Twenty of them, free. Everybody else gates their practice questions behind a signup or publishes flashcards with no explanation. These are on the page with the reasoning, because the reasoning is the part worth having.
A practice question you get right by luck teaches you nothing, and one you get wrong without an explanation teaches you less. So what follows is twenty questions with the complete working, the traps named, and the reason each one matters in the field as well as in the exam room. Nothing is gated and no email is required.
They are weighted toward design and safety, which is where the marks and the expensive field mistakes both are. If they are useful, there are another 120 in the question bank at $210 CAD, each with a video of somebody working through it.
A module has an open-circuit voltage of 41.0 V at standard test conditions and a temperature coefficient of Vₒₐ of −0.28 %/°C. Fourteen modules are wired in series. The record low temperature at the site is −24 °C. Will the string exceed an inverter rated at 600 V maximum input?
Working. Standard test conditions are 25 °C, so the temperature difference is −24 − 25 = −49 °C. Voltage rises as temperature falls, and a negative coefficient multiplied by a negative difference gives a positive change: −0.28 %/°C × −49 °C = +13.72 %. So each module reaches 41.0 × 1.1372 = 46.6 V. Fourteen in series gives 14 × 46.6 = 652 V.
Answer: yes, it exceeds 600 V, and the string is too long. Thirteen modules would give 606 V, which still exceeds it. Twelve gives 559 V, which does not.
Why this question matters more than any other on the paper. A string sized at 25 °C looks fine and destroys an inverter on the first cold bright morning. If you can do this calculation without hesitating you have dealt with the hardest reliable source of marks in the examination, and the most expensive mistake in the field.
A household uses 900 kWh a month. The site receives an average of 4.2 peak sun hours a day. Ignoring losses, roughly what array size is needed?
Working. 900 kWh a month is about 900 ÷ 30 = 30 kWh a day. Peak sun hours is the useful fiction that turns a day's irradiance into a number of hours at 1,000 W/m², so an array of P kilowatts produces roughly P × 4.2 kilowatt-hours a day. Setting P × 4.2 = 30 gives P ≈ 7.1 kW.
Then apply losses — soiling, temperature, wiring, inverter efficiency, typically 20 to 25 % combined — and you need nearer 9 kW.
The trap. Candidates lose this one on units, not on method. A watt is a rate; a watt-hour is a quantity. If the answer you reach has the wrong one, the arithmetic was never the problem.
A crew needs to work on the DC side of an operating array at midday. What is the correct statement about isolating it?
Working. The DC conductors from the array can be switched, but the array itself cannot be de-energised while light falls on it. Covering modules reduces output; it does not reliably make them safe. So the correct approach treats the array as a live source at all times during daylight: isolate downstream, verify absence of voltage at the point of work, and expect voltage to be present upstream regardless of switch position.
Answer: there is no switch that makes an illuminated array dead.
Why experienced trades get this wrong. An AC habit — open the breaker and the conductors are dead — does not transfer. The sun is not a breaker, and DC arcs do not self-extinguish at a zero crossing because there is no zero crossing. This is the single most important difference between PV work and general electrical work.
A PV source circuit carries a short-circuit current of 9.6 A. Code requires the continuous current to be taken as 125 % of short-circuit current, and the conductor to be sized at 125 % of that. What minimum ampacity must the conductor have before any temperature or conduit-fill correction?
Working. 9.6 × 1.25 = 12 A continuous. Then 12 × 1.25 = 15 A.
Answer: 15 A — and note it is the same as 9.6 × 1.56.
The trap. Applying 125 % once instead of twice gives 12 A, which will be one of the four options. The two factors are doing different jobs: the first turns an instantaneous rating into a continuous-duty value, the second is the conductor-sizing allowance for continuous loads.
An inverter has a minimum MPPT input voltage of 200 V. Modules have a maximum-power voltage of 31.2 V at standard test conditions and a temperature coefficient of −0.38 %/°C. The hottest expected cell temperature is 70 °C. What is the shortest string that still works?
