Chapter 13.3 - Primary, Secondary and Complementary colours

In the previous section we saw recombination of colours. We also saw persistence of vision. In this section, we will see primary, secondary and complementary colours.

Let us do an activity. We will write it in steps:
1. Take three torches.  Cover the front ends with red, blue and green glass papers.
2. Take the red and blue torches.
• Turn them on and direct the lights towards a white wall. 
• We can see a red circle and a blue circle. This is shown in fig.13.8(a) below:

Fig.13.8

3. Now move the torches in such a way that, the red circle and blue circle overlap.
This is shown in fig.13.8(b) above. 
• When they overlap, there is a 'region common to both circles'.  
• This common region is magenta in colour. 
4. Take the red and green torches.
• Turn them on and direct the lights towards the white wall. 
• We can see a red circle and a green circle. This is shown in fig.13.8(c) above.  
5. Now move the torches in such a way that, the red circle and green circle overlap.
This is shown in fig.13.8(d) above.  
• When they overlap, there is a 'region common to both circles'.  
• This common region is yellow in colour. 
6. Take the blue and green torches.
• Turn them on and direct the lights towards the white wall. 
• We can see a blue circle and a green circle. This is shown in fig.13.9(a) below:

Fig.13.9

7. Now move the torches in such a way that, the blue circle and green circle overlap.
This is shown in fig.13.9(b) 
• When they overlap, there is a 'region common to both circles'.  
• This common region is cyan in colour. 
8. Now take all the three torches together. 
• Turn them on and direct the lights towards the white wall
• Move the torches in such a way that the the three circles overlap. This is shown in fig.13.9(d)
    ♦ We can see a common region between red and blue. It is magenta as before 
    ♦ We can see a common region between red and green. It is yellow as before 
    ♦ We can see a common region between green and blue. It is cyan as before 
• But there is something new this time:
■ The region which is common to all colours is white
■ So we can write:
White light can be obtained by combining red, blue and green colours.
9. Remove the red blue and green glass papers from the torch  
• Cover them with other coloured glass papers.  
• Now try to make new colours by different combinations.  
• After some trials, we will find that, no matter how many ways we try, it is not possible to make red, blue or green. 

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■ By suitably superimposing red, blue and green, it is possible to make white light or other colours.
■ But it is impossible to make red, blue and green by combining other colours.  
■ Hence red, blue and green are the primary colours of light.
■ The colour obtained by combining any two primary colours of the same intensity is a secondary colour of light.  
Note that, 'same intensity' is specially mentioned. 

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From the above activity we can write the following 3 points about secondary colours: 
(i) When the primary colours blue and red are combined we get the secondary colour magenta 
(ii) When the primary colours green and red are combined we get the secondary colour yellow 
(iii) When the primary colours blue and green are combined we get the secondary colour cyan

■ An easy method to remember primary and secondary colours:
Step 1: Remember the two combinations 'RGB' and 'CMY'
'RGB' stands for the primary colours red, green and blue
'CMY' stands for the secondary colours cyan, magenta and yellow
Step 2: Write them vertically downwards, side by side as shown in fig. 13.10 (a) below:

Fig.13.10

• R must correspond to C
• G must correspond to M
• B must correspond to Y
Step 3: If we want to know how C is obtained:
    ♦ Strike out the primary colour 'R' corresponding to C
    ♦ Then C is obtained by the other two primary colours G and B
    ♦ This is shown in fig.13.10(b) above
• If we want to know how M is obtained:
    ♦ Strike out the primary colour 'G' corresponding to M
    ♦ Then M is obtained by the other two primary colours R and B
    ♦ This is shown in fig.13.10(c) above
• If we want to know how Y is obtained:
    ♦ Strike out the primary colour 'B' corresponding to Y
    ♦ Then Y is obtained by the other two primary colours R and G
    ♦ This is shown in fig.13.10(d) above

Complementary colours

Let us do another activity:
From the previous activity, we already have three torches: red, blue and green
1. Take three more torches.  Cover the front ends with the secondary colours: cyan, magenta and yellow glass papers.
2. Take the red and cyan torches.
• Turn them on and direct the lights towards the white wall. 
• We can see a red circle and a cyan circle. This is shown in fig.13.11(a) below:

Fig.13.11

3. Now move the torches in such a way that, the red circle and cyan circle overlap.
This is shown in fig.13.11(b) above. 
• When they overlap, there is a 'region common to both circles'.  
• We can see that this common region is white in colour. 
4. Take the Green and magenta torches.
• Turn them on and direct the lights towards the white wall. 
• We can see a green circle and a magenta circle. This is shown in fig.13.11(c) above
5. Now move the torches in such a way that, the green circle and magenta circle overlap.
This is shown in fig.13.11(d) above. 
• When they overlap, there is a 'region common to both circles'.  
• We can see that this common region is white in colour. 
6. Take the blue and yellow torches.
• Turn them on and direct the lights towards the white wall. 
• We can see a blue circle and a yellow circle. This is shown in fig.13.11(e) above
7. Now move the torches in such a way that, the blue circle and yellow circle overlap.
This is shown in fig.13.11(f) above. 
• When they overlap, there is a 'region common to both circles'.  
• We can see that this common region is white in colour.

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• From the above activity, we get three new methods to obtain white colour.  
• In each of those methods, only two colours are mixed.  
• Note that, in the previous activity be mixed three colours together to obtain white light.  
■ If white light is formed by combining just two colours, each one of those two colour is said to be complementary to the other. 

case 1: 
(i) We mixed red and cyan together.  Red is a primary colour while cyan is a secondary colour.  
(ii) When cyan is included, we are automatically including blue and green. 
(iii) So in effect, we are mixing red, blue and green together. Thus we get white colour.
case 2: 
(i) We mixed green and magenta together. Green is a primary colour while magenta is a secondary colour.  
(ii) When magenta is included, we are automatically including red and blue. 
(iii) So in effect, we are mixing red, blue and green together. Thus we get white colour.
case 3: 
(i) We mixed blue and yellow together.  Blue is a primary colour while yellow is a secondary colour.  
(ii) When yellow is included, we are automatically including red and green. 
(iii) So in effect, we are mixing red, blue and green together. Thus we get white colour.

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If white light is formed by combining a primary colour and secondary colour, two are mutually complementary.

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■ An easy method to remember complementary colours:
Step 1: Remember the two combinations 'RGB' and 'CMY'
'RGB' stands for the primary colours red, green and blue
'CMY' stands for the secondary colours cyan, magenta and yellow
Step 2: Write them vertically downwards, side by side as shown in fig. 13.10 (a) above.
• That is it. We have three pairs: [R,C], [G,Y], [B,M]
• Consider any one pair. Each member of that pair will be the complement colour of the other member

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In the next section, we will see colour of opaque objects.

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Chapter 13.2 - Recombination of Colours

In the previous section we saw the formation of rainbow. In this section, we will see recombination of colours.
Consider figure 13.6 below:

Fig.13.6

1. White light is falling on the prism on the left.  
• As a result, dispersion occurs.  
• Seven colours emerged from this prism. 
2. These 7 colours fall on the surface of the prism on the right.  
• This prism is similar to the first prism. But it is kept with its base on the top.
• We can see that the light emerging from the second prism is a single beam of white light.  
3. So what happens inside the second prism?
Ans: Each colour emerging from the first prism is bent by the glass material of the second prism.
• This 'bending' occurs in a direction opposite to 'the bending that occurred in the first prism'  
• The result is that, the seven rays converge at a point on the opposite face of the second prism.  
4. When the seven colours converge at a point, we get white light.  
■ This experiment proves that, if the seven colours are combined together, we will get white light. 

Newtons' Colour Disc

1. Consider the disk shown in fig.13.7 below.

Fig.13.7

• Seven colours of the rainbow are painted on it.  
2. The area of each colour in the disk should be the same.  
■ How can we achieve the same area?
Ans: 
(i) We know that, the full circle has a central angle of 360^o.  
(ii) If we divide this 360 by 7, we will get 51.42.
• So the central angle of each sector on the disk is 51.42^o
(iv) We have learned about sectors in our maths classes. (Details here)
3. Once we divide the circle into 7 equal sectors,  we can give each sector a colour.
• The colouring should be done in the same order as in the rainbow. That is., the order in the word VIBGYOR  
• Such a disc is called Newton's colour disc.  
4. If we spin this disc about it's centre, we can see that the colours gradually fade.  
• If we spin it very fast, we will see only white colour.
5. We want to know the reason for obtaining white.
We will write it in steps:
(i) Consider any one colour in the disc, say blue.  
(ii) Consider the instant at which the ray from the blue colour falls on our eye.  
(iii) Just after that instant, the ray from the next colour green will reach the eye. This is because, the disc is spinning fast
(iii) Even when the green colour has reached the eye, the previous blue ray will be still remaining in the retina
■ This is because, a ray will remain in the retina for (¹⁄₁₆) seconds. This is called persistence of vision
(iv) In fact, if the wheel spins very fast, the rays from all 7 colours will reach the retina within (¹⁄₁₆) seconds
(v) Since all the rays are together at a point within such a small interval of time, we will not be able to distinguish between the various colours
(vi) We will see only the resulting white colour
■ This proves that, white colour is a combination of seven colours.

