High school Physics Lessons: step up transformer

In the previous section we saw the basics about production of electricity. In this section, we will see it's transmission and distribution.
• We have seen that electricity is obtained from a generator.
• What is the energy conversion taking place in an AC generator?
Ans: We have seen that, mechanical energy is required to rotate the armature. (Details here).
• So we can say: Mechanical energy is converted into electrical energy.

But where does this mechanical energy come from?
There are several sources. Let us see some of them:
Method 1:
1. Water can be allowed to flow from a height.
2. This flowing water will rotate a turbine, which in turn will rotate the armature.
• When the armature rotates, we get electricity
3. The required water is stored in huge dams. These dams are constructed at elevated areas.
4. So the water stored will have potential energy.
5. This potential energy is converted into kinetic energy when the 'flow of water' takes place.
6. This kinetic energy causes the turbine to rotate.
7. A power station which uses this method is called hydroelectric power station.
8. So, in a hydroelectric power station, the energy change taking place is:
Potential energy ➡️ kinetic energy ➡️ Mechanical energy ➡️ Electrical energy

Method 2:
1. Water can be heated to make steam.
2. Steam under high pressure can be used to rotate the turbine.
3. A power station which uses this method is called thermal power station.
4. Coal can be used as a fuel to heat water.
5. But burning coal can cause environmental pollution.
6. In coal, energy is stored as chemical energy
7. So, in a thermal power station, the energy change taking place is:
Chemical energy ➡️ heat energy ➡️ Mechanical energy ➡️ Electrical energy

Method 3:
1. Instead of coal, nuclear energy can be used to heat water.
2. A power station which uses this method is called a nuclear power station.
7. So, in a nuclear power station, the energy change taking place is:
Nuclear energy ➡️ heat energy ➡️ Mechanical energy ➡️ Electrical energy

■ Electricity produced in a power station is transmitted to distant places. This process is called power transmission.
■ Special conducting wires are used for this purpose. Such wires are called transmission lines.

The various stages of power transmission and distribution are shown schematically in the fig.12.1 below:

Fig.12.1

1. We can see that the generator produces electricity of 11 kilo volts
2. This is sent to a power transformer
• So the input into this Power transformer is 11 kV.
3. The output from the power transformer is shown as 220 kV.
• That means, the 11 kV is increased 20 times to become 220 kV.
4. Why do we need such an increase?
We will write the answer in steps:
(i) When current flows through a conductor, heat is generated.
• To reach the consumer, from the power station, current has to flow through large distances
• So a large quantity of heat will be generated
■ We can say:
When current flows from the power station to the consumers, a portion of electrical energy is lost in the form of heat.
• We have to find ways to reduce this energy loss.
(ii) We have seen from Joules law that: Heat generated H = I²Rt (Details here)
• So, when resistance increases, heat also increases.
(iii) If we can reduce the 'resistance of the conducting wires', heat can be reduced
• To reduce the 'resistance of the conducting wires', we have to increase the 'area of cross section' of those wires. (We have seen the reason here)
(iv) But when the area of cross section is increased, two disadvantages occur:
(a) The conducting wires will become very expensive
(b) The conducting wires will become very heavy. Then, 'very strong and huge supports' will be required to keep them in position
• So 'increasing the area of cross section' is not an option here. We have to look for other methods
(v) Consider again the equation H = I²Rt
• We see that, the heat generated depends upon the square of the current (I)
• So, if we can reduce the current, heat generated will be low.
(vi) Consider the equation for power: P = VI (Details here)
• We do not want any 'energy loss'. That is., we do not want any 'power loss'
• So the left side 'P' must remain the same
• At the same time, the current 'I' should decrease
• There is only one way to satisfy both the conditions:
'Increase the voltage (V)'
An example:
• Let P₁ = V₁I₁ and P₂ = V₂I₂
• But we want: P₁ = P₂
• So we get: V₁I₁ = V₂I₂
• If the current is reduced to one tenth, we get: I₂ = ^I₁⁄₁₀
• Substituting this, we get:
V₁I₁ = V₂×^I₁⁄₁₀⟹ V₁ = V₂×¹⁄₁₀ ⟹ V₂ = 10V₁
■ That is: The voltage should be increased 10 times
Thus we can write:
• If the voltage is increased 10 times, we can reduce the current to one tenth
• When the current is thus reduced, heat generated will be low
• That is the reason for stepping up the voltage from 11 kV to 220 kV in the 'stage 1' shown in fig.12.1 above
5. Now we can get back to our main discussion:
The current at 220 kV travels long distances through 'transmission towers' (images here).
• Then they reach a substation. The name of this substation is '220 kV substation'
• Here the current is split into different branches
• Each branch enters into it's own designated step down transformer
• The outputs from these different step down transformers are:
110 kV, 66 kV, 33 kV, 11 kV etc.,
• The voltages 66 kV, 33 kV, 11 kV etc., are given to large scale industries because, they require such high voltages
6. The 110 kV continues the journey and reach another substation. The name of this substation is '110 kV substation'
• Here, all the current enters into a step down transformer.
• The output from this transformer is 11 kV
• This 11 kV current is split into two parts.
♦ One 11 kV part is given to small scale industries
7. The other 11 kV part continues it's journey and reach a transformer.
• It is a step down transformer. It's name is 'distribution transformer'
•The output from this transformer is 230 V
• 230 V is just what is required for house hold appliances

• This completes the description about the schematic diagram shown in figure 12.1.
• So now we know the basics about power transmission.
■ Consider the last transformer in figure 12.1.
• We see that 3 lines go into that transformer.
• But there are 4 lines coming out. why is that so?
We will see the reason in the next section

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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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High school Physics Lessons

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Wednesday, June 13, 2018

Chapter 12 - Power Transmission and Distribution

Thursday, June 7, 2018

Chapter 11.5 - Mutual Induction and Transformer