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NABCEP PV practice questions

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Question 1Site Assessment

A roof plane faces true south. What azimuth value is assigned to this plane?

  1. 0°
  2. 180° Correct answer.
  3. 90°
  4. 270°

Explanation

Azimuth is measured clockwise from true north, so true south corresponds to 180°. East is 90°, west is 270°, and north is 0° or 360°.

Question 2Site Assessment

For a fixed-tilt array in the northern hemisphere, a common starting-point tilt angle is:

  1. 90° minus the site latitude
  2. equal to the site longitude
  3. equal to the site latitude Correct answer.
  4. 45° regardless of location

Explanation

A common rule of thumb for fixed-tilt annual energy in the northern hemisphere is to set tilt near the site latitude. Roof constraints and seasonal snow or summer production goals can justify other angles.

Question 3Site Assessment

A compass reads 170° magnetic at a site where magnetic declination is 10° east. What is the true azimuth?

  1. 170°
  2. 190°
  3. 180° Correct answer.
  4. 160°

Explanation

East declination means magnetic north lies east of true north, so true azimuth = magnetic azimuth + declination. Here 170° + 10° = 180°.

Question 4Site Assessment

A roof rises 6 inches over a horizontal run of 12 inches. What is the pitch angle?

  1. 26.6° Correct answer.
  2. 45.0°
  3. 30.0°
  4. 18.4°

Explanation

Pitch angle = arctan(rise/run) = arctan(6/12) = 26.6°. The 6/12 pitch is a 1:2 rise-to-run ratio, not 30° or 45°.

Question 5Site Assessment

At solar noon in the northern hemisphere, the sun is:

  1. lowest, due north
  2. due west
  3. due east
  4. highest, due south Correct answer.

Explanation

At solar noon the sun crosses the local meridian and reaches its highest elevation of the day. For sites north of the Tropic of Cancer, it lies due south at that time.

Question 6Site Assessment

During a site survey, a solar access tool is used primarily to:

  1. plot the sun path and obstructions Correct answer.
  2. verify transformer rating
  3. calculate conductor ampacity
  4. calculate the structural span of the rafters

Explanation

A solar access or shading tool records the sun path and nearby obstructions so the designer can estimate shade at the array location. Structural and electrical ratings are assessed with other methods.

Question 7Site Assessment

When laying out modules on a roof, which module datasheet value is most directly needed to determine row-to-row and edge spacing?

  1. module length and width Correct answer.
  2. temperature coefficient of power
  3. maximum series fuse rating
  4. NOCT

Explanation

Module length and width define the physical footprint, which sets row-to-row and edge clearances. Electrical ratings do not determine the layout geometry.

Question 8Site Assessment

In the northern hemisphere, which roof slope receives the least annual solar irradiance?

  1. west-facing
  2. east-facing
  3. north-facing Correct answer.
  4. south-facing

Explanation

In the northern hemisphere, north-facing planes have the lowest annual irradiance because they receive mostly indirect light. South-facing planes generally receive the most.

Question 9Site Assessment

A chimney located south of a proposed array is most likely to cause:

  1. increased wind uplift on the array
  2. reduced module efficiency at all times
  3. partial shading at times Correct answer.
  4. higher utility voltage

Explanation

An obstruction south of the array can shade it near solar noon and during lower-sun seasons. This reduces output from the affected modules during those periods.

Question 10Site Assessment

The most direct way to determine the available rectangular area of a roof plane is to:

  1. estimate from the building footprint only
  2. measure plane dimensions minus obstructions Correct answer.
  3. measure the gutter length at the eave
  4. count the shingle courses on the roof

Explanation

Usable area is the gross roof-plane dimensions minus areas occupied by vents, chimneys, walkways, and required setbacks. Building footprint estimates and eave measurements do not capture the actual plane geometry.

