22 Site Analysis & System Design Practice Questions & Answers

1. 1. A homeowner's twelve monthly utility bills total 14,400 kWh for the year, with no month below 900 kWh and none above 1,500 kWh. For a preliminary system estimate, what does this record establish?

   • A.The customer's monthly demand charge, billed on the single highest hour of use
   • B.The customer's minimum battery capacity, set by the lowest month of recorded use
   • C.The customer's peak instantaneous demand, which works out to roughly 39 kW
   • D.The customer's average daily energy use, which works out to roughly 39 kWh per day✓ Answer

   Billed kilowatt-hours record energy, not power. Dividing 14,400 kWh by 365 days gives about 39 kWh per day, which is the consumption figure a preliminary array size is built on. Demand is a rate of use measured over short intervals and cannot be extracted from monthly energy totals.

   Source: NABCEP PV Associate Job Task Analysis, preliminary estimate from utility usage history; energy in kilowatt-hours versus power in kilowattsReport a problem with this question

2. 2. A small commercial customer's monthly bill shows 6,000 kWh of consumption plus a separate charge based on a 45 kW figure recorded during the month. What does the 45 kW figure represent?

   • A.The average rate at which the facility drew power across the billing period
   • B.The highest rate at which the facility drew power during any billing interval✓ Answer
   • C.The lowest rate at which the facility drew power during any billing interval
   • D.The total quantity of energy the facility consumed over the entire billing period

   A demand charge is billed on kilowatts, the peak rate of energy use recorded in any short metering interval during the month, while the kilowatt-hour line bills the total quantity used. Because the two are billed separately, reducing total consumption does not automatically reduce the peak that sets the demand charge.

   Source: Utility demand-charge billing practice; peak demand in kilowatts versus consumption in kilowatt-hoursReport a problem with this question

3. 3. A home's usage history shows about 600 kWh in each spring and autumn month, rising to roughly 1,600 kWh in July and in August. What does this pattern most likely indicate?

   • A.A large summer cooling load concentrated in the hottest months of the year✓ Answer
   • B.A large electric heating load concentrated in the coldest months of the year
   • C.A metering error that overstates consumption during the summer read cycle
   • D.A steady base load that is essentially unaffected by outdoor temperature

   A usage curve that nearly triples in the two hottest months and falls back in the mild months tracks outdoor temperature, which is the signature of air conditioning. Recognising the season in which demand peaks matters because a south-facing fixed array produces most in the same summer months, so production and load line up well.

   Source: Solar site assessment practice: seasonal load profile analysis from utility usage historyReport a problem with this question

4. 4. During a site visit an assessor records a plumbing vent about three feet from the proposed array and a mountain ridgeline roughly a mile to the southwest. How are these two obstructions classified?

   • A.The vent is far shading and the ridgeline is near shading
   • B.Both are far shading because each lies outside the array itself
   • C.Both are near shading because each lies within the solar window
   • D.The vent is near shading and the distant ridgeline is far shading✓ Answer

   Near shading comes from objects close to the array, such as vents, chimneys and nearby trees, and casts a sharp-edged shadow that sweeps across individual modules. Far shading comes from distant features such as ridgelines and tall buildings, and typically removes production from the whole array at once during the early or late part of the day.

   Source: Standard PV site-assessment terminology: near shading versus far shading obstructionsReport a problem with this question

5. 5. A chimney casts no shadow on the proposed array at midday in June, yet shades several modules at midday in December. What explains the difference?

   • A.The sun's output is weaker in winter, so shadows spread wider across the roof
   • B.The sun's midday azimuth swings to due east in winter, moving the shadow over
   • C.The sun's midday altitude is lower in winter, so the chimney's shadow is longer✓ Answer
   • D.The sun's midday altitude is higher in winter, so the chimney's shadow is longer

   Shadow length depends on the sun's altitude above the horizon, and in the northern hemisphere the midday sun sits much lower in December than in June. The same object therefore throws a far longer shadow in winter, which is why a shading survey must consider the whole year rather than one visit.

   Source: Solar geometry: seasonal variation in solar altitude and resulting shadow lengthReport a problem with this question

6. 6. One module in a fourteen-module series string has a single cell covered by a fallen leaf. Why can the string's output fall by far more than the shaded fraction of area?

   • A.Series-connected cells reverse polarity, so the module drives current backwards
   • B.Series-connected cells lose their bypass paths, so the module opens the circuit
   • C.Series-connected cells share the same voltage, so one cell halves the string
   • D.Series-connected cells carry the same current, so the weakest cell limits all✓ Answer

   In a series circuit the same current flows through every cell, so a shaded cell that can only pass a small current throttles the current available to the entire string. Bypass diodes limit the damage by routing current around the affected cell group, but the string still loses the output of that group, which is far more than the leaf's area.

