In the previous section we saw the basics about three phase generator. In this section, we will see moving coil microphone.
• The moving coil microphone (see images here) is a device which works on the principle of electromagnetic induction.
• That is., this devices utilises the current formed in a coil due to the relative motion of the coil and a magnet.
Let us see it's working. We will write the steps:
1. Consider fig.11.25 below:
• The sound waves made by the speaker falls on a diaphragm, causing it to vibrate.
• The vibrations will be in accordance with the sound falling on it.
♦ If the sound is loud, vibrations will be more.
♦ If the sound is soft, vibrations will be less.
2. The diaphragm is attached to a coil.
• So the vibrations are passed on to the coil and it also begins to vibrate.
3. This coil is placed in the strong magnetic field of a permanent magnet.
So we can say: The vibrations cause the coil to move inside a magnetic field.
4. Due to electromagnetic induction, current will begin to flow in the coil.
• This current is called electrical signals.
• It has all the characteristics of the sound made by the speaker.
• These electrical signals flow out from the coil through lead wires.
5. Then they reach an amplifier.
• Here the signals are 'strengthened'.
6. After amplification, the signals are sent to the loudspeaker.
• We have already seen the working of a loudspeaker (Details here).
• If the signals are not amplified, they will not be able to move the paper cone of the loudspeaker.
7. In a microphone, first we have a mechanical energy in the form of 'vibrations of the diaphragm'.
• We effectively convert it into 'electrical signals'. So we can write:
■ In a microphone, mechanical energy is converted into electrical energy.
Now we will learn about mutual induction. Let us do an activity:
1. Consider the arrangement in fig.11.26 below:
• At each end of a soft iron core, finely insulated copper wire is wound. They are named as P and Q
♦ The ends of the copper wire of P are connected to a cell and a switch.
♦ The ends of the copper wire of Q are connected to a bulb
2. Now we can begin the trials:
Trial 1:
(i) Turn the switch on and off continuously
(ii) Note down the observation:
• The bulb glows
3. Trial 2:
(i) Turn on the switch and keep it in on position
(ii)Note down the observation:
• The bulb does not glow
4. The trials are complete. Let us do the analysis:
(i) When we turn on the switch, a magnetic field is formed around P
(ii) Since Q is close by, it will have access to the field produced in P
(iii) But 'having access' is not enough. There must be a 'change in flux'
(iv) When the switch is turned on and off continuously, the flux connected with Q changes.
• So a current will be produced in it. Thus the bulb glows
(v) But when the switch is kept in on position, there is no 'change in flux'.
• So current will not be produced and hence the bulb does not glow.
■ So we can write:
Due to a 'changing magnetic field' produced in a coil, an emf is induced in another coil which is kept nearby
• The coil to which we give current for the production of magnetic field is called the primary coil
• The coil in which induced emf is generated is called the secondary coil
• It would be very convenient if we can cause the 'change in flux' with out turning the switch on and off continuously
• Let us do another activity
1. Consider fig.11.27 below
• The arrangements are the same as in fig.11.26 except for the power source:
• The cell is replaced by a 6 V AC source
1. Now we can begin the trials. This activity has only one trial
Trial 1:
(i) Turn on the switch
(ii) Note down the observation:
• The bulb glows continuously
2. Let us do the analysis:
(i) When the AC passes through the primary, the direction of current changes continuously
• So the polarity of the magnetic field produced in the primary changes continuously
• We can say: A varying magnetic field is produced in the primary
(ii) The secondary is situated in this varying magnetic field
• That is., the magnetic flux connected to the secondary changes continuously
• This situation is similar to 'moving a magnet inside the secondary coil
[Recall that, moving a magnet inside a coil causes a change in flux]
(iii) The flux change produced in the secondary, induces an emf in it.
• So a current begins to flow in the secondary. Thus the bulb glows
• Consider two coils of wire kept side by side.
• When the strength or direction of current in one coil changes, the magnetic flux around it changes.
• As a result, an emf is induced in the secondary coil.
