Ohm's Law, Watt's Law, and the AC Power Triangle

The 12 Relationships You Need to Know

Ohm's Law and Watt's Law combine to produce 12 formulas that relate four quantities: voltage (V or E), current (I), resistance (R), and power (P). Arranged in a wheel, they let you solve for any quantity given two others.

Solving for Voltage: V = I x R, V = P / I, V = sqrt(P x R). Solving for Current: I = V / R, I = P / V, I = sqrt(P / R). Solving for Resistance: R = V / I, R = V^2 / P, R = P / I^2. Solving for Power: P = V x I, P = I^2 x R, P = V^2 / R.

These 12 formulas are exact for DC circuits and for purely resistive AC loads (electric heaters, incandescent lamps). They work because in a resistive circuit, voltage and current are in phase: they rise and fall together. Power is simply the product of the two at any instant, and the average power equals V_rms times I_rms.

The practical value of these relationships is enormous. If a 240V heater draws 20A, you know it consumes P = 240 x 20 = 4,800 watts and has a resistance of R = 240 / 20 = 12 ohms. If the heater element partially shorts and resistance drops to 8 ohms, current rises to I = 240 / 8 = 30A, which will trip a 20A breaker. This kind of quick mental math is how experienced electricians diagnose problems in the field.

When you see "E" instead of "V" in older references, they mean the same thing: electromotive force in volts. The symbol convention varies between textbooks and trade publications, but the math is identical.

Why AC Circuits Need Power Factor

In an AC circuit with motors, transformers, or other inductive loads, current does not stay in phase with voltage. The magnetic field in the motor windings causes current to lag behind voltage by some angle, typically 15 to 40 degrees for loaded motors. This phase shift means that at some instants, voltage is positive while current is negative, and the product (instantaneous power) is negative. Energy is flowing backward from the load to the source during those instants.

The result is that the simple formula P = V x I overestimates the actual useful power. The product V x I gives you apparent power (measured in volt-amperes, VA), but the actual work-producing power is less. The ratio between real power and apparent power is the power factor (PF).

Real Power (P) is measured in watts (W) or kilowatts (kW). It represents actual energy conversion: shaft work in a motor, heat in a resistance heater, light output from a lamp. This is what does useful work and what the utility meter charges you for on the energy (kWh) portion of your bill.

Apparent Power (S) is measured in volt-amperes (VA) or kilovolt-amperes (kVA). It is the product of RMS voltage and RMS current: S = V x I. This determines conductor sizing, transformer ratings, and generator capacity because the conductors carry the full current regardless of phase angle.

Reactive Power (Q) is measured in volt-amperes reactive (VAR) or kilovolt-amperes reactive (kVAR). It represents energy that oscillates between the source and the magnetic field of the load. It does no useful work but it does require current to flow, which means larger conductors, larger transformers, and higher losses.

The relationship is: PF = P / S = cos(theta), where theta is the phase angle between voltage and current. A purely resistive load has PF = 1.0 (theta = 0). A typical loaded induction motor has PF = 0.80 to 0.90. A lightly loaded motor can drop to PF = 0.40 to 0.60. Welders and fluorescent lighting without correction typically run PF = 0.50 to 0.70.

The Power Triangle: Real, Reactive, and Apparent

The power triangle is a right triangle where real power (P, in kW) is the horizontal leg, reactive power (Q, in kVAR) is the vertical leg, and apparent power (S, in kVA) is the hypotenuse. The angle between P and S is the power factor angle (theta). This geometric relationship comes from the Pythagorean theorem: S^2 = P^2 + Q^2.

This triangle is not just a teaching tool. It is the working model for every power system calculation involving mixed loads. When you have multiple loads on a panel or feeder, you add the real powers (kW) arithmetically and add the reactive powers (kVAR) arithmetically, but you cannot add the apparent powers (kVA) arithmetically. You must calculate the combined S from the combined P and Q using S = sqrt(P^2 + Q^2).

