2.4 Electrical Theory for the Exam: Ohm's Law, Power & Three-Phase
Key Takeaways
Ohm's law and the power formulas cover most single-phase theory items: E = I x R and P = E x I, with P = I squared x R for conductor heating.
Single-phase apparent power is VA = E x I; three-phase apparent power is VA = 1.732 x E-line x I-line, so a three-phase line current is kVA x 1000 divided by (1.732 x line voltage).
Power factor is real power divided by apparent power, so line current rises as power factor falls for the same kilowatt load.
In a balanced wye system, line voltage equals 1.732 times phase voltage while line current equals phase current; in a delta system, line current equals 1.732 times phase current while line voltage equals phase voltage.
The single-phase voltage-drop formula is VD = 2 x K x I x L divided by CM; the three-phase version substitutes 1.732 for the 2, using K of about 12.9 for copper and 21.2 for aluminum.
The Formulas Worth Owning
The NEC contains rules, not physics. Theory questions and the arithmetic inside code questions come from a short list of relationships. Write these on the inside cover of your working notes and drill until they are automatic.
Ohm's Law and Power
| Quantity | Formula |
|---|---|
| Voltage | E = I x R |
| Current | I = E / R |
| Resistance | R = E / I |
| Power (resistive) | P = E x I |
| Power from current and resistance | P = I squared x R |
| Power from voltage and resistance | P = E squared / R |
P = I squared x R is the one that explains the Code. Conductor heating rises with the square of current, which is why ampacity tables exist, why bundling forces adjustment, and why a 20 percent overload is far worse than it sounds.
Single-Phase Current
I = VA / E (apparent power)
I = (kW x 1000) / (E x PF) (real power with power factor)
A 9.6 kW, 240 V single-phase resistance heater at unity power factor draws 9600 / 240 = 40 A.
Three-Phase Current
I(line) = VA / (1.732 x E-line)
I(line) = (kW x 1000) / (1.732 x E-line x PF)
The 1.732 is the square root of 3. A 30 kW, 480 V three-phase heater at unity power factor draws 30,000 / (1.732 x 480) = 30,000 / 831.4 = 36.1 A.
A 75 kVA, 208 V three-phase transformer secondary delivers 75,000 / (1.732 x 208) = 75,000 / 360.3 = 208.2 A. That number is worth remembering as a sanity check: a 75 kVA 208Y/120 V transformer is roughly a 200 A secondary.
Power Factor
PF = P (watts) / S (volt-amperes)
Power factor is the ratio of real power doing work to apparent power the system must carry. Purely resistive loads — heaters, incandescent lamps — run at unity. Motors and transformers run lagging because magnetizing current does no work.
The practical consequence is that current increases as power factor falls. A 20 kW three-phase 480 V load at unity PF draws 24.1 A. The same 20 kW at 0.80 PF draws 20,000 / (1.732 x 480 x 0.80) = 30.1 A — the conductors, raceway, and overcurrent device must all be sized for the larger figure.
The power triangle relates the three: apparent power S in kVA is the hypotenuse, real power P in kW is the horizontal leg, and reactive power Q in kVAR is the vertical leg, with S squared = P squared + Q squared.
Wye and Delta Relationships
These two sets of relationships are frequently reversed in distractor answers:
| System | Voltage | Current |
|---|---|---|
| Wye (star) | E-line = 1.732 x E-phase | I-line = I-phase |
| Delta | E-line = E-phase | I-line = 1.732 x I-phase |
A 208Y/120 V system is a wye: 120 V phase-to-neutral, 120 x 1.732 = 208 V phase-to-phase. A 480Y/277 V system is the same relationship scaled: 277 x 1.732 = 480 V.
The high-leg delta. A 240/120 V three-phase four-wire delta has a centre tap on one winding, giving 120 V to neutral on two phases and 208 V to neutral on the third — the "high leg" or "wild leg." 110.15 requires the high leg to be durably and permanently marked by an orange outer finish or other effective means at each point where a connection is made if the grounded conductor is present, and 408.3(E)(1) requires it to be phase B in switchboards and panelboards. Never land a 120 V load on the high leg.
Voltage Drop
Voltage drop is a performance issue, not a mandatory rule for most conductors. The 3 percent branch-circuit and 5 percent total figures live in Informational Notes to 210.19(A) and 215.2(A), which 90.5(C) makes non-enforceable. Some specific rules do mandate voltage-drop limits — 647.4(D) for sensitive electronic equipment, 695.7 for fire pumps, and Article 690 for PV — but for ordinary branch circuits and feeders the 3/5 figures are recommendations.
The Circular-Mil Formulas
Single-phase: VD = (2 x K x I x L) / CM
Three-phase: VD = (1.732 x K x I x L) / CM
Solve for wire size:
Single-phase: CM = (2 x K x I x L) / VD
Three-phase: CM = (1.732 x K x I x L) / VD
Where K is the resistivity constant — approximately 12.9 for copper and 21.2 for aluminum — I is load current in amperes, L is the one-way length in feet, CM is the conductor's circular-mil area from Chapter 9, Table 8, and VD is the voltage drop in volts. The 2 in the single-phase formula accounts for the out-and-back path, which is why L is one-way.
Useful circular-mil values from Table 8: 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 = 105,600.
Worked Example 1 — Find the Drop
A 240 V single-phase circuit carries 30 A to a load 150 feet away on 10 AWG copper.
VD = (2 x 12.9 x 30 x 150) / 10,380
= 116,100 / 10,380
= 11.19 V
That is 11.19 / 240 = 4.7 percent — beyond the 3 percent recommendation, so upsize if performance matters.
Worked Example 2 — Find the Wire Size
A 480 V three-phase load draws 50 A over 400 feet, and you want to hold the drop to 3 percent (0.03 x 480 = 14.4 V) using copper.
CM = (1.732 x 12.9 x 50 x 400) / 14.4
= 446,856 / 14.4
= 31,032 circular mils
From Table 8, 6 AWG is 26,240 CM — too small. 4 AWG at 41,740 CM is the smallest conductor that satisfies the target. Always verify separately that the chosen conductor also meets its ampacity requirement; voltage drop can only drive you larger, never smaller.
What is the line current of a 45 kVA, 208 V three-phase transformer secondary at full load?
374.8 A
144.3 A
216.3 A
125.0 A
A 20 kW three-phase load operates at 480 V with a power factor of 0.80. What is the line current?
24.1 A
41.7 A
30.1 A
52.1 A
In a balanced three-phase delta system, what is the relationship between line current and phase current?
Line current equals 1.732 times phase current
Line current equals phase current
Line current equals phase current divided by 1.732
Line current equals three times phase current
A 120 V single-phase branch circuit carries 16 A over a one-way distance of 100 feet using 12 AWG copper with a circular-mil area of 6,530. What is the approximate voltage drop?
3.2 V
4.9 V
9.8 V
6.3 V
Sections you finish are checked off in the contents.