• This phenomenon is called mutual induction
Now we will see a practical application of mutual induction
Consider the arrangement in fig.11.28(a) below:
1. We have seen that, when an AC is applied to the primary, an emf will be induced in the secondary
• We also know that, when the number of turns in a coil increases, greater emf is induced in it
2. The number of turns in the secondary coil in fig.11.28(a) is greater than that in the primary.
• So we get a higher voltage in the secondary.
• This arrangement can be used when we want to increase the voltage
■ It is called a Step up transformer
3. The number of turns in the secondary coil in fig.11.28(b) is lesser than that in the primary.
• So we get a lower voltage in the secondary.
• This arrangement can be used when we want to decrease the voltage
■ It is called a Step down transformer
4. Let the emf in each turn of the primary be e1
• Let the emf in each turn of the secondary be e2
5. These two values will always be the same. That is., e1 = e2
• Let us denote it as 'e'. So we can write: e1 = e2 = e
6. If the number of turns in the primary is Np, then The voltage in primary Vp = Npe
• If the number of turns in the secondary is Ns, then The voltage in primary Vs = Nse
7. Taking ratios, we get:
■ Thus we get a relation between the four items:
(i) Voltage in primary (ii) Voltage in secondary
(iii) Number of turns in primary (iv) Number of turns in secondary
Let us see some solved examples:
Solved example 11.2
(a) In a transformer,
Number of turns in the primary is 500
Voltage in the primary is 10 V
Number of turns in the secondary is 2500
What is the voltage in the secondary?
Solution:
1. Given that:
Number of turns in the primary is 500. So Np = 500
Voltage in the primary is 10 V. So Vp = 10 V
Number of turns in the secondary is 2500. So Ns = 2500
2. We have: Vs⁄Vp = Ns⁄Np
• Substituting the known values, we get:
Vs⁄10 = 2500s⁄500 ⟹ Vs⁄10 = 5 ⟹ Vs = 50 V
(b) In a transformer,
Voltage in the primary is 100 V
Number of turns in the secondary is 800
Voltage in the secondary is 25 V
What is the number of turns in the primary?
Solution:
1. Given that:
Voltage in the primary is 100 V. So Vp = 100 V
Number of turns in the secondary is 800. So Ns = 800
Voltage in the secondary is 25 V. So Vs = 25 V
2. We have: Vs⁄Vp = Ns⁄Np
• Substituting the known values, we get:
25⁄100 = 800⁄Np ⟹ 1⁄4 = 800⁄Np ⟹ Np = 800 × 4 = 3200
(c) In a transformer,
Number of turns in the primary is 600
Number of turns in the secondary is 1800
Voltage in the secondary is 120 V
What is the voltage in the primary?
Solution:
1. Given that:
Number of turns in the primary is 600. So Np = 600
Number of turns in the secondary is 1800. So Ns = 1800
Voltage in the secondary is 120 V. So Vp = 120 V
2. We have: Vs⁄Vp = Ns⁄Np
• Substituting the known values, we get:
120⁄Vp = 1800⁄600 ⟹ 120⁄Vp = 3 ⟹ Vp = 120⁄3 = 40 V
(d) In a transformer,
Number of turns in the primary is 12000
Voltage in the primary is 240 V
Voltage in the secondary is 12 V
What is the number of turns in the secondary?
Solution:
1. Given that:
Number of turns in the primary is 12000. So Np = 12000
Voltage in the primary is 240 V. So Vp = 240 V
Voltage in the secondary is 12 V. So Vs = 12 V
2. We have: Vs⁄Vp = Ns⁄Np
• Substituting the known values, we get:
12⁄240 = Ns⁄12000 ⟹ Ns⁄12000 = 1⁄20 ⟹ Ns = 12000⁄20= 600
Solved example 11.3
A transformer working on a 240 V AC supplies a voltage of 8 V to an electric bell. The number of turns in the primary coil is 4800. Calculate the number of turns in the secondary coil.
Solution:
1. Given that:
Voltage in the primary is 240 V. So Vp = 240 V
Voltage in the secondary is 8 V. So Vs = 8 V
Number of turns in the primary is 4800. So Np = 4800
2. We have: Vs⁄Vp = Ns⁄Np
• Substituting the known values, we get:
8⁄240 = Ns⁄4800 ⟹ Ns⁄4800 = 1⁄30 ⟹ Ns = 4800⁄30 = 160
Solved example 11.4
The input voltage of a transformer is 240 V AC. There are 80 turns in the secondary coil and 800 turns in the primary. What is the output voltage of the transformer?
