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SOLAR PV PRACTICE TEST QUESTIONS WITH 100% CORRECT ANSWERS

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SOLAR PV PRACTICE TEST QUESTIONS WITH 100% CORRECT ANSWERS

Institution
Solar Energy Equipment Installer
Course
Solar Energy Equipment Installer

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A 50 kw interactive PV system has 15% system losses in a location getting 5.4 peak sun
hours per day average throughout the year. What would be the expected annual
energy putout?


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83,768 kWh




What is the largest array size if you are using a 24V battery bank and 20A charge
controller?


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280W

,What is the solar altitude angle?


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The angle between the horizon and the sun




Which type of PV system can dissimilar PV module types be used with little or no
system loss?


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A system that uses Ac PV modules that are coupled only to the Ac system




Which is the best way to determine the state of charge of a typical flooded lead acid
battery?


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Hydrometer testing




A 1MW inverter with 1200kW of PV is operating on a sunny day with 780W per square
meter
of irradiance and a cell temperature of 45oC and dc to ac derating of 95%. What is
the expected
inverter output power? Assume a typical temperature coefficient of power of
-0.45%/C
a. 1MW

, b. 1.1MW
c. 908kW
d. 809kW


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D 809kW

Explanation:
We are looking for different derating factors and we can see three of them
here:
1. irradiance below 1000W/m2
2. cell temperature above 25C
3. given derating of 95%.

We need to get derating numbers for these factors:
For irradiance if we have 780W/m2 out of 1000W/m2
:
= 0.78 derating for irradiance


For temperature we are going to lose some power.
First, we are going to determine our difference in temperature from how
the modules were tested,
which was 25C.
45C - 25C = 20C

Then we multiply our difference in temperature (delta T) by the temperature
coefficient for power.
20C × -0.45%/C = -9% change in power

Since we lose 9% out of 100% then
100% - 9% = 91%
Change to decimal 91%/100% = 0.91
0.91 derating for temperature; our last derating was a given 95% or 0.95

Now we multiply our PV capacity of 1200kW by the derating factors:
1200kW × 0.78 × 0.91 × 0.95 = 809kW

We should double check that the 1MW inverter can handle the output of
809kW, which it can
without clipping.

Therefore, we would expect this system to get somewhere near 809kW

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Institution
Solar Energy Equipment Installer
Course
Solar Energy Equipment Installer

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Uploaded on
July 17, 2026
Number of pages
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Written in
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Subjects

  • solar pv
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