losses are minimized: (i) by using thicker copper wires in the windings. Less heat is generated in the wires. (ii) by using a laminated iron core. 65 Power Loss in Cables 66 Heating Heatingpower powerininthe thetransmitting transmittingcables cables== I2? R P=VI 2,000000 = 10000 x I I = 200 A Heating power in cables = 2002 x 40 = 1,600000 W = 1.6 MW A huge loss 2MW 10KV P.S. 0V Power transmission R = 20 R = 20 0.4MW L N 67 Heating Heatingpower powerininthe thetransmitting transmittingcables cables== I2? R P=VI 2,000000 = 20000 x I I = 100 A Heating power in cables = 1002 x 40 = 400,000 W = 0.4 MW A smaller loss 2MW 20KV P.S. 0V Power transmission R = 20 R = 20 1.6MW L N 68