Question 11Site Assessment

A solar pathfinder image shows an obstruction crossing the December sun-path band between 9 a.m. and 11 a.m. This indicates:

  1. no shading in December
  2. shading only at summer noon
  3. no usable solar window
  4. winter morning shade there Correct answer.

Explanation

The December sun-path band shows where the sun travels during that month. An obstruction crossing the 9–11 a.m. segment shades the array location during those hours.

Question 12Site Assessment

For a fixed-tilt ground mount, increasing the tilt angle while keeping row length and module height the same generally requires:

  1. no change in spacing
  2. less row-to-row spacing
  3. more row-to-row spacing Correct answer.
  4. smaller modules

Explanation

A steeper row casts a longer shadow, so rows must be spaced farther apart to maintain the same shade-free window. Row length and module size do not change that geometric effect.

Question 13Site Assessment

In the northern hemisphere, a fixed array tilted at a lower angle than latitude will typically produce relatively more energy in:

  1. summer Correct answer.
  2. winter
  3. night
  4. annual total only

Explanation

Lower tilt favors the high summer sun and reduces the angle of incidence then. Steeper tilt favors the low winter sun.

Question 14Site Assessment

During a roof assessment for a retrofit PV system, the most important roofing consideration is:

  1. the color of the roof shingles
  2. the brand of the gutter
  3. remaining roof service life Correct answer.
  4. the number of roof vents

Explanation

PV arrays are long-lived and costly to remove, so the roof covering should have enough remaining service life to avoid removal and reinstallation. Color, gutter brand, and vent count are secondary site details.

Question 15Site Assessment

A site assessment finds a 200 A main service panel with a 200 A main breaker and a 40 A PV backfeed breaker. The busbar rating must be at least:

  1. 200 A Correct answer.
  2. 150 A
  3. 100 A
  4. 125 A

Explanation

Under the 120% rule, main breaker plus PV backfeed breaker must not exceed 120% of the busbar rating. Here (200 A + 40 A) / 1.2 = 200 A.

Question 16Site Assessment

A deciduous tree currently casts no shadow on a roof at solar noon but is expected to grow 1 m per year. The best assessment practice is to:

  1. ignore the tree because it is deciduous
  2. check winter shade over system life Correct answer.
  3. assume the tree will be removed
  4. measure shade only at the current date

Explanation

Shading changes as trees grow, so the design should evaluate winter-solstice shade over the system life. Low winter sun is most easily blocked by future tree growth.

Question 17Site Assessment

For a ground-mounted array, a geotechnical assessment is mainly used to determine:

  1. the local utility net-metering rate
  2. module frame material
  3. inverter efficiency
  4. soil bearing and foundation design Correct answer.

Explanation

Geotechnical testing characterizes soil strength, bearing capacity, and groundwater conditions that drive foundation design. It does not set module materials or utility rates.

Question 18Site Assessment

Which roof characteristic most directly affects the need for snow guards above a PV array?

  1. roof slope and shedding Correct answer.
  2. roof color
  3. number of occupants
  4. module wattage

Explanation

Snow guards are selected based on how snow sheds from the roof. Slope and shedding behavior determine the risk of sliding snow above walkways, doors, and gutters.

Question 19Site Assessment

The distance from the array to the inverter is being assessed. A long DC run is most likely to require:

  1. larger conductors for voltage drop Correct answer.
  2. a higher module tilt
  3. smaller conductor size
  4. a larger roof setback

Explanation

Long DC conductors have higher resistance and voltage drop, so larger conductors may be needed to keep drop within acceptable limits. Tilt and roof setback do not address conductor voltage drop.

Question 20Site Assessment

At a given time, the sun elevation is 30° and a 10 m tree stands due south of an array. Approximately how long is the tree's shadow on level ground?

  1. 20.0 m
  2. 17.3 m Correct answer.
  3. 5.0 m
  4. 8.7 m

Explanation

Shadow length = height / tan(elevation) = 10 m / tan 30° = 17.3 m. The shadow is longer than the tree because the sun is fairly low.