Example: A panel feeds a 50 kW heater bank (PF = 1.0) and a 30 kW motor load (PF = 0.85 lagging). The heater has Q = 0 kVAR because PF = 1.0. The motor has S = 30 / 0.85 = 35.3 kVA, and Q = sqrt(35.3^2 - 30^2) = 18.6 kVAR. Combined: P_total = 50 + 30 = 80 kW, Q_total = 0 + 18.6 = 18.6 kVAR, S_total = sqrt(80^2 + 18.6^2) = 82.1 kVA. The combined PF = 80 / 82.1 = 0.974.

Notice that the combined PF (0.974) is much better than the motor alone (0.85) because the resistive heater load "dilutes" the reactive component. This is why facilities with large resistive loads (furnaces, heaters) often have acceptable power factors even without correction capacitors.

When a utility charges a power factor penalty (typically below PF = 0.90), the solution is to add capacitor banks that supply reactive power locally. Capacitors generate leading reactive power that cancels the lagging reactive power from motors. The sizing calculation uses the triangle: Q_cap = P x (tan(theta_old) - tan(theta_new)), where theta values come from the arccos of the respective power factors.

Single-Phase vs Three-Phase Power Formulas

The power formulas change depending on whether the circuit is single-phase or three-phase. Getting the formula wrong is one of the most common errors in electrical calculations.

Single-phase: S = V x I, so I = S / V. For a 10 kVA load at 240V single-phase: I = 10,000 / 240 = 41.7A. Real power: P = V x I x PF.

Three-phase: S = V x I x sqrt(3), so I = S / (V x sqrt(3)). For a 10 kVA load at 480V three-phase: I = 10,000 / (480 x 1.732) = 12.0A. The sqrt(3) factor (1.732) accounts for the phase relationship between the three conductors. Real power: P = V x I x sqrt(3) x PF.

The voltage in three-phase formulas is always the line-to-line voltage (480V, 208V, 600V), not the line-to-neutral voltage. This is a frequent source of confusion on 208/120V systems. If someone tells you the load is "208V three-phase," use 208V in the formula with the sqrt(3) factor. If the load is connected line-to-neutral (120V on a 208/120V system), it is a single-phase load and you use 120V without the sqrt(3) factor.

A practical example: What is the line current for a 25 HP motor at 460V, 3-phase, with a power factor of 0.87 and an efficiency of 0.91? First, convert HP to watts: 25 x 746 = 18,650 watts input to the shaft. Account for efficiency: electrical input = 18,650 / 0.91 = 20,495 watts. Now solve for current: I = P / (V x sqrt(3) x PF) = 20,495 / (460 x 1.732 x 0.87) = 29.6A. This matches closely with NEC Table 430.250, which lists 34A for a 25 HP, 460V motor. The table value is higher because NEC uses conservative assumptions to ensure conductors are adequately sized.

Worked Example: Sizing a Feeder for a Known Motor Load

Problem: Size the feeder conductors and overcurrent protection for a 50 HP, 460V, 3-phase motor. The feeder runs 200 feet from the MCC to the motor.

Step 1: Find FLA from NEC Table 430.250. For a 50 HP, 460V, 3-phase motor, the table value is 65A. Use this value, not the nameplate current, for all conductor and protection sizing per NEC 430.6(A)(1).

Step 2: Size the conductors. NEC 430.22 requires conductor ampacity of at least 125% of FLA: 65 x 1.25 = 81.25A. From NEC Table 310.16, 4 AWG copper THHN is rated 85A at 75 degrees C column (the termination temperature rating governs per 110.14(C)). 4 AWG copper is adequate.

Step 3: Check voltage drop. At 200 feet, using the standard voltage drop formula: VD = (sqrt(3) x I x rho x L) / (CM), where rho = 12.9 for copper, L = 200 ft, CM = 41,740 for 4 AWG. VD = (1.732 x 65 x 12.9 x 200) / 41,740 = 6.97V. Percentage: 6.97 / 460 = 1.5%. This is well under the NEC 210.19(A) informational note recommendation of 3% for branch circuits (5% total including feeder).

Step 4: Size the overload relay. NEC 430.32(A)(1) requires the overload device to trip at no more than 125% of the motor nameplate FLA (use nameplate here, not table value). If the nameplate reads 62A: 62 x 1.25 = 77.5A overload heater or electronic overload setting.

