High school Physics Lessons: electromotive force

Showing posts with label electromotive force. Show all posts

Thursday, June 7, 2018

Chapter 11.5 - Mutual Induction and Transformer

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:

Fig.11.25

• 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:

Fig.11.26

• 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

Fig.11.27

• 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:

Fig.11.28

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 N_p, then The voltage in primary V_p = N_pe
• If the number of turns in the secondary is N_s, then The voltage in primary V_s = N_se
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 N_p= 500
Voltage in the primary is 10 V. So V_p= 10 V
Number of turns in the secondary is 2500. So N_s = 2500
2. We have: ^V_s⁄_{V_p} = ^N_s⁄_{N_p}
• Substituting the known values, we get:
^V_s⁄₁₀ = ^2500_s⁄₅₀₀⟹ ^V_s⁄₁₀ = 5⟹ V_s= 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 V_p= 100 V
Number of turns in the secondary is 800. So N_s = 800
Voltage in the secondary is 25 V. So V_s= 25 V
2. We have: ^V_s⁄_{V_p} = ^N_s⁄_{N_p}
• Substituting the known values, we get:
²⁵⁄₁₀₀ = ⁸⁰⁰⁄_{N_p}⟹ ¹⁄₄ = ⁸⁰⁰⁄_{N_p}⟹ N_p= 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 N_p= 600
Number of turns in the secondary is 1800. So N_s = 1800
Voltage in the secondary is 120 V. So V_p= 120 V
2. We have: ^V_s⁄_{V_p} = ^N_s⁄_{N_p}
• Substituting the known values, we get:
¹²⁰⁄_{V_p}= ¹⁸⁰⁰⁄₆₀₀⟹ ¹²⁰⁄_{V_p} = 3⟹ V_p= 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 N_p= 12000
Voltage in the primary is 240 V. So V_p= 240 V
Voltage in the secondary is 12 V. So V_s= 12 V
2. We have: ^V_s⁄_{V_p} = ^N_s⁄_{N_p}
• Substituting the known values, we get:
¹²⁄₂₄₀ = ^N_s⁄₁₂₀₀₀⟹ ^N_s⁄₁₂₀₀₀ =  ¹⁄₂₀⟹ N_s= 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 V_p= 240 V
Voltage in the secondary is 8 V. So V_s= 8 V
Number of turns in the primary is 4800. So N_p= 4800
2. We have: ^V_s⁄_{V_p} = ^N_s⁄_{N_p}
• Substituting the known values, we get:
⁸⁄₂₄₀ = ^N_s⁄₄₈₀₀⟹ ^N_s⁄₄₈₀₀ =  ¹⁄₃₀⟹ N_s= 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 V_p= 240 V
Number of turns in the secondary is 80. So N_s= 80
Number of turns in the primary is 800. So N_p= 800
2. We have: ^V_s⁄_{V_p} = ^N_s⁄_{N_p}
• Substituting the known values, we get:

^V_s⁄₂₄₀ = ⁸⁰⁄₈₀₀⟹ ^V_s⁄₂₄₀ =  ¹⁄₁₀⟹ V_s= 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: P_p = P_s
2. We have seen in the previous chapter that: Power (P) = Voltage (V) × Current (I) [Details here]
• So we can write: P_p= V_p× I_pand P_s= V_s× I_s
3.But P_p = P_s
• So we can write:  V_p× I_p= V_s× I_s
• Rearranging the above, we get: ^V_s⁄_{V_p} = ^I_p⁄_{I_s}
4. But earlier in this section, we saw that '^V_s⁄_{V_p}' is equal to ^N_s⁄_{N_p}
• So we can write them together:
^V_s⁄_{V_p} = ^N_s⁄_{N_p} = ^I_p⁄_{I_s}.
5. From the above relation, we get an important information:
• Consider the first and last ratios together: ^V_s⁄_{V_p} =  ^I_p⁄_{I_s}
• Rearranging, we get: V_sI_s= V_pI_p
• That means, the two quantities on either sides of the '=' sign will always be the same.
6. So, if the secondary voltage V_sbecomes high, the secondary current I_swill 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: I_s< I_p.
■ 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 V_sbecomes low, the secondary current I_swill 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: I_p< I_s.
■ 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 N_p= 5000
Number of turns in the secondary is 250. So N_s= 250
Voltage in the primary is 120 V. So V_p= 120 V
2. We have: ^V_s⁄_{V_p} = ^N_s⁄_{N_p}
• Substituting the known values, we get:
^V_s⁄₁₂₀ = ²⁵⁰⁄₅₀₀₀⟹ ^V_s⁄₁₂₀ =  ¹⁄₂₀⟹ V_s= ¹²⁰⁄₂₀= 6 V
3. We have: ^V_s⁄_{V_p} = ^I_p⁄_{I_s}
• Substituting the known values, we get:
⁶⁄₁₂₀ = ^0.1⁄_{I_s}⟹ ¹⁄₂₀ = ^0.1⁄_{I_s}⟹ I_s= 0.1 × 20= 2 A

Solved example 11.6
(a) Given that, in a transformer, V_s> V_p
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, I_s< I_p
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, ^N_s⁄_{N_p}< 1
Is it a step up or step down transformer?
Solution:
^N_s⁄_{N_p}< 1 ⟹ N_s< N_p
Since Number of turns in secondary is less than that in primary, it is a step down transformer
(d) Given that, in a transformer, V_s< V_p
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, I_s> I_p
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, ^N_s⁄_{N_p}> 1
Is it a step up or step down transformer?
Solution:
^N_s⁄_{N_p}> 1 ⟹ N_s> N_p
Since Number of turns in secondary is greater than that in primary, it is a step up transformer

In the next section we will see self induction.