Question 21Site Assessment

A roof faces 45° east of true south. Compared with true south at the same tilt, annual production in a sunny climate is generally:

  1. identical
  2. much higher
  3. zero
  4. slightly lower Correct answer.

Explanation

Azimuth deviation from true south reduces annual irradiance modestly in a sunny climate. A 45° offset is not zero production, but it is not identical to true south either.

Question 22Site Assessment

When measuring roof azimuth with a smartphone compass, the phone should be held:

  1. pointing directly at the sun
  2. vertical against the fascia board
  3. flat on roof, aligned with slope Correct answer.
  4. parallel to the ridge line

Explanation

To measure roof azimuth, hold the device level and align its axis with the downslope direction. A tilted phone or an axis along the ridge can give an erroneous heading.

Question 23Site Assessment

A sun-path chart shows that an array location is unshaded from 9 a.m. to 3 p.m. on the winter solstice. The solar window is best described as:

  1. 6 hours on the winter solstice Correct answer.
  2. 9 hours year-round
  3. the total annual sunshine hours
  4. the time between sunrise and sunset

Explanation

The solar window is the unshaded interval within the sun path, not the full sunrise-to-sunset day. Here it is 9 a.m. to 3 p.m. on the winter solstice, or 6 hours.

Question 24Site Assessment

For a row height difference H and a required minimum solar elevation angle α above the horizon, the minimum row-to-row spacing to avoid shading is approximately:

  1. H × cos α
  2. H × tan α
  3. H × sin α
  4. H / tan α Correct answer.

Explanation

At solar elevation α, a height H casts a horizontal shadow of H / tan α. Row spacing must be at least that length to avoid shading at the design angle.

Question 25Site Assessment

A string of 10 series modules has one module fully shaded while the other nine are unshaded. With no bypass diodes, the string current is:

  1. increased by the unshaded modules
  2. equal to the inverter maximum current
  3. unchanged at the unshaded module current
  4. limited to the shaded module's current Correct answer.

Explanation

Series-connected devices carry the same current, so a shaded module with no bypass diode limits the entire string to its reduced current. The unshaded modules cannot push more current through the shaded device.

Question 26Site Assessment

During a rooftop site survey, the quickest way to directly measure the pitch angle of a roof plane is to use:

  1. a digital inclinometer Correct answer.
  2. a tape measure and the plans
  3. a laser distance meter
  4. a compass against the fascia

Explanation

A digital inclinometer placed directly on the roof surface displays the slope angle without any calculation. Tape-and-rise methods require a separate run measurement and trigonometry, and plans may not reflect as-built conditions. A compass measures azimuth, not slope.

Question 27Site Assessment

At a site where magnetic declination is 10° east, a compass held against a roof edge reads an azimuth of 180° magnetic. The true azimuth of that edge is:

  1. 190° Correct answer.
  2. 180°
  3. 200°
  4. 170°

Explanation

True azimuth equals magnetic azimuth plus east declination (or minus west declination). With 10° east declination, 180° magnetic becomes 190° true. Ignoring declination would leave a systematic orientation error in the production estimate.

Question 28Site Assessment

A site receives an average of 5.0 peak sun hours per day. This is equivalent to an average daily insolation of:

  1. 5.0 W/m² averaged over 24 hours
  2. 5.0 kW/m² of irradiance sustained all day
  3. 5.0 kWh/m² per day Correct answer.
  4. 5.0 hours of cloud-free daylight

Explanation

One peak sun hour is defined as 1 kWh/m² of insolation, so 5.0 peak sun hours equals 5.0 kWh/m² per day. It is an energy quantity, not a constant power level, and it is not the same as the number of daylight hours.

Question 29Site Assessment

In the northern hemisphere, which period is most critical when evaluating how a distant obstruction will affect annual production?

  1. any day at solar noon
  2. March 21, 6 a.m. to 9 a.m.
  3. June 21, from noon until 3 p.m.
  4. December 21, 9 a.m.–3 p.m. Correct answer.

