In the previous section we saw the production of alternating current in an AC generator. In this section, we will see some properties of the alternating current.
• We saw that, the armature ABCD rotates in the space between a N-pole and a S-pole.
♦ That is., the armature rotates inside a magnetic field
• The magnetic force between the poles has a definite direction: From N-pole to S-pole
• We can represent this force by drawing arrows from the N-pole to S-pole
• We have seen that, when a conductor moves in the magnetic field, current will be induced in that conductor
• But to achieve 'maximum production of current', the conductor should move in a 'particular direction'
■ What is this 'perpendicular direction'?
To find the answer, we must analyse fig.11.14 below:
We will write the steps:
1. Consider fig.11.14(a)
(i) The conductor AB is moving vertically downwards.
(ii) When it moves in such a manner, it cuts the blue arrows (the field lines) perpendicularly
• For our present discussion, we will call it 'perfect cutting'
(iii) Maximum current is induced when 'perfect cutting' occurs
2. Consider fig.b
(i) The conductor AB is moving horizontally towards the right.
(ii) When it moves in such a manner, it does not cut any of the blue arrows
• For our present discussion, we will call it 'no cutting'
(iii) No current is induced when 'no cutting' occurs
3. Consider fig.c
(i) The conductor AB is moving along a sloping line.
(ii) When it moves in such a manner, it does cut the blue arrows. But the cutting is not perfect
• For our present discussion, we will call it 'intermediate cutting'
(iii) Some current is induced when 'intermediate cutting' occurs. But maximum current will not be obtained as in a 'perfect cutting'
• When the armature rotates in an AC generator, all the three types of cutting will occur
• They continue to occur one after the other, in a 'to and fro' manner. That is:
♦ First a 'no cutting' occurs
♦ Then an 'intermediate cutting' occurs
♦ After that, a 'perfect cutting' occurs
♦ Then again an 'intermediate cutting' occurs
♦ Then a 'no cutting' occurs
- - -
- - -
so on . . .
• How does this happen?
• To find the answer, we must analyse fig.11.15 below:
We will write the steps:
1. We know that the armature is continuously rotating in the space between the poles.
• Let us stop the rotation at the very instant when the armature is exactly vertical
(with AB at top and CD at bottom).
• This is shown in fig.11.15(a)
2. Now there is no current in the circuit.
• This is our initial point. We start our analysis from here.
• So we will call this point as 'stage 1'
3. Now we begin the rotation. The armature begins to move.
• At the instant when rotation begins, the segment AB of the armature has a horizontal motion.
♦ This is indicated by the red horizontal arrow
4. Note that, this horizontal motion of AB is 'instantaneous'.
• That is., it is valid for a very short time only. After that instant, the movement is not horizontal.
5. But however small it may be, this 'instantaneous horizontal motion' is valuable for our analysis.
• Because, during the horizontal motion, there is 'no cutting' of the magnetic field
• Since there is 'no cutting', there is no current at that instant
6. The same happens at the bottom segment CD also.
• It also has an 'instantaneous horizontal movement' (in the opposite direction)
• Hence there is no current in CD also.
7. There is no current in AB as well as CD.
• As a result, there is zero current in the armature as a whole
• We can indeed call that instant as 'stage 1'
8. So what is the next stage?
• As the rotation continues, current begins to flow.
9. The armature becomes less and less vertical
• It becomes more and more horizontal
• During this process, the cuttings are 'intermediate cuttings'
10. Consider the very instant when the armature is exactly horizontal.
This is shown in fig.11.15(b)
• At this instant, the segment AB has a vertical motion. This is indicated by the red arrow.
• Note that, this vertical motion of AB is 'instantaneous'.
• That is., it is valid for a very short time only. After that instant, the movement is not vertical.
11. But however small it may be, this 'instantaneous vertical motion' is valuable for our analysis. • Because, due to the vertical motion, there is 'perfect cutting' of the magnetic field
• Since there is 'perfect cutting', there is 'maximum current' at that instant
12. The same happens at the other segment CD also.
• It also has an 'instantaneous vertical motion' (in the opposite direction)
• Hence there is 'maximum current' in CD also.
13. There is 'maximum current' in AB as well as CD.
• As a result, there is 'maximum current' in the armature as a whole
• We can call that instant as 'stage 2'
14. Now, how did the armature reach 'stage 2' from 'stage 1'
• Ans: Stage 1 is vertical and stage 2 is horizontal
• So, to reach stage 2 from stage 1, the armature rotated through an angle of 90o.