   Source: Series-circuit behaviour in PV source circuits; current mismatch and bypass diode functionReport a problem with this question

7. 7. A dormer unavoidably shades the lower corner of a planned array for part of every morning. Which design response most limits the resulting production penalty?

   • A.Wiring each module to its own power electronics so one module cannot drag others✓ Answer
   • B.Wiring every module into one long series string so healthy modules average it out
   • C.Wiring the array at a lower tilt so the dormer's shadow clears the modules sooner
   • D.Wiring the shaded modules first in the string so their loss occurs before the rest

   Microinverters and DC optimizers give each module its own maximum power point tracking, so a shaded module produces less while its unshaded neighbours keep operating at their own best point. With a single long series string the shaded module instead limits the current of every module wired with it, which is what makes the loss disproportionate.

   Source: Module-level power electronics (microinverters and DC optimizers) as a partial-shading mitigation strategyReport a problem with this question

8. 8. Standing on a roof, an assessor reads 180 degrees on a magnetic compass at a location where magnetic declination is 12 degrees east. What is the roof's true azimuth, and on what convention?

   • A.192 degrees, on the convention where south is 0 and north is 180
   • B.168 degrees, on the convention where south is 0 and north is 180
   • C.168 degrees, on the convention where north is 0 and south is 180
   • D.192 degrees, on the convention where north is 0 and south is 180✓ Answer

   Solar azimuth is stated with north at 0, east at 90, south at 180 and west at 270 degrees. An easterly declination is added to a magnetic reading to obtain the true bearing, so 180 plus 12 gives a true azimuth of 192 degrees, meaning the roof faces slightly west of true south.

   Source: Solar azimuth convention (north 0, east 90, south 180, west 270) and magnetic declination correction to true bearingReport a problem with this question

9. 9. For a fixed roof-mounted array facing true south in the northern hemisphere, which statement about tilt angle and annual energy yield is correct?

   • A.A tilt of nearly zero degrees tends to maximise annual production at any site
   • B.A tilt of about double the site's latitude maximises annual production
   • C.A tilt much steeper than the site's latitude maximises annual production
   • D.A tilt roughly equal to the site's latitude tends to maximise annual production✓ Answer

   Setting the tilt near the site's latitude keeps the array closest to perpendicular to the sun's rays averaged over the year, which is the rule of thumb for maximum annual energy. Tilting lower than latitude shifts production toward summer and tilting higher shifts it toward winter, so the choice follows the seasonal goal.

   Source: Fixed-tilt array design rule of thumb: tilt near site latitude for maximum annual yieldReport a problem with this question

10. 10. The roof covering beneath a proposed array has roughly five years of service life left, while the array is expected to operate for decades. What should the assessor recommend?

    • A.Replace the roof covering first, so the array is not removed and reset later✓ Answer
    • B.Replace the mounting method with ballast so the array never pierces the roof
    • C.Replace only the shingles directly under each mount and leave the rest as is
    • D.Replace the roof covering later and mount the array over it now to save cost

    When the roof covering will fail long before the array does, re-roofing first avoids paying later to remove the array, re-roof, and reinstall it, and it avoids leaving a failing surface under new penetrations. Matching the remaining roof life to the expected system life is a standard part of site assessment.

    Source: PV site assessment practice: roof covering condition and remaining service life evaluated before array installationReport a problem with this question

11. 11. A designer must leave clear pathways and setbacks around a rooftop array. Where do the required dimensions come from?

    • A.From the module manufacturer's installation manual for the racking hardware
    • B.From the utility's interconnection agreement for that class of customer service
    • C.From the array itself, allowing one module width along every unattached edge
    • D.From the fire and building codes adopted by the authority having jurisdiction✓ Answer

    Access pathways and setbacks exist so firefighters can reach the roof and ventilate it, and the specific dimensions are set by the fire and building codes that the local authority having jurisdiction has adopted, including any local amendments. Because adopted editions and amendments differ from place to place, the designer applies the requirements in force at that site rather than a remembered number.

    Source: Fire and building codes as adopted by the authority having jurisdiction; rooftop access pathways and setbacks for PV arraysReport a problem with this question

12. 12. A homeowner asks whether the existing rafters can carry the added weight of the array plus snow load. What should an associate-level assessor do?