Solution:
1. Given that:
Voltage in the primary is 240 V. So Vp = 240 V
Number of turns in the secondary is 80. So Ns = 80
Number of turns in the primary is 800. So Np = 800
2. We have: Vs⁄Vp = Ns⁄Np
• Substituting the known values, we get:
Vs⁄240 = 80⁄800 ⟹ Vs⁄240 = 1⁄10 ⟹ Vs = 240⁄10 = 24 V
Another useful relation:
1. In an ideal transformer, we can assume that, there is no loss of energy in the form of heat.
■ In such a transformer, we can write:
Power in the primary and secondary are equal. That is: Pp = Ps
2. We have seen in the previous chapter that: Power (P) = Voltage (V) × Current (I) [Details here]
• So we can write: Pp = Vp × Ip and Ps = Vs × Is
3.But Pp = Ps
• So we can write: Vp × Ip = Vs × Is
• Rearranging the above, we get: Vs⁄Vp = Ip⁄Is
4. But earlier in this section, we saw that 'Vs⁄Vp' is equal to Ns⁄Np
• So we can write them together:
Vs⁄Vp = Ns⁄Np = Ip⁄Is.
5. From the above relation, we get an important information:
• Consider the first and last ratios together: Vs⁄Vp = Ip⁄Is
• Rearranging, we get: VsIs = VpIp
• That means, the two quantities on either sides of the '=' sign will always be the same.
6. So, if the secondary voltage Vs becomes high, the secondary current Is will have to become low
• Otherwise the equality will not be maintained
• That is., in a step up transformer, the secondary current will be lower than the primary current
■ In other words: In a step up transformer: Is < Ip.
■ In a step up transformer, the coil in the primary has to carry greater current. So it is made of thicker wires
7. In a similar way, if the secondary voltage Vs becomes low, the secondary current Is will have to become high
• Otherwise the equality will not be maintained
• That is., in a step down transformer, the secondary current will be higher than the primary current
■ In other words: In a step down transformer: Ip < Is.
■ In a step down transformer, the coil in the secondary has to carry greater current. So it is made of thicker wires
Solved example 11.5
In a transformer without any loss in power, there are 5000 turns in the primary and 250 turns in the secondary. The primary voltage is 120 V and the primary current is 0.1 A. Find the voltage and current in the secondary.
Solution:
1. Given that:
Number of turns in the primary is 5000. So Np = 5000
Number of turns in the secondary is 250. So Ns = 250
Voltage in the primary is 120 V. So Vp = 120 V
2. We have: Vs⁄Vp = Ns⁄Np
• Substituting the known values, we get:
Vs⁄120 = 250⁄5000 ⟹ Vs⁄120 = 1⁄20 ⟹ Vs = 120⁄20 = 6 V
3. We have: Vs⁄Vp = Ip⁄Is
• Substituting the known values, we get:
6⁄120 = 0.1⁄Is ⟹ 1⁄20 = 0.1⁄Is ⟹ Is = 0.1 × 20 = 2 A
Solved example 11.6
(a) Given that, in a transformer, Vs > Vp
Is it a step up or step down transformer?
Solution:
Since secondary voltage is greater than primary voltage, it is a step up transformer
(b) Given that, in a transformer, Is < Ip
Is it a step up or step down transformer?
Solution:
Since secondary current is less than primary current, it is a step up transformer
(c) Given that, in a transformer, Ns⁄Np < 1
Is it a step up or step down transformer?
Solution:
Ns⁄Np < 1 ⟹ Ns < Np
Since Number of turns in secondary is less than that in primary, it is a step down transformer
(d) Given that, in a transformer, Vs < Vp
Is it a step up or step down transformer?
Solution:
Since secondary voltage is less than primary voltage, it is a step down transformer
(e) Given that, in a transformer, Is > Ip
Is it a step up or step down transformer?
Solution:
Since secondary current is greater than primary current, it is a step down transformer
(f) Given that, in a transformer, Ns⁄Np > 1
Is it a step up or step down transformer?