• The various angles between horizontal and vertical planes is shown in the fig.11.16 below:
15. We can write:
• During a rotation from 0o to 90o, the current reached 'maximum current' from 'zero current'
16. Current is directly related to voltage (emf)
[Recall the relation: V = IR]
♦ When emf increases, current increases
♦ When emf decreases, current decreases
• In other words:
♦ 'maximum current' is caused due to 'maximum emf'
♦ 'zero current' is caused due to 'zero emf'
17. So we can write:
• During a rotation from 0o to 90o, the emf reached 'maximum emf' from 'zero emf'
• In other words:
♦ Stage 1 → 0o → zero emf
♦ Stage 2 → 90o → maximum emf
18. This variation from 'zero' to 'maximum' can be represented graphically as shown in fig.11.17 below:
• Angle is plotted along the x axis
• emf is plotted along the y axis
• The green curve is the required graph
• Note that, it is a 'rising curve'. So it indicates the rise of emf from zero to maximum
• Consider the cyan coloured vertical dashed lines in the above graph
• They start from the center of the armature. Why is that so?
• Ans: The graph represents the 'emf in the armature as a whole'. Not the 'emf in individual segments'
• So it is appropriate to use the center point of the armature
19. Now we move on to see stage 3
• The armature is continuing it's rotation
• It becomes less and less horizontal and more and more vertical
• During this process, the cuttings are 'intermediate cuttings'
• The angle is increasing from 90o
20. Consider the very instant when the armature is exactly vertical again. The angle now is 180o.
• This is a situation very similar to 'stage 1'
• The only difference is that, AB is now at the bottom and CD is at the top
21. At this instant, both the segments AB and CD has a horizontal motion.
• The cuttings at this instant are 'zero cuttings'
• So, at this instant, there is 'zero emf' in the armature
• We can call this instant 'stage 3'
22. We can write:
• During a rotation from 90o to 180o, the emf reached 'zero emf' from 'maximum emf'
• In other words:
♦ Stage 2 → 90o → maximum emf
♦ Stage 3 → 180o → zero emf
23. This variation from 'maximum' to 'zero' can also be represented graphically.
• We will add it to the previous graph. The new graph is shown in fig.11.18 below:
• The green curve from stage 2 to stage 3 is the required graph
• Note that, it is a 'falling curve'. So it indicates the fall of emf from maximum to zero
24. Now we move on to see stage 4
• The armature is continuing it's rotation
• It becomes less and less vertical and more and more horizontal
• During this process, the cuttings are 'intermediate cuttings'
• The angle is increasing from 180o
25. Consider the very instant when the armature is exactly horizontal again. The angle now is 270o.
• This is a situation very similar to 'stage 2'
• The only difference is that, AB is now at the left side and CD is at the right side
26. At this instant, both the segments AB and CD has a vertical motion.
• The cuttings at this instant are 'perfect cuttings'
• So, at this instant, there is 'maximum emf' in the armature
• We can call this instant 'stage 4'
27. We can write:
• During a rotation from 90o to 180o, the emf reached 'zero emf' from 'maximum emf'
• In other words:
♦ Stage 3 → 180o → zero emf
♦ Stage 4 → 270o → maximum emf
28. This variation from 'zero' to 'maximum' can also be represented graphically.
• We will add it to the previous graph. The new graph is shown in fig.11.19 below:
• The green curve from stage 3 to stage 4 is the required graph
• Note that, though the emf increases from zero to maximum, we draw it on the negative side of the y axis
• This is to indicate that, the direction of current changes after one 'half rotation'
29. Now we move on to see stage 5
• The armature is continuing it's rotation
• It becomes less and less horizontal and more and more vertical
• During this process, the cuttings are 'intermediate cuttings'
• The angle is increasing from 270o
30. Consider the very instant when the armature is exactly vertical again. The angle now is 360o.
• This is a situation very similar to 'stage 1'. In fact, it is exactly same as 'stage 1'
31. At this instant, both the segments AB and CD has a horizontal motion.
• The cuttings at this instant are 'zero cuttings'
• So, at this instant, there is 'zero emf' in the armature
• We can call this instant 'stage 5'
32. We can write:
• During a rotation from 270o to 360o, the emf reached 'zero emf' from 'maximum emf'
• In other words:
♦ Stage 4 → 270o → maximum emf
♦ Stage 5 → 360o → zero emf
33. This variation from 'maximum' to 'zero' can also be represented graphically.
• We will add it to the previous graph. The new graph is shown in fig.11.20 below:
• In stage 5, we reach back where we began. So 'one cycle' can be considered to be complete
Thus we can write:
■ 'One cycle is complete' when 'the armature completes on rotation'.
• The second cycle begins when the armature begins to rotate after stage 5. It is a continuous process.
• We get electricity as long as we repeat the cycles
• Graphs can be drawn for those cycles also
• The beginning of the 'graph of the second cycle' is shown by the dotted green curve
• Note that, the intervals between any two consecutive cyan vertical lines are the same. It is 90o
• We saw that, the armature ABCD rotates in the space between a N-pole and a S-pole.
