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Category: Sample Data

The watt hour is a unit of energy equal to 3600 joules

Examples
  • A heater, rated at 1000 watts (1 kilowatt), operating for one hour uses one kilowatt hour 
  • Using a 60 watt light bulb for one hour consumes 0.06 kilowatt hours of electricity. Using a 60 watt light bulb for one houndres hours consumes 6 kilowatt hours of electricity.
  • If a 100 watt light bulb is on for one hour per day for 30 days, the energy used is

100 W × 30 h = 3000 W/h = 3 kW/h.The International System of Units (SI) 
1000 Watt = 1 Kilowatt
1000 Kilowatt = 1 Megawatt
1000 Megawat = 1 Gigawatt
1000 Gigawatt = 1 Terawatt

The Watt, kiloWatt and kiloWatt/hour explained 

 

The Watt kiloWatt explainedEnergy released by electricity is measured in watts. In a home every appliance has an electricity rating. This, usually found on the back of the appliance, tells the user how many watts that particular appliance uses in an hour. (If, for example, a lamp uses 100 Watts, it will use this amount of electricity in an hour.)
 
One-thousand Watts (1 000 W) equals one kiloWatt (1kW). When you are charged for electricity, your bill will usually tell you how many units were used during the month. One unit of electricity is exactly equal to 1 000 watts of power used for 1 hour- or kWh, which suppliers use to determine your costs. (Your lamp using 100 Watts will therefore use 0.1 kW in an hour when it’s usage is monitored by your home’s electricity meter.)
 
To determine what the lamp costs you to run, you simply perform the following calculation: unit cost x power consumption in Watts divided by 1 000A 100 watt lamp in a house paying 85.39 cents per unit (kWh) would cost:85.39 cents x 100w, divided by 1 000 = 8.54 cents an hour.

Volts, Amps and Ohms – Measuring Electricity 

 

OhmThe three most basic units in electricity are voltage (V), current (I) and resistance (r).
 
Voltage is measured in volts, current is measured in amps and resistance is measured in ohms. A neat analogy to help understand these terms is a system of plumbing pipes. The voltage is equivalent 
to the water pressure, the current is equivalent to the flow rate, and the resistance is like the pipe size.

I = V/r

 

Voltage is measured in volts, current is measured in amps and resistance is measured in ohms. The ohm (symbol: Ω) is the SI derived unit of electrical resistance
Let's see how this relation applies to the plumbing system. 
Let's say you have a tank of pressurized water connected to a hose that you are using to water the garden.What happens if you increase the pressure in the tank? You probably can guess that this makes more water come out of the hose. The same is true of an electrical system: Increasing the voltage will make more current flow.Let's say you increase the diameter of the hose and all of the fittings to the tank. You probably guessed that this also makes more water come out of the hose. This is like decreasing the resistance in an electrical system, which increases the current flow.

 

Electrical power is measured in watts.

 

In an electrical system power (P) is equal to the voltage multiplied by the current.P = VI
The water analogy still applies. Take a hose and point it at a waterwheel like the ones that were used to turn grinding stones in watermills. You can increase the power generated by the waterwheel in two ways. If you increase the pressure of the water coming out of the hose, it hits the waterwheel with a lot more force and the wheel turns faster, generating more power. If you increase the flow rate, the waterwheel turns faster because of the weight of the extra water hitting it.

Electrical Efficiency 

 

flowElectrical systems are more efficient when a higher voltage is used to reduce current.
In an electrical system, increasing either the current or the voltage will result in higher power. Let's say you have a system with a 6-volt light bulb hooked up to a 6-volt battery. The power output of the light bulb is 100 watts. Using the equation above, we can calculate how much current in amps would be required to get 100 watts out of this 6-volt bulb.You know that P = 100 W, and V = 6 V. So you can rearrange the equation to solve for I and substitute in the numbers.
I = P/V = 100 W / 6 V = 16.66 amps
What would happen if you use a 12-volt battery and a 12-volt light bulb to get 100 watts of power?
100 W / 12 V = 8.33 amps