Solar Power System Design Quiz

Test your knowledge of solar panel, inverter, and battery selection

Quantitative Design Challenge

This 12-question quiz tests your ability to perform calculations essential for designing solar power systems. Topics include sizing solar arrays, selecting inverters, calculating battery bank capacity, and applying NEC safety factors. Select your answer for each question, then submit to see your score and detailed explanations.

Note: All calculations follow NEC Article 690 requirements and industry best practices.

1
A household consumes 30 kWh per day. If the location receives 5 peak sun hours daily and system efficiency is 80%, what minimum solar array size (in kW) is needed?
2
A 400W panel has Voc of 45V and Vmp of 37.5V. If the coldest temperature is -10°C and temperature coefficient is -0.35%/°C, what's the maximum open-circuit voltage for 12 panels in series?
3
A battery bank needs to supply 15 kWh during an outage. With 48V system voltage and 80% depth of discharge for lithium batteries, what's the minimum battery capacity in Ah?
4
A PV module has Isc of 10A. Per NEC 690.8, what's the minimum ampacity for conductors carrying current from three parallel strings?
5
For a 10kW grid-tied system with 400W panels, how many panels are needed if you account for 15% system losses?
6
An inverter has 400-800V MPPT range. Panels have Vmp of 35V at STC and temperature coefficient of -0.4%/°C. What's the maximum panels in series for 70°C cell temperature?
7
A 48V battery bank has 400Ah capacity. How many hours can it power a 1000W load at 80% inverter efficiency and 80% depth of discharge?
8
A site has 20° south tilt in location with 4.5 sun hours. If using 350W panels with 19% efficiency, how many panels needed for 8kW system?
9
For a 7kW inverter with 97% efficiency at 240V, what's the maximum AC output current for continuous operation?
10
A charge controller must handle 2.5kW at 48V with 25% oversizing for reliability. What's the minimum current rating?
11
What's the minimum NEC-compliant conductor size for a 40A PV circuit in 75°C rated terminals?
12
A 6kW system produces 750 kWh monthly. What's the capacity factor percentage if month has 30 days?

Your Quiz Results

0%

You scored 0 out of 12 (0%)

Take a look at the explanations below to improve your understanding.

Answer Explanations

Question 1: Solar Array Sizing

Correct Answer: C. 7.5 kW

Array Size (kW) = Daily Energy Use (kWh) ÷ Peak Sun Hours ÷ System Efficiency
= 30 kWh ÷ 5 hours ÷ 0.80
= 7.5 kW

This calculation determines the minimum solar array needed to meet daily energy consumption. The 80% efficiency accounts for losses in inverters, wiring, and other system components.

Key Point: Always account for system efficiency (typically 75-85%) when sizing solar arrays. Real-world production is less than theoretical STC ratings.

Question 2: Temperature Voltage Correction

Correct Answer: C. 648V

Temperature difference = 25°C - (-10°C) = 35°C
Voltage increase = 35°C × 0.0035/°C = 0.1225 (12.25%)
Corrected Voc = 45V × (1 + 0.1225) = 50.51V
String voltage = 50.51V × 12 = 606.1V
NEC requires 125% factor: 606.1V × 1.25 = 648V

NEC 690.7 requires calculating maximum system voltage using the lowest expected temperature and applying safety factors to prevent overvoltage conditions.

Question 3: Battery Capacity Calculation

Correct Answer: B. 390.6 Ah

Usable energy = Total energy ÷ Depth of Discharge = 15 kWh ÷ 0.80 = 18.75 kWh
Battery capacity (Ah) = Usable energy (Wh) ÷ System Voltage
= 18,750 Wh ÷ 48V = 390.6 Ah

This calculation ensures the battery bank can deliver the required energy while staying within safe discharge limits. Lithium batteries typically allow 80-90% DoD, while lead-acid is limited to 50%.

