Voltage Drop Practice Problems
These 34 voltage drop practice problems cover DC basics, single-phase and three-phase circuits, conductor sizing, maximum distance, and applied calculation concepts. Every problem gives you the values you need; all are original, unofficial practice, not real exam questions.
A. DC and circuit basics
1. Wire loop resistance
A 12-V DC circuit carries 7.5 A. The outgoing and return wires together have 0.16 Ω of resistance (the load is not included). What is the voltage drop across the wires?
- A. 0.60 V
- B. 1.20 V
- C. 2.40 V
- D. 10.80 V
Show answer and worked solution
Answer: B — 1.20 V
Work it: VD = I × R = 7.5 × 0.16 = 1.20 V.
Why: The 0.16 Ω already covers both wires, so there is nothing to double.
The other choices: A — Halves a resistance that already includes both wires; C — Doubles the combined resistance again; D — That is the voltage left for the load (12 − 1.20), not the drop
Reference: OpenStax College Physics 2e §20.2; OpenStax College Physics 2e §21.1
For the remaining calculations, unless a problem says otherwise: copper conductors; unity power factor; reactance and AC-resistance corrections neglected; L = one-way length in feet; K = 12.9 for copper or 21.2 for aluminum, in Ω·cmil/ft; and √3 ≈ 1.732. Circuit voltages are source values; three-phase voltages and drops are line-to-line. Use the stated operating current and keep intermediate values unrounded.
Resistance values and percentage targets are supplied exercise inputs, not a claim about a current code-table entry or a universal installation limit. Conductor-sizing answers address voltage drop only.
Quick formula card (full formula reference below the problems)
| Find | Single-phase or DC (two-wire) | Balanced three-phase |
|---|---|---|
| Voltage drop, K method | VD = 2 × K × I × L ÷ CM | VD = 1.732 × K × I × L ÷ CM |
| Voltage drop, Ω per 1,000 ft | VD = 2 × L × R × I ÷ 1,000 | VD = 1.732 × L × R × I ÷ 1,000 |
| Conductor area | CM = 2 × K × I × L ÷ VD | CM = 1.732 × K × I × L ÷ VD |
Choose the smallest listed conductor area at or above the calculated requirement. Round maximum distances, currents and kVA down to the increment requested.
Jump to: B. Single-phase · C. Three-phase · D. Conductor sizing · E. Maximum distance and load · F. Applied targets and concepts
2. Wire resistance changes the current
An ideal 36-V DC source supplies an 8.40-Ω resistive load through wires with 0.60 Ω of total loop resistance. Ignore all other resistance. What voltage reaches the load? Round to 0.01 V.
- A. 33.43 V
- B. 33.60 V
- C. 35.40 V
- D. 36.00 V
Show answer and worked solution
Answer: B — 33.60 V
Work it: Total resistance = 8.40 + 0.60 = 9.00 Ω. I = 36 ÷ 9.00 = 4.00 A. Wire drop = 4.00 × 0.60 = 2.40 V. Load voltage = 36 − 2.40 = 33.60 V (check: 4.00 × 8.40 = 33.60 V).
Why: The wires sit in series with the load, so they reduce the current as well as taking their own share of the voltage.
The other choices: A — Finds current as 36 ÷ 8.40 (about 4.29 A), ignoring the wire resistance, then subtracts that larger drop; C — Subtracts 0.60 as if it were volts; D — Ignores the wires entirely
Reference: OpenStax College Physics 2e §20.2; OpenStax College Physics 2e §21.1
3. Percent drop from measurements
At the same moment, a meter reads 120.0 V at the source and 117.6 V at the load. What is the voltage drop as a percentage of the source voltage?
- A. 2.00%
- B. 2.04%
- C. 2.40%
- D. 98.00%
Show answer and worked solution
Answer: A — 2.00%
Work it: Drop = 120.0 − 117.6 = 2.4 V. Percent = 2.4 ÷ 120 × 100 = 2.00%.
Why: Percent voltage drop is measured against the source voltage.
The other choices: B — Divides by the load voltage (117.6 V); C — Writes the 2.4-V drop as a percentage; D — That is the share of voltage that arrives, not the share lost
Reference: Schneider Electric Electrical Installation Guide, “Calculation of voltage drop in steady load conditions,” Figure G29
4. Low-voltage DC run
A load on a 24-V DC supply draws 10 A. It is 50 ft (one way) from the supply on 10 AWG stranded copper; use the supplied resistance of 1.24 Ω per 1,000 ft. What is the percent voltage drop? Round to 0.01%.
- A. 1.24%
- B. 2.58%
- C. 4.47%
- D. 5.17%
Show answer and worked solution
Answer: D — 5.17%
Work it: VD = 2 × L × R × I ÷ 1,000 = 2 × 50 × 1.24 × 10 ÷ 1,000 = 1.24 V. Percent = 1.24 ÷ 24 × 100 ≈ 5.17%.
Why: Same two-wire math as any other circuit. At low voltage, the same few volts become a much bigger percentage.
The other choices: A — Reports the 1.24-V drop as a percentage; B — Counts only one conductor; C — Uses the three-phase multiplier (1.732) on a DC circuit
Reference: OpenStax College Physics 2e §20.2; OpenStax College Physics 2e §20.3
B. Single-phase voltage drop
5. K method, single-phase
A 120-V, two-wire circuit supplies an 18-A resistive load 90 ft (one way) from the panel on 12 AWG copper (6,530 cmil). Using K = 12.9, what is the voltage drop? Round to 0.01 V.
- A. 3.20 V
- B. 5.54 V
- C. 12.80 V
- D. 6.40 V
Show answer and worked solution
Answer: D — 6.40 V
Work it: VD = 2 × K × I × L ÷ CM = 2 × 12.9 × 18 × 90 ÷ 6,530 ≈ 6.40 V (about 5.33% of 120 V).
Why: Current flows out on one wire and back on the other. The 2 covers that return trip, so L stays one-way.