    • A.Refer the question to a licensed structural professional for a written evaluation✓ Answer
    • B.Refer the question to the module datasheet and accept it because modules are light
    • C.Refer the question to the homeowner's account and accept it if nothing looks bent
    • D.Refer the question to the racking manual's span table and accept a matching spacing

    Judging whether a roof structure can carry dead load plus snow, wind and seismic load is engineering work that falls outside an associate-level assessor's scope. The correct action is to document the framing observed and hand the determination to a licensed structural professional, whose evaluation the permitting authority will normally require anyway.

    Source: NABCEP PV Associate scope of practice; structural adequacy determinations referred to a licensed structural professionalReport a problem with this question

13. 13. A south-facing roof plane measures 900 square feet, but vents, a skylight and required access pathways take up 300 square feet. Each module occupies about 20 square feet. How many modules fit?

    • A.45 modules, because the entire area of the roof plane counts
    • B.60 modules, because modules may be placed over obstructions
    • C.30 modules, because only the unobstructed area of the plane counts✓ Answer
    • D.15 modules, because only half the unobstructed area counts

    Usable area is the roof plane minus everything that cannot hold modules, so 900 minus 300 leaves 600 square feet, and 600 divided by 20 gives 30 modules. Counting the gross roof area overstates the array because obstructions and required pathways must stay clear.

    Source: Array layout from usable roof area after deducting obstructions and required access pathwaysReport a problem with this question

14. 14. Three PV source circuits, each of ten modules in series, are combined in parallel. Each module is rated 40 V and 10 A at maximum power. What are the array's maximum-power voltage and current?

    • A.400 V and 30 A✓ Answer
    • B.400 V and 10 A
    • C.40 V and 300 A
    • D.1,200 V and 10 A

    Series connections add voltage while current stays the same, so ten modules in series give 10 times 40 volts, or 400 volts, at 10 amperes. Parallel connections add current while voltage stays the same, so three such circuits in parallel give 400 volts at 3 times 10 amperes, or 30 amperes.

    Source: Series and parallel circuit rules applied to PV source circuits and array configurationReport a problem with this question

15. 15. A design uses 24 modules, each with a nameplate rating of 400 W at standard test conditions. What is the array's DC capacity?

    • A.9,600 kW DC
    • B.0.96 kW DC
    • C.96 kW DC
    • D.9.6 kW DC✓ Answer

    Array DC capacity is simply the module count times the nameplate rating, so 24 times 400 watts equals 9,600 watts. Dividing by 1,000 converts that to 9.6 kilowatts DC, the figure used for permitting, interconnection paperwork and production estimates.

    Source: Array DC capacity from module nameplate rating at standard test conditionsReport a problem with this question

16. 16. An 8.0 kW DC array is installed where the site receives 5.0 peak sun hours per day. Applying a system derate factor of 0.80, what daily AC energy should be expected?

    • A.50 kWh per day
    • B.40 kWh per day
    • C.6.4 kWh per day
    • D.32 kWh per day✓ Answer

    Daily energy equals the array's DC rating multiplied by peak sun hours and then by the derate factor: 8.0 times 5.0 gives 40 kWh before losses, and 40 times 0.80 gives 32 kWh. The derate factor accounts for temperature, soiling, wiring, mismatch and inverter losses, which is why the answer is below the ideal 40 kWh.

    Source: Energy estimate: kilowatt-hours equal array kilowatts DC times peak sun hours times the system derate factorReport a problem with this question

17. 17. Modules are rated 41.0 V open-circuit at standard test conditions, with a temperature coefficient of open-circuit voltage of -0.30 %/degC. The site's record low temperature is -10 degC and the inverter datasheet lists a maximum DC input voltage of 600 V. What is the greatest number of modules permitted in one series string?

    • A.12 modules
    • B.14 modules
    • C.13 modules✓ Answer
    • D.16 modules

    Open-circuit voltage has a negative temperature coefficient, so voltage rises as cells get colder and the record low is the worst case. The site is 35 degrees C below the 25 degree C standard condition, giving a rise of 35 times 0.30 percent, or 10.5 percent: 41.0 times 1.105 is about 45.3 V, and 600 divided by 45.3 is 13.2, which must be rounded down to 13 modules.

    Source: Cold-temperature open-circuit voltage correction checked against the inverter's maximum DC input voltageReport a problem with this question

18. 18. A design pairs a 12 kW DC array with a 10 kW AC inverter, and during a few clear midday hours each year the inverter holds output at its AC rating. How should this behaviour be characterised?