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Some examples of persistence of vision:
• When a lighted torch is rotated fast, it appears as an illuminated circle
• Continuously falling rain drops appear as glass rods
• The leaves of a rotating fan cannot be distinguished from one another

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In the next section, we will see primary and secondary colours.

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Chapter 13.1 - Rainbow formation

In the previous section we saw dispersion of light. In this section, we will see formation of rainbow.

• For the formation of a rainbow in the sky, three conditions are to be satisfied. Let us see them first:
1.A rainbow is formed when the following two items are present at the same time in the atmosphere:
(i) sun light (ii) water particles 
2. A rainbow is formed either in the morning or in the evening. 
• Note that, at these times, the light from the sun are nearly parallel to the surface of the earth
3. When the sun is in the east, the rainbow is in the west 
• When the sun is in the west, the rainbow is in the east 
• So in the morning time, we must look towards the west to see the rainbow
• And in the evening time, we must look towards the east to see the rainbow
• In other words, to see the rainbow, our backs must be facing the sun

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Above three are the conditions. Now we will see two properties of the rainbow
1. When viewed from the ground, a rainbow is always seen in the form of an arc. See image below:

Source: By Eric Rolph at English Wikipedia - English Wikipedia, CC BY-SA 2.5, https://commons.wikimedia.org/w/index.php?curid=2406447

• But when viewed from an aeroplane, a rainbow can be seen as a full circle. Some images can be seen here.
2. In a rainbow, red will be always at the outer edge
• Voilet will always be at the inner edge 

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Now we will discuss how a rainbow is formed. During this discussion, we will see the explanations for the above 3 conditions and the 2 properties. We will write the steps:
1. The water particles in the atmosphere will be spherical in shape. 
• They can be considered as perfect spheres. 
• Note that, they are not bubbles with air trapped inside. They are solid spheres, made up of water. 
2. One such sphere is shown in fig.13.3(a) below:

Fig.13.3

• P is a point on the sphere. A white ray of light is incident at P
• A ray of light can fall on the surface of a sphere in two ways:
♦ It can fall obliquely on the surface
♦ It can fall normally on the surface
3. If a ray falls 'normally on the surface of a water sphere', 
♦ it will pass (with out any deviation) through the centre of the sphere. 
♦ also it will emerge without any deviation from the other side of the sphere. 
■ When do we say that 'a ray falls normally on the surface of a sphere'?
Ans: 
(i) Consider the green line drawn at 'P'
(ii) It is a radial line. That is., if the green line is extended downwards, it will pass through the centre of the sphere
(iii) We can easily draw this green line by joining 'P' and the 'centre'
(iv) This green line is the normal to the surface of the sphere at p.
(v) If a light ray falls along this normal, we say: 'that ray falls normally on the surface of a sphere' 
4. Thus, in our present case, it is obvious that, the ray under consideration is falling obliquely at P. 
• Because it is not falling along the normal at P
• Since it falls obliquely, refraction takes place. We have seen the result of such a refraction in fig.13.1 in the previous section
5. So the white light is split into it's component colours. 
• We will take the two extreme colours: Red and violet. This is shown in fig.13.3(b)
♦ The red ray falls at Q on the inside surface of the sphere
♦ The violet ray falls at R on the inside surface of the sphere
6. Now consider fig.13.4(a) below:

Fig.13.4

• Green normals are drawn at Q and R
(i) Carefully consider the angle between the following two lines:
♦ The green normal at Q
♦ The line PQ
• The angle between the above two lines, is the 'angle of incidence at Q'
• This angle is greater than the critical angle 
♦ So ray PQ cannot emerge out of the sphere. It will be reflected back
(We have learned about critical angle and total internal reflection here)
(ii) Again carefully consider the angle between the following two lines:
♦ The green normal at R
♦ The line PR
• The angle between the above two lines, is the 'angle of incidence at R'
• This angle is also greater than the critical angle 
♦ So ray PR cannot emerge out of the sphere. It will also be reflected back
• The other colours which fall in between red and violet will also be reflected back
7. After reflection at Q, the red ray falls at S
• After reflection at R, the violet ray falls at T
• This time the angle of incidence at S is less than the critical angle
♦ So the red ray emerges out
• The angle of incidence at T is also less than the critical angle
♦ So the violet ray also emerges out
8. Thus, in effect, we have these:
• A white ray going into a water drop
• It comes out of the water drop as seven different colours. 
• Red at one edge and violet at the other edge. 
This is shown in fig.13.4(b)
• The other 5 colours fall in between red and violet in regular order.
9. Note that, in the case of a glass prism, we get the component colours on the other side of the prism
♦ With red at top and violet at bottom 
• But in the case of a water drop, we get the component colours on the same side 
♦ With red at bottom and violet at top
10. The resulting red ray will be always at an 'angle of deviation' of 42 from the original white ray 
• The resulting violet ray will be always at an 'angle of deviation' of 40 from the original white ray 
This is shown in fig.13.4(b)
■ 42 is greater than 40. So, when the refraction is done by a water drop in this way:
• Red will always be at bottom and 
• Violet will always be at top
11. Now consider a person viewing the rainbow. This is shown in fig.13.5 below:

Fig.13.5

• Consider the line joining the eye of the viewer to the top most red arc of the rainbow
• One water drop along that line is shown in the fig.
• All seven colours will be emerging from that drop
• But red is the bottom most colour. Since all the other six colours are above red, they will travel above the head of the observer
• All water drops along that line will be producing all the seven colours
• But, since red is at the bottom, the colours other than red, from all of those drops, will pass above the head
• The observer will not see those colours
• That is the reason why red is always seen at the outer most arc in the rainbow
12. Now consider the line joining the eye of the viewer to the bottom most violet arc of the rainbow
• One water drop along that line is shown in the fig.
• All seven colours will be emerging from that drop
• But violet is the top most colour. Since all the other six colours are below violet, they will travel towards the feet of the observer
• All water drops along that line will be producing all the seven colours
• But, since violet is at the top, the colours other than violet, from all of those drops, will travel towards the feet
• The observer will not see those colours
• That is the reason why violet is always seen at the inner most arc in the rainbow
13. There are large numbers of water drops. All of them are dispersing light in different directions
• But those 'drops capable of sending particular colours to the viewer' lies in a circle around the viewer
• So we see a circular shape
14. The line from the eye to the red arc (in fig.13.5) will be a sloping line because red arc will be above the eye level
• The line from the eye to the violet arc (in fig.13.5) will also be a sloping line because violet arc will also be above the eye level
• But the line from the eye to the point 'O' will be a horizontal line.  
15. Part of the circle below the horizon will not be visible. Also, there are no water particles below the horizon to disperse the light
• So we see only part of the circle. That is., an arc.
• But when viewed from an aeroplane, water particles are available below the horizontal line. So the full circle will be visible

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In the next section, we will see recombination of light.

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Chapter 13 - Colours of Light

In the previous chapter we completed a discussion on power transmission and distribution. In this section, we will see colours of light.

• We have learned about refraction of light in a previous chapter. Based on that discussion, we will now see some advanced topics.
• Consider fig.13.1 below. A ray of white light falls on a glass prism.

