In the previous section we saw the graph of an alternating current. In this section, we will see some more properties.
• Consider fig.11.21 below. It is the same fig.11.20 that we saw at the end of the previous section.
• The only difference is that, some fractions are written at the bottom
Let us see what those fractions are:
1. We have:
• 'zero' written at the bottom of the y axis
2. This 't' indicates time.
• We have seen that, the fourth vertical dashed line (which corresponds to stage 5) marks the end of one cycle
• So it is clear that 't' is the time required to complete one cycle.
■ That is:
A person is rotating the armature in such a way that, 't' seconds is required for one rotation of the armature
OR
Water is falling on the turbine in such a way that, 't' seconds is required for one rotation of the armature. (see fig.11.13)
■ The time required for one complete rotation of the armature is called period
It's symbol is 'T'
■ So, for our present case, the period (T) = t seconds
3. The four cyan vertical dashed lines (together with the y axis), divide this total 't' into 4 equal parts.
• So each part is 't⁄4'
• This 't⁄4' is added successively at each cyan line.
• Thus we get: 0, t⁄4, t⁄2, 3t⁄4 and t
♦ The emf is maximum at t⁄4 and 3t⁄4
♦ The emf is minimum at 0, t⁄2, and t
4. For an armature,
If Period (T) = 1 s, it means that the armature requires 1 second to complete one rotation
If Period (T) = 2 s, it means that the armature requires 2 seconds to complete one rotation
If Period (T) = 3 s, it means that the armature requires 3 seconds to complete one rotation
- - -
- - -
so on . . .
5. But what if the armature does not require even one second to complete one rotation?
• For example, an armature may rotate at a 'higher speed' so that, it requires only 0.5 seconds to complete one rotation.
• That is, for that armature, T = 0.5 s
• Then number of rotations that can be made in 1 second = 1⁄0.5 = 2
6. Another example:
• An armature may rotate at a higher speed so that, it requires only 0.2 seconds to complete one rotation.
• That is, for that armature, T = 0.2 s
• Then number of rotations that can be made in 1 second = 1⁄0.2 = 5
7. One more example:
• An armature may rotate at a very high speed so that, it requires only 0.025 seconds to complete one rotation.
• That is, for that armature, T = 0.025 s
• Then number of rotations that can be made in 1 second = 1⁄0.025 = 40
8. We have seen that, 'one rotation' is called 'one cycle'
• So 'number of rotations' in one second
is same as:
'Number of cycles' in one second
■ 'Number of cycles that the armature completes in one second' is called frequency of that AC
■ The unit of frequency is hertz. It is named after the German scientist Heinrich Hertz.
• It's symbol is 'Hz'
• One Hz means that: 'an event occurs once per second'
• In fact, a previous name of this unit was 'cycles per second (cps)'
9. For our present case of AC, 'Frequency of AC is one Hz' means that, 'the armature rotates once in every second'
■ The 'frequency of the AC' that is generated for distribution in India is 50 Hz
• That means, the armature has to rotate 50 times in one second
• There are practical difficulties involved in rotating the armature 50 times in just one second. To overcome those difficulties, scientists and engineers use some special techniques. We shall see some basic principles of those techniques later in this discussion
Solved example 11.1
The frequency of the AC from a power station is 50 Hz. How many times will the direction of current change in a circuit which use this AC?
Solution:
1. Given that frequency of the AC is 50 Hz
• That means, the armature makes 50 rotations in one second
2. Imagine that, the galvanometer (in fig.11.11 that we saw earlier) is replaced by the circuit mentioned in this problem.
• In each rotation, the current flows in two different directions:
• One direction in first 'half rotation'
• Opposite direction in second 'half rotation'
3. So in the circuit in which that AC is used, in one second:
• Current flows in one direction 50 times
• Current flows in the opposite direction 50 times
■ These two flows occur alternately
■ Power stations are centres that generate and distribute large quantities of electricity
■ The generators used here are called power generators
• Power generators have the same structure as AC generators
The main parts of an AC generator are:
• Field magnet • Armature • Brushes • Slip rings
• We have seen that, in an AC generator, electricity is generated when the armature rotates in a magnetic field.