Solution:
Ns⁄Np > 1 ⟹ Ns > Np
Since Number of turns in secondary is greater than that in primary, it is a step up transformer
• The moving coil microphone (see images here) is a device which works on the principle of electromagnetic induction.
• That is., this devices utilises the current formed in a coil due to the relative motion of the coil and a magnet.
Let us see it's working. We will write the steps:
1. Consider fig.11.25 below:
 |
| Fig.11.25 |
• The vibrations will be in accordance with the sound falling on it.
♦ If the sound is loud, vibrations will be more.
♦ If the sound is soft, vibrations will be less.
2. The diaphragm is attached to a coil.
• So the vibrations are passed on to the coil and it also begins to vibrate.
3. This coil is placed in the strong magnetic field of a permanent magnet.
So we can say: The vibrations cause the coil to move inside a magnetic field.
4. Due to electromagnetic induction, current will begin to flow in the coil.
• This current is called electrical signals.
• It has all the characteristics of the sound made by the speaker.
• These electrical signals flow out from the coil through lead wires.
5. Then they reach an amplifier.
• Here the signals are 'strengthened'.
6. After amplification, the signals are sent to the loudspeaker.
• We have already seen the working of a loudspeaker (Details here).
• If the signals are not amplified, they will not be able to move the paper cone of the loudspeaker.
7. In a microphone, first we have a mechanical energy in the form of 'vibrations of the diaphragm'.
• We effectively convert it into 'electrical signals'. So we can write:
■ In a microphone, mechanical energy is converted into electrical energy.
Now we will learn about mutual induction. Let us do an activity:
1. Consider the arrangement in fig.11.26 below:
 |
| Fig.11.26 |
♦ The ends of the copper wire of P are connected to a cell and a switch.
♦ The ends of the copper wire of Q are connected to a bulb
2. Now we can begin the trials:
Trial 1:
(i) Turn the switch on and off continuously
(ii) Note down the observation:
• The bulb glows
3. Trial 2:
(i) Turn on the switch and keep it in on position
(ii)Note down the observation:
• The bulb does not glow
4. The trials are complete. Let us do the analysis:
(i) When we turn on the switch, a magnetic field is formed around P
(ii) Since Q is close by, it will have access to the field produced in P
(iii) But 'having access' is not enough. There must be a 'change in flux'
(iv) When the switch is turned on and off continuously, the flux connected with Q changes.
• So a current will be produced in it. Thus the bulb glows
(v) But when the switch is kept in on position, there is no 'change in flux'.
• So current will not be produced and hence the bulb does not glow.
■ So we can write:
Due to a 'changing magnetic field' produced in a coil, an emf is induced in another coil which is kept nearby
• The coil to which we give current for the production of magnetic field is called the primary coil
• The coil in which induced emf is generated is called the secondary coil
• It would be very convenient if we can cause the 'change in flux' with out turning the switch on and off continuously
• Let us do another activity
1. Consider fig.11.27 below
 |
| Fig.11.27 |
• The cell is replaced by a 6 V AC source
1. Now we can begin the trials. This activity has only one trial
Trial 1:
(i) Turn on the switch
(ii) Note down the observation:
• The bulb glows continuously
2. Let us do the analysis:
(i) When the AC passes through the primary, the direction of current changes continuously
• So the polarity of the magnetic field produced in the primary changes continuously
• We can say: A varying magnetic field is produced in the primary
(ii) The secondary is situated in this varying magnetic field
• That is., the magnetic flux connected to the secondary changes continuously
• This situation is similar to 'moving a magnet inside the secondary coil
[Recall that, moving a magnet inside a coil causes a change in flux]
(iii) The flux change produced in the secondary, induces an emf in it.
• So a current begins to flow in the secondary. Thus the bulb glows
• Consider two coils of wire kept side by side.
• When the strength or direction of current in one coil changes, the magnetic flux around it changes.
• As a result, an emf is induced in the secondary coil.
• This phenomenon is called mutual induction
Now we will see a practical application of mutual induction
Consider the arrangement in fig.11.28(a) below:
 |
| Fig.11.28 |
• We also know that, when the number of turns in a coil increases, greater emf is induced in it
2. The number of turns in the secondary coil in fig.11.28(a) is greater than that in the primary.
• So we get a higher voltage in the secondary.