Fig.13.1 Source: https://en.m.wikipedia.org/wiki/File:Dispersion_prism.jpg

• The ray passes through the prism and emerges from the other side. 
• But the emerging ray is not white. It is composed of different colours. 
• We want to know the reason for the splitting up of white light into different colours. Let us write an analysis. We will write it in steps:
1. First of all, the incident ray should fall on the prism obliquely. 
• 'Obliquely' means, 'at an angle'. This can be explained based on fig.13.2 below:

Fig.13.2

• The magenta line is drawn perpendicular to the surface of the prism. 
• Such a perpendicular is called the 'normal of the surface'. 
• The ray of light should make an angle with the normal. We shall call this angle as 'θ'. This 'θ' can take any value.
• If there is no angle (that is., if θ = 0), then the ray will be passing exactly along the normal. In that case, we will not get different colours.
2. Now consider the black dashed line in the fig.13.2.
• It is the original path of the light. The light should have passed along that line. 
• But it cannot do so. This is because, the glass material of the prism bends the light. 
• We have seen the details when we learned about refraction. 
3. But the glass is not able to bend all lights to the same extent
• Violet is bent more
• Red is bent less
4. The other colours that fall in between are: indigo, blue, green, yellow and orange
• The order of bending can be arranged in decreasing order:
■ Violet, Indigo, Blue, Green, Yellow, Red
■ Because of this difference in bending, components of the white light gets separated from each other.

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■ Any light that is composed of more than one colour is called a composite light
■ So we can write: White light is a composite light. 
■ A composite light will undergo dispersion when it passes through a prism.
■ Dispersion is the phenomenon of splitting up of a composite light into it's constituent colours. 
■ The components obtained by dispersion will appear as a regular array. 
■ This regular array of colours is called visible spectrum

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Now we will see the reason for 'violet being bent more than red' 
1. Consider the table given below:

• It gives the Wave lengths of different colours.
• We have seen some basics about waves in an earlier chapter (Details here)
2. Wave lengths are arranged in ascending order. 
• We can see that, the violet which undergoes most bending is at the top of the list. 
• Red which undergoes least bending is at the bottom of the list.
3. So we can write:
• The deviation from the original path depends on the wave length of the colour
• The colour which has more wave length deviates less
• The colour which has less wave length deviates more
• The other colours fall in between in regular order.

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Light undergoes refraction when it enters the prism obliquely and when it comes out of the prism.  The extent of deviation depends on the wavelength.  Therefore waves undergo deviation at different angles and get separated.  This is the reason for dispersion.

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The formation of rainbows in the sky can be explained based on dispersion. We will see it in the next section.

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Chapter 12.3 - Three pin Plug and Safety

In the previous section we saw the details about household circuits. We also saw the kwh meter. In this section, we will see the details about 3 pin plug.


1. Consider the fig.12.8 shown below. It shows the inside of an electric iron.  

Fig.12.8

• The current enters the heating coil from the phase line shown in red colour.  
• It leaves the coil through the neutral line shown in black colour.  
• Due to this flow of current, the required heat is produced in the coil. 
2. If there is any fault in the insulation,  the current will enter the body of the iron.  
• In such a situation,  a person who touches the iron will get an electric shock.
3. Such an accident can be avoided by using a 3 pin plug.  
Let us see how the three pin plug helps to avoid the accident:  
• Fig 12.9 below shows different types of 3 pin plugs available in the market

Fig.12.9 Source: https://commons.wikimedia.org/wiki/File:BS-546-3-pin-plugs.jpg 

4. The basic functioning of all of them are the same.  
• There are two small pins and one large pin.  
    ♦ The large pin is called the earth pin 
    ♦ Among the two smaller pins, one is for phase line and the other is for the neutral line.  
5. Now consider figure 12.8 again.  
• A green wire starts from the body of the iron.  It enters into the three pin plug.  
• Inside the three pin plug, this green wire is connected to the earth pin. This is shown in fig.12.10 below:

Fig.12.10 Source: https://commons.wikimedia.org/wiki/File:Three_pin_mains_plug_(UK).svg

6. So the current which has entered the body of the iron has two options:
(i) Flow into the body of the person who touches the iron
(ii) Flow into the earth through the green earth line
• Current will choose the second option because, it is a 'path of less resistance'
■ To make this flow more easier, the earth wire is always given a greater thickness
■ The earth pin also has a greater thickness
7. When the current thus begins to flow into the earth, more current will be drawn into the appliance
• Such a higher current will increase the temperature in the fuse wire. 
• As a result, the fuse wire will melt and thus the circuit will become open. Danger is thus averted.
8. We can see that, the length of the earth pin is greater than the other two pins. Why is that so?
Ans: Consider the instant at which a person inserts the 3 pin plug into the socket
• Two things happen at that instant:
(i) The earth pin comes into contact with the 'earth line connecting the socket to the earth'
(ii) The phase pin and neutral pin come into contact with their respective lines, and current begins to flow
• Since the earth pin is longer, (i) happens before (ii)
• So the 'whole earth line from the appliance to the earth' will be ready before the flow of current begins. 
• If any current enters the body of the appliance, 'a path is already waiting' for it's easy flow into the earth.
9. Consider the instant at which the person pulls the plug out of the socket 
• Two things happen at that instant:
(i) The earth pin loses it's contact with the 'earth line connecting the socket to the earth'
(ii) The phase pin and neutral pin lose their contact with their respective lines, and current is cut off
• Since the earth pin is longer, (ii) happens before (i)
• So the 'earth line is fully capable to carry any current' even after the instant at which current from the phase line is stopped.