Working. Temperature difference 70 − 25 = +45 °C. Change −0.38 %/°C × 45 = −17.1 %. So Vₘₚ falls to 31.2 × 0.829 = 25.9 V per module. Required: 200 ÷ 25.9 = 7.7, so 8 modules minimum.
Why this is the question people forget. Everybody checks the cold limit, because a dead inverter is dramatic. The hot limit is quieter: a string that drops below the MPPT window on the hottest afternoon of the year simply stops producing, at the exact hour it should be producing most, and the owner never finds out unless somebody reads the data.
Three strings, each of twelve modules, are wired in parallel into one inverter input. Each module has Iₛℂ of 9.1 A and Vₒₐ of 40.2 V. What are the array's open-circuit voltage and short-circuit current at standard test conditions?
Working. In series, voltage adds and current does not: each string is 12 × 40.2 = 482.4 V at 9.1 A. In parallel, current adds and voltage does not: the array is 482.4 V at 3 × 9.1 = 27.3 A.
Answer: 482.4 V, 27.3 A.
The trap. Multiplying everything by everything gives 1,447 V, which appears as an option and would be an unlawful and lethal array. Series adds volts; parallel adds amps. If you can hold that one sentence you will not lose a mark anywhere in this topic.
A string carries 8.5 A over a one-way distance of 40 m. The conductor has a resistance of 0.00328 Ω per metre. The string operates at 420 V. What is the percentage voltage drop?
Working. Current travels out and back, so the circuit length is 2 × 40 = 80 m. Resistance = 80 × 0.00328 = 0.262 Ω. Drop = I × R = 8.5 × 0.262 = 2.23 V. As a percentage: 2.23 ÷ 420 = 0.53 %.
Answer: about 0.5 %, comfortably inside normal design limits.
The trap. Forgetting the return path halves the answer. It is the single most common arithmetic slip in voltage-drop questions, and the examiners know it.
Worth noticing: this is why strings are run at high voltage. At 48 V the same 2.23 V drop would be 4.6 % and you would be buying much heavier cable.
A 9.6 kW array is connected to a 7.6 kW inverter. Is this a design error?
Working. The ratio is 9.6 ÷ 7.6 = 1.26.
Answer: no. It is deliberate, and usually correct.
Why. An array almost never produces its nameplate output. The rating is measured at 1,000 W/m² and 25 °C, and a real roof is rarely at both at once — when irradiance is high the modules are hot, and hot modules produce less. Oversizing the array relative to the inverter means the inverter runs nearer its efficient range for more of the day. The cost is a small amount of clipping at the few brightest moments of the year, which is cheaper than buying a larger inverter that idles.
Ratios of roughly 1.15 to 1.35 are ordinary. Beyond that, clipping losses start to matter, and manufacturer limits apply.
One module in a twelve-module series string is 50 % shaded by a vent pipe. Roughly what happens to the string's output, ignoring bypass diodes?
Working. In a series circuit the current is the same through every element, so the whole string is limited by the module passing the least current. A module at half irradiance passes roughly half the current. So the string delivers roughly half its output — not one twelfth less.
Answer: about 50 % of the string, from one shaded module out of twelve.
What changes it in practice. Bypass diodes inside the module let current route around the affected cell group, so a real module loses a third or two thirds of itself rather than all of it, and the string loses correspondingly less. Module-level power electronics — optimisers or microinverters — remove the problem almost entirely by letting each module find its own operating point.
This is why a shading assessment is not paperwork, and why "it is only a small vent pipe" is a sentence that costs money.
A system shuts down briefly most sunny afternoons and restarts by itself. Output is otherwise normal. What is the most likely cause?
Working. Two candidates fit a pattern that is daily, brief, self-clearing and weather-correlated: grid over-voltage, and inverter over-temperature.
Grid over-voltage is the more common. An inverter must push its output slightly above the grid voltage to export at all. If the service voltage is already high — a long service run, a lightly loaded transformer, several neighbours exporting at once — the inverter reaches its statutory upper limit and must disconnect. It returns when the voltage settles.
Answer: most likely grid over-voltage, with inverter over-temperature the second candidate.