The other choices: A — Leaves out the 2; B — Uses 1.732, the three-phase multiplier; C — Doubles the length and also uses 2, counting the return path twice
Reference: Mike Holt, “Voltage Drop Calculations,” definitions of K, I, D and CM; NBS Handbook 100, Copper Wire Tables, Table 5 (circular-mil areas), printed p. 16
6. Resistance per 1,000 feet
A 240-V single-phase load draws 40 A. It is 150 ft (one way) away on 8 AWG stranded copper. Use 0.778 Ω per 1,000 ft and neglect reactance. What is the voltage drop? Round to 0.01 V.
- A. 3.89 V
- B. 4.67 V
- C. 8.08 V
- D. 9.34 V
Show answer and worked solution
Answer: D — 9.34 V
Work it: VD = 2 × L × R × I ÷ 1,000 = 2 × 150 × 0.778 × 40 ÷ 1,000 ≈ 9.34 V (about 3.89% of 240 V).
Why: Find the resistance of the 300 ft of conductor carrying current, then multiply by current.
The other choices: A — That is the percentage (3.89%) written as volts; B — Counts only one conductor; C — Uses the three-phase multiplier
Reference: OpenStax College Physics 2e §20.2; OpenStax College Physics 2e §20.3
7. The length already includes both wires
A 120-V, two-wire single-phase circuit carries 10 A to a resistive load. The total length of the outgoing plus return conductors is 160 ft, and the conductor resistance is 1.60 Ω per 1,000 ft. Neglect reactance. A student calculates 2 × 10 × 1.60 × 160 ÷ 1,000 = 5.12 V. What is the correct voltage drop?
- A. 1.28 V
- B. 2.56 V
- C. 5.12 V
- D. 16.00 V
Show answer and worked solution
Answer: B — 2.56 V
Work it: The 160 ft is already the full loop. R = 1.60 × 160 ÷ 1,000 = 0.256 Ω. VD = 10 × 0.256 = 2.56 V.
Why: Use 2 × one-way length, or the total loop length once — never both.
The other choices: A — Uses only half the loop, dropping the return conductor; C — The student’s answer: counts the loop twice; D — Treats 1.60 Ω per 1,000 ft as the resistance of this run (10 × 1.60)
Reference: OpenStax College Physics 2e §20.2; OpenStax College Physics 2e §21.1
8. Voltage at a motor (use running current)
A motor on a 240-V single-phase supply draws 24 A while running. It is 125 ft (one way) from its panel on 10 AWG copper (10,380 cmil), K = 12.9. For this simplified exercise, use unity power factor and neglect reactance. Using the running current, what voltage reaches the motor? Round to 0.1 V.
- A. 7.5 V
- B. 232.5 V
- C. 236.3 V
- D. 247.5 V
Show answer and worked solution
Answer: B — 232.5 V
Work it: VD = 2 × 12.9 × 24 × 125 ÷ 10,380 ≈ 7.46 V. Motor voltage ≈ 240 − 7.46 ≈ 232.5 V.
Why: Voltage drop uses the current the load actually draws — 24 A here, not 24 × 125%. This is a running-current calculation, not a starting-current or conductor-ampacity calculation.
The other choices: A — That is the drop, not the voltage at the motor; C — Counts one conductor (half the drop); D — Adds the drop instead of subtracting it
Reference: Mike Holt, “Voltage Drop Calculations,” definitions of K, I, D and CM; Schneider Electric Electrical Installation Guide, “Calculation of voltage drop in steady load conditions,” Figure G29; NBS Handbook 100, Copper Wire Tables, Table 5 (circular-mil areas), printed p. 16
9. Continuous load: 100%, not 125%
An EV charger draws 48 A continuously from a 240-V single-phase supply, 140 ft (one way) from the panel on 6 AWG copper (26,240 cmil), K = 12.9. What is the percent voltage drop at full charging current? Round to 0.01%.
- A. 1.38%
- B. 2.38%
- C. 2.75%
- D. 3.44%
Show answer and worked solution
Answer: C — 2.75%
Work it: VD = 2 × 12.9 × 48 × 140 ÷ 26,240 ≈ 6.61 V. Percent = 6.61 ÷ 240 × 100 ≈ 2.75%.
Why: Voltage drop depends on the current that actually flows. The stated charging current is 48 A, not 60 A; this question does not determine conductor ampacity or overcurrent protection.
The other choices: A — Leaves out the 2; B — Uses 1.732; D — Uses 60 A (48 × 125%)
Reference: Mike Holt, “Voltage Drop Calculations,” definitions of K, I, D and CM; NBS Handbook 100, Copper Wire Tables, Table 5 (circular-mil areas), printed p. 16
C. Three-phase voltage drop
10. Three-phase basics
A balanced 480-V three-phase load draws 60 A per line. It is 250 ft (one way) away on 4 AWG copper (41,740 cmil), K = 12.9. What is the line-to-line voltage drop? Round to 0.01 V.
- A. 4.64 V
- B. 8.03 V
- C. 9.27 V
- D. 24.09 V
Show answer and worked solution
Answer: B — 8.03 V
Work it: VD = 1.732 × K × I × L ÷ CM = 1.732 × 12.9 × 60 × 250 ÷ 41,740 ≈ 8.03 V (about 1.67%).
Why: For a balanced three-phase circuit, 1.732 (√3) replaces the single-phase 2. L is still one-way.
The other choices: A — No multiplier at all; C — Uses the single-phase 2; D — Uses 1.732 and also triples the length
Reference: Schneider Electric Electrical Installation Guide, “Calculation of voltage drop in steady load conditions,” Figure G29; Mike Holt, “Voltage Drop Calculations,” definitions of K, I, D and CM; NBS Handbook 100, Copper Wire Tables, Table 5 (circular-mil areas), printed p. 16
11. Three-phase percent drop at 208 V
A balanced 208-V three-phase load draws 32 A. It is 150 ft (one way) away on 10 AWG copper (10,380 cmil), K = 12.9. What is the percent voltage drop? Round to 0.01%.
- A. 4.97%
- B. 2.87%
- C. 5.74%
- D. 14.90%
Show answer and worked solution
Answer: A — 4.97%
Work it: VD = 1.732 × 12.9 × 32 × 150 ÷ 10,380 ≈ 10.33 V. Percent = 10.33 ÷ 208 × 100 ≈ 4.97%.
Why: The drop from this formula is line-to-line, so divide by the line-to-line voltage.