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First aid to be given in the case of electric shock
■ First Aid should be given only after disconnecting the victim from the electric wire
• As a result of electric shock,
♦ The body temperature of the victim decreases 
♦ Viscosity of blood increases
♦ Clotting of blood occurs
♦ Muscles of the body contract
■ How to provide first aid:
• Raise the temperature of the body by massaging 
• Give artificial respiration 
• Massage the muscles and bring them to the original condition 
• Start first aid for the functioning of the heart (apply pressure on the chest regularly)
• Take the person to the nearest hospital immediately

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• Electricity is an integral part of everyday life. This energy is to be conserved for tomorrow and so its consumption should be reduced to the minimum possible level.
■ Saving electricity is equivalent to generating electricity
• Electricity is highly useful. And at the same time, it is a dangerous form of energy. Hence electrical appliances should be handled with extreme care 

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In the next chapter, we will see colors of light

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Chapter 12.2 - Household Electric Circuits

In the previous section we saw that household circuits should be in parallel mode. In this section, we will see the details of an actual household circuit.

We will write it in steps:
1. Consider fig.12.6 below:

Fig.12.6

• 4 lines are passing through the electric post. We can obtain electrical energy from them. 
2. For household purposes, we need only two lines:
♦ Any one phase line and 
♦ The neutral line.  
• Here we have chosen the red phase line. 
3. Both the lines first enters the kwh metre. 
• From this metre, we get 'the amount of energy used by the consumer'. We will see details later in this section. 
4. The main line then enters the main fuse box.  
• This is for safety purposes.  
• When current flows above the accepted level,  the fuse wire melts and thus, the circuit is broken.
• We have already seen the working of the safety fuse.     
5. Then the lines enter the main switch. 
• The current flow into the house can be stopped or resumed at any time by using the main switch.
6. After that, both the lines enter the ELCB. (Earth Leakage Circuit Breaker)
• This device breaks the circuit (and thus stops the flow of current) if it detects any flow of current in the 'earth wire'. 
• If even small quantities of current flows through the earth wire, it will be due to faults in the circuit. 
• So it is essential to break the circuit immediately
• We will learn about earth wire in the next section
• However, devices which are even more advanced than the ELCB, are available today.
7. After that, both the lines enter the MCB distribution board. We have already learned about MCB. 
• It is a safety device which will break the circuit if there is any fault in the circuit. Details here.
• What we have now, is a 'MCB distribution board'. This board combines two functions:
(i) It helps to take out branches from the phase line
♦ Each of those branch will go into it's own circuit
♦ One such circuit will be available for each room in the house
(ii) It provides the 'circuit breaking safety' for each of those circuits
8. So, after branching at the 'MCB distribution board', we will get several 'live lines'. In our present case, we have 3 'live lines'.
• Each live line goes into it's own circuit.
• One such circuit is shown in detail. This circuit has 3 appliances:
a bulb, a fan and a 3 pin socket
9. These 3 appliances should be connected in parallel
• In the fig.12.6, they are indeed connected in parallel. 
• If we have more space to draw, that circuit can be drawn as shown in fig.12.7 below: 

Fig.12.7

• Now the 'parallel mode' is more clear
• The reader may compare the switch board in fig.12.7 with that in fig.12.6 and verify that, they are the same.
• In fig.12.7, notice how 'an additional bulb taken from the 3 pin socket' will effectively complete the flow of current through the socket. The green line is the 'earth line'. We will see it's details in the next section.
■ The system shown in fig.12.6 is called the tree system.

