Voltage Drop Calculations for the C-10 Exam
Voltage drop separates the people who memorized a formula from the people who understand what the formula is doing. The math isn't hard. The traps live in the setup: one-way versus round-trip distance, single-phase versus three-phase, and knowing that the code recommends limits rather than mandating them. Get those three straight and these questions turn into free points.
These concepts are explained here to help you study. Always defer to the current adopted code and confirm anything that changes with the CSLB at cslb.ca.gov.
What the code actually says (and doesn't)
Here's the part that trips people up: the NEC does not flat-out require you to limit voltage drop on general branch circuits and feeders. The familiar 3% and 5% figures live in informational notes to 210.19(A) for branch circuits and 215.2(A) for feeders. Informational notes are guidance, not enforceable text.
The recommendation breaks down like this:
| Segment | Recommended max drop |
|---|---|
| Branch circuit only | 3% |
| Feeder only | 3% |
| Feeder + branch combined | 5% total |
On a 120V circuit, 3% is 3.6 volts on the branch. Across a feeder and branch together, 5% of 120V is 6 volts total from the service to the farthest outlet. When a question says "per NEC recommendation" or "to keep voltage drop within recommended limits," that 3%/5% pairing is what they want.
There are places where voltage drop becomes mandatory — sensitive electronic equipment, fire pump feeders, and a handful of specific-purpose articles carry hard limits. But for standard branch and feeder questions, treat 3%/5% as the target and know it's advisory. Knowing what's actually on the C-10 electrical exam helps you predict which flavor of question you'll get.
The two formulas you have to know cold
Both formulas use K, the resistivity constant — the resistance of one circular mil-foot of conductor. Use roughly 12.9 for copper and 21.2 for aluminum. Some references round copper to 12 or 11; the exam usually hands you the value to use, so read the problem.
Single-phase:
VD = (2 × K × I × D) / CM
Three-phase:
VD = (1.732 × K × I × D) / CM
Where:
- VD = voltage drop in volts
- K = resistivity constant (12.9 Cu / 21.2 Al)
- I = current in amps
- D = one-way distance in feet
- CM = conductor cross-section in circular mils (Chapter 9, Table 8)
The only difference between the two is the multiplier out front: 2 for single-phase, 1.732 (the square root of 3) for three-phase. The 2 accounts for current traveling out on the hot and back on the neutral — the full round trip. Three-phase circuits don't double, because of how the phases share the return path, so you use 1.732 instead. That's the single most common mistake on these problems: doubling a three-phase run, or forgetting to double a single-phase one.
The other constant killer is D. D is one-way distance in the standard formula — panel to load, not there-and-back. The round trip is already baked into that leading 2. If a question hands you total conductor length, don't double it again.
Worked example: single-phase
A 120V, single-phase circuit carries 16 amps to a load 150 feet away. The conductor is #12 copper (6,530 circular mils per Table 8). What's the voltage drop?
VD = (2 × 12.9 × 16 × 150) / 6,530
VD = 61,920 / 6,530
VD = 9.5 volts
That's 9.5V on a 120V circuit — about 7.9%. Way past the 3% branch target of 3.6V. This circuit needs a bigger wire, which brings us to the reverse calculation.
Solving for wire size
Most real exam questions don't ask "what's the drop." They ask "what size conductor keeps the drop under the recommended limit." Flip the formula to solve for CM:
CM = (2 × K × I × D) / VD_allowed
Same numbers, capping the drop at 3.6V (3% of 120V):
CM = (2 × 12.9 × 16 × 150) / 3.6
CM = 61,920 / 3.6
CM = 17,200 circular mils
Now go to Chapter 9, Table 8 and round up to the next standard size. #8 copper is 16,510 CM — just under, so it won't cut it. #6 copper is 26,240 CM — that clears it. You'd run #6 copper to hold the drop under 3%. Always round up; the next size down defeats the whole point.
This is exactly where voltage drop overrides ampacity. A #12 copper conductor carries the 16-amp load safely all day per conductor ampacity and wire sizing and Table 310.16. The load isn't the problem — the distance is. You upsize for voltage drop even though the smaller wire handles the current fine.
When to upsize conductors
Three conditions push you past what ampacity alone requires:
- Long runs. Distance is linear in the formula. Double the run, double the drop. Anything past 100 feet on a 120V branch deserves a check.
- High current. Current is linear too. A heavily loaded circuit drops more voltage over the same distance.
- Continuous and motor loads. Motors draw hard on startup and run continuously, so sagging voltage at the terminals causes overheating and nuisance trips. If you're working through motor and transformer calculations, voltage drop shows up there too.
A quick gut check before you commit to the full formula: on 120V circuits, drop climbs fast. On 240V or 480V three-phase, the same load over the same distance is a smaller percentage because the base voltage is higher. Raising system voltage is the cheapest fix for a long run when the design allows it.
Exam-day habits that save points
Read the question twice and lock down four things before you touch the calculator: single- or three-phase (sets your 2 vs. 1.732), one-way or total distance (don't double twice), the K value they want, and whether they're asking for volts dropped or a percentage. Percentage means one more step — divide your VD by the source voltage and multiply by 100.
If you're folding this into a study routine, working timed problems until the setup is automatic is what gets you through the electrical load calculations and voltage drop items without burning clock. Grinding practice questions and full timed exams is how this material actually sticks.
Voltage drop rewards a calm, checklist approach more than raw math speed. Nail the setup — phase, distance, constant, target — and the arithmetic is the easy part.