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.
• 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.
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
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
• 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
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
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
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
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.
• 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.
• 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.
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
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
• 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
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
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
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
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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