NABCEP PV ASSOCIATE
CERTIFICATION SCRIPT 2026
QUESTIONS WITH SOLUTIONS
GRADED A+
◍ An off-grid installation has a 1HP pump that is on 10% of the time and a
20W light that is on 100% of the time. What is the monthly energy use?.
Answer: 68KWh1HP = 746W746W 0.1 24hours * 30days = 53712Wh20W
24hrs 30days = 14400Wh14400 + 53712 = 68112Wh68kWh
◍ What kind of ground faults do most inverters detect?.
Answer: DCInverters detect dc ground faults. AC ground faults are detected
on the AC side of the inverter at the distribution paned (MSP, load center).
Most inverters are not connected to high or medium voltage. Medium
voltage is thousands of volts, large utility scale inverters are connected to
medium voltage via a transformer
◍ 16. A 48 volt battery bank is used to provide power for critical loads
requiring 7458 Wh/day. Three days of autonomy are required. What is the
required capacity of the battery bank?.
Answer: 16. Answer: b. 466 Ah 7458 Wh ÷ 48V = 155.4 Ah/day 155.4
Ah/day x 3 days autonomy = 466 Ah
◍ 1. NEC section ________ shows the requirements for working spaces
around live electrical equipment?.
Answer: 110.26
◍ 2. What is the maximum latitude at which the sun can achieve a 90º altitude
angle?.
Answer: 2. Answer: b. Approx. 23.45º north and south of the equator, but
, like magnetic declination, the latitude of the Tropics varies over time.
◍ 3. If the open circuit voltage of a polycrystalline silicon PV module is
37.0V, the module Vmp is 29.9V, the inverter max voltage is 600VDC and
its MPPT voltage range is 300 to 480VDC, and the minimum temperature is
-24°
C. What is the maximum number of modules per source circuit according to
the NEC? List the NEC section where the answer is found..
Answer: 3. Answer C: 37V x 1.20 = 44.4V 600V max ÷ 44.4V = 13.4 → 13
modules NEC Table 690.7 - 1.20 correction factor
◍ Current carrying ability of a conductor (wire) is called.
Answer: Ampacity
◍ 4. A PV array of Suniva 300 Watt modules consists of 3 rows and 10
columns of racked modules mounted in landscape and facing south at
latitude 30°. The modules are tilted at 20⁰. The mounting posts are installed
3 ft. deep. How long must the posts be? Module dimensions are 77.6" x
38.7"..
Answer: Answer: C Sin(angle) = opposite ÷ hypotenuse Sin(20⁰) = Height ÷
38.7" x 3 Height = sin(20⁰) x 116.1" Height = 0.342 x 116.1" = 39.7 " 39.7"
+ 36" = 75.7
◍ An array located at 30⁰N latitude consists of two rows racked facing south.
Both rows are on a level surface and the height from the ground to the
highest point on the module is 39.7". Calculate the minimum distance in feet
needed between rows so the modules will not be shaded at 9AM on
December 21. Use the sun chart provided..
Answer: Answer: b Tan(angle) = opposite ÷ adjacent where opposite is array
height and adjacent is row-to-row spacing Tan(21°) = 39.7" ÷ Row-to-row
Row-to-row = 39.7" ÷ 0.384 Row-to-row = 103.4" or 8.6
◍ 6. At 43⁰ North latitude on the winter solstice, the solar altitude angle at
noon is____..
Answer: d. 23.55⁰ 90° - 43° (latitude) + (-23.45°) = 23.55° For the Summer
Solstice, the third number is +23.45°C, resulting in 70.45°. For either
, equinox, the last number is zero.
◍ 7. An array is comprised of 22 modules. Each module is 64.5" x 38.7" and
weighs 44.1 lbs. The site will experience 50 psf. of uplift force. What is the
approximate total uplift on the array?.
Answer: 7. Answer: a. 19,000 lbs. 64.5" x 38.7" ÷ 144 sq. in/sq. ft. = 17.33
sq. ft. 17.33 sq. ft. x 22 modules x 50 psf = 19063 lbs.
◍ According to ohms law and voltage drop, the relationship between voltage
drop and current is.
Answer: Proportional. Ohms law: V = I*RMore current makes more voltage
drop. This is a reason why we need a larger wire with more current, so the
wire resistance is less with a larger wire and there will be less voltage drop.
◍ 8. What is the temperature correction factor if the module correction factor
is -0.335 %/⁰C and the cell temperature is 54⁰C?.
Answer: 8. Answer: a. 0.903 1 + (-0.00335 x (54 - 25)) = 0.903
◍ 9. A module has dimensions of 64.5" x 38.7" and is in a landscape
orientation on a flat roof. The position of the sun at 9am on Dec 21 is 11°
elevation and 130° azimuth. What is the maximum tilt angle the modules
can have so that there is no inter-row shading. A 2 foot walkway is required
between adjacent rows?.
Answer: 9. Answer: a. 10° is the closest to 10.8°Module width x SIN(tilt
angle) = module height Module height x COS(azimuth angle) ÷
TAN(elevation angle) = minimum row spacing The azimuth angle is 180 -
130 = 50° The height is 24" x TAN(11) ÷ COS (50) = 7.26" SIN-1(7.26" ÷
38.7") = 10.8° tilt angle
◍ 10. Where no overcurrent protection is provided for the PV dc circuit, an
assumed overcurrent device rated at the PV circuit Isc is used to size the
equipment grounding conductor in accordance with NEC ____..
