Chapter 10.3 - Electric Power

In the previous section we completed a discussion on the 'lighting effect of electric current'. In this section, we will see 'electric power'.

• We have seen in earlier classes that, power is: Work done in unit time
• But 'work done' is same as 'energy consumed'. Both have the same unit joule
• So another way of defining power is: Energy consumed in unit time
We have seen those details here.

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• Now, in our present discussion, consider any electric appliance. 
• We know that, it will consume electric energy. By consuming energy, it does work. 
■ How much work does it produce? OR, How much energy does it consume? 
We cannot give an exact answer. Because, it depends on time. 
• If it works for a long time, we say: It does a lot of work for us
    ♦ Same as: It consumes a lot of energy from us
• If it works for a short time, we say: It does only a little work for us 
    ♦ Same as: It consumes only a little energy from us
■ So time is an important factor.
• 'Work done per unit time' is what matters more.
• Some heavy appliances can do more work per unit time. 
    ♦ Unit time can be taken as one second, one minute, one hour etc.,
    ♦ Usually we take  'one second'
• Some lighter appliances can do only a less work per unit time. 

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Based on the above, we can derive an equation:
• Let an appliance consume electric energy for 't' seconds
• As a result, let it produce a work of 'H' joules
■ Then the power of that appliance is given by: P = ^H⁄_{t
• The unit of power is Watt}

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We will apply this to a heating coil. We will write it in steps:
1. Let us give electrical energy to that heating coil, for 't' seconds.
• Then, based on joules law, it will produces a work (in the form of heat) given by H = I²Rt
[Here, 'heat energy produced' can be considered as 'work'. Because, that heat can be used to produce steam, which can turn a small turbine for example] 
2. So in 1 second, it produces a work of ^I2Rt⁄_t 
• 't' in numerator and denominator cancels each other. Thus we can write: 
Eq.10.4: P = I²R

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• The equation P = I²R gives us the relation between power and the quantities: [current, resistance]
• Can we relate power to voltage? Let us try:
• From ohm's law, we have: V = IR (Details here)
• From this we get: I = ^V⁄_R
• So we can use (^V⁄_R) instead of I
• Thus we get: P = (^V⁄_R)²R ⟹ P = (^V2⁄_R²)R =  ^V2⁄_R
So we can write:
Eq.10.5: P = ^V2⁄_R 

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Another derivation:
• From ohm's law, we have: V = IR
• From this we get: R = ^V⁄_I
• So we can use (^V⁄_I) instead of R
• Thus we get: P = I²(^V⁄_I) ⟹ P = VI
So we can write:
Eq.10.6: P = VI

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Now we will see a solved example:
Solved example 10.3:
A heating appliance has a resistance of 115 Ω. If 2 A current flows through it, what is the power of the appliance?
Solution:
1. Given: R = 115 Ω, I = 2 A
2. We want a relation that will connect the power (P) to: [R, I]
• So we will use Eq.10.4: P = I²R
3. Substituting the values, we get: P = 2² × 115 = 460 W

Solved example 10.4
An appliance of power 540 W is used in a branch circuit. If the voltage is 230 V, What is the amperage?
Solution:
1. Given: P = 540 W, V = 230 volts
2. We are asked to find the amperage (I). So we want a relation that will connect the power (P) to: [V, I]
• We will use Eq.10.6: P = VI
3. Substituting the values, we get: 540 = 230 × I ⟹ I = ⁵⁴⁰⁄₂₃₀ = 2.3478 A
• The next whole number should be taken as the amperage. So we get:
Amperage = 3 A

