10 problems, all seven types in every set. Work each one before you open its steps.
Problem 1: Single-phase power
A single-phase load uses 2,400 W of true power on 208 V at a power factor of 0.9. What current does it draw, in amperes?
Show the answer and steps
Answer: 12.82 A
- Rearrange P = E × I × PF for current: I = P ÷ (E × PF).
- E × PF = 208 × 0.9 = 187.2.
- I = 2,400 ÷ 187.2 = 12.82 A.
- Sanity check: a power factor below 1 always means more current than P ÷ E alone.
Problem 2: Continuous load at 125%
A 240 V single-phase branch circuit serves a continuous load of 2,400 VA and 4 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: 16.5 A
- Continuous current: 2,400 VA ÷ 240 V = 10 A.
- At 125%: 10 × 1.25 = 12.5 A.
- Add the noncontinuous part at 100%: 12.5 + 4 = 16.5 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 3: Ohm's law
A resistive heater is rated 2,000 W at 240 V. What is its resistance, in ohms?
Show the answer and steps
Answer: 28.8 Ω
- With voltage and power known, use R = E² ÷ P.
- R = 240² ÷ 2,000 = 57,600 ÷ 2,000 = 28.8 Ω.
- Check: I = 2,000 ÷ 240 = 8.33 A, and 8.33 × 28.8 = 240 V.
Problem 4: Series and parallel resistance
A 60 Ω resistor is in series with a parallel pair of 24 Ω and 15 Ω, across a 24 V source. What is the total current, in amperes?
Show the answer and steps
Answer: 0.35 A
- Reduce the pair first, product over sum: (24 × 15) ÷ (24 + 15) = 9.23 Ω.
- Add the series resistor: 60 + 9.23 = 69.23 Ω.
- Ohm's law: I = 24 ÷ 69.23 = 0.35 A.
Problem 5: Transformer full-load current
A 100 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: 208.33 A
- Single-phase: I = kVA × 1,000 ÷ E.
- I = 100 × 1,000 ÷ 480 = 100,000 ÷ 480 = 208.33 A.
- The other winding has the same kVA, so its current changes in inverse proportion to its voltage.
Problem 6: Voltage drop
A 480 V three-phase circuit carries 80 A to a load 250 ft away (one way). The copper conductors are 1/0 AWG, 105,535 circular mils. Use K = 12.9 Ω·cmil/ft. What is the voltage drop as a percentage of the source voltage?
Show the answer and steps
Answer: 0.88%
- Three-phase: VD = 1.732 × K × I × L ÷ cmil, with L the one-way length.
- VD = 1.732 × 12.9 × 80 × 250 ÷ 105,535 = 4.23 V.
- Percent drop = 4.23 ÷ 480 × 100 = 0.88%.
Problem 7: Three-phase power
A balanced three-phase load uses 25 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: 37.59 A
- Rearrange P = 1.732 × E × I × PF for current: I = P ÷ (1.732 × E × PF).
- Convert kW to watts: 25 × 1,000 = 25,000 W.
- 1.732 × 480 × 0.8 = 665.09.
- I = 25,000 ÷ 665.09 = 37.59 A.
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: Voltage drop
A 120 V single-phase, two-wire circuit carries 40 A to a load 100 ft away (one way). The copper conductors are 6 AWG, 26,251 circular mils. Use K = 12.9 Ω·cmil/ft. What is the voltage drop as a percentage of the source voltage?
Show the answer and steps
Answer: 3.28%
- Single-phase: VD = 2 × K × I × L ÷ cmil, with L the one-way length.
- VD = 2 × 12.9 × 40 × 100 ÷ 26,251 = 3.93 V.
- Percent drop = 3.93 ÷ 120 × 100 = 3.28%.
Problem 10: Series and parallel resistance
Resistors of 60 Ω, 5 Ω and 5 Ω are connected in series. What is the total resistance, in ohms?
Show the answer and steps
Answer: 70 Ω
- In series, resistances add: R = R1 + R2 + R3.
- R = 60 + 5 + 5 = 70 Ω.
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.