Answer: 10. Answer: c.Table 250.122 NEC Article 690.45(A) refers to
Table 250.122
◍ 11. There are to be two critical loads on a PV system. Your analysis shows
CERTIFICATION SCRIPT 2026
QUESTIONS WITH SOLUTIONS
GRADED A+
◍ An off-grid installation has a 1HP pump that is on 10% of the time and a
20W light that is on 100% of the time. What is the monthly energy use?.
Answer: 68KWh1HP = 746W746W 0.1 24hours * 30days = 53712Wh20W
24hrs 30days = 14400Wh14400 + 53712 = 68112Wh68kWh
◍ What kind of ground faults do most inverters detect?.
Answer: DCInverters detect dc ground faults. AC ground faults are detected
on the AC side of the inverter at the distribution paned (MSP, load center).
Most inverters are not connected to high or medium voltage. Medium
voltage is thousands of volts, large utility scale inverters are connected to
medium voltage via a transformer
◍ 16. A 48 volt battery bank is used to provide power for critical loads
requiring 7458 Wh/day. Three days of autonomy are required. What is the
required capacity of the battery bank?.
Answer: 16. Answer: b. 466 Ah 7458 Wh ÷ 48V = 155.4 Ah/day 155.4
Ah/day x 3 days autonomy = 466 Ah
◍ 1. NEC section ________ shows the requirements for working spaces
around live electrical equipment?.
Answer: 110.26
◍ 2. What is the maximum latitude at which the sun can achieve a 90º altitude
angle?.
Answer: 2. Answer: b. Approx. 23.45º north and south of the equator, but
, like magnetic declination, the latitude of the Tropics varies over time.
◍ 3. If the open circuit voltage of a polycrystalline silicon PV module is
37.0V, the module Vmp is 29.9V, the inverter max voltage is 600VDC and
its MPPT voltage range is 300 to 480VDC, and the minimum temperature is
-24°
C. What is the maximum number of modules per source circuit according to
the NEC? List the NEC section where the answer is found..
Answer: 3. Answer C: 37V x 1.20 = 44.4V 600V max ÷ 44.4V = 13.4 → 13
modules NEC Table 690.7 - 1.20 correction factor
◍ Current carrying ability of a conductor (wire) is called.
Answer: Ampacity
◍ 4. A PV array of Suniva 300 Watt modules consists of 3 rows and 10
columns of racked modules mounted in landscape and facing south at
latitude 30°. The modules are tilted at 20⁰. The mounting posts are installed
3 ft. deep. How long must the posts be? Module dimensions are 77.6" x
38.7"..
Answer: Answer: C Sin(angle) = opposite ÷ hypotenuse Sin(20⁰) = Height ÷
38.7" x 3 Height = sin(20⁰) x 116.1" Height = 0.342 x 116.1" = 39.7 " 39.7"
+ 36" = 75.7
◍ An array located at 30⁰N latitude consists of two rows racked facing south.
Both rows are on a level surface and the height from the ground to the
highest point on the module is 39.7". Calculate the minimum distance in feet
needed between rows so the modules will not be shaded at 9AM on
December 21. Use the sun chart provided..
Answer: Answer: b Tan(angle) = opposite ÷ adjacent where opposite is array
height and adjacent is row-to-row spacing Tan(21°) = 39.7" ÷ Row-to-row
Row-to-row = 39.7" ÷ 0.384 Row-to-row = 103.4" or 8.6
◍ 6. At 43⁰ North latitude on the winter solstice, the solar altitude angle at
noon is____..
Answer: d. 23.55⁰ 90° - 43° (latitude) + (-23.45°) = 23.55° For the Summer
Solstice, the third number is +23.45°C, resulting in 70.45°. For either
, equinox, the last number is zero.
◍ 7. An array is comprised of 22 modules. Each module is 64.5" x 38.7" and
weighs 44.1 lbs. The site will experience 50 psf. of uplift force. What is the
approximate total uplift on the array?.
Answer: 7. Answer: a. 19,000 lbs. 64.5" x 38.7" ÷ 144 sq. in/sq. ft. = 17.33
sq. ft. 17.33 sq. ft. x 22 modules x 50 psf = 19063 lbs.
◍ According to ohms law and voltage drop, the relationship between voltage
drop and current is.
Answer: Proportional. Ohms law: V = I*RMore current makes more voltage
drop. This is a reason why we need a larger wire with more current, so the
wire resistance is less with a larger wire and there will be less voltage drop.
◍ 8. What is the temperature correction factor if the module correction factor
is -0.335 %/⁰C and the cell temperature is 54⁰C?.
Answer: 8. Answer: a. 0.903 1 + (-0.00335 x (54 - 25)) = 0.903
◍ 9. A module has dimensions of 64.5" x 38.7" and is in a landscape
orientation on a flat roof. The position of the sun at 9am on Dec 21 is 11°
elevation and 130° azimuth. What is the maximum tilt angle the modules
can have so that there is no inter-row shading. A 2 foot walkway is required
between adjacent rows?.
Answer: 9. Answer: a. 10° is the closest to 10.8°Module width x SIN(tilt
angle) = module height Module height x COS(azimuth angle) ÷
TAN(elevation angle) = minimum row spacing The azimuth angle is 180 -
130 = 50° The height is 24" x TAN(11) ÷ COS (50) = 7.26" SIN-1(7.26" ÷
38.7") = 10.8° tilt angle
◍ 10. Where no overcurrent protection is provided for the PV dc circuit, an
assumed overcurrent device rated at the PV circuit Isc is used to size the
equipment grounding conductor in accordance with NEC ____..
Answer: 10. Answer: c.Table 250.122 NEC Article 690.45(A) refers to
Table 250.122
◍ 11. There are to be two critical loads on a PV system. Your analysis shows
