2 The Watt is the SI unit of Electrical Power. It is the “work done” by electricity, and is a value of work generated by 1 Amp flowing at a voltage of.

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Presentation transcript:

2 The Watt is the SI unit of Electrical Power. It is the “work done” by electricity, and is a value of work generated by 1 Amp flowing at a voltage of 1 Volt. Therefore 1 Watt = 1 Volt multiplied by 1 Amp Or W = V I In mechanical terms, a Watt is the work required to move an object at 1 meter per second against a force of 1 Newton. W = V I

3 Now we know that Watts are a measurement of Electrical Power, we can note their uses everywhere. Such examples are lamps (light bulbs), motors and heaters which all have an electrical rating in Watts. The higher the wattage, the more electrical power the item will consume. One Kilowatt is One Thousand Watts, and is a typical rating for electric motors. 1 Kilowatt (KW) = 1.34 Horsepower (Hp) The consumption of Electricity is measured in KW/Hour. This is the amount of electrical power used in a specific hour.

4 Power and energy are frequently confused. Power is the rate at which energy is generated and consumed. For example, when a light bulb with a power rating of 100W is turned on for one hour, the energy used is 100 watt-hours (W·h), 0.1 kilowatt-hour, this is not the energy used to light the bulb, just the power consumed to keep the bulb lit for that period. This same amount of energy would light a 40-watt bulb for 2.5 hours, or a 50-watt bulb for 2 hours. 40 X 2.5 = 100 Watt hours and 50 X 2 = 100 Watt hours. A power station would be rated in multiples of watts, but its annual energy sales would be in multiples of watt-hours. A kilowatt-hour is the amount of energy equivalent to a steady power of 1 kilowatt running for 1 hour, the actual energy for this would be 3.6 Mega Joules.

5 A Simple battery torch Bulb = 6 Watt As we know Power (P) = V x I, We can also work out that from the triangle above, I =P/V and V = P/I Therefore the torch above would generate I = P/V = 6/3 = 2 amps in the circuit above.

6 Looking at the 2 triangles above we can see that V= I X R and P = V x I. So we can rightly assume P = (I x R) x I which = I x I x R or P = IR Also R = V/I and V = P/I, so R = P/ I x I or P/I and so on.

7

8 Power and Ohms law formulas are needed in a variety of everyday uses by electrical engineers. Examples of this include the calculation for over current and fuse protection, calculation for the size of wiring to use for new equipment, and for design and purchasing when you need the maximum allowed value of equipment for a given installation.