♦ That is., the armature rotates inside a magnetic field
• The magnetic force between the poles has a definite direction: From N-pole to S-pole
• We can represent this force by drawing arrows from the N-pole to S-pole
• We have seen that, when a conductor moves in the magnetic field, current will be induced in that conductor
• But to achieve 'maximum production of current', the conductor should move in a 'particular direction'
■ What is this 'perpendicular direction'?
To find the answer, we must analyse fig.11.14 below:
 |
| Fig.11.14 |
1. Consider fig.11.14(a)
(i) The conductor AB is moving vertically downwards.
(ii) When it moves in such a manner, it cuts the blue arrows (the field lines) perpendicularly
• For our present discussion, we will call it 'perfect cutting'
(iii) Maximum current is induced when 'perfect cutting' occurs
2. Consider fig.b
(i) The conductor AB is moving horizontally towards the right.
(ii) When it moves in such a manner, it does not cut any of the blue arrows
• For our present discussion, we will call it 'no cutting'
(iii) No current is induced when 'no cutting' occurs
3. Consider fig.c
(i) The conductor AB is moving along a sloping line.
(ii) When it moves in such a manner, it does cut the blue arrows. But the cutting is not perfect
• For our present discussion, we will call it 'intermediate cutting'
(iii) Some current is induced when 'intermediate cutting' occurs. But maximum current will not be obtained as in a 'perfect cutting'
• When the armature rotates in an AC generator, all the three types of cutting will occur
• They continue to occur one after the other, in a 'to and fro' manner. That is:
♦ First a 'no cutting' occurs
♦ Then an 'intermediate cutting' occurs
♦ After that, a 'perfect cutting' occurs
♦ Then again an 'intermediate cutting' occurs
♦ Then a 'no cutting' occurs
- - -
- - -
so on . . .
• How does this happen?
• To find the answer, we must analyse fig.11.15 below:
 |
| Fig.11.15 |
1. We know that the armature is continuously rotating in the space between the poles.
• Let us stop the rotation at the very instant when the armature is exactly vertical
(with AB at top and CD at bottom).
• This is shown in fig.11.15(a)
2. Now there is no current in the circuit.
• This is our initial point. We start our analysis from here.
• So we will call this point as 'stage 1'
3. Now we begin the rotation. The armature begins to move.
• At the instant when rotation begins, the segment AB of the armature has a horizontal motion.
♦ This is indicated by the red horizontal arrow
4. Note that, this horizontal motion of AB is 'instantaneous'.
• That is., it is valid for a very short time only. After that instant, the movement is not horizontal.
5. But however small it may be, this 'instantaneous horizontal motion' is valuable for our analysis.
• Because, during the horizontal motion, there is 'no cutting' of the magnetic field
• Since there is 'no cutting', there is no current at that instant
6. The same happens at the bottom segment CD also.
• It also has an 'instantaneous horizontal movement' (in the opposite direction)
• Hence there is no current in CD also.
7. There is no current in AB as well as CD.
• As a result, there is zero current in the armature as a whole
• We can indeed call that instant as 'stage 1'
8. So what is the next stage?
• As the rotation continues, current begins to flow.
9. The armature becomes less and less vertical
• It becomes more and more horizontal
• During this process, the cuttings are 'intermediate cuttings'
10. Consider the very instant when the armature is exactly horizontal.
This is shown in fig.11.15(b)
• At this instant, the segment AB has a vertical motion. This is indicated by the red arrow.
• Note that, this vertical motion of AB is 'instantaneous'.
• That is., it is valid for a very short time only. After that instant, the movement is not vertical.
11. But however small it may be, this 'instantaneous vertical motion' is valuable for our analysis. • Because, due to the vertical motion, there is 'perfect cutting' of the magnetic field
• Since there is 'perfect cutting', there is 'maximum current' at that instant
12. The same happens at the other segment CD also.
• It also has an 'instantaneous vertical motion' (in the opposite direction)
• Hence there is 'maximum current' in CD also.
13. There is 'maximum current' in AB as well as CD.
• As a result, there is 'maximum current' in the armature as a whole
• We can call that instant as 'stage 2'
14. Now, how did the armature reach 'stage 2' from 'stage 1'
• Ans: Stage 1 is vertical and stage 2 is horizontal
• So, to reach stage 2 from stage 1, the armature rotated through an angle of 90o.
