22 Electrical Fundamentals & PV Device Behavior Practice Questions & Answers

1. 1. A PV source-circuit conductor carries 9 A, and the conductor's total resistance is 0.4 ohm. Using Ohm's law, what is the voltage dropped across that conductor?

   • A.32.4 V
   • B.3.6 V✓ Answer
   • C.22.5 V
   • D.0.044 V

   Ohm's law states V = I x R, so 9 A x 0.4 ohm = 3.6 V. Dividing instead of multiplying produces 22.5 V or 0.044 V, and 32.4 is the I-squared-R heating power in watts, not a voltage.

   Source: NABCEP PV Associate Job Task Analysis, electrical concepts — Ohm's lawReport a problem with this question

2. 2. A module is operating at 38 V and 9.5 A. Using the power relationship P = V x I, what power is the module producing at that moment?

   • A.361 W✓ Answer
   • B.47.5 W
   • C.3,610 W
   • D.4.0 W

   Power is the product of voltage and current: 38 V x 9.5 A = 361 W. The 47.5 W option comes from adding the two figures, 4.0 W from dividing voltage by current, and 3,610 W from moving the decimal point one place.

   Source: NABCEP PV Associate Job Task Analysis, electrical concepts — power relationship P = V x IReport a problem with this question

3. 3. Current in a conductor doubles while the conductor's resistance stays the same. What happens to the power dissipated as heat in that conductor?

   • A.Stays the same as it was before
   • B.Rises to roughly two times the value
   • C.Falls to half the original value
   • D.Rises to about four times the value✓ Answer

   Resistive heating follows P = I squared x R, so the power varies with the square of the current. Doubling the current with resistance unchanged multiplies the heat by four, which is why an undersized conductor overheats quickly as current rises.

   Source: Basic electrical theory of resistive heating, P = I squared x RReport a problem with this question

4. 4. A PV array produces direct current while the utility grid distributes alternating current. Which statement explains why?

   • A.The junction reverses polarity sixty times a second, matching the grid's waveform
   • B.The junction drives current one way only, so an inverter makes the alternating waveform✓ Answer
   • C.The cell output alternates in bright sun and is direct current in low light
   • D.The cell makes alternating current that the array wiring rectifies to direct current

   Light absorbed at the cell's p-n junction separates charge carriers and drives them in one fixed direction, so the cell output is inherently DC. Transmission and distribution use AC, so an inverter must convert the array's DC into an alternating waveform synchronised with the grid.

   Source: PVEducation photovoltaics reference, the photovoltaic effect at the p-n junctionReport a problem with this question

5. 5. Ten identical modules are wired one after another in a single series string. Compared with one module on its own, what does the string produce?

   • A.The currents add and the voltage stays that of one module
   • B.Both the voltage and the current stay at one module's value
   • C.The voltages add and the current stays that of one module✓ Answer
   • D.Both the voltage and the current add across the modules

   In a series connection the same current must pass through every module in turn, so current is unchanged while the individual module voltages sum. Ten modules in series therefore give ten times the voltage of one module at the same current.

   Source: NABCEP PV Associate Job Task Analysis, electrical concepts — series and parallel circuitsReport a problem with this question

6. 6. Each of three identical strings produces 9 A at 360 V. The three strings are connected in parallel into one combiner. What does the combined output measure?

   • A.27 A at 360 V✓ Answer
   • B.9 A at 1,080 V
   • C.9 A at 360 V
   • D.27 A at 1,080 V

   Parallel branches share the same voltage while their currents add, so three 9 A strings give 27 A and the voltage stays at 360 V. Adding the voltages would only be correct for a series connection.

   Source: NABCEP PV Associate Job Task Analysis, electrical concepts — series and parallel circuitsReport a problem with this question

7. 7. A module is rated Voc 41 V and Isc 10 A. An array uses 12 of these modules per series string, with 4 such strings in parallel. What are the array's total Voc and Isc?

   • A.492 V and 40 A✓ Answer
   • B.492 V and 10 A
   • C.41 V and 480 A
   • D.164 V and 120 A

   Series connection multiplies voltage: 12 x 41 V = 492 V, with string current still 10 A. Parallel connection multiplies current: 4 x 10 A = 40 A, with voltage unchanged. The array is therefore 492 V and 40 A.

   Source: NABCEP PV Associate Job Task Analysis, electrical concepts — array voltage and current from series and parallel connectionsReport a problem with this question

8. 8. A DC circuit carries a fixed current between the array and the inverter. Which change would increase the percentage voltage drop on that circuit?

   • A.Increasing the conductor's cross-sectional area
   • B.Reducing the current that the circuit carries
   • C.Raising the operating voltage of the circuit
   • D.Lengthening the run from array to inverter✓ Answer

   Conductor resistance rises with the length of the run and falls as cross-sectional area grows, and the drop equals current times that resistance. Lengthening the run therefore raises the drop, while a larger conductor, less current, or a higher operating voltage all reduce it as a percentage.