The other choices: B — No multiplier; C — Uses 2 instead of 1.732; D — Triples the length on top of 1.732
Reference: Schneider Electric Electrical Installation Guide, “Calculation of voltage drop in steady load conditions,” Figure G29; Mike Holt, “Voltage Drop Calculations,” definitions of K, I, D and CM; NBS Handbook 100, Copper Wire Tables, Table 5 (circular-mil areas), printed p. 16
12. Convert kVA to amps first
A balanced three-phase circuit draws 45 kVA at its 208-V panel supply and runs 100 ft (one way) to the load on 1/0 copper (105,600 cmil), K = 12.9. What is the voltage drop? Round to 0.01 V.
- A. 2.64 V
- B. 1.53 V
- C. 3.05 V
- D. 4.58 V
Show answer and worked solution
Answer: A — 2.64 V
Work it: I = 45,000 ÷ (1.732 × 208) ≈ 124.9134 A. Using the unrounded current, VD = 1.732 × 12.9 × I × 100 ÷ 105,600 ≈ 2.64 V.
Why: Three-phase current needs √3 in the denominator. Get the current right before you touch the voltage-drop formula.
The other choices: B — Divides the VA by 3 × 208; C — Uses 2 in the voltage-drop formula; D — Finds current as VA ÷ 208, with no √3
Reference: U.S. DOE Fundamentals Handbook, Electrical Science Vol. 3, Module 9, printed p. 21; Schneider Electric Electrical Installation Guide, “Calculation of voltage drop in steady load conditions,” Figure G29; NBS Handbook 100, Copper Wire Tables, Table 5 (circular-mil areas), printed p. 16
13. Three-phase with resistance per 1,000 feet
A balanced 480-V three-phase load draws 150 A. It is 400 ft (one way) away on 3/0 copper; use the supplied resistance of 0.0766 Ω per 1,000 ft. Neglect reactance. What is the voltage drop? Round to 0.01 V.
- A. 4.60 V
- B. 7.96 V
- C. 9.19 V
- D. 23.88 V
Show answer and worked solution
Answer: B — 7.96 V
Work it: VD = 1.732 × L × R × I ÷ 1,000 = 1.732 × 400 × 0.0766 × 150 ÷ 1,000 ≈ 7.96 V (about 1.66%).
Why: Same structure as the K method: multiplier × one-way length × resistance per foot × current.
The other choices: A — No multiplier; C — Uses 2; D — Triples the length
Reference: Schneider Electric Electrical Installation Guide, “Calculation of voltage drop in steady load conditions,” Figure G29
14. Parallel conductor sets
A 480-V balanced three-phase feeder carries 400 A for 300 ft (one way) using two 250-kcmil copper conductors in parallel per phase. K = 12.9. Assume identical conductor lengths and resistances, equal current sharing, and no AC-resistance correction or reactance. What is the voltage drop? Round to 0.01 V.
- A. 2.68 V
- B. 5.36 V
- C. 6.19 V
- D. 10.72 V
Show answer and worked solution
Answer: B — 5.36 V
Work it: Area per phase = 2 × 250,000 = 500,000 cmil. VD = 1.732 × 12.9 × 400 × 300 ÷ 500,000 ≈ 5.36 V.
Why: Parallel sets share the current. Use the full current with the total area, or half the current with one set’s area — same answer either way.
The other choices: A — Halves the current and doubles the area, counting the split twice; C — Uses 2; D — Uses one set’s area with the full current
Reference: OpenStax College Physics 2e §21.1; Schneider Electric Electrical Installation Guide, “Calculation of voltage drop in steady load conditions,” Figure G29; OpenStax College Physics 2e §20.3
15. Match the voltage base
A balanced 480Y/277-V system has a calculated line-to-line voltage drop of 8.40 V. Which is the percentage drop on the matching base?
- A. 100 × 8.40 ÷ 480 = 1.75%
- B. 100 × 8.40 ÷ 277 ≈ 3.03%
- C. 100 × (2 × 8.40) ÷ 480 = 3.50%
- D. 8.40 ÷ 480 = 0.0175%
Show answer and worked solution
Answer: A — 100 × 8.40 ÷ 480 = 1.75%
Work it: 8.40 ÷ 480 × 100 = 1.75%.
Why: A line-to-line drop is compared with the line-to-line voltage.
The other choices: B — Mixes a line-to-line drop with the line-to-neutral voltage; C — Doubles a drop that is already complete; D — Finds the fraction 0.0175 but labels it as a percentage without multiplying by 100
Reference: Schneider Electric Electrical Installation Guide, “Calculation of voltage drop in steady load conditions,” Figure G29
16. Power factor and reactance
A balanced 480-V three-phase load draws 80 A over 200 ft (one way). Each conductor has a supplied AC resistance of 0.30 Ω and reactance of 0.08 Ω per 1,000 ft. The load power factor is 0.80 lagging (cos θ = 0.80, sin θ = 0.60). Use the specified effective-impedance approximation Z = R cos θ + X sin θ and VD = 1.732 × Z × I × L ÷ 1,000. What is the voltage drop? Round to 0.01 V.
- A. 7.98 V
- B. 8.31 V
- C. 8.60 V
- D. 9.22 V
Show answer and worked solution
Answer: A — 7.98 V
Work it: Z = 0.30 × 0.80 + 0.08 × 0.60 = 0.288 Ω per 1,000 ft. VD = 1.732 × 0.288 × 80 × 200 ÷ 1,000 ≈ 7.98 V (about 1.66%).
Why: This question specifies the effective-impedance approximation, so use its resistance, reactance and power factor rather than resistance alone. This is an approximation for balanced steady loads, not a full phasor solution.
The other choices: B — Uses resistance only and ignores the power factor and reactance the question supplied; C — Uses the full impedance magnitude √(0.30² + 0.08²) instead of the effective impedance at this power factor; D — Uses the single-phase 2
Reference: Schneider Electric Electrical Installation Guide, “Calculation of voltage drop in steady load conditions,” Figure G29
D. Size the conductor for voltage drop
These problems answer the voltage-drop target only. A real conductor must also meet ampacity, terminal-temperature and overcurrent-protection rules. Conductor sizing involves several separate checks.