• The various angles between horizontal and vertical planes is shown in the fig.11.16 below:
 |
| Fig.11.16 |
• During a rotation from 0o to 90o, the current reached 'maximum current' from 'zero current'
16. Current is directly related to voltage (emf)
[Recall the relation: V = IR]
♦ When emf increases, current increases
♦ When emf decreases, current decreases
• In other words:
♦ 'maximum current' is caused due to 'maximum emf'
♦ 'zero current' is caused due to 'zero emf'
17. So we can write:
• During a rotation from 0o to 90o, the emf reached 'maximum emf' from 'zero emf'
• In other words:
♦ Stage 1 → 0o → zero emf
♦ Stage 2 → 90o → maximum emf
18. This variation from 'zero' to 'maximum' can be represented graphically as shown in fig.11.17 below:
• Angle is plotted along the x axis
• emf is plotted along the y axis
 |
| Fig.11.17 |
• Note that, it is a 'rising curve'. So it indicates the rise of emf from zero to maximum
• Consider the cyan coloured vertical dashed lines in the above graph
• They start from the center of the armature. Why is that so?
• Ans: The graph represents the 'emf in the armature as a whole'. Not the 'emf in individual segments'
• So it is appropriate to use the center point of the armature
19. Now we move on to see stage 3
• The armature is continuing it's rotation
• It becomes less and less horizontal and more and more vertical
• During this process, the cuttings are 'intermediate cuttings'
• The angle is increasing from 90o
20. Consider the very instant when the armature is exactly vertical again. The angle now is 180o.
• This is a situation very similar to 'stage 1'
• The only difference is that, AB is now at the bottom and CD is at the top
21. At this instant, both the segments AB and CD has a horizontal motion.
• The cuttings at this instant are 'zero cuttings'
• So, at this instant, there is 'zero emf' in the armature
• We can call this instant 'stage 3'
22. We can write:
• During a rotation from 90o to 180o, the emf reached 'zero emf' from 'maximum emf'
• In other words:
♦ Stage 2 → 90o → maximum emf
♦ Stage 3 → 180o → zero emf
23. This variation from 'maximum' to 'zero' can also be represented graphically.
• We will add it to the previous graph. The new graph is shown in fig.11.18 below:
 |
| Fig.11.18 |
• Note that, it is a 'falling curve'. So it indicates the fall of emf from maximum to zero
24. Now we move on to see stage 4
• The armature is continuing it's rotation
• It becomes less and less vertical and more and more horizontal
• During this process, the cuttings are 'intermediate cuttings'
• The angle is increasing from 180o
25. Consider the very instant when the armature is exactly horizontal again. The angle now is 270o.
• This is a situation very similar to 'stage 2'
• The only difference is that, AB is now at the left side and CD is at the right side
26. At this instant, both the segments AB and CD has a vertical motion.
• The cuttings at this instant are 'perfect cuttings'
• So, at this instant, there is 'maximum emf' in the armature
• We can call this instant 'stage 4'
27. We can write:
• During a rotation from 90o to 180o, the emf reached 'zero emf' from 'maximum emf'
• In other words:
♦ Stage 3 → 180o → zero emf
♦ Stage 4 → 270o → maximum emf
28. This variation from 'zero' to 'maximum' can also be represented graphically.
• We will add it to the previous graph. The new graph is shown in fig.11.19 below:
 |
| Fig.11.19 |
• Note that, though the emf increases from zero to maximum, we draw it on the negative side of the y axis
• This is to indicate that, the direction of current changes after one 'half rotation'
29. Now we move on to see stage 5
• The armature is continuing it's rotation
• It becomes less and less horizontal and more and more vertical
• During this process, the cuttings are 'intermediate cuttings'
• The angle is increasing from 270o
30. Consider the very instant when the armature is exactly vertical again. The angle now is 360o.
• This is a situation very similar to 'stage 1'. In fact, it is exactly same as 'stage 1'
31. At this instant, both the segments AB and CD has a horizontal motion.
• The cuttings at this instant are 'zero cuttings'
• So, at this instant, there is 'zero emf' in the armature
• We can call this instant 'stage 5'
32. We can write:
• During a rotation from 270o to 360o, the emf reached 'zero emf' from 'maximum emf'
• In other words:
♦ Stage 4 → 270o → maximum emf
♦ Stage 5 → 360o → zero emf
33. This variation from 'maximum' to 'zero' can also be represented graphically.
• We will add it to the previous graph. The new graph is shown in fig.11.20 below:
 |
| Fig.11.20 |
Thus we can write:
■ 'One cycle is complete' when 'the armature completes on rotation'.
• The second cycle begins when the armature begins to rotate after stage 5. It is a continuous process.
• We get electricity as long as we repeat the cycles
• Graphs can be drawn for those cycles also
• The beginning of the 'graph of the second cycle' is shown by the dotted green curve
• Note that, the intervals between any two consecutive cyan vertical lines are the same. It is 90o
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