■ Can we use the same method in a power generator? Let us see:
1. In a power generator, we want to produce large quantities of current.
2. We know that when the number if turns in an armature increases, current also increases. So indeed we have to increase the number of turns.
3. But when we increase the number of turns, the weight of the armature as a whole will increase.
• This will make it difficult to rotate the armature.
4. The solution to this problem is to keep the armature stationary and rotate the magnet instead.
5. Making the armature stationary and magnet to rotate, has another advantage:
• The brushes can be avoided. This will eliminate any chances of sparks.
■ The rotating part of the generator is the rotor
■ The static part of the generator is the stator
■ So in a power generator, armature is the stator and field magnet is the rotor
6. We know that, a strong magnetic field is required for producing AC
• For that, we want a strong magnet
■ In power generators, permanent magnets are not used. Instead, they use electromagnets.
7. So a question arises:
■ From where does the 'electricity required for magnetising the electromagnets' come from?
[This question arises because, we can produce electricity only if magnetic field is available. But we are now saying that electromagnets are first magnetised to produce a magnetic field]
• The answer is that, an auxiliary generator is used to produce DC. This DC is used to magnetise the electromagnets. The auxiliary generator is called exciter.
• In modern generators, big batteries are used instead of exciter
8. The advantages of using electromagnets instead of permanent magnets are as follows:
• Strength of a permanent magnet decrease gradually. So the magnetic flux cannot be maintained at the required level. But an electromagnet does not have such a problem
• The strength of an electromagnet can be changed to required levels by changing the current flowing through it's coils. But such a change is not possible in the case of permanent magnets
• We cannot make permanent magnets of very high strength. But electromagnets can be suitably designed for required strength
1. Consider the circle in fig.11.22(a) below. It is divided into 'three equal sectors' by the three red lines.
• We can be sure that the sectors are equal because, the angle between the lines is 120o.
[When the total angle 360o of a circle is divided into 3 equal parts, we get: 360⁄3 = 120o]
2. Now consider fig.b. Three coils A, B and C are placed along the periphery of a circle
• The angles between the coils is 120o.
• So we can say: The three coils are distributed uniformly along the periphery of the circle.
3. A magnet is placed at the centre of the circle. This magnet is rotated as shown by the white arrows
• When the magnet rotates, the flux linked with the coils changes and so current will be produced in each of them.
4. Let us compare the three currents:
• Consider the instant at which coil A gets the north pole. At that instant:
♦ Coil B will not have access to the north pole
♦ Coil C will not have access to the north pole
5. As the magnet continues it's rotation, B will get access to the north pole.
• But this will happen only after a 120o rotation
• So there is a lag between the currents in coil A and coil B. A lag of 120o
6. Similarly, consider the instant at which coil B gets the north pole. At that instant:
♦ Coil C will not have access to the north pole
♦ Coil A will not have access to the north pole
7. As the magnet continues it's rotation, C will get access to the north pole. But this will happen only after a 120o rotation
• So there is a lag between the currents in coil B and coil C. A lag of 120o
• Also, the lag between A and C is (120 + 120) = 240
8. We can represent all the lags graphically. Consider fig.11.23 below:
• The yellow curve represents the current in armature A
• The magenta curve represents the current in armature B
• The cyan curve represents the current in armature C
9. Starting points:
• The yellow curve starts exactly at the origin, where angle = 0
• The magenta curve starts at angle = 120o (∵ there is a lag of 120o from A)
• The cyan curve starts at angle = 240o (∵ there is a lag of 240o from A)
10. Maximum +ve emf points:
• After the magnet begins the rotation from zero degrees, the armature A receives maximum emf after a rotation of 90o.