17. Minimum size for a 3% target
A 120-V single-phase, 20-A resistive load is 150 ft (one way) away on copper, K = 12.9. The target is 3% voltage drop. Which is the smallest listed conductor that meets it? Areas: 12 AWG = 6,530 cmil; 8 AWG = 16,510; 6 AWG = 26,240; 4 AWG = 41,740.
- A. 12 AWG
- B. 8 AWG
- C. 4 AWG
- D. 6 AWG
Show answer and worked solution
Answer: D — 6 AWG
Work it: Allowed drop = 0.03 × 120 = 3.6 V. CM = 2 × K × I × L ÷ VD = 2 × 12.9 × 20 × 150 ÷ 3.6 = 21,500 cmil → smallest listed area at or above this is 6 AWG. Check: 6 AWG gives 2.95 V; 8 AWG gives 4.69 V, over 3.6 V.
Why: Pick the first area at or above the calculated circular mils. This answers the voltage-drop target only — ampacity and overcurrent protection are separate checks.
The other choices: A — Its 6,530 cmil is below the required 21,500 cmil; it ignores the voltage-drop target; B — What you get using 5% (6 V) instead of 3% (12,900 cmil → 8 AWG); C — Works, but it is not the smallest size that meets 3%
Reference: Mike Holt, “Voltage Drop Calculations,” definitions of K, I, D and CM; NBS Handbook 100, Copper Wire Tables, Table 5 (circular-mil areas), printed p. 16; Schneider Electric, Conductor sizing: methodology and definition
18. Do not round down
A 240-V single-phase heat pump draws 45 A and is 190 ft (one way) from its panel. For this resistance-only exercise, use unity power factor, neglect reactance, and use copper, K = 12.9, with a 3% target. Smallest listed size? Areas: 8 AWG = 16,510 cmil; 6 AWG = 26,240; 4 AWG = 41,740; 3 AWG = 52,620.
- A. 8 AWG
- B. 6 AWG
- C. 4 AWG
- D. 3 AWG
Show answer and worked solution
Answer: C — 4 AWG
Work it: Allowed drop = 7.2 V. CM = 2 × 12.9 × 45 × 190 ÷ 7.2 = 30,637.5 cmil → 4 AWG. Check: 4 AWG gives 5.28 V; 6 AWG gives 8.41 V, over 7.2 V.
Why: Never round down to the closest size — the smaller conductor misses the target.
The other choices: A — Leaves out the 2 (15,319 cmil); B — Uses 5% (18,383 cmil), or rounds to the nearest size; D — Meets the target, but it is larger than needed
Reference: Mike Holt, “Voltage Drop Calculations,” definitions of K, I, D and CM; NBS Handbook 100, Copper Wire Tables, Table 5 (circular-mil areas), printed p. 16
19. Three-phase sizing
A balanced 208-V three-phase load draws 35 A and is 180 ft (one way) away. Copper, K = 12.9, target 3% of 208 V. Smallest listed size? Areas: 8 AWG = 16,510 cmil; 6 AWG = 26,240; 4 AWG = 41,740; 1 AWG = 83,690.
- A. 8 AWG
- B. 6 AWG
- C. 4 AWG
- D. 1 AWG
Show answer and worked solution
Answer: B — 6 AWG
Work it: Allowed drop = 0.03 × 208 = 6.24 V. CM = 1.732 × 12.9 × 35 × 180 ÷ 6.24 ≈ 22,558 cmil → 6 AWG. Check: 6 AWG gives 5.36 V; 8 AWG gives 8.53 V.
Why: Rearrange the three-phase formula the same way as single-phase, with 1.732 in place of 2.
The other choices: A — No multiplier (about 13,024 cmil); C — Takes 3% of 120 V instead of 208 V (about 39,100 cmil); D — Triples the length (about 67,673 cmil)
Reference: Mike Holt, “Voltage Drop Calculations,” definitions of K, I, D and CM; Schneider Electric Electrical Installation Guide, “Calculation of voltage drop in steady load conditions,” Figure G29; NBS Handbook 100, Copper Wire Tables, Table 5 (circular-mil areas), printed p. 16
20. Aluminum conductors
A 240-V single-phase, 60-A load is 150 ft (one way) away on aluminum conductors. Use K = 21.2 and a 3% target. Smallest listed size? Areas: 4 AWG = 41,740 cmil; 3 AWG = 52,620; 2 AWG = 66,360; 1 AWG = 83,690.
- A. 4 AWG
- B. 3 AWG
- C. 2 AWG
- D. 1 AWG
Show answer and worked solution
Answer: C — 2 AWG
Work it: Allowed drop = 7.2 V. CM = 2 × 21.2 × 60 × 150 ÷ 7.2 = 53,000 cmil. 3 AWG (52,620) is just short, so 2 AWG. Check: 3 AWG gives 7.25 V, over 7.2 V.
Why: Aluminum has more resistance than copper of the same area, so it needs a bigger K and a bigger conductor.
The other choices: A — Uses copper’s K = 12.9 (32,250 cmil); B — Rounds down to the closest area; D — Meets the target, but it is larger than needed
Reference: Mike Holt, “Voltage Drop Calculations,” definitions of K, I, D and CM; OpenStax College Physics 2e §20.3; NBS Handbook 100, Copper Wire Tables, Table 5 (circular-mil areas), printed p. 16
21. Splitting the 5% budget
For this exercise, the design targets are no more than 3% on a branch circuit and no more than 5% combined. The feeder already drops 1.5%; all three percentages use the same 120-V source base. A 120-V branch circuit serves a 20-A load 125 ft (one way) away. Copper, K = 12.9. Smallest listed branch conductor? Areas: 12 AWG = 6,530 cmil; 8 AWG = 16,510; 6 AWG = 26,240; 4 AWG = 41,740.
- A. 12 AWG
- B. 8 AWG
- C. 4 AWG
- D. 6 AWG
Show answer and worked solution
Answer: D — 6 AWG
Work it: Leftover total budget = 5% − 1.5% = 3.5%, but the branch cap is 3%. Use the smaller: 0.03 × 120 = 3.6 V. CM = 2 × 12.9 × 20 × 125 ÷ 3.6 ≈ 17,917 cmil → 6 AWG. Check: 8 AWG gives 3.26%.
