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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Chapter 4.4 - Relation between Energy and Power

In the previous section we saw the Law of conservation of energy. In this section we will see the basic details about Power.

■ Consider two labourers doing a certain work. 
• The first labourer has to load 50 bricks onto a platform at a height h. 
• The second labourer has to load another 50 bricks onto the same platform at height h.
• So the work to be done are same for both the labourers.
■ Now, out of the two labourers one is stronger than the other. 
• But based on what we have learned so far, there is no way to give any 'special consideration' for him. Because, both will be doing the same joules of work. 
■ So we introduce a new term power. 
• We have to calculate the power of each labourer separately. 
• Then we can give 'special consideration' for the labourer who shows greater power. 

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So we want to know how the power is calculated. Let us see:
1. Let one of the labourer start to load the bricks. 
• The moment when he starts to do the work, a stop watch should be started. 
2. Note the time when all 50 bricks are loaded. 
• Thus we get the time t1 required by the first labourer to do the work. 
3. Repeat the process for the second labourer. Let the time required by him be t2
4. We know that work done by both is the same. Let this work be W joules.
Then we get:
• Power of the first labourer = ^W⁄_t1
• Power of the second labourer = ^W⁄_t2

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So, to calculate the power, we are dividing the work by time. 

■ That means, power is the work done in unit time. 

We can write an equation:

Eq.4.3:
Power p = ^W⁄_t
• Note that, time t is in the denominator. So, if t decreases, power increases. 
• In the above example, the stronger labourer will most probably complete the work in a lesser time. 
• Thus we would be able to say that he has greater power.

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In the same way,

■ Consider a distance d
• A car having greater power travels this distance in time t1
• An ordinary car travels this same distance in time t2
• t1 will be less than t2
■ Consider two heaps of sand. Each having the same volume v
• An earth mover having greater power will carry away it's heap in t1
• An ordinary earth mover will carry away it's heap in time t2
• t1 will be less than t2

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Next we want a unit for power. 

■ We have seen that p = ^W⁄_t . 

• The unit of work is joules. 

• The unit of time is seconds. 

• So the unit of p is joules per second. It is written as J s^-1. 

• There is a special name for this J s^-1. It is watt. It's short form is W. This name is given in honour of the British scientist James Watt, who invented the Steam engine. 

■ We can say that power of an engine is 1 W, if it delivers 1 joule of work in every second. 
• We know that, to deliver work, an object has to consume energy. 
    ♦ The energy may be the chemical energy in petrol, coal etc., or 
    ♦ Electrical energy, heat energy etc., 
■ If an object consumes 1 joule of energy in every second, then also we say that, the power of that object is 1 W.

■ For expressing larger powers, we use larger forms of watt:

• 1 kilowatt = 1000 watts. In short form, it is: 1 kW = 1000 W
• If the power of an object is 1 kW, it consumes or delivers 1000 joules in every second.

Average power

• Consider an engine having a power of 500 W. Let it work for 2 hours. 
• We cannot expect it to give 500 joules every second. There may be variations. 
• It may give greater than 500 for some time duration, and less than 500 for some other time duration.
• In such cases, we first calculate the total work done in the duration of 2 hours. 
    ♦ Then we divide this total work by 2 × 3600 = 7200 seconds. 
• The result that we get is called the Average power. 
■ So average power can be defined as: ^{Total work}⁄_{Total time}

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Now we will see a solved example

Solved example 4.9
Two workers: A and B carry a weight of 350 N each to The second floor of a building, which is 6 m above the ground level. A takes 40 seconds and B takes 55 seconds to do the task. What is the power expended by each worker?
Solution:
1. Work done = Force × displacement = Weight × height
Weight = mg. It is given as 350 N
2. So work done = mgh = 350 × 6 = 2100 J
Both A and B do this same work. But the time taken is different
3. We have: Power p =  ^W⁄_t.
• So power expended by A = ²¹⁰⁰⁄₄₀ = 52.5 J s^-1 = 52.5 watts    
• Power expended by B = ²¹⁰⁰⁄₅₀ = 42 J s^-1 = 42 watts

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• So far in this section, we have been discussing about power. We saw the unit of power also. 
• Now we are going to revisit energy. In fact we are going to discuss about a unit of energy. It is called 'kilowatt hour'. It's symbol is kW h. Let us analyse this kW h:
1. We know that W is a unit for power. It is one joule work in one second. That is., one joule per second
2. So kiloWatt (kW) is 1000 joules in one second. That is 1000 joules per second
3. 'per second' means that the time (in seconds) is in the denominator
• That is., 1 kW = ^{1000 joules}⁄_{1 second}.
4. In kW h, we are not saying 'per hour'. That means the time (in hours) is in the numerator
5. So we can write: 1 kW h = [(^{1000 joules}⁄_{1 second} )×1 hour] = [(^{1000 joules}⁄_{1 second} )×3600 seconds]
• The seconds in numerator and denominator cancels out. What we get is:
1 kW h = 1000 × 3600 joules = 3600000 joules = 3.6 × 10⁶ J   
• So, kWh is a unit of energy. It is not a unit of power
■ 1 kW h is the total energy delivered by a machine in one hour, if it delivers 1000 joules in every second
OR
■ 1 kW h is the total energy consumed by a machine in one hour, if it consumes 1000 joules in every second

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So 1 kWh is a large quantity of energy. It is 36 followed by five zeroes. This large quantity of energy is used as '1 unit' for measuring electrical energy consumed in our homes, schools, offices etc.,

Let us see an example:

1. In an electricity bill, the following two details are written:

(a) Previous reading: 13541 in the electric meter

(b) Present reading: 13892

2. Calculate the difference: 

It is equal to: 13892 - 13541 = 351

3. So 351 kWh of electrical energy was consumed during the time between the following two points:

(a) The instant when previous reading was taken

(b) The instant when present reading is taken

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Now we will see some solved examples

Solved example 4.10

An electric bulb of 40 W is used for 5 hours per day. Calculate the ‘units’ of energy consumed in one day by the bulb.

Solution:

1. 'One W' is 'one joule' consumed every second. 

2. So 40 W is 40 joules consumed every second

3. So in one hour it will consume 40 × 60 × 60 = 144000 joules

4. So in 5 hours it will consume 144000 × 5 = 720000 joules

5. 1 unit = 1 kWh = 3600000 joules

6. So 'No. of units' contained inside 720000 joules =  ⁷²⁰⁰⁰⁰⁄_3600000 = 0.2

7. Thus we can write: When a 40 W bulb is used for 5 hours, 0.2 units of energy will be used up.

Solved example 4.11

Find the energy in kWh consumed in 6 hours by three devices of power 300 W each.

Solution:

Power of one device = 300 W

1. So it will consume 300 joules in every second

2. That means, it will consume 300 × 3600 = 1080000 joules in one hour 

3. So in 6 hours it will consume 6 × 1080000 = 6480000 joules

4. So three such devices will consume 3 × 6480000 = 19440000 joules in 6 hours

• One kWh = 3600000 joules.

5. So 19440000 joules = ^19440000⁄_3600000 = 5.4 kWh = 5.4 units

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In the next section, we will see Sound and Wave motion. 

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