# Electric Power and Efficiency | iCalculator™

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15.6Electric Power and Efficiency

In this Physics tutorial, you will learn:

• What is electric power and how is it related to the mechanical power?
• What is the difference between input and output power?
• How to calculate the efficiency of an electric circuit
• What does Joule's Law say on electric energy?
• What unit is used to measure the electric energy consumed?
• How to calculate the electric power and electric energy?

## Introduction

What does the number 1200W stamped on the plate of an electric appliance mean for you?

Do you think there is any relationship between mechanical power we have discussed in tutorial 5.5 "Power and Efficiency" and electric power?

Which electric devices are more powerful? Is it necessary for electric appliances to move objects in order to consider them as powerful? What other element determines the power of an appliance?

In this tutorial, we will discuss extensively the concept of electric power. Conceptually, it is similar to mechanical power; just the approach to find the formula is slightly different.

## Mechanical Power Recap. The Meaning of Electric Power

In tutorial 5.5 "Power and Efficiency" we have discussed the concept of mechanical power which is the work done by an object or system in the unit of time. Mathematically,

P = W/t

Given that work is the force F needed to move an object by Δx metres, we obtain for mechanical power

P = F ∙ ∆x/t
= F ∙ ∆x/t
= F ∙ v

where v is the velocity of the moving object.

However, work does not imply only being able to move an object by some linear distance. It also means delivering energy in the form of heat. This occurs in thermal systems where the power of such systems is evaluated by their ability to deliver heat energy in a given time.

The unit of power is [J/s] or watt [W]. However, when expressed in terms of fundamental SI units, the unit of power becomes

[W] = [J/s] = [N ∙ m/s] = [kg ∙ m/s2 ∙ m/s] = [kg ∙ m2/s3 ]

Likewise, electric power is the work done by an electric source during a given time. Given that the strength of an electric source is determined by its electromotive force, we must express the power delivered by such a source in terms of the work done by the source to make the charges move through a circuit. Obviously, this work does not depend only on the strength of the source but also on the length of the path along which the electric charges will move to do the required work. In other words, the role of displacement in the mechanical power here is played by the length of the circuit (the length of conducting wire), while the force necessary to move the charges is provided by the electric source, which has a certain electromotive force.

We have explained earlier that the electromotive force of an electric source represents the work done by this source to make the charges flow through the circuit, i.e.

emf = W/Q
= F ∙ L/Q

where L is the length of the conducting wire and Q is the charge. Therefore, the force provided by the source is

F = emf ∙ Q/L

Henceforth, we will denote the electromotive force by ε instead of emf. In this way, the formula of electric force provided by the source becomes

F = ε ∙ Q/L

Since the electric power is similar in concept to mechanical power, i.e. it represents the work done by the electric source in the unit of time, we obtain (after substituting Δx in the mechanical power with the length L of conductor):

P = F ∙ L/t
= ε ∙ Q/L ∙ L/t
= ε ∙ Q/t

From the definition of electric current I, we have

I= Q/t ⇒ Q = I ∙ t

Thus, we obtain after making these substitutions in the previous formula:

P = ε ∙ I ∙ t/t

or

P = ε ∙ I

Hence, the formula of electric power requires knowing only the electromotive force of the source and the electric current flowing through the circuit.

### Example 1

What is the power of the electric source shown in the figure? ### Solution 2

First, we find the equivalent resistance of the circuit. The resistance of the parallel branch is

1/Rp = 1/R1 + 1/R2
= 1/12 + 1/20
= 5/60 + 3/60
= 8/60

Thus,

Rp = 60/8 Ω
= 7.5 Ω

The equivalent resistance of the circuit therefore is

Req = Rp + R3
= 7.5Ω + 3Ω
= 10.5Ω

The current flowing in the circuit is

I= ε/Req
= 21 V/10.5 Ω
= 2 A

From here, we can calculate the power of the circuit:

P = I ∙ ε
= 2A ∙ 21 V
= 42 W

## Input and Output Electric Power. Joule's Law

Again, we must recall the concepts of output and input mechanical power explained in tutorial 5.5. We have explained that not all the energy produced by the source is used for doing work. Some part from the total energy is lost on the way, i.e. it converts into other forms of energy. This brings a decrease in the power of machine. We called the work done in a certain time as useful or output power. This is faction of the total or input power that is obtained by dividing total energy by time. We have

Poutput = Work/time

and

Pintput = Total Energy/time

The quality of a machine is largely determined by the part of total energy absorbed that it is able to convert into useful work. When expressed in percentage, this faction of total energy is known as efficiency (e) of machine, i.e.

e = W/Etot × 100%

This result, however, can be obtained by dividing the output and input power as well, and then write the result as a percentage, i.e.

e = Poutput/Pintput × 100%

The two above formulae are equivalent because

e = Poutput/Pintput × 100%
= W/t/Etot/t × 100%
= W/Etot × 100%

Similarly, when considering an electric source, the energy delivered by it represents the total or input energy. When dividing it to the time of operation, we obtain the total or input power we have discussed in the previous paragraph, i.e.

Pintput = Etot/t

Giving that

Pintput = I ∙ ε

we obtain

I ∙ ε = Etot/t

or

Etot = I ∙ ε ∙ t

where Etot is the total energy produced by the source.

The last formula is known as Joule's Law. It calculates the total energy delivered by an electric source during a given time. However, not all this energy goes for doing work. Some of the electricity is lost on the way to reach the appliance because of the existence of internal resistance of the source and conducting wire, which decreases the efficiency of circuit. An ideal circuit would have an efficiency of 100%, i.e. all the energy produced by the source was used by the appliance to do work. This is not possible, so we must account this element when calculating the work done by any circuit component.

The output power of an appliance is calculated by

Poutput = I ∙ ∆V

where ΔV is the potential difference across the terminals of appliance. We have explained earlier that ΔV < ε in every situation. Therefore, the efficiency of all electrical appliances is smaller than 100% as expected. We have

e = Poutput/Pintput × 100%
= I ∙ ∆V/I ∙ ε × 100%
= ∆V/ε × 100%

Also, we can write Joule's Law for work done by electrical appliances:

W = I ∙ ∆V ∙ t

For electric heaters in which the electric energy is converted into heat, we replace W by Q (do not confuse the symbol Q used for heat with Q used for the electric charge in the previous chapter; they express different things). Therefore, Joule's Law for electric heaters is written as

Q = I ∙ ∆V ∙ t

### Example 2

An electric circuit powered by a 221 V source containing two resistors, R1 = 15 Ω and R2 = 30 Ω connected in parallel is shown in the figure below. The internal resistance of the power source is r = 1 Ω while the resistance of the conducting wire of the circuit is Rw = 2 Ω. Calculate:

1. Total power delivered by the source
2. Useful power of the circuit
3. Electric energy produced by the source in 2 minutes
4. Heat energy dissipated by the resistors during this time
5. Efficiency of the circuit

### Solution 2

First, we must find the equivalent resistance of the parallel setup. We have

1/Rp = 1/R1 + 1/R2
= 1/15 + 1/30
= 2 + 1/30
= 3/30
= 1/10

Therefore, the equivalent resistance of the parallel branch is

Rp = 10/1 Ω
= 10 Ω

The total resistance in the circuit therefore is

Rtot = Req + Rw + r
= 10Ω + 2Ω + 1Ω
= 13Ω

The current in the main branch is

I = ε/Rtot
= 221 V/13 Ω
= 17 A
1. The total power delivered by the source is
Ptot = I ∙ ε
= 17A ∙ 221V
= 3757 W
2. To find the useful power we must find the potential difference of the parallel branch. We have
∆V = ε - I ∙ Rw - I ∙ r
= 221V - 17A ∙ 2Ω - 17A ∙ 1Ω
= 221V - 34V - 17V
= 170V
Therefore, the useful power of the circuit is
Puseful = I ∙ ∆V
= 17A ∙ 170V
= 2890 W
3. Two minutes are equal to 120 seconds. Thus, the electric energy produced by the source in 2 minutes is
Etot = Ptot ∙ t
= 3757 W ∙ 120 s
= 450 860 J
4. Heat energy dissipated by the resistors during the 120 seconds is the useful energy of the circuit. It is
Q = W = Poutput ∙ t
= 2890 W ∙ 120 s
= 346 800 J
5. Efficiency of the circuit is calculated using any of the formulae provided earlier. Thus,
e = Poutput/Pintput × 100%
= I ∙ ∆V/I ∙ ε × 100%
= ∆V/ε × 100%
= 170 V/221 V × 100%
= 10/13 × 100%
= 76.923%

This result means about 23% of the energy produced by the source is wasted during the way to the resistors.