Why: Both limits apply. The branch gets whichever allowance is smaller.
The other choices: A — Its 6,530 cmil is below the required 17,916.67 cmil; B — Uses the 3.5% leftover (15,357 cmil) and breaks the 3% branch limit; C — Uses the feeder’s 1.5% as the branch limit (35,833 cmil)
Reference: Mike Holt, “Voltage Drop Calculations,” definitions of K, I, D and CM; Schneider Electric Electrical Installation Guide, “Calculation of voltage drop in steady load conditions,” Figure G29; NBS Handbook 100, Copper Wire Tables, Table 5 (circular-mil areas), printed p. 16
E. Maximum distance and maximum load
22. Maximum one-way distance
A 240-V single-phase, 30-A load uses 10 AWG copper (10,380 cmil), K = 12.9. With a 3% limit, what is the longest one-way run, rounded down to the whole foot?
- A. 48 ft
- B. 193 ft
- C. 111 ft
- D. 96 ft
Show answer and worked solution
Answer: D — 96 ft
Work it: L = CM × VD ÷ (2 × K × I) = 10,380 × 7.2 ÷ (2 × 12.9 × 30) ≈ 96.56 ft → 96 ft. Check: 96 ft gives 7.16 V; 97 ft gives 7.23 V, over the 7.2-V limit.
Why: A maximum rounds down. Rounding 96.56 up to 97 ft breaks the limit.
The other choices: A — Divides by 4 instead of 2; B — Leaves out the 2; C — Uses 1.732
Reference: Mike Holt, “Voltage Drop Calculations,” definitions of K, I, D and CM; NBS Handbook 100, Copper Wire Tables, Table 5 (circular-mil areas), printed p. 16
23. Maximum load current
A 120-V single-phase circuit runs 80 ft (one way) on 12 AWG copper (6,530 cmil), K = 12.9. With a 3% limit, what is the largest load current, rounded down to 0.1 A?
- A. 5.6 A
- B. 11.3 A
- C. 13.1 A
- D. 22.7 A
Show answer and worked solution
Answer: B — 11.3 A
Work it: I = CM × VD ÷ (2 × K × L) = 6,530 × 3.6 ÷ (2 × 12.9 × 80) ≈ 11.39 A → 11.3 A. Check: 11.3 A gives about 3.572 V; 11.4 A gives about 3.603 V, above 3.6 V.
Why: Solve the same formula for I. As a maximum, it rounds down.
The other choices: A — Divides by 4; C — Uses 1.732; D — Leaves out the 2
Reference: Mike Holt, “Voltage Drop Calculations,” definitions of K, I, D and CM; NBS Handbook 100, Copper Wire Tables, Table 5 (circular-mil areas), printed p. 16
24. Maximum distance, three-phase
A balanced 480-V three-phase, 55-A load uses 6 AWG copper (26,240 cmil), K = 12.9. With a 3% limit, what is the longest one-way run, rounded down to the whole foot?
- A. 102 ft
- B. 266 ft
- C. 307 ft
- D. 532 ft
Show answer and worked solution
Answer: C — 307 ft
Work it: Allowed drop = 14.4 V. L = 26,240 × 14.4 ÷ (1.732 × 12.9 × 55) ≈ 307.5 ft → 307 ft. Check: 307 ft gives about 14.377 V; 308 ft gives about 14.424 V, above 14.4 V.
Why: Rearrange with 1.732 in place of 2.
The other choices: A — Divides by an extra 3; B — Uses 2; D — Leaves out the multiplier
Reference: Schneider Electric Electrical Installation Guide, “Calculation of voltage drop in steady load conditions,” Figure G29; Mike Holt, “Voltage Drop Calculations,” definitions of K, I, D and CM; NBS Handbook 100, Copper Wire Tables, Table 5 (circular-mil areas), printed p. 16
25. Maximum load in kVA
A balanced 208-V three-phase circuit runs 150 ft (one way) on 3 AWG copper (52,620 cmil), K = 12.9. With a 3% limit, what is the largest apparent power at the 208-V source, rounded down to 0.1 kVA?
- A. 20.3 kVA
- B. 30.5 kVA
- C. 61.1 kVA
- D. 35.2 kVA
Show answer and worked solution
Answer: D — 35.2 kVA
Work it: Allowed drop = 6.24 V. I = 52,620 × 6.24 ÷ (1.732 × 12.9 × 150) ≈ 97.9735 A. Using the unrounded current, kVA = 1.732 × 208 × I ÷ 1,000 ≈ 35.2954 kVA → 35.2 kVA. Check: 35.2 kVA gives about 6.22314 V; 35.3 kVA gives about 6.24082 V, above 6.24 V.
Why: Solve for current first, then convert to three-phase kVA with √3. Keep the unrounded current and round the final maximum down.
The other choices: A — Finds kVA as volts × amps with no √3; B — Uses 2 in the voltage-drop step; C — Multiplies by 3 instead of √3
Reference: U.S. DOE Fundamentals Handbook, Electrical Science Vol. 3, Module 9, printed p. 21; Schneider Electric Electrical Installation Guide, “Calculation of voltage drop in steady load conditions,” Figure G29; NBS Handbook 100, Copper Wire Tables, Table 5 (circular-mil areas), printed p. 16
F. Applied targets and concepts
26. Feeder plus branch
A 240-V feeder drops 4.8 V to a subpanel. A 240-V branch circuit from that subpanel drops 6.0 V at its farthest outlet. What is the combined percent drop, and how does it compare with the exercise targets (3% branch, 5% total), both measured on the 240-V source base?
- A. 2.25% — within both
- B. 2.5% — within both
- C. 4.5% — within both
- D. 10.8% — over 5%
Show answer and worked solution
Answer: C — 4.5% — within both
Work it: Feeder = 4.8 ÷ 240 × 100 = 2.0%. Branch = 6.0 ÷ 240 × 100 = 2.5%. Total = 10.8 ÷ 240 × 100 = 4.5%.
Why: Drops on the same voltage base add. The branch (2.5%) stays under 3%, and the total (4.5%) stays under 5%.