♦ So it's peak point is (0+90) = 90o
♦ This can be seen in fig.11.23
• After the magnet begins the rotation from zero degrees, the armature B begins to receive emf only after 120o. Only after a further 90o, will B receive it's maximum.
♦ So B's maximum point is (120+90) = 210o
♦ This can be seen in fig.11.23
• Consider fig.11.21 below. It is the same fig.11.20 that we saw at the end of the previous section.
• The only difference is that, some fractions are written at the bottom
 |
| Fig.11.21 |
1. We have:
• 'zero' written at the bottom of the y axis
• 't⁄4' written at the bottom of first cyan vertical dashed line
• 't⁄2' written at the bottom of second vertical dashed line
• '3t⁄4' written at the bottom of third vertical dashed line
• 't' written at the bottom of fourth vertical dashed line 2. This 't' indicates time.
• We have seen that, the fourth vertical dashed line (which corresponds to stage 5) marks the end of one cycle
• So it is clear that 't' is the time required to complete one cycle.
■ That is:
A person is rotating the armature in such a way that, 't' seconds is required for one rotation of the armature
OR
Water is falling on the turbine in such a way that, 't' seconds is required for one rotation of the armature. (see fig.11.13)
■ The time required for one complete rotation of the armature is called period
It's symbol is 'T'
■ So, for our present case, the period (T) = t seconds
3. The four cyan vertical dashed lines (together with the y axis), divide this total 't' into 4 equal parts.
• So each part is 't⁄4'
• This 't⁄4' is added successively at each cyan line.
• Thus we get: 0, t⁄4, t⁄2, 3t⁄4 and t
♦ The emf is maximum at t⁄4 and 3t⁄4
♦ The emf is minimum at 0, t⁄2, and t
4. For an armature,
If Period (T) = 1 s, it means that the armature requires 1 second to complete one rotation
If Period (T) = 2 s, it means that the armature requires 2 seconds to complete one rotation
If Period (T) = 3 s, it means that the armature requires 3 seconds to complete one rotation
- - -
- - -
so on . . .
5. But what if the armature does not require even one second to complete one rotation?
• For example, an armature may rotate at a 'higher speed' so that, it requires only 0.5 seconds to complete one rotation.
• That is, for that armature, T = 0.5 s
• Then number of rotations that can be made in 1 second = 1⁄0.5 = 2
6. Another example:
• An armature may rotate at a higher speed so that, it requires only 0.2 seconds to complete one rotation.
• That is, for that armature, T = 0.2 s
• Then number of rotations that can be made in 1 second = 1⁄0.2 = 5
7. One more example:
• An armature may rotate at a very high speed so that, it requires only 0.025 seconds to complete one rotation.
• That is, for that armature, T = 0.025 s
• Then number of rotations that can be made in 1 second = 1⁄0.025 = 40
8. We have seen that, 'one rotation' is called 'one cycle'
• So 'number of rotations' in one second
is same as:
'Number of cycles' in one second
■ 'Number of cycles that the armature completes in one second' is called frequency of that AC
■ The unit of frequency is hertz. It is named after the German scientist Heinrich Hertz.
• It's symbol is 'Hz'
• One Hz means that: 'an event occurs once per second'
• In fact, a previous name of this unit was 'cycles per second (cps)'
9. For our present case of AC, 'Frequency of AC is one Hz' means that, 'the armature rotates once in every second'
■ The 'frequency of the AC' that is generated for distribution in India is 50 Hz
• That means, the armature has to rotate 50 times in one second
• There are practical difficulties involved in rotating the armature 50 times in just one second. To overcome those difficulties, scientists and engineers use some special techniques. We shall see some basic principles of those techniques later in this discussion
Solved example 11.1
The frequency of the AC from a power station is 50 Hz. How many times will the direction of current change in a circuit which use this AC?