How you tell them apart. Read the inverter's event log, which records the fault code. If it is voltage, the fix is usually reducing the AC conductor's voltage drop or a conversation with the utility, not replacing the inverter — which is what often gets replaced first, expensively and pointlessly.
An off-grid cabin uses 4.8 kWh a day. The site gets 3.5 peak sun hours in December. System losses are 25 %. What array size is needed to carry the load in December?
Working. Energy needed from the array after losses: 4.8 ÷ 0.75 = 6.4 kWh a day. Array size = 6.4 ÷ 3.5 = 1.83 kW.
Answer: about 1.8 kW, rounded up in practice to 2 kW.
The trap, and it is a big one. Candidates divide by 3.5 first and then apply losses, which gives 1.37 kW — a system that fails every winter. Losses increase the array you need; they do not reduce it. If your answer came out smaller than the no-loss case, you applied the factor the wrong way round.
And the design point: off-grid sizing uses the worst month, not the average. An average-month design leaves somebody cold in January, which is when they are using the cabin's heating most.
A battery bank is rated 10 kWh with a maximum depth of discharge of 80 % and a round-trip efficiency of 90 %. How much energy can actually be delivered to a load per cycle?
Working. Usable before losses: 10 × 0.80 = 8 kWh. After round-trip efficiency: 8 × 0.90 = 7.2 kWh.
Answer: 7.2 kWh from a “10 kWh” battery.
Why it matters outside the exam. A customer told they are buying 10 kWh and delivered 7.2 has been misled, even if nobody intended to mislead them. Quoting usable capacity rather than nameplate is the difference between a referral and a complaint, and it is a habit worth forming before you ever sell one.
A module is 1.95 m by 1.10 m and rated 430 W. What is its efficiency?
Working. Area = 1.95 × 1.10 = 2.145 m². At standard test conditions the incoming power is 1,000 W/m², so the module receives 2.145 × 1,000 = 2,145 W. Efficiency = 430 ÷ 2,145 = 20.0 %.
Answer: 20 %.
What the number is good for, and what it is not. Efficiency tells you how much power you get per square metre, so it matters when roof area is the constraint. It tells you nothing about value: a less efficient module at a lower price per watt is the better buy on a roof with room to spare. Efficiency is a packaging statistic, not a quality statistic.
Why must module frames and racking be bonded together and connected to the equipment grounding conductor, even though the array is nowhere near 120 V?
Working. Two reasons, and both matter.
Fault current needs a path. If a conductor's insulation fails against a metal frame, the frame becomes energised. A bonded, grounded path gives that current somewhere to go and lets protective devices operate. Without it, the frame stays live and waits for a person.
Ground-fault detection depends on it. An inverter's ground-fault protection works by detecting current where there should be none. If the array's metalwork is not referenced, there is nothing to detect against and the protection is decorative.
Answer: to provide a fault-current path and to make ground-fault protection functional.
The field point. Anodised aluminium is an insulator. Rails and frames that look metal-to-metal are often not electrically continuous, which is why bonding hardware bites through the coating — and why a bond that was never tested is a bond you do not have.
What problem is rapid shutdown intended to solve, and what does it actually do?
Working. The problem is that an array cannot be switched off. Opening the DC disconnect de-energises the conductors below it, but the conductors on the roof, between the modules and that disconnect, stay at full voltage for as long as light falls on the array. A firefighter cutting through a roof meets live DC at hundreds of volts.
Rapid shutdown requires that, once initiated, conductors outside the array boundary drop to a low voltage within a short time, and in the stricter case that voltage inside the array boundary is limited too — which in practice means electronics at or near each module.
Answer: it limits conductor voltage on and near the roof after initiation, so emergency responders are not working beside an energised DC source.
Worth stating plainly: rapid shutdown does not make the modules themselves safe. They still generate. What it does is stop the wiring from being lethal.
A string of fourteen modules, each with Vₒₐ of 41.0 V, is measured at 15 °C ambient and reads 548 V open-circuit. Calculated expectation at that temperature is about 588 V. What has gone wrong?
Working. The shortfall is 588 − 548 = 40 V, which is almost exactly one module's open-circuit voltage.