• This arrangement can be used when we want to increase the voltage
■ It is called a Step up transformer
3. The number of turns in the secondary coil in fig.11.28(b) is lesser than that in the primary.
• So we get a lower voltage in the secondary.
• This arrangement can be used when we want to decrease the voltage
■ It is called a Step down transformer
4. Let the emf in each turn of the primary be e1
• Let the emf in each turn of the secondary be e2
5. These two values will always be the same. That is., e1 = e2
• Let us denote it as 'e'. So we can write: e1 = e2 = e
6. If the number of turns in the primary is Np, then The voltage in primary Vp = Npe
• If the number of turns in the secondary is Ns, then The voltage in primary Vs = Nse
7. Taking ratios, we get:
(i) Voltage in primary (ii) Voltage in secondary
(iii) Number of turns in primary (iv) Number of turns in secondary
Let us see some solved examples:
Solved example 11.2
(a) In a transformer,
Number of turns in the primary is 500
Voltage in the primary is 10 V
Number of turns in the secondary is 2500
What is the voltage in the secondary?
Solution:
1. Given that:
Number of turns in the primary is 500. So Np = 500
Voltage in the primary is 10 V. So Vp = 10 V
Number of turns in the secondary is 2500. So Ns = 2500
2. We have: Vs⁄Vp = Ns⁄Np
• Substituting the known values, we get:
Vs⁄10 = 2500s⁄500 ⟹ Vs⁄10 = 5 ⟹ Vs = 50 V
(b) In a transformer,
Voltage in the primary is 100 V
Number of turns in the secondary is 800
Voltage in the secondary is 25 V
What is the number of turns in the primary?
Solution:
1. Given that:
Voltage in the primary is 100 V. So Vp = 100 V
Number of turns in the secondary is 800. So Ns = 800
Voltage in the secondary is 25 V. So Vs = 25 V
2. We have: Vs⁄Vp = Ns⁄Np
• Substituting the known values, we get:
25⁄100 = 800⁄Np ⟹ 1⁄4 = 800⁄Np ⟹ Np = 800 × 4 = 3200
(c) In a transformer,
Number of turns in the primary is 600
Number of turns in the secondary is 1800
Voltage in the secondary is 120 V
What is the voltage in the primary?
Solution:
1. Given that:
Number of turns in the primary is 600. So Np = 600
Number of turns in the secondary is 1800. So Ns = 1800
Voltage in the secondary is 120 V. So Vp = 120 V
2. We have: Vs⁄Vp = Ns⁄Np
• Substituting the known values, we get:
120⁄Vp = 1800⁄600 ⟹ 120⁄Vp = 3 ⟹ Vp = 120⁄3 = 40 V
(d) In a transformer,
Number of turns in the primary is 12000
Voltage in the primary is 240 V
Voltage in the secondary is 12 V
What is the number of turns in the secondary?
Solution:
1. Given that:
Number of turns in the primary is 12000. So Np = 12000
Voltage in the primary is 240 V. So Vp = 240 V
Voltage in the secondary is 12 V. So Vs = 12 V
2. We have: Vs⁄Vp = Ns⁄Np
• Substituting the known values, we get:
12⁄240 = Ns⁄12000 ⟹ Ns⁄12000 = 1⁄20 ⟹ Ns = 12000⁄20= 600
Solved example 11.3
A transformer working on a 240 V AC supplies a voltage of 8 V to an electric bell. The number of turns in the primary coil is 4800. Calculate the number of turns in the secondary coil.
Solution:
1. Given that:
Voltage in the primary is 240 V. So Vp = 240 V
Voltage in the secondary is 8 V. So Vs = 8 V
Number of turns in the primary is 4800. So Np = 4800
2. We have: Vs⁄Vp = Ns⁄Np
• Substituting the known values, we get:
8⁄240 = Ns⁄4800 ⟹ Ns⁄4800 = 1⁄30 ⟹ Ns = 4800⁄30 = 160
Solved example 11.4
The input voltage of a transformer is 240 V AC. There are 80 turns in the secondary coil and 800 turns in the primary. What is the output voltage of the transformer?