Solution:
1. Given that frequency of the AC is 50 Hz
• That means, the armature makes 50 rotations in one second
2. Imagine that, the galvanometer (in fig.11.11 that we saw earlier) is replaced by the circuit mentioned in this problem.
• In each rotation, the current flows in two different directions:
• One direction in first 'half rotation'
• Opposite direction in second 'half rotation'
3. So in the circuit in which that AC is used, in one second:
• Current flows in one direction 50 times
• Current flows in the opposite direction 50 times
■ These two flows occur alternately
Power generator
■ Power stations are centres that generate and distribute large quantities of electricity
■ The generators used here are called power generators
• Power generators have the same structure as AC generators
The main parts of an AC generator are:
• Field magnet • Armature • Brushes • Slip rings
• We have seen that, in an AC generator, electricity is generated when the armature rotates in a magnetic field.
■ Can we use the same method in a power generator? Let us see:
1. In a power generator, we want to produce large quantities of current.
2. We know that when the number if turns in an armature increases, current also increases. So indeed we have to increase the number of turns.
3. But when we increase the number of turns, the weight of the armature as a whole will increase.
• This will make it difficult to rotate the armature.
4. The solution to this problem is to keep the armature stationary and rotate the magnet instead.
5. Making the armature stationary and magnet to rotate, has another advantage:
• The brushes can be avoided. This will eliminate any chances of sparks.
■ The rotating part of the generator is the rotor
■ The static part of the generator is the stator
■ So in a power generator, armature is the stator and field magnet is the rotor
6. We know that, a strong magnetic field is required for producing AC
• For that, we want a strong magnet
■ In power generators, permanent magnets are not used. Instead, they use electromagnets.
7. So a question arises:
■ From where does the 'electricity required for magnetising the electromagnets' come from?
[This question arises because, we can produce electricity only if magnetic field is available. But we are now saying that electromagnets are first magnetised to produce a magnetic field]
• The answer is that, an auxiliary generator is used to produce DC. This DC is used to magnetise the electromagnets. The auxiliary generator is called exciter.
• In modern generators, big batteries are used instead of exciter
8. The advantages of using electromagnets instead of permanent magnets are as follows:
• Strength of a permanent magnet decrease gradually. So the magnetic flux cannot be maintained at the required level. But an electromagnet does not have such a problem
• The strength of an electromagnet can be changed to required levels by changing the current flowing through it's coils. But such a change is not possible in the case of permanent magnets
• We cannot make permanent magnets of very high strength. But electromagnets can be suitably designed for required strength
Three phase generator
Now we will learn some basics about the 'three phase generator'. We will write it in steps:1. Consider the circle in fig.11.22(a) below. It is divided into 'three equal sectors' by the three red lines.
 |
| Fig.11.22 |
[When the total angle 360o of a circle is divided into 3 equal parts, we get: 360⁄3 = 120o]
2. Now consider fig.b. Three coils A, B and C are placed along the periphery of a circle
• The angles between the coils is 120o.
• So we can say: The three coils are distributed uniformly along the periphery of the circle.
3. A magnet is placed at the centre of the circle. This magnet is rotated as shown by the white arrows
• When the magnet rotates, the flux linked with the coils changes and so current will be produced in each of them.
4. Let us compare the three currents:
• Consider the instant at which coil A gets the north pole. At that instant:
♦ Coil B will not have access to the north pole
♦ Coil C will not have access to the north pole
5. As the magnet continues it's rotation, B will get access to the north pole.
• But this will happen only after a 120o rotation
• So there is a lag between the currents in coil A and coil B. A lag of 120o
6. Similarly, consider the instant at which coil B gets the north pole. At that instant:
♦ Coil C will not have access to the north pole
♦ Coil A will not have access to the north pole
7. As the magnet continues it's rotation, C will get access to the north pole. But this will happen only after a 120o rotation
• So there is a lag between the currents in coil B and coil C. A lag of 120o
• Also, the lag between A and C is (120 + 120) = 240
8. We can represent all the lags graphically. Consider fig.11.23 below:
 |
| Fig.11.23 |
• The magenta curve represents the current in armature B
• The cyan curve represents the current in armature C
9. Starting points:
• The yellow curve starts exactly at the origin, where angle = 0
• The magenta curve starts at angle = 120o (∵ there is a lag of 120o from A)
• The cyan curve starts at angle = 240o (∵ there is a lag of 240o from A)
10. Maximum +ve emf points:
• After the magnet begins the rotation from zero degrees, the armature A receives maximum emf after a rotation of 90o.
