2.4 Electrical Theory for the Exam: Ohm's Law, Power & Three-Phase

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

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:

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.