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Tuesday, April 17, 2018

Chapter 8 - Current Electricity

In the previous section we completed a discussion on light rays. In this section, we will learn about Current electricity.
1. Consider the positively charged electroscope in fig 8.1 below.

Fig.8.1

• The above fig. is obtained from wikipedia.
2. This electroscope can be connected to the earth by a conductor through a switch.
• When the switch is turned on, charge will flow through the conductor.
3. Why do we say that there is 'flow of charge'?
• When the switch is 'turned on', electrons flow from the earth towards the electroscope.
• These electrons will neutralise the positive charge.
• So we say that there is a 'flow of charge'
■ But this 'flow of charge' will stop within a short time.

An activity:
1. Consider the circuit shown in fig.8.2 below:

Fig.8.2

• It consists of a cell, a bulb and a switch
2. When the switch is turned on, the bulb will glow.
■ The bulb will glow continuously.
• This indicates that unlike in the electroscope, the 'flow of charge' in this case is continuous.

• A 'flow of charge' creates a current.
■ In conductors, current is formed by the flow of electrons
• But in gases and electrolytes, electrons cannot flow freely
■ So in gases and electrolytes, the current is formed by the flow of ions
• For many practical applications, we want a continuous flow of charge as we saw in fig.8.2
• An instantaneous flow as in the case of the electroscope in fig.8.1 is not sufficient

Let us see a comparison between 'flow of charge' and 'flow of water'. We will write the comparison in steps:
1. Consider fig.8.3(a) below:

Fig.8.3

• In the arrangement in fig.a, water at P is at a higher level than Q.
2. When we open the valve, water will flow from P to Q
• As a result of this flow, the wheel will spin
3. But after some time, the levels at P and Q become the same.
• Then water will no longer flow from P to Q.
• So the wheel will no longer spin. This is shown in fig.b
• So we can write: The flow of water was due to the 'difference in water levels' between P and Q
4. We can write it in a more scientific way:
(i) Water at both P and Q have potential energy. We have seen potential energy here.
(ii) But water at P is at a higher level. So it has more potential energy than at Q
(iii) In other words. there is a 'difference in potential energy' between P and Q.
(iv) It is this 'difference in potential energy' that causes the water to flow.
(v) In fig.b, both P and Q are at the same level. So there is no difference in potential energy between P and Q. So the water will not flow.
5. In the case of electric circuits also, a similar case as in fig.8.3(b) occurs. Consider fig.8.4 below:

Fig.8.4

• A bulb is connected to a conductor through a switch.
6. There is no 'difference in potential' between X and Y.
• So when the switch is turned on, there will not be any flow of charge and so, the bulb will not glow.

7. The flow of water in fig.8.3(a) soon comes to an end.
■ Can we achieve a continuous flow? Let us try:
• In fig.8.5 below, a pump is introduced. All other arrangements are the same as in fig.8.3

Fig.8.5

8. Now, what does this pump do?
(i) The pump has an inlet pipe and an outlet pipe
♦ The inlet pipe takes in water from Q
♦ The outlet pipe delivers the water to P
(ii) The blue portion attached to the pump is like a closed vessel.
• Inside this vessel, there are rotating blades.
• Rotation of the blades is achieved with the help of the motor.
(iii) Due to the rotation of those blades, water will be sucked from Q.
(iv) Due to the continuing rotation of the blades, this sucked up water will be sent upwards into the outlet pipe of the pump. Thus the water will reach P
9. The pump has it's own definite 'rate of flow'.
• That is., in one second, it will deliver a 'fixed litres' of water. We cannot change it
• But we can adjust the valve at the bottom. By adjusting this valve, we can ensure that the following two values are the same:
(a) The litres of water reaching P from Q (through the pump) in one second
(b) The litres of water reaching Q from P (through the value) in one second
If the above two items are equal, we will achieve the following conditions:
(i) Level of water at P does not change
(ii) Level of water at Q does not change
10. In other words, the 'difference in water levels' between P and Q does not change
• If this 'difference in water levels' is maintained in this way, there will be continuous flow of water from P to Q, and thus the wheel at the bottom will spin continuously.
11. In the same way, we want the bulb in fig.8.4 to glow continuously
• The wheel in fig.8.5 spins continuously because there is a 'difference in energy level is maintained continuously'
♦ The quantity of energy at P is E_P
♦ The quantity of energy at Q is E_Q
12. Since E_P ≠ E_Q, there is a 'difference in energy level' between P and Q
■ What type of energy is E_Pand E_Q?
• Is it Potential energy?
• Is it Kinetic energy?
• Is it Chemical energy?
• Is it Solar energy?
- - -
- - -
• Obviously, it is potential energy.
13. In the same way,
• We want some energy E_X at X in the fig.8.4
• We want some energy E_Y at Y in the fig.8.4
• Also we want E_X≠ E_Y
14. But what type of energy is E_Xand E_Y?
• Is it Potential energy?
• Is it Kinetic energy?
• Is it Chemical energy?
• Is it Solar energy?
- - -
- - -
■ In this case, the energy is called electric potential
• E_X is a particular value of 'electric potential' at X
• E_Y is a particular value of 'electric potential' at Y
■ This is similar to saying:
• E_P is a particular value of 'potential energy' at P
• E_Q is a particular value of 'potential energy' at Q

