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Question 1 of 10
1. Question
What limits the production of a supply side inverter?
Correct
The answer is service entrance conductors. It is not load side breaker because as the name suggests, it is used for load side connections. It is not the ampere capacity of combiner box either because the combiner box’s purpose is to combine outputs of multiple PVs.
Incorrect
The answer is service entrance conductors. It is not load side breaker because as the name suggests, it is used for load side connections. It is not the ampere capacity of combiner box either because the combiner box’s purpose is to combine outputs of multiple PVs.

Question 2 of 10
2. Question
Express in watt hours the quarterly production of a 9kW PV placed in a spot with 4.5 PSH and generates 17% losses. Assume that a month is 30 days.
Correct
Solution: Get the derating factor (1 – 17% = 0.83). In a year, there are 12 months, so a quarter would be 3 months or 30 days. After getting the derating factor and days, multiply ((9kW)(0.83)(4.5 PSH)(90 days) = 3025kWh). To convert kWh to Wh, multiply by 1000 ((3025kWh)(1000) = 3025000Wh).
Incorrect
Solution: Get the derating factor (1 – 17% = 0.83). In a year, there are 12 months, so a quarter would be 3 months or 30 days. After getting the derating factor and days, multiply ((9kW)(0.83)(4.5 PSH)(90 days) = 3025kWh). To convert kWh to Wh, multiply by 1000 ((3025kWh)(1000) = 3025000Wh).

Question 3 of 10
3. Question
How much 20% efficient PV will fit in a 6m by 3m roof?
Correct
Solution: Get irradiance ((20%)(1000W/m^2) = 200W/m^2). Multiply length by width to get area ((6m)(3m) = 18m^2). Multiply area and irradiance ((18m^2)( 200W/m^2) = 3600W) then convert W to kW (3600W/1000 = 3.6kW)
Incorrect
Solution: Get irradiance ((20%)(1000W/m^2) = 200W/m^2). Multiply length by width to get area ((6m)(3m) = 18m^2). Multiply area and irradiance ((18m^2)( 200W/m^2) = 3600W) then convert W to kW (3600W/1000 = 3.6kW)

Question 4 of 10
4. Question
Why is a water pump commonly powered by a direct PV system?
Correct
A direct PV system does not have batteries which means that it will only generate electricity during the day when water needs to be pumped the most.
Incorrect
A direct PV system does not have batteries which means that it will only generate electricity during the day when water needs to be pumped the most.

Question 5 of 10
5. Question
Calculate the current of an 800W microwave oven plugged in a 220V socket.
Correct
Solution: 800W/220V = 3.636A
Incorrect
Solution: 800W/220V = 3.636A

Question 6 of 10
6. Question
How much power does a 4HP motor generate?
Correct
Note that 1HP is equal to 746W. Solution: ((746W)(4))/1000 = 2.984kW
Incorrect
Note that 1HP is equal to 746W. Solution: ((746W)(4))/1000 = 2.984kW

Question 7 of 10
7. Question
Which of the following is true about concentrated PV?
Correct
A concentrated PV is a type of PV that works with 2axis trackers and direct sunlight. It does not use steam nor will it work during cloudy weather.
Incorrect
A concentrated PV is a type of PV that works with 2axis trackers and direct sunlight. It does not use steam nor will it work during cloudy weather.

Question 8 of 10
8. Question
Which type of PV has the least components?
Correct
Among the choices, the system with the least components is the direct coupled which is composed of a PV and a load. The others have inverters, batteries, regulators, and other parts.
Incorrect
Among the choices, the system with the least components is the direct coupled which is composed of a PV and a load. The others have inverters, batteries, regulators, and other parts.

Question 9 of 10
9. Question
Which would be the most probable reason why a ground mount PV may be preferred than rooftop PV?
Correct
The airflow of a ground mount PV is better, making it cooler and more efficient.
Incorrect
The airflow of a ground mount PV is better, making it cooler and more efficient.

Question 10 of 10
10. Question
Which of the following is true about bypass diodes?
Correct
The answer is that it limits shading effects which increases efficiency. It’s also wired parallel and is in the module’s junction box.
Incorrect
The answer is that it limits shading effects which increases efficiency. It’s also wired parallel and is in the module’s junction box.