Explanation

On the winter solstice the sun reaches its lowest path, so shadows from obstructions are longest and the affected solar window is largest. The standard solar window of roughly 9 a.m. to 3 p.m. solar time is used because hours outside it contribute little to annual insolation.

Question 30Site Assessment

A gable roof ridge runs true north–south. The two roof planes face approximately:

  1. north and south
  2. southeast and northwest
  3. northeast and southwest
  4. east and west Correct answer.

Explanation

Roof planes face perpendicular to the ridge line, so a ridge running north–south produces one plane facing east and one facing west. Each plane will produce a strongly time-of-day-dependent output that should be modeled rather than assumed to behave like a south-facing array.

Question 31Site Assessment

Which information is obtained by inspecting the attic space beneath a proposed roof-mounted array?

  1. rafter size, spacing and condition Correct answer.
  2. the roof azimuth and tilt
  3. the shingle manufacturer, color and warranty terms
  4. the utility service amperage

Explanation

Attic access reveals the framing that mounting hardware must attach to, including member size, spacing, condition and any obstructions such as ducts or wiring. Roof covering details come from the exterior, geometry from field measurement, and service size from the electrical panel.

Question 32Site Assessment

The generally accepted attachment approach for mounting PV racking on a standing-seam metal roof is:

  1. non-penetrating clamps that grip the standing seam Correct answer.
  2. lag bolts driven through the standing seam
  3. self-tapping screws into the flat pan between seams
  4. adhesive-only mounting pads

Explanation

Manufacturer-approved seam clamps transfer load to the structural seam without creating penetrations, preserving the roof's weathertightness and warranty. Fastening through the pan or seam introduces leak paths, and adhesive-only methods are not rated for PV wind uplift loads.

Question 33Site Assessment

A ground-mounted array will be installed on an exposed ridge with no windbreaks. Which site characteristic most directly drives the racking and foundation design?

  1. the distance from the array to the inverter
  2. the module's temperature coefficient
  3. the soil color at the site
  4. the site's basic wind speed and exposure Correct answer.

Explanation

Structural design loads for exposed sites come from the code-based basic wind speed adjusted by exposure category, which reflects terrain roughness and lack of shielding. Those values set pile embedment, rack member sizing and module clamping requirements.

Question 34Site Assessment

For a bifacial array mounted above open ground, which surface beneath and around the array produces the largest rear-side irradiance gain?

  1. weathered dark gravel
  2. bare, dark soil
  3. fresh snow Correct answer.
  4. mowed green grass

Explanation

Rear-side gain scales with ground albedo, and fresh snow has the highest albedo of the listed surfaces. Dark soil and dark gravel absorb most incident light and yield little reflected irradiance, while grass is intermediate.

Question 35Site Assessment

A string inverter is mounted outdoors. Placing it on a wall that receives strong afternoon sun is poor practice mainly because:

  1. it voids the module manufacturer's warranty
  2. it raises the DC array voltage above the inverter maximum input
  3. it changes the array's effective azimuth
  4. heat causes the unit to derate or shut down Correct answer.

Explanation

Inverter output is limited by internal component temperature, so solar gain plus high ambient temperature pushes the unit toward derating or thermal shutdown. Shading the unit or mounting it on a shaded wall keeps it within its rated ambient range.

Question 36Site Assessment

On a clear day, the most reliable field method for establishing true south at a site is to:

  1. point a handheld GPS unit along the roof ridge
  2. use the direction of the prevailing winter wind
  3. read the azimuth from a smartphone compass without correction
  4. mark the plumb-line shadow at local solar noon Correct answer.

Explanation

At local solar noon the sun lies on the observer's meridian, so the shadow of a vertical plumb line aligns with the true north–south line regardless of magnetic declination. The opposite direction of that shadow line gives true south.