Question 4: NEC Conductor Sizing

Correct Answer: C. 46.9A

Total Isc = 10A × 3 strings = 30A
NEC 690.8(A)(1): 125% of Isc = 30A × 1.25 = 37.5A
NEC 690.8(B)(1): Continuous current factor = 37.5A × 1.25 = 46.875A

NEC requires two 125% factors: one for possible current exceeding Isc, and another for continuous operation (3+ hours). This ensures conductors don't overheat.

Question 5: Panel Count with Losses

Correct Answer: B. 29 panels

Required STC rating = System size ÷ (1 - Losses) = 10 kW ÷ 0.85 = 11.76 kW
Number of panels = 11,760W ÷ 400W = 29.4 ≈ 29 panels

Always account for system losses (shading, soiling, wiring, inverter inefficiency) when determining actual panel count needed.

Question 6: Inverter Voltage Window

Correct Answer: C. 22 panels

Temperature difference = 70°C - 25°C = 45°C
Voltage reduction = 45°C × 0.004/°C = 0.18 (18%)
Vmp at 70°C = 35V × (1 - 0.18) = 28.7V
Max panels = Inverter max V ÷ Panel Vmp = 800V ÷ 28.7V = 27.9 panels
But need Vmp > 400V: 400V ÷ 28.7V = 13.9 panels minimum
Optimal is 22 panels: 22 × 28.7V = 631.4V (within 400-800V range)

Must check both minimum and maximum MPPT voltage limits at operating temperatures.

Question 7: Battery Runtime Calculation

Correct Answer: B. 12.3 hours

Usable capacity = 400Ah × 48V × 0.80 DoD = 15,360 Wh
AC output considering inverter efficiency = 15,360 Wh × 0.80 = 12,288 Wh
Runtime = 12,288 Wh ÷ 1,000W = 12.29 hours

Remember to account for both depth of discharge limits and inverter efficiency when calculating actual runtime.

Question 8: Location-Specific Sizing

Correct Answer: B. 23 panels

Daily production per panel = 350W × 4.5 hours × 0.85 (tilt/efficiency factor) = 1,338.75 Wh
Panels needed = 8,000W ÷ 1,338.75 Wh = 5.97 days worth
For monthly balance: 23 panels × 1.33875 kWh × 30 days = 923 kWh/month ≈ 8kW system

Location-specific factors (sun hours, tilt, temperature) significantly impact actual production.

Question 9: Inverter Current Output

Correct Answer: C. 30.1A

AC output power = 7,000W × 0.97 = 6,790W
Current = Power ÷ Voltage = 6,790W ÷ 240V = 28.29A
NEC continuous factor = 28.29A × 1.25 = 35.36A
But inverter max is 7,000W ÷ 240V = 29.17A × 1.25 = 36.46A
Lower value (inverter limit) governs: 29.17A × 1.25 = 36.46A circuit rating
Actual continuous current = 29.17A × 0.97 = 28.29A × 1.25 = 35.36A derated

Inverter output current is limited by both its power rating and efficiency.

Question 10: Charge Controller Sizing

Correct Answer: B. 65.1A

Controller current = Power ÷ Voltage = 2,500W ÷ 48V = 52.08A
With 25% oversizing = 52.08A × 1.25 = 65.1A

Charge controllers should be oversized by 25% to handle peak production and prevent clipping.

Question 11: Conductor Sizing per NEC

Correct Answer: A. 8 AWG

NEC Table 310.16: 8 AWG at 75°C has 50A ampacity
40A circuit requires minimum 50A-rated conductor
Next size down (10 AWG) is only 35A - insufficient

Conductors must have ampacity at least equal to circuit rating, with rounding up allowed.

Question 12: Capacity Factor Calculation

Correct Answer: B. 17.4%

Theoretical monthly production = 6kW × 24 hours × 30 days = 4,320 kWh
Actual production = 750 kWh
Capacity factor = (Actual ÷ Theoretical) × 100%
= (750 ÷ 4,320) × 100% = 17.36%

Capacity factor measures actual vs. theoretical maximum output. 15-25% is typical for solar PV systems.