Physics: Solar Panels: Energy, Power, and Angle Data
Calculating solar energy, power output, efficiency, and angle effects
Physics: Solar Panels: Energy, Power, and Angle Data
Calculating solar energy, power output, efficiency, and angle effects
Physics - Grade 9-12
- 1
A solar panel produces an average power of 120 W for 4.0 hours. How much electrical energy does it produce in watt-hours and in joules?
Use E = Pt, then convert watt-hours to joules.
The panel produces 480 Wh because E = Pt = 120 W x 4.0 h. Since 1 Wh = 3600 J, the energy is 480 x 3600 = 1,728,000 J. - 2
A 2.0 m² solar panel receives sunlight with an intensity of 850 W/m². If the panel is 18% efficient, what electrical power output should it produce?
First find the incoming power on the panel, then multiply by efficiency.
The incoming solar power is P = IA = 850 W/m² x 2.0 m² = 1700 W. The electrical output is 0.18 x 1700 W = 306 W. - 3
A student measures the voltage across a solar panel as 18 V and the current through a load as 2.5 A. What is the electrical power delivered to the load?
The electrical power is P = IV = 2.5 A x 18 V = 45 W. The panel delivers 45 W to the load. - 4
The table shows solar panel power output at different tilt angles: 0 degrees: 68 W, 15 degrees: 82 W, 30 degrees: 95 W, 45 degrees: 88 W, 60 degrees: 70 W. Which tilt angle gives the maximum power output, and what is that output?
Look for the largest power value in the data table.
The maximum output occurs at 30 degrees. At that angle, the panel produces 95 W, which is greater than the power at the other listed angles. - 5
A solar panel is rated at 300 W under standard test conditions. During a cloudy period, it produces 90 W. What percent of its rated power is it producing?
Use percent = part divided by whole times 100%.
The percent of rated power is 90 W divided by 300 W times 100%. This equals 30%, so the panel is producing 30% of its rated power. - 6
A solar array produces 1.8 kWh of energy in 6.0 hours. What was its average power output in kilowatts and watts?
The average power is P = E/t = 1.8 kWh divided by 6.0 h = 0.30 kW. This is equal to 300 W. - 7
Sunlight strikes a flat solar panel at an angle, so the effective intensity is I_effective = I cos theta. If the sunlight intensity is 1000 W/m² and theta = 40 degrees from perpendicular, what is the effective intensity on the panel? Use cos 40 degrees = 0.766.
Multiply the sunlight intensity by the cosine of the angle from perpendicular.
The effective intensity is I_effective = 1000 W/m² x 0.766 = 766 W/m². The panel receives an effective intensity of 766 W/m². - 8
A 1.5 m² solar panel receives an effective intensity of 700 W/m² and produces 160 W of electrical power. What is the efficiency of the panel?
Efficiency = useful output power divided by input power times 100%.
The incoming power is 700 W/m² x 1.5 m² = 1050 W. The efficiency is output divided by input, so 160 W divided by 1050 W = 0.152, or about 15.2%. - 9
A solar panel charges a battery by supplying a current of 3.0 A at 12 V for 2.0 hours. How much energy is transferred to the battery in watt-hours?
The power is P = IV = 3.0 A x 12 V = 36 W. The energy is E = Pt = 36 W x 2.0 h = 72 Wh. - 10
A graph of power output versus time shows these average power values during each hour: 8-9 AM: 80 W, 9-10 AM: 140 W, 10-11 AM: 210 W, 11 AM-12 PM: 260 W. Estimate the total energy produced over these four hours.
For each one-hour interval, energy in Wh equals power in W times 1 hour.
For one-hour intervals, add the energy from each hour: 80 Wh + 140 Wh + 210 Wh + 260 Wh = 690 Wh. The panel produces about 690 Wh over the four hours. - 11
Two identical solar panels are tested at noon. Panel A is aimed directly at the Sun and produces 200 W. Panel B is at an angle where cos theta = 0.60. If all other conditions are the same, estimate the power output of Panel B.
When other factors are the same, power output is approximately proportional to effective sunlight intensity.
Panel B receives 60% as much effective sunlight as Panel A. Its estimated power output is 0.60 x 200 W = 120 W. - 12
A home uses 12 kWh of electrical energy per day. A solar system produces an average of 3.0 kWh per day. What fraction and percent of the home's daily energy use is supplied by the solar system?
The fraction supplied is 3.0 kWh divided by 12 kWh = 0.25, which is 1/4. The percent supplied is 25%. - 13
A solar panel has an open-circuit voltage of 21 V, but when connected to a load it operates at 17 V and 4.0 A. Why should the load values be used to calculate useful power, and what is the useful power?
Open-circuit voltage is measured when no current flows.
The load values should be used because useful electrical power is delivered only when current flows through a connected circuit. The useful power is P = IV = 4.0 A x 17 V = 68 W. - 14
A student records power output at three times during a lab: 10 AM: 150 W, 12 PM: 240 W, 2 PM: 170 W. Explain one physical reason the noon value may be the largest.
The noon value may be largest because the Sun is higher in the sky, so sunlight strikes the panel more directly. A more direct angle increases the effective intensity on the panel and can increase electrical output. - 15
A 250 W solar panel costs $200. A 400 W solar panel costs $360. Which panel has the lower cost per watt, and by how much?
Cost per watt equals price divided by rated power.
The 250 W panel costs $200 divided by 250 W = $0.80 per watt. The 400 W panel costs $360 divided by 400 W = $0.90 per watt. The 250 W panel has the lower cost per watt by $0.10 per watt.