Photovoltaic solar panels provide a means of setting up a system on your own property to meet some or all of your electrical demand. However, to determine how effective a solar power system is, you must first convert its output into the same language used on your electric bill -- kilowatt hours, or the use of 1,000 watts for one hour. By making this calculation, you can determine how effective a prospective solar system will be simply by comparing it to a recent electricity bill.

Add up the output of all the panels and/or cells in the solar power system if you are building an array from a set number of panels. A six-panel array with each panel producing 60 watts puts out 360 watts. Alternately, if you are fitting solar panels to a set area, such as your roof, multiply the average output of solar panels (between 8-10 watts per square foot) by the surface area to be used. If you cover a 250-square-foot roof with panels averaging 9 watts per square foot, the total output is 2,250 watts.

Determine the amount of peak solar energy the panels receive each day. For an unobstructed set of solar panels that can be tilted to the appropriate seasonal angle, this is usually five hours of peak energy. However, if a nearby building obstructs your panels with shadow for 20 per cent of the day, the figure is reduced to four hours.

Multiply the total output from Step 1 by the peak hours from Step 2 to determine the total daily output. If the six-panel array described in Step 1 is placed in a prime location, it could put out 1,800 watts every day.

Divide the result of Step 3 by 1,000 to determine kWh. The six-panel array produces 1.8 kWh every day.

Factor the efficiency of your control system by multiplying the result in Step 4 by 0.7, which is typical of home solar power systems. Your power system loses some electricity in converting, storing and transmitting power, and you need to factor this into your raw number. The six-panel system's useful output is, therefore, reduced to 1.26 kWh.