15. But in problems, we will be more interested in the difference: (E_P- E_Q)
• For example, using (E_P- E_Q), scientists and engineers can calculate the speed at which the wheel will spin
16. In the same way, for our present problem in fig.8.4, we are more interested in (E_X- E_Y)
• For example, using (E_X- E_Y), scientists and engineers can calculate the brightness with which the bulb will glow

■ This difference in electric potential is called potential difference between X and Y

17. We know that the unit of potential energy is joules
• So the 'difference between two potential energies' will also be measured in joules
■ The unit of 'electric potential' is volt
■ So the 'difference between two electric potentials' will also be measured in volts
For example:
• If E_P = 14 joules and E_Q= 5 joules, then (E_P- E_Q) = 9 joules
• If E_X = 8 volts and E_Y= 3 volts, then (E_X- E_Y) = 5 volts
[Note that 'unit' is just a way for measuring. There can be several ways for measuring a quantity
For example: A certain length can be measured in cm or inch. They will give different values. But the actual length does not change.]
• For measuring 'electric potential', the unit 'volt' is universally accepted
■ If the potential difference between X and Y is 5 volts, we say this:
'A voltage of 5 volts is available between X and Y'

18. So our next task is to apply a potential difference between X and Y in fig.8.4.
• Also to make the bulb glow continuously, we have to maintain that potential difference
19. In the case of the spinning wheel, we used a pump, which is an 'external source of energy' to maintain the difference in water levels
• For our circuit in fig.8.4 also, we will need an 'external source of energy'.
• This 'external source of energy' that we will use is called a cell.
20. In fig.8.6 below, a cell is connected to the circuit. It is a 6 V cell.
• That means, it is able to maintain a potential difference of 6 volts.

Fig.8.6

• Note that X is connected to the positive terminal of the cell. So X will get a potential of 6 volts.
• Y is connected to the negative terminal of the cell. So Y will get a potential of zero volts.
• Thus the potential difference between X and Y is (6-0) = 6 volts.
21. The cell is an external source of energy. It continuously supplies 6 volts of energy.
• So the 'potential difference' between X and Y is maintained as 6 volts.
• Thus, when the switch is turned on, the bulb will glow continuously.

How is the cell able to supply this potential difference?
• Inside the cell there are chemical substances.
• When we connect a cell to a circuit, and the switch is turned on, chemical reactions will begin between these substances.
• As a result, there will be a flow of electrons.
■ But is the cell able to maintain the 'flow of electrons' for very long periods?
• The answer is 'no'. We know that, when chemical reactions takes place, the reactants will be converted into products.
• So when all the reactants stored in the cell are converted into products, the reactions will no longer take place.
• So there will not be any further flow of electrons. We will learn more details in higher classes.
• At present, all we need to know is that, a cell is an 'external source of energy'.
• It can help to maintain a 'potential difference'.

Electromotive force

1. In the fig.8.4 that we saw previously above, we wanted a potential difference between X and Y.
• But how much potential difference do we need?
♦ If it is a large bulb, we will need a large potential difference
♦ If it is a small bulb, we will need only a small potential difference
2. So before buying a cell we must decide how much energy we will need to make the bulb glow.
• This 'required energy' is specified in terms of electromotive force.
■ Electromotive force is defined as the energy provided by a cell which gives us the ability to maintain the potential difference between the ends of a conductor. It is measured in volts.
• In short form, the electromotive force is written as emf

3. If a cell is marked as 1.5 V, then it's emf is 1.5 volts.
• That is., if we use that cell, we will be able to maintain a potential difference of 1.5 volts between the ends of a conductor. This is shown in fig.8.7 below:

Fig.8.7

4. But when the electrons flow through the conductor some energy will be lost as heat.
• So if we want to check the emf of a cell using a 'measuring device', we must connect the terminals of the 'measuring device' directly to the terminals of the cell.
5. The voltmeter is such a 'measuring device'.
• The positive terminal of the cell should be connected to the positive terminal of the voltmeter.
• The negative terminal of the cell should be connected to the negative terminal of the voltmeter
• This is shown in fig.8.8 below:

Fig.8.8

In the next section, we will see combination of cells.