Question 37Site Assessment

A DC home run carries 10 A at 400 V over a one-way distance of 50 m in 6 mm² copper conductors (resistivity 0.0172 Ω·mm²/m). The approximate voltage drop is:

  1. 1.4 V
  2. 5.7 V
  3. 0.3 V
  4. 2.9 V Correct answer.

Explanation

Voltage drop uses the total circuit length, which is twice the one-way run: R = 0.0172 × 100 / 6 ≈ 0.287 Ω. Multiplying by 10 A gives about 2.9 V, roughly 0.7% of 400 V, which is well within typical design limits.

Question 38Site Assessment

A site's record low ambient temperature is −15 °C. A string has a Voc of 400 V at 25 °C with a Voc temperature coefficient of −0.30%/°C. The expected string Voc at the record low is approximately:

  1. 424 V
  2. 448 V Correct answer.
  3. 412 V
  4. 460 V

Explanation

The temperature difference from 25 °C to −15 °C is 40 °C, so Voc rises by 0.30% × 40 = 12%: 400 V × 1.12 ≈ 448 V. This cold-condition value must be checked against the inverter's maximum input voltage and the modules' maximum system voltage.

Question 39Site Assessment

A shade study shows an array fully clear from 8 a.m. to 4 p.m. every day of the year, with shading only outside that window. Compared with an otherwise identical unshaded array, the expected annual energy loss is:

  1. about 25%, because the shade covers two hours of the day
  2. roughly low single-digit percent Correct answer.
  3. essentially 0%
  4. about 50%, because morning and evening are half the day

Explanation

Annual insolation is heavily weighted toward midday and summer, when the sun is high and irradiance is greatest. Shading limited to early morning and late afternoon removes only a small share of the yearly resource, so the loss is typically a few percent rather than a proportion of clock hours.

Question 40Site Assessment

For a ground-mounted array, the clearance between the bottom of the modules and grade is primarily set by:

  1. expected maximum snow depth and vegetation growth Correct answer.
  2. the inverter's rated ambient temperature operating range
  3. the string voltage
  4. the module width

Explanation

Low module edges can be buried, damaged or shaded by drifted snow and by growing vegetation, so design clearance reflects site records for snow depth and maintenance practices. Electrical and module dimensions influence structure and wiring, not the height above grade.

Question 41System Design

Per NEC requirements for PV systems, the maximum system voltage used to select conductors and equipment is determined using which temperature condition?

  1. Standard test conditions of 25°C cell temperature
  2. The highest expected ambient temperature at the installation site
  3. The lowest expected ambient temperature at the installation site Correct answer.
  4. The nominal operating cell temperature of the selected module

Explanation

Module open-circuit voltage rises as cell temperature falls, so the worst-case voltage stress occurs on the coldest day. The Code therefore requires the maximum voltage to be calculated from the lowest expected ambient temperature at the site using the module's voltage temperature coefficient. Using 25°C or a high temperature would understate the true maximum voltage.

Question 42System Design

What is the primary function of a bypass diode in a PV module?

  1. Provide an alternate path around a shaded or failed cell group to limit hot-spot heating Correct answer.
  2. Disconnect the module from the inverter during a grid outage
  3. Limit the open-circuit voltage of the module to a safe level
  4. Block reverse current from flowing into the module at night

Explanation

When cells are shaded while the rest of the module conducts, the shaded cells can be forced into reverse bias and overheat. A bypass diode provides a parallel path so string current flows around the affected cell group instead of through it. This limits both power loss and hot-spot damage.

Question 43System Design

An array rated 8.0 kW at STC is connected to a 6.0 kW inverter. What is the DC-to-AC ratio of this system?

  1. 2.00
  2. 0.75
  3. 1.33 Correct answer.
  4. 1.00

Explanation

The DC-to-AC ratio is the array STC rating divided by the inverter's AC rating, or 8.0 kW ÷ 6.0 kW = 1.33. Ratios above 1.0 mean the array can exceed the inverter rating at peak irradiance, which is a deliberate design choice in many systems.