The other choices: A — Averages 2.0% and 2.5% instead of adding them; B — Counts the branch only; D — Writes 10.8 V as a percentage
Reference: Schneider Electric Electrical Installation Guide, “Calculation of voltage drop in steady load conditions,” Figure G29; OpenStax College Physics 2e §21.1
27. What the calculation proves
A supplied-data calculation gives a branch-circuit voltage drop of 2.8% against a stated 3% target. No ampacity, protection or installation checks have been performed. Which conclusion is supported?
- A. The entire installation is code-compliant
- B. The conductor is suitable for any load current
- C. The voltage-drop result meets the stated target; the other checks remain separate
- D. The equipment will operate correctly under every starting condition
Show answer and worked solution
Answer: C — The voltage-drop result meets the stated target; the other checks remain separate
Work it: 2.8% ≤ 3%, so this voltage-drop result meets the exercise target. That comparison does not evaluate the other conditions.
Why: Passing one calculation is not the same as completing the conductor and protection design.
The other choices: A — Claims compliance checks that were not performed; B — A voltage-drop percentage does not establish current-carrying capacity; D — The stem gives no starting-condition result or equipment operating limit
Reference: Schneider Electric, Conductor sizing: methodology and definition
28. The tighter remaining allowance
An exercise sets a 1.5% branch-circuit target and a 2.5% combined feeder-plus-branch target. The feeder already drops 1.4%. All percentages use the same source-voltage base. What is the largest remaining branch allowance?
- A. 1.1%
- B. 1.4%
- C. 1.5%
- D. 2.5%
Show answer and worked solution
Answer: A — 1.1%
Work it: Remaining combined allowance = 2.5% − 1.4% = 1.1%. The branch allowance is the smaller of 1.5% and 1.1%: 1.1%.
Why: Both targets must hold. Here the remaining combined allowance is tighter than the branch-only target.
The other choices: B — Uses the feeder drop as the branch allowance; C — Ignores the combined target; 1.4% + 1.5% would be 2.9%; D — Gives the entire combined allowance to the branch without subtracting the feeder
Reference: Schneider Electric Electrical Installation Guide, “Calculation of voltage drop in steady load conditions,” Figure G29
29. A stated starting-voltage target
For a hypothetical motor-starting exercise, a 480-V source is used and the stated maximum drop is 15% of that source voltage. What is the minimum terminal voltage that meets this exercise target?
- A. 72 V
- B. 432 V
- C. 456 V
- D. 408 V
Show answer and worked solution
Answer: D — 408 V
Work it: Allowed drop = 480 × 0.15 = 72 V. Minimum terminal voltage = 480 − 72 = 408 V.
Why: The question gives a percentage loss; subtract that loss from the source voltage to find what must remain. The 15% target is supplied for this exercise, not a fire-pump or general motor-code rule.
The other choices: A — Reports the allowed drop instead of the voltage remaining; B — Subtracts 10% instead of the stated 15%; C — Subtracts 5% instead of the stated 15%
Reference: Schneider Electric Electrical Installation Guide, “Calculation of voltage drop in steady load conditions,” Figure G29
30. Where K comes from
For this exercise, a 1/0 copper conductor has a supplied resistance of 0.122 Ω per 1,000 ft and an area of 105,600 cmil. What K, in Ω·cmil/ft, do those values imply? Round to 0.1.
- A. 1.3
- B. 12.6
- C. 21.2
- D. 12.9
Show answer and worked solution
Answer: D — 12.9
Work it: K = resistance per foot × area = (0.122 ÷ 1,000) × 105,600 ≈ 12.88 → 12.9.
Why: K has units of Ω·cmil/ft: resistance per foot multiplied by conductor area. Because it is rounded, the K method and a calculation using the supplied resistance can differ slightly in the last digit.
The other choices: A — A decimal-place slip; B — Does not follow from the supplied 0.122 Ω per 1,000 ft and 105,600 cmil; those inputs give 12.8832; C — Aluminum’s K
Reference: OpenStax College Physics 2e §20.3; NBS Handbook 100, Copper Wire Tables, Table 5 (circular-mil areas), printed p. 16
31. Resistance at a hotter temperature
For this exercise, 12 AWG stranded copper has a supplied resistance of 1.98 Ω per 1,000 ft at 75 °C. Use R₂ = R₁ × [1 + 0.00323 × (T₂ − 75)], where the supplied coefficient is 0.00323 per °C relative to 75 °C. What is the resistance per 1,000 ft at 90 °C? Round to 0.01 Ω per 1,000 ft.
- A. 1.88 Ω per 1,000 ft
- B. 2.08 Ω per 1,000 ft
- C. 2.28 Ω per 1,000 ft
- D. 2.56 Ω per 1,000 ft
Show answer and worked solution
Answer: B — 2.08 Ω per 1,000 ft
Work it: 1.98 × [1 + 0.00323 × (90 − 75)] = 1.98 × 1.04845 ≈ 2.08 Ω per 1,000 ft.
Why: Copper resistance rises with temperature. Here the starting resistance and the correction coefficient are supplied at a 75 °C reference temperature.
The other choices: A — Subtracts the correction; C — Multiplies by 1.15; D — Uses 90 instead of the 15-degree difference
Reference: OpenStax College Physics 2e §20.3
32. Double the length, preserve the drop
A resistance-only two-wire circuit uses 10 AWG copper (10,380 cmil) over 125 ft one way. The run is extended to 250 ft one way, with the same current, material, conductor temperature and maximum voltage drop. What is the smallest listed conductor area that keeps the drop no higher than before? Areas: 10 AWG = 10,380 cmil; 8 AWG = 16,510; 6 AWG = 26,240; 3 AWG = 52,620.
- A. 10 AWG
- B. 8 AWG
- C. 6 AWG
- D. 3 AWG
Show answer and worked solution
Answer: C — 6 AWG
Work it: VD is proportional to L ÷ CM when current and K stay fixed. Required area = 10,380 × (250 ÷ 125) = 20,760 cmil. The smallest listed area at or above 20,760 is 26,240 cmil → 6 AWG.
Why: Double the length and you need twice the area to preserve the resistance and voltage drop. Select the first listed area that meets or exceeds the result.