   Source: Voltage-drop principle: conductor resistance varies with length and inversely with cross-sectional areaReport a problem with this question

9. 9. Excessive voltage drop is found on the conductors between a PV array and its inverter. What is the practical consequence for the system?

   • A.Short-circuit current rises to make up for the lost voltage
   • B.The inverter raises its own output voltage to compensate
   • C.Power is lost as heat in the wiring, so less reaches the inverter✓ Answer
   • D.Current falls to zero once the drop passes a few percent

   The voltage missing at the far end of the run has been dropped across the resistance of the conductors themselves, and that lost voltage times the circuit current is power turned into heat. The result is less energy delivered, warmer conductors, and an operating voltage that may sit below where the inverter tracks best.

   Source: Voltage-drop principle and resistive power loss in PV circuit conductorsReport a problem with this question

10. 10. Under a steady light level, and operating below its maximum power point, how does a photovoltaic cell behave electrically?

    • A.A fixed resistance whose value falls as irradiance rises
    • B.A constant-voltage source whose current follows the load
    • C.A battery that stores charge by day and releases it later
    • D.A light-driven current source that tracks the irradiance✓ Answer

    The number of charge carriers a cell can deliver is set by the number of photons it absorbs, so its current stays nearly constant across a wide range of terminal voltage below the knee of the curve. A cell stores no energy of its own; it converts light only while light falls on it.

    Source: PVEducation photovoltaics reference, solar cell operation as a light-generated current sourceReport a problem with this question

11. 11. On a module's current-voltage curve, where are short-circuit current (Isc) and open-circuit voltage (Voc) read?

    • A.Isc is read at zero volts and Voc at zero current✓ Answer
    • B.Both are read at the maximum power point of the curve
    • C.Isc and Voc are both read with the module under load
    • D.Isc is read at zero current and Voc at zero volts

    Isc is the current that flows when the terminals are shorted together, so the voltage between them is zero; Voc is the voltage across open terminals, where no current flows. Between those two endpoints lies the knee, where the product of voltage and current is greatest.

    Source: PVEducation photovoltaics reference, IV curve parameters — short-circuit current and open-circuit voltageReport a problem with this question

12. 12. A module datasheet lists Voc 41 V, Isc 11 A, Vmp 34 V and Imp 10.5 A. What is the module's maximum power?

    • A.374 W
    • B.451 W
    • C.357 W✓ Answer
    • D.430.5 W

    Maximum power is the product of the voltage and current at the knee of the curve: Vmp x Imp = 34 V x 10.5 A = 357 W. Multiplying Voc by Isc gives 451 W, a figure no module delivers, because the real power at each of those two endpoints is zero.

    Source: PVEducation photovoltaics reference, maximum power point of the IV curveReport a problem with this question

13. 13. Fill factor is quoted for a module's current-voltage curve. What does fill factor express?

    • A.The ratio of maximum power to the product of Voc and Isc✓ Answer
    • B.The fraction of sunlight converted to electricity
    • C.The ratio of rated power to the module's total area
    • D.The share of module surface that active cells cover

    Fill factor is (Vmp x Imp) divided by (Voc x Isc), a measure of how square the curve is near its knee. It falls as series resistance rises, for example with corroded or loose terminations, so a declining fill factor is a useful maintenance signal.

    Source: PVEducation photovoltaics reference, fill factor of a solar cell IV curveReport a problem with this question

14. 14. Irradiance on an array falls from 1,000 W/m² to 500 W/m² while cell temperature is unchanged. What happens to the current the modules deliver?

    • A.Rises a little as the cells run cooler
    • B.Drops to about one quarter of the value
    • C.Falls to roughly half, tracking irradiance✓ Answer
    • D.Stays nearly constant, as the voltage does too

    Photocurrent is set by the number of photons absorbed, so short-circuit and operating current scale almost linearly with irradiance: half the light gives roughly half the current. Open-circuit voltage changes only weakly with light level, so it is the voltage, not the current, that stays nearly constant.

    Source: PVEducation photovoltaics reference, effect of light intensity on solar cell IV characteristicsReport a problem with this question

15. 15. A partly shaded string measures an open-circuit voltage close to the expected value, yet its operating current is far below normal. What explains this pattern?

    • A.Voltage rises under shade because less current flows
    • B.Current tracks irradiance closely; voltage barely moves✓ Answer
    • C.Cell voltage is fixed by the cell count and never varies
    • D.Shade affects only wiring resistance, not the cells

    Current is very nearly proportional to irradiance, while open-circuit voltage changes only logarithmically with light level and so barely moves until the light is extremely low. A near-normal Voc with badly reduced current is therefore the signature of shading or soiling rather than an open connection, which would show little or no voltage.

    Source: PVEducation photovoltaics reference, effect of light intensity on short-circuit current and open-circuit voltageReport a problem with this question

16. 16. Cell temperature rises well above 25 degC while irradiance holds steady. Which set of changes describes what happens to a crystalline silicon module?

