10 problems, all seven types in every set. Work each one before you open its steps.
Problem 1: Continuous load at 125%
A branch circuit serves 8 A of continuous load and 10 A of noncontinuous load. What minimum ampere rating does the 125% rule give, before rounding up to a standard size?
Show the answer and steps
Answer: 20 A
- Continuous part at 125%: 8 × 1.25 = 10 A.
- Add the noncontinuous part at 100%: 10 + 10 = 20 A.
- In 2023 NEC numbering the rule is at 210.19(A)(1) for the conductor and 210.20(A) for the overcurrent device; confirm the numbers in your 2026 book. Rounding up to a standard device rating is a separate step this drill leaves out.
Problem 2: Three-phase power
A balanced three-phase load uses 30 kW on 480 V line-to-line at a power factor of 0.8. What is the line current, in amperes?
Show the answer and steps
Answer: 45.11 A
- Rearrange P = 1.732 × E × I × PF for current: I = P ÷ (1.732 × E × PF).
- Convert kW to watts: 30 × 1,000 = 30,000 W.
- 1.732 × 480 × 0.8 = 665.09.
- I = 30,000 ÷ 665.09 = 45.11 A.
Problem 3: Series and parallel resistance
A 15 Ω resistor is in series with a parallel pair of 10 Ω and 60 Ω, across a 24 V source. What is the total current, in amperes?
Show the answer and steps
Answer: 1.02 A
- Reduce the pair first, product over sum: (10 × 60) ÷ (10 + 60) = 8.57 Ω.
- Add the series resistor: 15 + 8.57 = 23.57 Ω.
- Ohm's law: I = 24 ÷ 23.57 = 1.02 A.
Problem 4: Single-phase power
A single-phase load on 277 V draws 8 A at a power factor of 0.85. What is its true power, in watts?
Show the answer and steps
Answer: 1,884 W
- Single-phase true power: P = E × I × PF.
- Apparent power first: 277 × 8 = 2,216 VA.
- True power: 2,216 × 0.85 = 1,884 W.
- Sanity check: true power can never be more than the volt-amperes.
Problem 5: Ohm's law
A heating element of 4 Ω carries 12 A. How much power does it use, in watts?
Show the answer and steps
Answer: 576 W
- With current and resistance known, use P = I² × R and skip the voltage step.
- P = 12² × 4 = 144 × 4 = 576 W.
- Check with E × I: E = 12 × 4 = 48 V, and 48 × 12 = 576 W.
Problem 6: Voltage drop
A 120 V single-phase, two-wire circuit carries 12 A to a load 150 ft away (one way). The copper conductors are 12 AWG, 6,530 circular mils. Use K = 12.9 Ω·cmil/ft. What is the voltage drop, in volts?
Show the answer and steps
Answer: 7.11 V
- Single-phase: VD = 2 × K × I × L ÷ cmil, with L the one-way length.
- VD = 2 × 12.9 × 12 × 150 ÷ 6,530 = 7.11 V.
- The 2 is there because the current travels out and back.
Problem 7: Transformer full-load current
A 25 kVA single-phase transformer has a 480 V winding. What is the full-load current of that winding, in amperes?
Show the answer and steps
Answer: 52.08 A
- Single-phase: I = kVA × 1,000 ÷ E.
- I = 25 × 1,000 ÷ 480 = 25,000 ÷ 480 = 52.08 A.
- The other winding has the same kVA, so its current changes in inverse proportion to its voltage.
Problem 8: Transformer full-load current
A 75 kVA single-phase transformer has a 240 V winding. What is the full-load current of that winding, in amperes?
Show the answer and steps
Answer: 312.5 A
- Single-phase: I = kVA × 1,000 ÷ E.
- I = 75 × 1,000 ÷ 240 = 75,000 ÷ 240 = 312.5 A.
- The other winding has the same kVA, so its current changes in inverse proportion to its voltage.
Problem 9: Three-phase power
A balanced three-phase load on 480 V line-to-line draws 20 A per line at a power factor of 0.85. What is its true power, in watts?
Show the answer and steps
Answer: 14,134 W
- Three-phase true power: P = 1.732 × E × I × PF, with line voltage and line current.