    • A.A wiring error that appears whenever array voltage leaves the MPPT window
    • B.A deliberate trade-off that lifts annual yield despite brief losses at the peak✓ Answer
    • C.A design fault that must be cured by fitting a larger inverter to the system
    • D.A listing violation that voids the inverter's approval under its safety standard

    Arrays rarely reach their nameplate DC rating because of temperature, soiling and imperfect sun angles, so oversizing the array relative to the inverter keeps the inverter working nearer its efficient range for far more hours of the year. The limited output during the few hours of clipping costs less energy than is gained the rest of the time, which makes the ratio a design choice rather than a fault.

    Source: DC-to-AC ratio and inverter clipping as an intentional array-to-inverter sizing trade-offReport a problem with this question

19. 19. A grid-tied PV system without energy storage is running on a sunny afternoon when the utility loses power. What happens to the home's loads?

    • A.The loads stay powered, because the inverter switches to an islanded output mode
    • B.The loads stay powered, because the array feeds the panel around the inverter
    • C.The loads go dark, because the inverter must disconnect when the grid is absent✓ Answer
    • D.The loads go dark, because sunlight alone cannot produce usable household voltage

    An interactive inverter is required to detect the loss of utility voltage and stop producing, so that it cannot energise lines that utility crews believe are dead. This anti-islanding behaviour is why a grid-tied system without storage provides no backup power, even in full sun, and customers who want outage coverage need storage with a protected-loads subpanel.

    Source: IEEE 1547 and UL 1741 anti-islanding requirements for utility-interactive invertersReport a problem with this question

20. 20. Which statement correctly distinguishes a stand-alone off-grid system from a grid-tied system with energy storage?

    • A.The off-grid system has a utility connection used only when the battery runs down
    • B.The off-grid system has no battery and runs its loads straight from the array
    • C.The off-grid system has no utility connection and meets all load from array and battery✓ Answer
    • D.The off-grid system has anti-islanding controls that stop it when utility power fails

    A stand-alone system has no utility service at all, so its battery must be sized on days of autonomy to carry every load through cloudy weather and darkness. A grid-tied system with storage keeps the utility connection and uses the battery mainly to serve a protected-loads subpanel during outages or to shift energy, so anti-islanding still applies to it.

    Source: PV system architectures: stand-alone (off-grid) versus utility-interactive systems with energy storageReport a problem with this question

21. 21. A customer asks how much PV capacity the serving utility will permit to be interconnected at their service. What is the correct response?

    • A.The limit comes from the array's DC rating, never from the AC output rating
    • B.The limit comes from the utility's rules and the authority having jurisdiction✓ Answer
    • C.The limit comes from the inverter manufacturer's listing for that equipment
    • D.The limit comes from a national cap set as a fixed share of annual consumption

    Interconnection limits are set by the serving utility's own tariff and interconnection rules together with the electrical requirements enforced by the authority having jurisdiction, and both vary from one territory to the next. The honest answer is that the applicable rules must be looked up for that utility and that jurisdiction rather than quoted from memory as a national figure.

    Source: Utility interconnection requirements and the role of the authority having jurisdiction in PV permittingReport a problem with this question

22. 22. A site is described as receiving 5.5 peak sun hours per day. What does that figure express?

    • A.Daily hours of unshaded operation, counted from 9 a.m. through 3 p.m.
    • B.Daily solar power arriving per unit area, equal to 5.5 kW per square metre
    • C.Daily solar energy arriving per unit area, equal to 5.5 kWh per square metre✓ Answer
    • D.Daily hours of visible daylight, counted from sunrise to sunset each day

    Peak sun hours restate daily insolation, an amount of energy per square metre, as the number of hours the sun would have to shine at the reference irradiance of 1,000 watts per square metre to deliver the same total. So 5.5 peak sun hours means 5.5 kWh per square metre per day, which is why irradiance in watts per square metre and insolation in kilowatt-hours per square metre must not be confused.

    Source: Peak sun hours as daily insolation in kilowatt-hours per square metre versus irradiance in watts per square metreReport a problem with this question

Practice questions based on the published knowledge domains of the NABCEP PV Associate Job Task Analysis and on standard photovoltaic engineering and safety references. NABCEP is not affiliated with this site and does not endorse it. Answers here deliberately avoid code dimensions, manufacturer specifications, incentive rules and prices, all of which change and vary by jurisdiction — always apply the electrical and building codes adopted by the authority having jurisdiction, the equipment manufacturer's instructions, and your employer's safety program. Confirm current exam requirements with NABCEP before testing. About the NABCEP Associate program →