Solution:
1. Given that:
Voltage in the primary is 240 V. So Vp = 240 V
Number of turns in the secondary is 80. So Ns = 80
Number of turns in the primary is 800. So Np = 800
2. We have: Vs⁄Vp = Ns⁄Np
• Substituting the known values, we get:
Vs⁄240 = 80⁄800 ⟹ Vs⁄240 = 1⁄10 ⟹ Vs = 240⁄10 = 24 V
Another useful relation:
1. In an ideal transformer, we can assume that, there is no loss of energy in the form of heat.
■ In such a transformer, we can write:
Power in the primary and secondary are equal. That is: Pp = Ps
2. We have seen in the previous chapter that: Power (P) = Voltage (V) × Current (I) [Details here]
• So we can write: Pp = Vp × Ip and Ps = Vs × Is
3.But Pp = Ps
• So we can write: Vp × Ip = Vs × Is
• Rearranging the above, we get: Vs⁄Vp = Ip⁄Is
4. But earlier in this section, we saw that 'Vs⁄Vp' is equal to Ns⁄Np
• So we can write them together:
Vs⁄Vp = Ns⁄Np = Ip⁄Is.
5. From the above relation, we get an important information:
• Consider the first and last ratios together: Vs⁄Vp = Ip⁄Is
• Rearranging, we get: VsIs = VpIp
• That means, the two quantities on either sides of the '=' sign will always be the same.
6. So, if the secondary voltage Vs becomes high, the secondary current Is will have to become low
• Otherwise the equality will not be maintained
• That is., in a step up transformer, the secondary current will be lower than the primary current
■ In other words: In a step up transformer: Is < Ip.
■ In a step up transformer, the coil in the primary has to carry greater current. So it is made of thicker wires
7. In a similar way, if the secondary voltage Vs becomes low, the secondary current Is will have to become high
• Otherwise the equality will not be maintained
• That is., in a step down transformer, the secondary current will be higher than the primary current
■ In other words: In a step down transformer: Ip < Is.
■ In a step down transformer, the coil in the secondary has to carry greater current. So it is made of thicker wires
Solved example 11.5
In a transformer without any loss in power, there are 5000 turns in the primary and 250 turns in the secondary. The primary voltage is 120 V and the primary current is 0.1 A. Find the voltage and current in the secondary.
Solution:
1. Given that:
Number of turns in the primary is 5000. So Np = 5000
Number of turns in the secondary is 250. So Ns = 250
Voltage in the primary is 120 V. So Vp = 120 V
2. We have: Vs⁄Vp = Ns⁄Np
• Substituting the known values, we get:
Vs⁄120 = 250⁄5000 ⟹ Vs⁄120 = 1⁄20 ⟹ Vs = 120⁄20 = 6 V
3. We have: Vs⁄Vp = Ip⁄Is
• Substituting the known values, we get:
6⁄120 = 0.1⁄Is ⟹ 1⁄20 = 0.1⁄Is ⟹ Is = 0.1 × 20 = 2 A
Solved example 11.6
(a) Given that, in a transformer, Vs > Vp
Is it a step up or step down transformer?
Solution:
Since secondary voltage is greater than primary voltage, it is a step up transformer
(b) Given that, in a transformer, Is < Ip
Is it a step up or step down transformer?
Solution:
Since secondary current is less than primary current, it is a step up transformer
(c) Given that, in a transformer, Ns⁄Np < 1
Is it a step up or step down transformer?
Solution:
Ns⁄Np < 1 ⟹ Ns < Np
Since Number of turns in secondary is less than that in primary, it is a step down transformer
(d) Given that, in a transformer, Vs < Vp
Is it a step up or step down transformer?
Solution:
Since secondary voltage is less than primary voltage, it is a step down transformer
(e) Given that, in a transformer, Is > Ip
Is it a step up or step down transformer?
Solution:
Since secondary current is greater than primary current, it is a step down transformer
(f) Given that, in a transformer, Ns⁄Np > 1
Is it a step up or step down transformer?
Solution:
Ns⁄Np > 1 ⟹ Ns > Np
Since Number of turns in secondary is greater than that in primary, it is a step up transformer
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