- Apparent power: 1.732 × 480 × 20 = 16,627 VA.
- True power: 16,627 × 0.85 = 14,133 W.
- With the exact square root of 3 the key prints as 14,134 W. Both are inside the self-check tolerance.
- The common miss is using 3 in place of 1.732, or leaving out the power factor.
Problem 10: Voltage drop
A 120 V single-phase, two-wire circuit carries 20 A to a load 250 ft away (one way). The copper conductors are 10 AWG, 10,383 circular mils. Use K = 12.9 Ω·cmil/ft. What is the voltage drop, in volts?
Show the answer and steps
Answer: 12.42 V
- Single-phase: VD = 2 × K × I × L ÷ cmil, with L the one-way length.
- VD = 2 × 12.9 × 20 × 250 ÷ 10,383 = 12.42 V.
- The 2 is there because the current travels out and back.
Score yourself
Give yourself one point for each answer within 0.5% of the key. That leaves room for 1.732 against the exact square root of 3 and for rounding once at the end. Rework every miss by hand from its steps before you start the next set.
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How the practice set works
Each set has 10 problems. The first seven cover every type once in a shuffled order, and the last three repeat types at random. Every problem gives you all its numbers, so you need a calculator and nothing else. Open Show the answer and steps only after you have an answer written down.
The set number at the top picks the numbers. The same set number always gives the same ten problems, so quote it with the problem number if you report an error. New set moves to the next set number.
The formulas behind each type
- Ohm's law. E = I × R, with P = I² × R and P = E² ÷ R when resistance is known.
- Single-phase power. P = E × I × PF. E × I alone is apparent power in volt-amperes.
- Three-phase power. P = 1.732 × E × I × PF, with line-to-line voltage and line current.
- Transformer full-load current. I = kVA × 1,000 ÷ E for single-phase, and kVA × 1,000 ÷ (1.732 × E) for three-phase.
- Voltage drop. VD = 2 × K × I × L ÷ cmil for single-phase and 1.732 × K × I × L ÷ cmil for three-phase, with L the one-way length. K and the circular mils are given in each problem.
- Series and parallel resistance. Series resistances add. In parallel, the reciprocal of the total is the sum of the reciprocals.
- Continuous load at 125%. 1.25 × the continuous load plus the noncontinuous load gives the minimum before you round up to a standard rating. The rule is at 210.19(A)(1) and 210.20(A) in 2023 numbering, so confirm the numbers in your 2026 book.
The first six are standard physics in our own words. The method for the table-driven calculations, by NEC article and table number, is in the Calculations portion guide, and the master Calculations guide works the longer version.
Worked example: a three-phase line current
A balanced 30 kW load runs on 480 V, three-phase, at a power factor of 0.9. Step 1: convert to watts, 30 × 1,000 = 30,000 W. Step 2: the divisor, 1.732 × 480 × 0.9 = 748.22. Step 3: I = 30,000 ÷ 748.22 = 40.09 A. Leaving out the power factor gives 36.09 A, and using 3 in place of 1.732 gives 23.15 A. Both are the kind of wrong option a careful setup avoids. This is an original example, not an exam question.
Calculations practice FAQ
Where do these practice problems come from?
Every problem is original and written to the published PSI outline for the Texas electrician exams, which lists Calculations and Theory, Transformers and Branch Circuit Calculations in the Calculations portion. The numbers are made up and drawn from a seed, and every key comes from the same math as the calculators on this site.
Why are there no ampacity, motor or conduit fill problems?
Those answers come from NEC tables, and we never reproduce code-book values. Every problem here gives you all its inputs, including K and circular mils for voltage drop. Work the table problems from your own NEC with the method in our Calculations portion guide.
Does this help with the Texas master electrician exam too?
Yes. The PSI outline lists Calculations and Theory, Transformers and Branch Circuit Calculations in the master Calculations portion as well, and these formulas are the same at both levels. The master portion adds more service and motor work, which needs your NEC tables.
What calculator can I bring to the exam?
PSI allows only non-programmable calculators that are silent, battery-operated, have no paper tape printing and have no alphabet keyboard. Practice these sets on the calculator you will carry, so the key order is familiar.