Answer: the string has thirteen modules connected, not fourteen — or one module is bypassed, disconnected, or has a failed connection.
Why this question is on the exam. It is the entire reason commissioning measurements are compared against a calculation rather than just recorded. A number that is merely "about right" hides a missing module. A number checked against a prediction finds it in two minutes, on the day, while the crew and the ladder are still there.
The habit: calculate first, measure second, and investigate any disagreement greater than a few per cent.
A 7 kW system costs $21,000 installed and produces 8,400 kWh a year. Electricity costs $0.16/kWh. What is the simple payback?
Working. Annual saving = 8,400 × 0.16 = $1,344. Payback = 21,000 ÷ 1,344 = 15.6 years.
Answer: about 15 to 16 years.
Everything that number leaves out, and you should say so to a customer before they work it out themselves: electricity prices rise, which shortens it; module output degrades slowly, which lengthens it; an inverter will probably need replacing within the period, which lengthens it; and any incentive or tax measure available to that customer changes it materially. A simple payback is a conversation starter, not an answer, and presenting it as an answer is how installers acquire unhappy customers in year three.
A site has 14 hours of daylight in June and 4.6 peak sun hours. Why are those numbers so different?
Working. Daylight hours count the time the sun is above the horizon. Peak sun hours express the total energy arriving over the day as an equivalent number of hours at full strength, 1,000 W/m².
Early and late in the day the sun is low, the light passes through more atmosphere and strikes the array at a poor angle, so irradiance is a fraction of full strength. Sum the whole day's energy and it is equivalent to about four and a half hours at full strength — even though the sun was up for fourteen.
Answer: peak sun hours measures energy, not time.
The practical use. It lets you multiply an array's kilowatt rating by one number to estimate a day's kilowatt-hours, which is the single most useful shortcut in solar design. Using daylight hours instead overstates production by roughly three times.
A crew arrives to work on an existing array at 7 am in October. It is barely light. What is the correct assumption about the DC conductors?
Working. That they are energised. Low light produces reduced current, but voltage rises quickly with even modest irradiance and is near its full value long before the array produces useful power. Open-circuit voltage is close to maximum at light levels where output is almost nil — and voltage is what injures you.
Answer: treat them as live, in any daylight, at any time of year.
Why this catches experienced people and not beginners. An electrician's habit is lock out, prove dead, work. It is a good habit and it does not transfer, because there is no upstream device that makes an illuminated array dead. The people who get hurt on arrays are usually the ones with the most confidence in a procedure that does not apply here.
A 7.6 kW inverter is to be connected to a 200 A service panel with a 200 A main breaker, by a back-fed breaker. The busbar is rated 200 A. Using the common 120 % allowance, what is the largest PV breaker permitted?
Working. The busbar may be loaded to 120 % of its rating from both ends combined: 200 × 1.20 = 240 A. Subtract the main breaker: 240 − 200 = 40 A.
Answer: a 40 A PV breaker — which comfortably carries a 7.6 kW inverter at 240 V (about 32 A, requiring 40 A at the 125 % continuous allowance).
Why this is the question that kills jobs. It is the calculation that determines whether a system can connect to an existing panel at all. Find out at the survey and you price the job correctly, including a service upgrade or a supply-side connection if one is needed. Find out on installation day, with a crew standing in the driveway, and you have lost the margin on the job and possibly the customer.
Check the current code edition and your local amendments for the applicable allowance before relying on any figure, including this one.
140 questions across all six job-task domains, each with the working shown on video — the twenty above and 120 more. Weighted toward the design domain, because that is where the calculations are and where the marks go. The question bank, $210 CAD →
More on how the questions are constructed, and how to read a four-option question quickly: multiple-choice technique →
No, and nobody honest would publish those. These are written to NABCEP's published job task analysis, testing the same things in the same way. Reproducing live examination questions would cost a provider its registration, and rightly.
No. They are on the page. That is the point of them.
Another 120 of them, with a video of the working for every one, weighted toward the design domain where the calculations are.
Temperature-corrected open-circuit voltage and string length. It is the most reliable source of marks and the most expensive mistake to make in the field.
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