♦ So it's peak point is (0+90) = 90o
♦ This can be seen in fig.11.23
• After the magnet begins the rotation from zero degrees, the armature B begins to receive emf only after 120o. Only after a further 90o, will B receive it's maximum.
♦ So B's maximum point is (120+90) = 210o
♦ This can be seen in fig.11.23
• After the magnet begins the rotation from zero degrees, the armature C begins to receive emf only after 240o. Only after a further 90o, will C receive it's maximum.
♦ So C's maximum point is (240+90) = 330o
♦ So C's maximum point is (240+90) = 330o
♦ This can be seen in fig.11.23
11. Maximum -ve emf points:
• Once we find the 'maximum +ve emf points', we can easily find the 'maximum -ve emf points'.
• Because they are separated by 180o (this 180o corresponds to 'half a rotation')
• The maximum +ve point for armature A is 90o
♦ So the maximum -ve point is (90+180) = 270o
♦ This can be seen in fig.11.23
• The maximum +ve point for armature B is 210o
♦ So the maximum -ve point is (210+180) = 390o
♦ This can be seen in fig.11.23
• The maximum +ve point for armature C is 330o
♦ So the maximum -ve point is (330+180) = 510o
♦ This can be seen in fig.11.23
12. Now we know the salient points in all the three graphs
• However, the 'rotation of the armature' is a continuous process. The exact beginning of the rotation does not have much significance
• We can discard the 'first complete rotation'
• One complete rotation is 360o. So we need to consider only the portion to the right of 360o
• This will give a general picture of the three currents produced from the generator. It is shown in fig.11.24 below
13. When the field magnet rotates, 3 different alternating currents are generated
• In each armature, maximum and minimum emfs are generated at different instances
• In other words, no two currents have maximum or minimum at a same instant
• In other words, the three currents are in 'three different phases'
■ Such generators are three phase generators
14. In the AC generator that we saw in fig.11.21 at the beginning of this section, there is only one set of coil. Such generators are called single phase generators.
11. Maximum -ve emf points:
• Once we find the 'maximum +ve emf points', we can easily find the 'maximum -ve emf points'.
• Because they are separated by 180o (this 180o corresponds to 'half a rotation')
• The maximum +ve point for armature A is 90o
♦ So the maximum -ve point is (90+180) = 270o
♦ This can be seen in fig.11.23
• The maximum +ve point for armature B is 210o
♦ So the maximum -ve point is (210+180) = 390o
♦ This can be seen in fig.11.23
• The maximum +ve point for armature C is 330o
♦ So the maximum -ve point is (330+180) = 510o
♦ This can be seen in fig.11.23
12. Now we know the salient points in all the three graphs
• However, the 'rotation of the armature' is a continuous process. The exact beginning of the rotation does not have much significance
• We can discard the 'first complete rotation'
• One complete rotation is 360o. So we need to consider only the portion to the right of 360o
• This will give a general picture of the three currents produced from the generator. It is shown in fig.11.24 below
 |
| Fig.11.24 |
• In each armature, maximum and minimum emfs are generated at different instances
• In other words, no two currents have maximum or minimum at a same instant
• In other words, the three currents are in 'three different phases'
■ Such generators are three phase generators
14. In the AC generator that we saw in fig.11.21 at the beginning of this section, there is only one set of coil. Such generators are called single phase generators.
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