Question 44System Design

Which set of conditions defines Standard Test Conditions for rating a PV module?

  1. 800 W/m² irradiance, 20°C cell temperature, air mass 1.0
  2. 1000 W/m² irradiance, 20°C ambient temperature, air mass 1.0
  3. 1000 W/m² irradiance, 25°C cell temperature, air mass 1.0
  4. 1000 W/m² irradiance, 25°C cell temperature, air mass 1.5 Correct answer.

Explanation

STC is defined as 1000 W/m² irradiance, a 25°C cell temperature, and an air mass of 1.5. These fixed conditions allow modules from different manufacturers to be compared on a common basis. Nameplate power and voltage ratings are all referenced to STC.

Question 45System Design

What is the primary purpose of a rapid shutdown system on a PV array?

  1. Disconnect the array from the grid during periods of high utility demand
  2. Prevent the inverter from islanding the array during a grid outage
  3. Reduce shock hazard for responders by limiting conductor voltage near and outside the array Correct answer.
  4. Discharge the stored energy held in the inverter's DC bus capacitors

Explanation

Rapid shutdown limits the voltage present on conductors within and at the boundary of the array so firefighters and other responders can work safely around a building. Anti-islanding and capacitor discharge are separate functions handled by the inverter. The system is initiated at a clearly marked switch.

Question 46System Design

In the northern hemisphere, which array azimuth generally produces the greatest annual energy yield?

  1. Due north (0°)
  2. Due south (180°) Correct answer.
  3. Due east (90°)
  4. Due west (270°)

Explanation

In the northern hemisphere the sun's path lies to the south, so a surface facing due south collects the most annual insolation at a fixed tilt. East and west orientations shift production toward morning or afternoon hours. A north-facing array in this hemisphere loses substantial annual yield.

Question 47System Design

Four identical PV modules are connected in series. Compared with a single module, the resulting string has:

  1. Twice the voltage and twice the current
  2. The same voltage and the same current
  3. Four times the voltage at the same current Correct answer.
  4. Four times the current at the same voltage

Explanation

In a series circuit the voltages add while the current through every element is identical, so four modules produce four times the voltage at the single-module current. Parallel connections do the opposite, adding current at a fixed voltage. This relationship drives both string voltage limits and conductor ampacity calculations.

Question 48System Design

In the northern hemisphere, how does mounting modules at a steeper tilt than latitude affect seasonal production?

  1. It increases winter yield and decreases summer yield Correct answer.
  2. It increases annual yield in every season
  3. It increases summer yield and decreases winter yield
  4. It has essentially no effect on seasonal yield

Explanation

A steeper tilt points the array more directly at the low winter sun, which improves cold-season production. The trade-off is reduced collection during high summer sun angles. Latitude tilt is a compromise that balances the seasons for year-round output.

Question 49System Design

A module has a Voc of 41.2 V at STC and a Voc temperature coefficient of −0.30%/°C. What is the module's open-circuit voltage at a design low temperature of −10°C?

  1. 41.2 V
  2. 45.5 V Correct answer.
  3. 44.3 V
  4. 37.9 V

Explanation

The cell is 35°C colder than STC, so Voc rises by 0.30%/°C × 35°C = 10.5%. Multiplying 41.2 V by 1.105 gives approximately 45.5 V. This cold-temperature value is the one used for maximum voltage checks.

Question 50System Design

A 24 V nominal array circuit carries 10 A through 50 ft of 4 AWG copper with a resistance of 0.308 ohms per 1000 ft. What is the approximate voltage drop on this circuit?

  1. 0.15 V
  2. 0.31 V Correct answer.
  3. 0.62 V
  4. 1.54 V

Explanation

The current travels out and back, so the effective length is 100 ft: 100 ft × 0.000308 ohms/ft = 0.0308 ohms. Multiplying by 10 A gives about 0.31 V, roughly 1.3% of 24 V. The return conductor must always be included in DC voltage drop calculations.