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Electric Power and Efficiency

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15.6Electric Power and Efficiency


In these revision notes for Electric Power and Efficiency, we cover the following key points:

  • 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?

Electric Power and Efficiency Revision Notes

Electric power is the work done by an electric source in the unit time.

There are two types of electric power: (1) the total (or input) power, i.e. the power generated by the source; it is calculated by

Pintput = I ∙ ε

where ε is the electromotive force of the source and I is the current in the circuit, and (2) the useful (or output) power, i.e. the power generated by an operating appliance; it is calculated by

Poutput = I ∙ ∆V

where ΔV is the potential difference between the terminals of the appliance.

The unit of power is Watt. 1 Watt = 1 Joule/second. In terms of fundamental SI units, the unit of power is:

1[W] = 1[kg ∙ m2/s3 ]

Since Useful (output) Power = Work/Time and Total (input) Power = Total Energy/Time, we obtain

W = I ∙ ∆V ∙ t

and

Etot = I ∙ ε ∙ t

The last formula is known as Joule's Law. It calculates the total energy delivered by an electric source during a given time.

Not all the electric energy produced by the source goes for doing work. Some of this 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.

Therefore, the efficiency of all electrical appliances is smaller than 100% as expected. We have

e = Poutput/Pintput × 100%

Joule's Law for electric heaters is written as

Q = I ∙ ∆V ∙ t

where Q is the heat energy delivered by the heater's resistor.

Joule is not suitable to use as a unit of electrical energy s it is too small. 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, another unit of electrical energy known as kilowatt - hour (kW-h) is used. 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=3 600 000 J

Other forms of writing the electric power formula written when combining Ohm's Law and the meaning of electric power include:

Poutput = I2 ∙ R
Pintput = I2 ∙ Rtot
Poutput = ∆V2/R
Pintput = ε2/R

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

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