    • A.Voc and Isc both fall, and Pmax stays constant
    • B.Voc rises, Isc falls slightly, and Pmax rises
    • C.Voc falls, Isc rises slightly, and Pmax falls✓ Answer
    • D.Voc and Isc both rise, and Pmax rises as well

    Heat raises the cell's reverse saturation current, pushing open-circuit voltage down by roughly three tenths of a percent per degree C, while the narrowing bandgap lifts short-circuit current only very slightly, on the order of five hundredths of a percent per degree C. The voltage loss dominates, so maximum power falls as the cells get hotter.

    Source: PVEducation photovoltaics reference, effect of temperature on solar cell IV characteristicsReport a problem with this question

17. 17. A module has an open-circuit voltage of 40.0 V at 25 degC and a Voc temperature coefficient of -0.30 %/degC. On a cold morning the cell temperature is -10 degC. What is the module's open-circuit voltage?

    • A.35.8 V
    • B.44.2 V✓ Answer
    • C.47.2 V
    • D.40.0 V

    The cell is 35 degC below the rating temperature, and -0.30 %/degC x -35 degC is a gain of 10.5 %, so 40.0 V x 1.105 = 44.2 V. Applying the sign the wrong way round gives 35.8 V, and that is the mistake that lets a cold-weather string exceed an inverter's maximum input voltage.

    Source: PVEducation photovoltaics reference, temperature coefficient of open-circuit voltageReport a problem with this question

18. 18. When during the year does a fixed PV array reach its highest DC voltage?

    • A.On a clear, very cold morning just after sunlight reaches the array✓ Answer
    • B.On a mild overcast day when diffuse light is spread evenly
    • C.On the hottest summer afternoon, when irradiance is at its peak
    • D.During a warm, humid evening while the array is still working

    Module voltage rises as cell temperature falls, and it takes only modest light to bring a cell close to its open-circuit voltage. The coldest bright moment of the year therefore sets the highest voltage the wiring and the inverter will ever see, while a hot afternoon gives the lowest voltage and the lowest power.

    Source: PVEducation photovoltaics reference, temperature dependence of module voltage; Sandia PV Performance Modeling CollaborativeReport a problem with this question

19. 19. One module in a twelve-module series string is heavily soiled and can pass less current than its neighbours. What happens to the string?

    • A.Only one twelfth of the string output is lost, in proportion
    • B.The other modules raise their current to cover the shortfall
    • C.String voltage drops to roughly that of the weakest module
    • D.String current drops to roughly that of the weakest module✓ Answer

    The same current must flow through every module in a series string, so the whole string is held down to what its weakest member can pass. That is why one shaded or soiled module costs far more than its own share of the array, and why module-level electronics are used where mismatch cannot be avoided.

    Source: PVEducation photovoltaics reference, module mismatch in series-connected cells and modulesReport a problem with this question

20. 20. A bypass diode is fitted across a group of cells inside a module. What does that diode do when part of the group is shaded?

    • A.It disconnects the shaded module until the shade has passed
    • B.It blocks current from flowing back into the array from a battery
    • C.It routes current around a shaded cell group, limiting heating✓ Answer
    • D.It restores a shaded module's voltage to its unshaded value

    A shaded cell in a series group becomes reverse biased and dissipates the string current as heat, creating a hot spot. The bypass diode conducts and gives that current a path around the affected group, protecting the cells at the cost of losing that group's contribution. Preventing reverse current from a battery is instead the job of a blocking diode.

    Source: PVEducation photovoltaics reference, bypass diodes and hot-spot protectionReport a problem with this question

21. 21. How do bonding and grounding differ in a PV installation?

    • A.Grounding joins metal parts to a common potential; bonding connects it to earth
    • B.Both terms describe the same connection and are fully interchangeable
    • C.Bonding carries the normal operating current; grounding carries fault current
    • D.Bonding joins metal parts to a common potential; grounding ties it to earth✓ Answer

    Bonding connects metal enclosures, racking and frames together so they sit at the same potential and fault current has a low-impedance path back to its source; grounding is the separate act of referencing the system to earth. The equipment grounding conductor exists to carry that fault current and open the protective device, and it carries no current in normal operation.

    Source: NFPA 70 definitions of bonding, grounding and the equipment grounding conductor; NABCEP PV Associate Job Task Analysis, grounding and bondingReport a problem with this question

22. 22. Why is a ground fault on the DC side of a PV array especially hazardous?

    • A.The lit array keeps feeding the fault, and a DC arc will not self-extinguish✓ Answer
    • B.A ground fault matters only once the inverter has begun exporting
    • C.Direct current cannot sustain an arc, so only the AC side is risky
    • D.Opening the DC disconnect removes all voltage from the faulted wires

    An illuminated module cannot be switched off, so opening a disconnect does not de-energise the conductors on the array side of it and fault current keeps flowing. Direct current also has no zero crossing to interrupt an arc, so a DC arc persists once struck. Treat array conductors as live whenever there is light, and have a qualified person clear the fault.

    Source: NABCEP PV Associate Job Task Analysis, PV electrical safety — energised array conductors and DC arcingReport 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 →