## 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

**To determine what the lamp costs you to run, you simply perform the following calculation:**

**unit cost x power consumption in Watts**divided by

**1 000**A 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

to the water pressure, the current is equivalent to the flow rate, and the resistance is like the pipe size.

**I = V/r**

**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**.

**P**) is equal to the voltage multiplied by the current.

**P = VI**

## Electrical Efficiency

**Electrical systems are more**

**efficient when a higher voltage****is used to reduce current.****P**= 100 W, and

**V**= 6 V. So you can rearrange the equation to solve for

**and substitute in the numbers.**

**I****I = P/V = 100 W / 6 V = 16.66 amps**

**100 W / 12 V = 8.33 amps**