The other choices: A — Leaves the area unchanged, so doubling the length doubles the drop; B — Rounds down; 16,510 is less than 20,760; D — Meets the target, but it is larger than the smallest listed size needed
Reference: OpenStax College Physics 2e §20.3; OpenStax College Physics 2e §20.2; NBS Handbook 100, Copper Wire Tables, Table 5 (circular-mil areas), printed p. 16
33. Raise the voltage, same source power
A resistive single-phase circuit draws 9.6 kW at its 240-V source (40 A) and has an 8.0-V wire drop (3.33%). A second configuration draws the same 9.6 kW at a 480-V source (20 A), using the same conductors, length and conductor temperature. What is the new percent drop? Round to 0.01%.
- A. 0.83%
- B. 1.67%
- C. 3.33%
- D. 6.67%
Show answer and worked solution
Answer: A — 0.83%
Work it: Current halves (40 A → 20 A), so the drop halves to 4.0 V. 4.0 ÷ 480 × 100 ≈ 0.83%.
Why: For the same source-input power, doubling the voltage cuts the percent drop to one quarter: half the volts lost, on twice the base.
The other choices: B — Halves only once (4.0 V ÷ 240); C — Assumes nothing changes; D — Doubles instead of reducing
Reference: OpenStax College Physics 2e §20.2; OpenStax College Physics 2e §20.4
34. What a 3% drop costs a heater
A 4,800-W, 240-V resistance heater receives only 232.8 V (a 3% drop). Assuming its resistance stays constant, how much power does it deliver? Round to the nearest watt.
- A. 4,516 W
- B. 4,656 W
- C. 4,800 W
- D. 4,944 W
Show answer and worked solution
Answer: A — 4,516 W
Work it: R = 240² ÷ 4,800 = 12 Ω. P = 232.8² ÷ 12 ≈ 4,516 W (about 94.1% of rated).
Why: Power in a resistance varies with the square of voltage, so a 3% drop costs about 6% of the heat.
The other choices: B — Reduces the power by 3% as if it were linear; C — Assumes no change; D — Increases the power
Reference: OpenStax College Physics 2e §20.4
Check your results
Count your correct answers in each section. Count skipped problems separately and keep them in the total.
| Section | Problems | Your correct | Skipped |
|---|---|---|---|
| A. DC and circuit basics | 1–4 | ___ / 4 | ___ |
| B. Single-phase voltage drop | 5–9 | ___ / 5 | ___ |
| C. Three-phase voltage drop | 10–16 | ___ / 7 | ___ |
| D. Size the conductor | 17–21 | ___ / 5 | ___ |
| E. Maximum distance and load | 22–25 | ___ / 4 | ___ |
| F. Applied targets and concepts | 26–34 | ___ / 9 | ___ |
| Total | 34 | ___ / 34 | ___ |
Your total shows how you did on these 34 problems. It isn't a scaled score, and it can't tell you whether you'll pass a licensing exam. Four to nine problems per section is a small sample, so treat a low section as a place to look, not a verdict.
When you miss one, find the first line where your setup differs from the worked answer. The table below points you to the problems that drill each mistake. Tomorrow, hide the solution and redo the setup with units before you touch the calculator.
| If you missed… | The usual cause | Rework |
|---|---|---|
| A single-phase drop | Left out the 2, or used 1.732 | 4, 5, 6 |
| A three-phase drop | Used 2, or multiplied the length by 3 | 10, 11, 13 |
| A "total length" problem | Doubled a length that already included both wires | 1, 7 |
| A continuous or motor load | Used 125% of the current instead of 100% | 8, 9 |
| A percentage | Divided by the load voltage, or mixed line-to-line and line-to-neutral | 3, 11, 15 |
| A sizing problem | Rounded down to the closest area | 17, 18, 20 |
| A maximum distance | Rounded the length up | 22, 24 |
| An aluminum problem | Used copper's K | 20 |
| A three-phase current | Found amps from kVA without √3 | 12, 25 |
Formula reference
Use the supplied resistance-only model unless the question specifies otherwise; problem 16 explicitly includes reactance and power factor.
| Calculation | Formula | Notes |
|---|---|---|
| Ohm's law | VD = I × R | R is the resistance of the wires carrying current |
| Two-wire (DC or single-phase), K method | VD = 2 × K × I × L ÷ CM | L = one-way feet; CM = area in circular mils |
| Balanced three-phase, K method | VD = 1.732 × K × I × L ÷ CM | Result is line-to-line |
| Two-wire, resistance method | VD = 2 × L × R × I ÷ 1,000 | R in ohms per 1,000 ft |
| Balanced three-phase, resistance method | VD = 1.732 × L × R × I ÷ 1,000 | Same R units |
| Minimum area | CM = (2 or 1.732) × K × I × L ÷ VD | VD in volts, not percent |
| Maximum one-way length | L = CM × VD ÷ [(2 or 1.732) × K × I] | Round down |
| Maximum current | I = CM × VD ÷ [(2 or 1.732) × K × L] | Round down |
| Percent drop | VD ÷ source voltage × 100 | Match the voltage base |
| Allowed drop in volts | Source voltage × percent ÷ 100 | Convert before using the area formula |
| Three-phase current | I = VA ÷ (1.732 × line-to-line volts) | Balanced loads |
| AC with power factor | Z = R cos θ + X sin θ, then use Z in place of R | Specified steady-load approximation; R and X have matching units |
| Temperature correction | R₂ = R₁ × [1 + α × (T₂ − T₁)] | Use the coefficient at the stated reference temperature; problem 31 supplies α = 0.00323 per °C at 75 °C |
| K from supplied resistance | K = (Ω per 1,000 ft ÷ 1,000) × CM | K is in Ω·cmil/ft |
What the letters mean. VD is voltage drop in volts. I is the current the load actually draws, in amps. L is the one-way distance from the source to the load, in feet. CM is the conductor's cross-sectional area in circular mils. K is resistivity expressed in Ω·cmil/ft. The supplied values 12.9 for copper and 21.2 for aluminum follow a commonly used 75 °C calculation convention (Mike Holt, definitions of K, I, D and CM).
