NABCEP PVIP practice questions
Original calculation questions of the kind the full pack drills, each with a worked solution. Every answer was checked by code. Pick an answer, then check it.
1. A PV module has Voc = 49.5 V at STC and a Voc temperature coefficient of −0.27 %/°C. The lowest expected ambient temperature at the site is −18 °C. Using the manufacturer coefficient method, what is the module's maximum Voc?
ΔT = −18 − 25 = −43 °C. Correction = 1 + (−0.0027 × −43) = 1.1161. Voc,max = 49.5 × 1.1161 = 55.2 V.
Reference: NEC 690.7
2. Using the module from the previous question (Voc,max = 55.2 V), what is the maximum number of modules in one series string if the inverter's maximum DC input voltage is 600 V?
600 ÷ 55.2 = 10.87. Always round down: 10 modules. Eleven would give 11 × 55.2 = 607 V, above the 600 V limit.
Reference: NEC 690.7
3. A module has Isc = 13.9 A. What is the maximum circuit current of a single PV source circuit?
Maximum circuit current = Isc × 1.25 = 13.9 × 1.25 = 17.4 A.
Reference: NEC 690.8(A)
4. For the same source circuit (maximum circuit current 17.4 A), what is the smallest standard overcurrent device that meets the continuous-current requirement?
OCPD ≥ 1.25 × 17.375 A = 21.7 A. The next standard size is 25 A. (Check that it does not exceed the module's maximum series fuse rating.)
Reference: NEC 690.9(B), 240.6(A)
5. A 2-wire DC circuit carries 10.5 A over a one-way length of 150 ft in #10 copper (use 1.24 Ω per 1,000 ft). The circuit operates at 400 V. What is the voltage drop as a percentage?
VD = 2 × L × I × R ÷ 1000 = 2 × 150 × 10.5 × 1.24 ÷ 1000 = 3.91 V. 3.91 ÷ 400 = 0.98 %. A common mistake is to forget the factor 2 (the current goes out and back).
6. A 7.6 kW single-phase inverter connects at 240 V. What is the minimum standard size of its output overcurrent device?
Continuous output current = 7,600 ÷ 240 = 31.7 A. × 1.25 = 39.6 A. The next standard size is 40 A.
Reference: NEC 705.30, 240.6(A)
7. A panelboard has a 200 A busbar and a 200 A main breaker. The PV backfed breaker will sit at the opposite end of the bus from the main. Under the 120 % rule, what is the most that 125 % of the inverter's rated output current may be?
120 % × 200 A = 240 A. 240 A − 200 A main = 40 A. So 125 % of the inverter output current may be at most 40 A, which allows an inverter of up to 32 A rated output (32 A is the trap answer).
Reference: NEC 705.12(B)
8. An 8.4 kWdc array sees an average of 5.2 peak sun hours per day. Total system derate is 0.80. Roughly how much energy will it produce in a year?
8.4 × 5.2 × 0.80 × 365 = 12,755 kWh ≈ 12,750 kWh. 15,940 kWh is the answer if you forget the derate.
9. Twenty 410 W modules feed a 6.0 kW (AC) inverter. What is the DC-to-AC ratio?
DC = 20 × 410 W = 8.2 kW. 8.2 ÷ 6.0 = 1.37.
10. A module has Vmp = 37.6 V at STC and a Vmp temperature coefficient of −0.35 %/°C. The highest expected cell temperature is 70 °C. The inverter's minimum MPPT voltage is 250 V. What is the minimum number of modules per string?
ΔT = 70 − 25 = 45 °C. Vmp,hot = 37.6 × (1 − 0.0035 × 45) = 31.7 V. 250 ÷ 31.7 = 7.9. Always round up: 8 modules.
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