Physics Lesson 15.6.2 - Input and Output Electric Power. Joule's Law

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Welcome to our Physics lesson on Input and Output Electric Power. Joule's Law, this is the second lesson of our suite of physics lessons covering the topic of Electric Power and Efficiency, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.

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

P_output = Work/time

and

P_intput = 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/E_tot × 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 = P_output/P_intput × 100%

The two above formulae are equivalent because

e = P_output/P_intput × 100%
= W/t/E_tot/t × 100%
= W/E_tot × 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.

P_intput = E_tot/t

Giving that

P_intput = I ∙ ε

we obtain

I ∙ ε = E_tot/t

E_tot = I ∙ ε ∙ t

where E_tot 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

P_output = 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 = P_output/P_intput × 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.

Physics Tutorials: This image provides visual information for the physics tutorial Electric Power and Efficiency

The internal resistance of the power source is r = 1 Ω while the resistance of the conducting wire of the circuit is R_w = 2 Ω. Calculate:

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

Solution 2

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

1/R^p = 1/R₁ + 1/R₂
= 1/15 + 1/30
= 2 + 1/30
= 3/30
= 1/10

Therefore, the equivalent resistance of the parallel branch is

R^p = 10/1 Ω
= 10 Ω

The total resistance in the circuit therefore is

R_tot = R_eq + R_w + r
= 10Ω + 2Ω + 1Ω
= 13Ω

The current in the main branch is

I = ε/R_tot
= 221 V/13 Ω
= 17 A

The total power delivered by the source is
P_tot = I ∙ ε
= 17A ∙ 221V
= 3757 W
To find the useful power we must find the potential difference of the parallel branch. We have
∆V = ε - I ∙ R_w - I ∙ r
= 221V - 17A ∙ 2Ω - 17A ∙ 1Ω
= 221V - 34V - 17V
= 170V
Therefore, the useful power of the circuit is
P_useful = I ∙ ∆V
= 17A ∙ 170V
= 2890 W
Two minutes are equal to 120 seconds. Thus, the electric energy produced by the source in 2 minutes is
E_tot = P_tot ∙ t
= 3757 W ∙ 120 s
= 450 860 J
Heat energy dissipated by the resistors during the 120 seconds is the useful energy of the circuit. It is
Q = W = P_output ∙ t
= 2890 W ∙ 120 s
= 346 800 J
Efficiency of the circuit is calculated using any of the formulae provided earlier. Thus,
e = P_output/P_intput × 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.

You have reached the end of Physics lesson 15.6.2 Input and Output Electric Power. Joule's Law. There are 4 lessons in this physics tutorial covering Electric Power and Efficiency, you can access all the lessons from this tutorial below.

More Electric Power and Efficiency Lessons and Learning Resources

Electrodynamics Learning Material
Tutorial ID	Physics Tutorial Title	Revision Notes	Revision Questions
15.6	Electric Power and Efficiency
Lesson ID	Physics Lesson Title	Lesson	Video Lesson
15.6.1	Mechanical Power Recap. The Meaning of Electric Power
15.6.2	Input and Output Electric Power. Joule's Law
15.6.3	Joule or kW-h?
15.6.4	Other Forms of Writing the Formula of Electric Power

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