Step 5: Size the branch circuit protection. NEC 430.52 and Table 430.52 allow up to 250% of FLA for an inverse-time circuit breaker with a standard motor: 65 x 2.5 = 162.5A. Round up to the next standard size: 175A breaker. If the motor does not start on 175A, NEC 430.52(C)(1) permits going to the next standard size, up to 400% for certain conditions.

This five-step process applies to every motor circuit. The key principle: NEC separates overload protection (protects the motor from sustained overcurrent) from short-circuit/ground-fault protection (protects the conductors from fault current). They are sized differently using different multipliers and different current values.

Power Factor Correction in Practice

Utilities penalize industrial customers with power factors below 0.85 to 0.90, depending on the rate schedule. The penalty takes several forms: a direct PF surcharge on the bill, billing based on kVA demand instead of kW demand, or a reactive power charge per kVAR. The economic incentive to correct power factor is real and often significant.

The correction method is to install capacitor banks at the service entrance, at distribution panels, or at individual motors. Capacitors supply the reactive current locally so it does not flow through the utility transformer and feeder. This reduces line current, reduces I^2R losses, and frees up transformer and conductor capacity.

Sizing example: A facility draws 500 kW at PF = 0.78. The utility requires PF = 0.95 to avoid penalties. Current reactive power: Q_old = 500 x tan(arccos(0.78)) = 500 x 0.8026 = 401.3 kVAR. Target reactive power: Q_new = 500 x tan(arccos(0.95)) = 500 x 0.3287 = 164.4 kVAR. Required capacitor bank: Q_cap = 401.3 - 164.4 = 236.9 kVAR. Select a 250 kVAR capacitor bank (standard size).

Capacitor placement matters. Installing capacitors at the service entrance corrects the power factor as seen by the utility but does not reduce losses inside the facility. Installing capacitors at individual motors corrects the power factor at every point upstream and reduces losses throughout the distribution system. The tradeoff is cost: one large bank at the entrance is cheaper to install than many small units at each motor.

A common hazard is overcorrection. If capacitors push the power factor above 1.0 (leading), the system becomes capacitive and can cause voltage rise, resonance with harmonic currents, and self-excitation of motors during power outages. Most installations target PF = 0.95 to 0.97, not 1.0, to provide a safety margin against overcorrection as loads vary throughout the day.

Common Calculation Mistakes and How to Avoid Them

Years of field experience and engineering reviews reveal the same mistakes appearing repeatedly in electrical calculations. Here are the most frequent ones.

Forgetting the sqrt(3) factor on three-phase circuits. This understates current by 42% on three-phase loads, leading to undersized conductors. If a load draws 100A per phase and you forget the sqrt(3), you calculate 100A instead of 173A for a kVA-based sizing. Always confirm: is this a single-phase or three-phase load?

Using watts instead of VA for conductor sizing. Conductors carry current, and current is determined by apparent power (VA), not real power (W). A 10 kW motor at PF = 0.80 draws 12.5 kVA, which means 12,500 / (480 x 1.732) = 15.0A, not 10,000 / (480 x 1.732) = 12.0A. Sizing conductors based on watts understates the required ampacity by the inverse of the power factor.

Confusing 208V and 240V. A 208V system is not the same as a 240V system. 208V comes from the line-to-line voltage of a 120/208V wye system (120 x sqrt(3) = 208). 240V comes from a center-tapped single-phase transformer. A motor rated for 230V will run on either, but it draws about 15% more current at 208V than at 240V because P = V x I x sqrt(3) x PF, and the lower voltage requires higher current for the same power. NEC Table 430.250 has separate columns for 200V and 230V motors for this reason.

Adding kVA values arithmetically for mixed loads. As explained in the power triangle section, kVA values from loads with different power factors cannot be simply added. You must convert to kW and kVAR, add those separately, then recalculate kVA from the totals. Adding kVA directly overstates the total by 5 to 15% for typical mixed loads.

Ignoring temperature derating. NEC Table 310.16 ampacities assume an ambient temperature of 30 degrees C (86 degrees F). In a hot attic, boiler room, or rooftop installation where ambient exceeds 30 degrees C, conductor ampacity must be derated using the correction factors at the bottom of Table 310.16. At 45 degrees C ambient, the 75 degrees C column ampacity drops to 87% of its 30 degrees C value.