## Joule or kW-h?

You may have realized in the above example that the values of energy are considerable. Only in two minutes of operation, the values of energy produced by the source were several hundreds of thousands of Joules. On the other hand, a home circuit contains a lot of electric appliances operating simultaneously and the electricity bill is paid once a month. If the electric energy consumed were calculated in Joules, the number showing the value of energy consumption would be too large. Therefore, it is better to use a more suitable unit of energy to make the work of the employees who deal with electricity bill easier but also to make the consumer to easily understand how much he has to pay. This unit is known as kilowatt - hour (kW-h). Kilowatt- hour means kilowatts multiplied by hour. Given that 1 kW = 1000 W and 1 hour = 60 ∙ 60 = 3600 s, the conversion factor between kW-h and Joule is

1 kWh = 1kW ∙ 1h
= 1000 W ∙ 3600s
= 3 600 000 J

A normal family consumes from a few hundreds to a few thousands kWh electrical energy in a month. This makes the electric board meter easy to read and you could complete these calculations for your own home.

### Example 3

A family electric board meter reads 034604.5 kWh on April 1 and 035179.3 kW-h on April 30. If the price is ₤0.2/kWh, what is the energy in Joules consumed during the month of April and what is the electricity bill the owner has to pay?

### Solution 3

First, we find the energy consumed in kW-h by subtracting the final and the initial value the electric board meter reads. We have

= 035179.3 kWh - 034604.5 kWh
= 574.8 kWh

When converted into Joules, this value becomes

E(J) = E(kWh) ∙ 3 600 000 J/kWh
= 574.8 kWh ∙ 3 600 000 J/kWh
= 2 069 280 000 J

The electricity bill is calculated by considering the electricity in kWh consumed. We have

Bill = Amount (in kWh) ∙ Price per kWh
= 574.8 kWh ∙ ₤0.2/kWh
= ₤114.96

## Other Forms of Writing the Formula of Electric Power

From Ohm's Law, we know that

∆V = I ∙ R

for a certain appliance and

ε = I ∙ Rtot

for the whole circuit. Substituting these two formulae in the equations of output and input power, we obtain

Poutput = I ∙ ∆V
= I ∙ I ∙ R
= I2 ∙ R

and

Pintput = I ∙ ε
= I ∙ I ∙ Rtot
= I2 ∙ Rtot

These formulae are used when current and resistances are given but not the potential difference of electromotive force.

On the other hand, when potential difference (or emf) and resistance (but not the current) are given, we write

Poutput = I ∙ ∆V
= ∆V/R ∙ ∆V
= ∆V2/R

for a certain appliance and

Pintput = I ∙ ε
= ε/R ∙ ε
= ε2/R

for the whole circuit.

Multiplying the above expressions by the time t, we obtain the work done (the useful energy consumed) by the appliance and the total energy delivered by the source in terms of resistance R:

W = I2 ∙ R ∙ t
W = ∆V2/R ∙ t
Etot = I2 ∙ Rtot ∙ t

and

Etot = ∆V2/Rtot ∙ t

### Example 4

An electric heater has 1200 W of output power. If the heater operates at 240 V, what is the resistance of heater?

### Solution 4

In this exercise, the current and resistance are not given, so we can use the formula

Poutput = ∆V2/R

to calculate the resistance, where ΔV = 240 V and Poutput = 1200 W.

Hence, we have

R = ∆V2/Poutput
= (240 V)2/1200 W
= 57600 V2/1200 W
= 48 Ω

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