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Now we will see the kwh meter. We will write it in steps:
1. We have learned about 'electric power' (Details here)
• Consider an appliance on which it is marked as '1000 watts' 
• That means, that appliance will consume an energy of 1000 joules in one second
2. If the consumer uses that appliance for one hour, how much energy will be used?
Ans: 1000 × 60 × 60 = 3600000 joules = 3600 kilo joules
• The consumer will have to pay money for this 3600 kilo joules
3. But the distribution companies do not measure energy in joules or kilo joules. 
■ They use another unit: kilowatt hour
4. Let us see how this unit is derived:
• We know that 'power' is the ratio of Energy to time. That is: 
Power = ^Energy⁄_time
• Multiplying both sides by 'time', we get:
Power × time = ^Energy⁄_time × time
5. But [^Energy⁄_time × time] = Energy
• Then (4) becomes: Power × time = Energy  
• So, to get 'amount of energy used', we can multiply the following two quantities:
(i) Power of the appliance  
(ii) Time for which the appliance is used
6. We can write:
Amount of energy used by an appliance 
= Power of the appliance × Time for which the appliance is used
7. Based on this, we can derive 'units':
• unit of energy = unit of power × unit of time
⟹ unit of energy = watts × sec
8. For large values, we can use:
• kilowatts instead of watts (∵ 1000 watts make up one kilo watt)
• hour instead of sec (∵ 3600 seconds make up one hour)
■ So we get:
Unit of energy = kilowatt hour (kwh)
• This unit is used by the distribution companies.
9. The companies use the simple term: 'units'
■ 1 unit = 1 kwh
An example:
• If a consumer uses '25 units' of electricity, it means that, he uses 25 kwh of electrical energy
10. On many occasions, we will want to know the 'number of units' consumed in our homes and offices. So let us derive an easy method:
• The two information that we will be having are:
♦ Power of the appliance (in watts)
♦ Time for which the appliance is used (in hours)
• Based on the above two, we must be able to quickly find the 'number of units'. Let us try:
We will write the steps:
(i) When 'power' is multiplied by 'time', we get 'energy'.
• So '(watt × hour) = watt hour' is energy
(ii) But we want 'kilowatt hour'
• So we must divide 'watt hour' by 1000
• Then we will get 'kwh' or the 'number of units' directly.
(iii) We can write the formula:

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Let us see an example:
Solved example 12.1
A grinder of power 750 W works for 2 hours. Calculate the energy consumed
Solution:
1. Given that, power = 750 W, Time = 2 hours
2. We have:

Substituting the values, we get:
Energy (kwh) = ^750×2⁄₁₀₀₀ = 1.5 kwh = 1.5 units
Another method:
1. Given that, power = 750 W
So the grinder consumes 750 joules every second
2. In 2 hours, there are (2 × 3600) seconds
So energy consumed in 2 hours = 750 × 2 × 3600 = 5400000 joules
3. We have to convert this into kwh:
(i) 1 kilowatt = 1000 watts = 1000 joules per second
• 1 hour = 3600 seconds
(ii) So energy of 1 kwh = (1000 3600) = 3600000 joules 
(iii) So 1 joule = ¹⁄_3600000 kwh
(iv) So 5400000 joules = (5400000 × ¹⁄_3600000) = (⁵⁴⁄₃₆) = 1.5 kwh

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• The kwh meter is installed by the distribution companies. 
• It directly shows the consumption in kwh
• So we need not calculate the consumption in joules
• Some images can be seen here.

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Solved example 12.2
In a house, in a day,  
• 5 CF lamps each of 20 W, work for 4 hours
• 4 fans each of 60 W, work for 5 hours    
• 1 TV of 100 W, works for 4 hours
What will be the consumption shown by the kwh meter per day?
Solution:
1. Power consumption of 1 CF lamp = ^20×4⁄₁₀₀₀ = 0.08 kwh
∴ Power consumption of 5 CF lamps = 0.08 × 5 = 0.4 kwh = 0.4 units
2. Power consumption of 1 fan = ^60×5⁄₁₀₀₀ = 0.3 kwh
∴ Power consumption of 4 fans = 0.3 × 4 = 1.2 kwh = 1.2 units
3. Power consumption of 1 TV = ^100×4⁄₁₀₀₀ = 0.4 kwh = 0.4 units
4. Total consumption = 0.4 + 1.2 + 0.4 = 2 units

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In the next section, we will see the details of a 3 pin plug

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