OP
amberleaf
AC motor theory
Question 1:
Electromechanical watt-hour meters use an aluminium disk that is spun by an electric motor. To generate a constant "drag" on the disk necessary to limit its rotational speed, a strong magnet is placed in such a way that its lines of magnetic flux pass perpendicularly through the disk's thickness:
the disk itself need not be made of a ferromagnetic material in order for the magnet to create a "drag" force. It simply needs to be a good conductor of electricity.
Explain the phenomenon accounting for the drag effect, and also explain what would happen if the roles of magnet and disk were reversed: if the magnet were moved in a circle around the periphery of a stationary disk.
This is an example of Lenz' Law. A rotating magnet would cause a torque to be generated in the disk.
Notes:
Mechanical speedometer assemblies used on many automobiles use this very principle: a magnet assembly is rotated by a cable connected to the vehicle's driveshaft. This magnet rotates in close proximity to a metal disk, which gets "dragged" in the same direction that the magnet spins. The disk's torque acts against the resistance of a spring, deflecting a pointer along a scale, indicating the speed of the vehicle. The faster the magnet spins, the more torque is felt by the disk.
Question 3:
Explain what slip speed is for an AC induction motor, and why there must be such as thing as slip" in order for an induction motor to generate torque.
The difference between the speed of the rotating magnetic field (fixed by line power frequency) and the speed of the rotor is called slip speed". Some amount of slip is necessary to generate torque because without it there would be no change in magnetic flux , seen by the rotor, and thus no induced currents in the rotor.
concept, because it is at the heart of induction motor operation.
Question 5:
What would we have to do in order to reverse the rotation of this three-phase induction motor?
Explain your answer. Describe how the (simple) solution to this problem works. Reverse any two lines. This will reverse the phase sequence (from ABC to CBA).
Notes:
One of the reasons three-phase motors are preferred in industry is the simplicity of rotation reversal. However, this is also a problem because when you connect a three-phase motor to its power source during maintenance or installation procedures, you often do not know which way it will rotate until you turn the power on!
Question 6:
If a copper ring is brought closer to the end of a permanent magnet, a repulsive force will develop between the magnet and the ring. This force will cease, however, when the ring stops moving. What is this effect called?
N .. Magnet .. S ( reaction → force ← Motion )
Also, describe what will happen if the copper ring is moved away from the end of the permanent magnet.
The phenomenon is known as Lenz' Law. If the copper ring is moved away from the end of the permanent magnet, the direction of force will reverse and become attractive rather than repulsive.
Follow-up question: trace the direction of rotation for the induced electric current in the ring necessary to produce both the repulsive and the attractive force.
Challenge question: what would happen if the magnet's orientation were reversed (south pole on left and north pole on right)?
Notes:
This phenomenon is difficult to demonstrate without a very powerful magnet. However, if you have such apparatus available in your lab area, it would make a great piece for demonstration!
One practical way I've demonstrated Lenz's Law is to obtain a rare-earth magnet (very powerful!), set it pole-up on a table, then drop an aluminium coin (such as a Japanese Yen) so it lands on top of the magnet. If the magnet is strong enough and the coin is light enough, the coin will gently come to rest on the magnet rather than hit hard and bounce off.
A more dramatic illustration of Lenz's Law is to take the same coin and spin it (on edge) on a table surface. Then, bring the magnet close to the edge of the spinning coin, and watch the coin promptly come to a halt, without contact between the coin and magnet.
Another illustration is to set the aluminium coin on a smooth table surface, then quickly move the magnet over the coin, parallel to the table surface. If the magnet is close enough, the coin will be "dragged" a short distance as the magnet passes over.
In all these demonstrations, it is significant to show to your students that the coin itself is not magnetic. It will not stick to the magnet as an iron or steel coin would, thus any force generated between the coin and magnet is strictly due to induced currents and not ferromagnetism.
Question 13:
Synchronous AC motors operate with zero slip, which is what primarily distinguishes them from induction motors. Explain what slip" means for an induction motor, and why synchronous motors do not have it.
Synchronous motors do not slip because their rotors are magnetized so as to always follow the rotating magnetic field precisely. Induction motor rotors are become magnetized by induction, necessitating a difference in speed (slip") between the rotating magnetic field and the rotor.
Step-up, step-down, and isolation transformers
Question 1:
Calculate the voltage output by the secondary winding of a transformer if the primary voltage is 35 volts, the secondary winding has 4500 turns, and the primary winding has 355 turns. V/secondary = ( V/ secondary = 443.7 volts )
Notes:
Transformer winding calculations are simply an exercise in mathematical ratios.
Question 3:
Calculate the number of turns needed in the secondary winding of a transformer to transform a primary voltage of 300 volts down to a secondary voltage of 180 volts, if the primary winding has 1150 turns of wire.