Solved example 10.5
A current of 0.4 A flows through an electric bulb working at 230 V. What is the power of the bulb?
Solution:
• Given: When the bulb is connected to a circuit and a voltage of 230 V is applied, a current of 0.4 A flows through it.
• If we apply less than 230 V, only a low current will flow. This will be less than the 'current specified by the manufacturer' of the bulb. 
• Because of such a low current, the filament will not get heated to the required temperature, and so we will get only a dim light.
• Thus, 0.4 A is the 'rated current' of the bulb. For this current, 230 V is essential.
• This current and voltage are related to the resistance of the bulb by the relation V = IR (Details here)
• For solving this problem however, we do not need to write the above details
1. Given: I = 0.4 A, V = 230 volts
2. We are asked to find the power. So we want a relation that will connect the power (P) to: [V, I]
• We will use Eq.10.6: P = VI
3. Substituting the values, we get: P = 230 × 0.4  = 92 W

Solved example 10.6
A device of resistance 690 Ω is working at 230 V. What is the power of the device?
Solution:
1. Given: R = 690 Ω, V = 230 volts
2. We are asked to find the power. So we want a relation that will connect the power (P) to: [R, V]
• So we will use Eq.10.5: P = ^V2⁄_R
3. Substituting the values, we get: P = ²³⁰²⁄₆₉₀ = 76.67 W

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Now we will see the 'changes in power' when changes occur in other factors: 
• resistance • current • voltage
1. First we will see the 'changes in power' when 'change in resistance and current' occur:
Consider a bulb filament which is broken at a point along it's length. 
• The filament is broken into two unusable pieces. 
2. If we join the two pieces together, current can flow again. 
• But when the joining is done, the length of the filament will decrease. 
3. 'Decrease in the length of a conductor' will decrease the resistance
• Because resistance is directly proportional to length (Details here)
4. Now, if the resistance decrease, the current will increase. 
• This is clear from the ohm's law: V = IR (Details here)
An example:
■ If the voltage is constant, calculate the current when resistance is halved
Solution: 
(i) Let the constant voltage be V
(ii) Let the initial current and resistance be I₁ and R₁ respectively 
• Then we can write: V = (I₁ × R₁)
(iii) Let the new current and resistance be I₂ and R₂ respectively 
• Then we can write: V = (I₂ × R₂)
(iv) Since voltage is constant at V, we can equate the results in (ii) and (iii):
(I₁ × R₁) = (I₂ × R₂)
(v) Given that R₂ = ^R₁⁄₂
• Substituting this in (iv), we get:
(I₁ × R₁) = (I₂ × ^R₁⁄₂) ⟹ (I₁) = (^I₂⁄₂) ⟹ I₂ = 2I₁
• We can write: When resistance is halved, current becomes double  
5. We also have the Eq.10.4: P = I²R, which relates power to current
■ From this equation, we see that, when current increases, power increases   
So we can write:
■ When resistance decreases, current increases and so, power also increases
6. Next we will see the 'changes in power' when 'change in voltage' occurs:
• We have the Eq.10.5: P = ^V2⁄_R, which relates power to voltage
■ From this equation, we see that, when voltage increases, power increases

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Solved example 10.7
An instrument is marked 150 W, 230 V. If the voltage is lowered to 110 V, what will be it's power?
Solution:
1. Given: P = 150 W, V = 230 volts
2. The manufacturer knows that, if the customer supplies a voltage of 230 V, that instrument will do a work of 150 joules in one second (∵ power is marked as 150 W)
• How is the manufacturer so sure about it?
• Ans: Because he has fitted a fixed resistance (R) in that instrument
3. So we want an equation which relates power to [voltage, resistance]
• We can use Eq.10.5: P = ^V2⁄_R
• Substituting the values, we get: 150 = ²³⁰²⁄_R ⟹ R = ²³⁰²⁄₁₅₀ = 352.67 Ω 
4. So the manufacturer has fitted a resistance of 352.67 Ω in that instrument. This resistance value will not change.
• Now, if we supply only 110 volts (instead of 230), the instrument will not give us a work of 150 joules in one second 
• It will be less than 150. How much will that be?
• Ans: we again use Eq.10.5: P = ^V2⁄_R
• We get: P = ¹¹⁰²⁄_352.67 = 34.31 W
5. Now we know the reason why some instruments slow down when voltage is low.

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In the next section we will see a few more solved examples.

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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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