Solar Exam Math II: Voltage Drop, the 120% Rule & Energy Production Estimates

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Alright, Solar Exam Math II. Don't let the plain title fool you. Here's the part that actually matters on the job: voltage drop, busbar limits, and energy estimates are the calculations that decide whether a design works for the customer and the utility — and they're exam favorites because they mix formulas, code tables, and judgment. Let's work each one forward and backward.

Educational content only — not official NABCEP material, electrical design, or a substitute for a licensed electrician. Eligibility rules, exam content, fees, and renewal requirements are set by NABCEP and change; always confirm them in NABCEP's current candidate handbook and exam blueprint. Real installations must follow the NEC (NFPA 70) edition adopted by your Authority Having Jurisdiction, the manufacturers' instructions, and your utility's rules.

Voltage drop

For a two-wire DC circuit or a single-phase AC circuit: VD = (2 x L x I x R) / 1,000

• L = one-way length in feet
• I = current in amps (operating current, such as Imp, for energy loss)
• R = conductor resistance in ohms per 1,000 feet — from NEC Chapter 9, Table 8 for DC (and as a close approximation for small AC conductors; Table 9 gives AC impedance)

Percent VD = VD / circuit voltage x 100

Typical Table 8 values for uncoated stranded copper (check your edition): 12 AWG ≈ 1.98, 10 AWG ≈ 1.24, 8 AWG ≈ 0.778, 6 AWG ≈ 0.491 ohms per 1,000 ft.

Example 1 (forward). A string with Imp = 10.5 A, Vmp = 400 V, 10 AWG copper, 120 ft one way. VD = 2 x 120 x 10.5 x 1.24 / 1,000 = 3.12 V; 3.12 / 400 = 0.78%.

Example 2 (backward — choose a wire). An inverter output circuit at 240 V, 32 A, 100 ft one way. Target: 2% or less. Allowed VD = 0.02 x 240 = 4.8 V. Maximum R = VD x 1,000 / (2 x L x I) = 4.8 x 1,000 / (2 x 100 x 32) = 0.75 ohm per 1,000 ft. 8 AWG (0.778) is slightly too high; 6 AWG (0.491) meets the target. (The conductor must also meet ampacity requirements.)

Voltage drop limits are mostly informational notes in the NEC, not mandatory rules; designers and exam questions commonly use targets around 1–3%. Read the question for the target it gives.

The 120% rule (NEC 705.12)

For a load-side backfed breaker in a panel: (125% x inverter continuous output current) + main breaker rating ≤ 120% x busbar rating ...with the PV breaker at the opposite end of the busbar from the main breaker.

Example 3. 200 A busbar, 200 A main. 240 - 200 = 40 A available → inverter current ≤ 40 / 1.25 = 32 A → 32 x 240 V = 7,680 W max inverter.

Example 4. 225 A busbar, 200 A main. 270 - 200 = 70 A → 70 / 1.25 = 56 A of inverter output.

Example 5 (two inverters). 200 A busbar, 200 A main, two inverters at 16 A each. 125% x 32 A total = 40 A; 200 + 40 = 240 ≤ 240. Passes — exactly at the limit.

Example 6 (main derate). 200 A busbar, inverter output 40 A (needs 50 A). What main size works? Main ≤ 240 - 50 = 190 A → next standard size down, 175 A — only if the load calculation allows.

Also know the 100% rule (125% inverter current + main ≤ busbar, any position) and the supply-side connection (705.11), which avoids the busbar limit.

Energy production estimates

A simple annual estimate: Annual kWh ≈ array kW (DC, STC) x peak sun hours per day x 365 x derate factor

• Peak sun hours (PSH) = average daily insolation in kWh/m²/day at the array's tilt and orientation — one PSH equals one hour at 1,000 W/m².
• Derate factor combines losses: temperature, soiling, wiring, inverter efficiency, mismatch, shading, and availability. Exam problems often give it, commonly somewhere around 0.75 to 0.85.

Example 7. 6 kW array, 5.0 PSH, derate 0.80. 6 x 5.0 x 365 x 0.80 = 8,760 kWh/year.

Example 8 (backward — size the array). Customer uses 10,000 kWh/year and wants to offset 100%. Site: 4.5 PSH, derate 0.80. kW = 10,000 / (4.5 x 365 x 0.80) = 10,000 / 1,314 = 7.6 kW DC. With 400 W modules: 7,600 / 400 = 19 modules.

Example 9 (specific yield). A 7.6 kW system producing 10,000 kWh/year has a specific yield of 10,000 / 7.6 ≈ 1,316 kWh/kW/year — a handy number for comparing sites and spotting underperformance.

Professional estimates use modeling software with hourly weather data, but the exam expects the hand method and an understanding of what each loss means.

Going Deeper (Intermediate)

Exam questions often combine steps: find the string current, compute voltage drop, then decide whether to upsize; or find the inverter size, check the 120% rule, then choose the breaker. Break combined problems into labeled steps.

Remember inverter efficiency when going from DC to AC energy, if the question separates it from the derate. And check whether a question's PSH is for horizontal or plane-of-array — tilt matters.

Advanced / Pro-Level

Know how to work each formula backward: maximum length for a given voltage drop (L = VD x 1,000 / (2 x I x R)), maximum inverter size for a given panel, required array size for a given usage. Many exam questions are written in reverse. Estimate before you calculate: if a 7 kW system in a sunny climate comes out to 50,000 kWh a year, you made a mistake.

Practice Challenge

A home with a 200 A busbar and 150 A main uses 12,000 kWh per year. The site has 4.8 PSH and a 0.78 derate. What array size offsets usage, and what is the largest inverter output current the 120% rule allows? (Answer: array = 12,000 / (4.8 x 365 x 0.78) = 12,000 / 1,366.6 ≈ 8.8 kW DC. Busbar: 240 - 150 = 90 A available; 90 / 1.25 = 72 A of inverter output — plenty of room for a typical inverter for an 8.8 kW array.)

In Practice

Run these numbers on every real job you touch — the panel label, the conduit run, the customer's bill. Numbers you've used in the field stick in a way flashcards never do.

Common Mistakes to Avoid

• Forgetting the 2 in the voltage drop formula (current goes out and back)
• Using diameter or area tables instead of Table 8 resistance
• Leaving out the 125% multiplier in the 120% rule
• Confusing kW (size) with kWh (energy)
• Forgetting the derate factor in production estimates

Takeaway: Voltage drop is 2 x L x I x R / 1,000; the 120% rule is 125% of inverter current plus the main within 120% of the busbar; annual energy is kW x PSH x 365 x derate — and every one of them must work backward as well as forward.