N/secondary = ( N/secondary = 690 turns )
Question 5:
Suppose 1200 turns of copper wire are wrapped around one portion of an iron hoop, and 3000 turns of wire are wrapped around another portion of that same hoop. If the 1200-turn coil is energized with 15 volts AC (RMS), how much voltage will appear between the ends of the 3000-turn coil ? ( 37.5 volts AC, RMS. )
Question 17:
In a typical step-up or step-down transformer, the higher-voltage winding usually uses finer gauge wire than the lower-voltage winding. Explain why this is.
The higher-voltage winding handles less current than the lower-voltage winding.
Question 24:
Explain how the construction of a step-down transformer differs from that of a step-up transformer.
Step-down transformers have fewer secondary turns than primary turns, while step-up transformers have more secondary turns than primary turns.
Question 26:
When calculating power in transformer circuits, how do the primary and secondary circuit powers (Pprimary = VprimaryIprimary and Psecondary = VsecondaryIsecondary) compare with each other? Is one greater than the other? If so, which one, and why?
Ideally, P/secondary = P/primary, although this equivalence is never quite exact. In practice, P/secondary will always be a little bit less than P/primary.
Question 27:
Explain why transformers are used extensively in long-distance power distribution systems. What advantage do they lend to a power system?
Transformers are used to step voltage up for efficient transportation over long distances, and used to step the high voltage down again for point-of-use circuits.
Question 31:
Suppose a step-down transformer fails due to an accidental short-circuit on the secondary (load) side of the circuit:
That the transformer actually failed as a result of the short is without any doubt: smoke was seen coming from it, shortly before current in the circuit stopped. A technician removes the burned-up transformer and does a quick continuity check of both windings to verify that it has failed open. What she finds is that the primary winding is open but that the secondary winding is still continuous. Puzzled at this finding, she asks you to explain how the primary winding could have failed open while the secondary winding is still intact, if indeed the short occurred on the secondary side of the circuit. What would you say? How is it possible that a fault on one side of the transformer caused the other side to be damaged?
A short-circuit would cause current in both windings of the transformer to increase.
Notes:
It is important for students to realize that a transformer "reflects" load conditions on the secondary side to the primary side, so that the source "feels" the load in all respects. What happens on the secondary (load) side will indeed be reflected on the primary (source) side.
Question 1:
Electromechanical watt-hour meters use an aluminium disk that is spun by an electric motor. To generate a constant "drag" on the disk necessary to limit its rotational speed, a strong magnet is placed in such a way that its lines of magnetic flux pass perpendicularly through the disk's thickness:
the disk itself need not be made of a ferromagnetic material in order for the magnet to create a "drag" force. It simply needs to be a good conductor of electricity.
Explain the phenomenon accounting for the drag effect, and also explain what would happen if the roles of magnet and disk were reversed: if the magnet were moved in a circle around the periphery of a stationary disk.
This is an example of Lenz' Law. A rotating magnet would cause a torque to be generated in the disk.
Notes:
Mechanical speedometer assemblies used on many automobiles use this very principle: a magnet assembly is rotated by a cable connected to the vehicle's driveshaft. This magnet rotates in close proximity to a metal disk, which gets "dragged" in the same direction that the magnet spins. The disk's torque acts against the resistance of a spring, deflecting a pointer along a scale, indicating the speed of the vehicle. The faster the magnet spins, the more torque is felt by the disk.
Question 3:
Explain what slip speed is for an AC induction motor, and why there must be such as thing as slip" in order for an induction motor to generate torque.
The difference between the speed of the rotating magnetic field (fixed by line power frequency) and the speed of the rotor is called slip speed". Some amount of slip is necessary to generate torque because without it there would be no change in magnetic flux , seen by the rotor, and thus no induced currents in the rotor.
concept, because it is at the heart of induction motor operation.
Question 5:
What would we have to do in order to reverse the rotation of this three-phase induction motor?
Explain your answer. Describe how the (simple) solution to this problem works. Reverse any two lines. This will reverse the phase sequence (from ABC to CBA).
Notes:
One of the reasons three-phase motors are preferred in industry is the simplicity of rotation reversal. However, this is also a problem because when you connect a three-phase motor to its power source during maintenance or installation procedures, you often do not know which way it will rotate until you turn the power on!
Question 6:
If a copper ring is brought closer to the end of a permanent magnet, a repulsive force will develop between the magnet and the ring. This force will cease, however, when the ring stops moving. What is this effect called?
N .. Magnet .. S ( reaction → force ← Motion )
Also, describe what will happen if the copper ring is moved away from the end of the permanent magnet.