Nominal areas used in these problems (AWG means American Wire Gauge; cmil means circular mils):
| Size | Area (cmil) |
|---|---|
| 14 AWG | 4,110 |
| 12 AWG | 6,530 |
| 10 AWG | 10,380 |
| 8 AWG | 16,510 |
| 6 AWG | 26,240 |
| 4 AWG | 41,740 |
| 3 AWG | 52,620 |
| 2 AWG | 66,360 |
| 1 AWG | 83,690 |
| 1/0 AWG | 105,600 |
These nominal areas are listed in the NBS copper wire tables (Handbook 100, Table 5). This is a selected area list, not a complete gauge series or an ampacity table. In problem 14, 1 kcmil means 1,000 cmil.
Why two methods give slightly different answers
Two calculations can both be right and still disagree. Before you change your arithmetic, check that both describe the same circuit:
- K method vs. supplied resistance. K = 12.9 is a rounded constant. In problem 30, 0.122 Ω per 1,000 ft and 105,600 cmil imply K = 12.8832, which rounds to 12.9. Use the method the question gives you (problem 30).
- One-way vs. total length. A formula with a 2 in it wants the one-way distance. A total loop length gets used once (problem 7).
- Temperature. Match the resistance and correction coefficient to their stated reference temperature. Hotter copper has more resistance in this model (problem 31) (OpenStax §20.3).
- DC resistance vs. AC impedance. A calculation that includes reactance and power factor can give a different result from the resistance-only model. Use the calculation method and assumptions specified in the question (problem 16) (Schneider Electric, Figure G29).
- Voltage base. A line-to-line drop goes over the line-to-line voltage (problem 15).
- Current. Use 100% of the load current, not 125% (Mike Holt, definition of I).
A quick way to check: write down the current, the length definition, the resistance basis, the phase and the voltage base. Match those first, then compare unrounded answers.
Use the target the problem gives you
The 3%, 5%, 1.5%, 2.5% and 15% figures used here are stated exercise targets. They are not a list of universal code requirements. When both a branch target and a combined target are given, use the smaller of the branch allowance and what remains after the feeder drop. Percentages can be added only when they use the same voltage base.
| What the problem supplies | What to do |
|---|---|
| A percent target | Convert it to volts using the stated source voltage |
| Branch and combined targets | Satisfy both; subtract the feeder drop from the combined allowance |
| A minimum conductor area | Choose the smallest listed area at or above the unrounded result |
| A maximum length, current or kVA | Round down to the requested increment, then back-check |
Meeting a voltage-drop target does not establish ampacity, protection or complete installation compliance. Schneider’s conductor-sizing methodology treats those as separate checks. These exercises are not installation instructions or a substitute for the requirements applicable to an actual job.
Common questions
Will voltage drop be on my exam? It depends on your state. Texas's current bulletin doesn't list voltage drop by name; its calculation portions group problems under areas such as branch-circuit calculations and conductors. The Texas Journeyman Calculations portion has 26 items (2 unscored) in 110 minutes, and the Master Calculations portion has 33 items (3 unscored) in 170 minutes, each requiring at least 70% correct on scored questions (PSI Texas electrician bulletin, printed pp. 10–13). Check your own licensing authority's candidate bulletin for its content outline.
Which NEC edition should I use? The one your licensing authority names for your exam date. For example, Texas exams reference the 2026 NEC from Sept. 1, 2026, and the bulletin tells candidates to answer without code exceptions or optional calculation methods unless the question directs them to (PSI Texas bulletin, printed p. 8). TDLR’s exam reference-material page also names the 2026 edition. An installation-code adoption list is not a substitute for your exam’s reference instructions. The supplied-data formulas here do not depend on a particular NEC edition.
Can I use a calculator? Use the type your exam bulletin permits. Texas allows silent, battery-operated, non-programmable calculators with no paper tape and no alphabetic keyboard (PSI Texas bulletin, printed p. 4). Unless a question says otherwise, the Texas calculation portions also treat all conductors as copper (PSI bulletin, printed pp. 10 and 13).
Keep practicing
Rework the problems tied to your missed setup: one-way versus total length, single-phase calculations, three-phase calculations, conductor sizing, or maximum distance and load. Hide the answer before trying again.
Sources and how we checked this page
- Circuit theory: OpenStax College Physics 2e §20.2 (Ohm’s law); §20.3 (resistivity and temperature); §20.4 (power); §21.1 (series and parallel circuits).
- K, one-way distance and operating current: Mike Holt, “Voltage Drop Calculations”, the K, I, D and CM definitions. Its older code references are not used as current installation rules here.
- Circular-mil areas: NBS Handbook 100, Copper Wire Tables, Table 5 (circular-mil areas), printed p. 16. The historical table supplies geometry, not a current ampacity or installation approval. Individual resistance values are supplied inputs in the problems.
- Three-phase formulas and the effective-impedance method: Schneider Electric Electrical Installation Guide, Figure G29; U.S. DOE Fundamentals Handbook, Electrical Science Vol. 3, Module 9, printed p. 21.
- Scope of conductor sizing: Schneider Electric, Conductor sizing: methodology and definition, Methodology.
- Texas exam details: PSI Texas Electrician Candidate Information Bulletin, cover and printed pp. 4, 8, 10 and 13; TDLR Electrician Exam Information, Reference Materials. These examples apply to Texas, not every electrician exam.
Source check: Sept. 29, 2026. Calculation and content checks: Sept. 30, 2026. We checked the cited technical principles, nominal area values, Texas exam details, and the arithmetic in all 34 solutions. This wasn't a review of every state's rules or the complete NEC.
How we made this page: These original problems were prepared with help from AI tools. We checked the source passages, recomputed the numerical solutions in code, and checked every answer choice for its reasoning, units and rounding. Source and arithmetic checks are not a licensed-electrician review; no professional-review credit is claimed.
By the Castleport Test Prep Editorial Team
Castleport Test Prep is an independent exam prep publisher. We aren't affiliated with, endorsed by, or approved by NFPA, PSI, the Texas Department of Licensing and Regulation, or any licensing authority or testing provider. NEC and National Electrical Code, and other exam and organization names, identify the subjects discussed; trademarks belong to their respective owners. These practice problems are original and aren't taken from any real exam. Using this page doesn't guarantee a score, a license, or a job.