The phenomenon is known as Lenz' Law. If the copper ring is moved away from the end of the permanent magnet, the direction of force will reverse and become attractive rather than repulsive.
Follow-up question: trace the direction of rotation for the induced electric current in the ring necessary to produce both the repulsive and the attractive force.
Challenge question: what would happen if the magnet's orientation were reversed (south pole on left and north pole on right)?
Notes:
This phenomenon is difficult to demonstrate without a very powerful magnet. However, if you have such apparatus available in your lab area, it would make a great piece for demonstration!
One practical way I've demonstrated Lenz's Law is to obtain a rare-earth magnet (very powerful!), set it pole-up on a table, then drop an aluminium coin (such as a Japanese Yen) so it lands on top of the magnet. If the magnet is strong enough and the coin is light enough, the coin will gently come to rest on the magnet rather than hit hard and bounce off.
A more dramatic illustration of Lenz's Law is to take the same coin and spin it (on edge) on a table surface. Then, bring the magnet close to the edge of the spinning coin, and watch the coin promptly come to a halt, without contact between the coin and magnet.
Another illustration is to set the aluminium coin on a smooth table surface, then quickly move the magnet over the coin, parallel to the table surface. If the magnet is close enough, the coin will be "dragged" a short distance as the magnet passes over.
In all these demonstrations, it is significant to show to your students that the coin itself is not magnetic. It will not stick to the magnet as an iron or steel coin would, thus any force generated between the coin and magnet is strictly due to induced currents and not ferromagnetism.
Question 13:
Synchronous AC motors operate with zero slip, which is what primarily distinguishes them from induction motors. Explain what slip" means for an induction motor, and why synchronous motors do not have it.
Synchronous motors do not slip because their rotors are magnetized so as to always follow the rotating magnetic field precisely. Induction motor rotors are become magnetized by induction, necessitating a difference in speed (slip") between the rotating magnetic field and the rotor.
Step-up, step-down, and isolation transformers
Question 1:
Calculate the voltage output by the secondary winding of a transformer if the primary voltage is 35 volts, the secondary winding has 4500 turns, and the primary winding has 355 turns. V/secondary = ( V/ secondary = 443.7 volts )
Notes:
Transformer winding calculations are simply an exercise in mathematical ratios.
Question 3:
Calculate the number of turns needed in the secondary winding of a transformer to transform a primary voltage of 300 volts down to a secondary voltage of 180 volts, if the primary winding has 1150 turns of wire.
N/secondary = ( N/secondary = 690 turns )
Question 5:
Suppose 1200 turns of copper wire are wrapped around one portion of an iron hoop, and 3000 turns of wire are wrapped around another portion of that same hoop. If the 1200-turn coil is energized with 15 volts AC (RMS), how much voltage will appear between the ends of the 3000-turn coil ? ( 37.5 volts AC, RMS. )
Question 17:
In a typical step-up or step-down transformer, the higher-voltage winding usually uses finer gauge wire than the lower-voltage winding. Explain why this is.
The higher-voltage winding handles less current than the lower-voltage winding.
Question 24:
Explain how the construction of a step-down transformer differs from that of a step-up transformer.
Step-down transformers have fewer secondary turns than primary turns, while step-up transformers have more secondary turns than primary turns.
Question 26:
When calculating power in transformer circuits, how do the primary and secondary circuit powers (Pprimary = VprimaryIprimary and Psecondary = VsecondaryIsecondary) compare with each other? Is one greater than the other? If so, which one, and why?
Ideally, P/secondary = P/primary, although this equivalence is never quite exact. In practice, P/secondary will always be a little bit less than P/primary.
Question 27:
Explain why transformers are used extensively in long-distance power distribution systems. What advantage do they lend to a power system?
Transformers are used to step voltage up for efficient transportation over long distances, and used to step the high voltage down again for point-of-use circuits.
Question 31:
Suppose a step-down transformer fails due to an accidental short-circuit on the secondary (load) side of the circuit:
That the transformer actually failed as a result of the short is without any doubt: smoke was seen coming from it, shortly before current in the circuit stopped. A technician removes the burned-up transformer and does a quick continuity check of both windings to verify that it has failed open. What she finds is that the primary winding is open but that the secondary winding is still continuous. Puzzled at this finding, she asks you to explain how the primary winding could have failed open while the secondary winding is still intact, if indeed the short occurred on the secondary side of the circuit. What would you say? How is it possible that a fault on one side of the transformer caused the other side to be damaged?
A short-circuit would cause current in both windings of the transformer to increase.
Notes:
It is important for students to realize that a transformer "reflects" load conditions on the secondary side to the primary side, so that the source "feels" the load in all respects. What happens on the secondary (load) side will indeed be reflected on the primary (source) side.
Last edited by a moderator: