Crash course: Ohm's Law for electricians with calculations (part 4)

Ohm's Law: the three equations you actually use

Voltage, current, resistance. Every troubleshooting call comes back to these three. Memorize the triangle or don't, but know the math cold:

• E = I x R (volts = amps times ohms)
• I = E / R (amps = volts divided by ohms)
• R = E / I (ohms = volts divided by amps)

Then the power wheel adds P = I x E, P = I squared x R, and P = E squared / R. That covers 95% of what you need on a service call. The other 5% is power factor and three-phase math, which we'll hit below.

Sizing conductors from load: a real calculation

Customer has a 5 kW resistive heater, 240V single-phase, dedicated circuit. Load current: I = P / E = 5000 / 240 = 20.83 amps. Continuous load per NEC 210.19(A)(1), so multiply by 125%: 20.83 x 1.25 = 26 amps minimum.

Pull Table 310.16. At 75C column, 10 AWG copper is rated 35 amps, 12 AWG is 25 amps. 12 AWG won't carry it at 125%, so 10 AWG THHN on a 30 amp breaker. Confirm termination rating at the breaker and heater lugs per 110.14(C). If either is 60C rated, drop to that column before sizing.

Field tip: if the equipment nameplate says "minimum circuit ampacity," use that number directly. The manufacturer already did the 125% math and any diversity calcs. Don't double-dip.

Voltage drop: when Ohm's Law bites you

NEC 210.19(A) Informational Note No. 4 recommends 3% drop on branch circuits, 5% total with feeders. It's not a hard rule, but it's what inspectors and engineers expect, and it's what keeps motors happy and LEDs from flickering.

Quick math for single-phase: VD = (2 x K x I x D) / CM. K is 12.9 for copper, 21.2 for aluminum. D is one-way distance in feet. CM is circular mils from Chapter 9 Table 8. For a 20 amp circuit at 120V running 150 feet on 12 AWG copper (6530 CM):

• VD = (2 x 12.9 x 20 x 150) / 6530 = 11.85 volts
• 11.85 / 120 = 9.9% drop. Way over. Upsize to 10 AWG (10380 CM), drop falls to 7.45V or 6.2%. Still over. Go 8 AWG and you're at 3.9%.

On long runs, voltage drop drives your conductor size, not ampacity. Calculate both and take the larger.

Three-phase: the square root of 3

Three-phase power: P = E x I x 1.732 x PF. For motors, use nameplate FLA, or pull Table 430.250 for standard values when sizing per Article 430. A 10 HP, 480V, three-phase motor: Table 430.250 gives 14 amps FLA.

Branch circuit conductors per 430.22: 14 x 1.25 = 17.5 amps. 12 AWG handles it. Overcurrent protection per 430.52 and Table 430.52: inverse time breaker at 250% of FLA = 35 amps, round up to next standard size if needed per 430.52(C)(1) Exception No. 1. Overload protection at the starter sized per 430.32.

Field tip: 1.732 is the square root of 3. Three-phase delivers more power per conductor than single-phase for the same current, which is why motors and big HVAC live on three-phase.

Troubleshooting with Ohm's Law

Receptacle reads 118V no load, 92V under a 1500W hair dryer. That's a 26V drop under roughly 12.5 amps of load. R = E / I = 26 / 12.5 = 2.08 ohms of loose or corroded connection somewhere in the circuit. A clean 12 AWG run of 50 feet should be around 0.08 ohms total. You're looking for a bad backstab, a loose wire nut, or a corroded neutral.

Same logic works for equipment grounding checks. Measure voltage drop on the EGC under fault simulation, calculate impedance, compare against what 250.4(A)(5) calls an effective ground-fault current path. High impedance means the breaker won't trip fast enough on a fault.

Quick reference for the truck

Stick these on the inside of your toolbox lid:

1. Single-phase amps from watts: I = W / E
2. Three-phase amps from watts: I = W / (E x 1.732 x PF)
3. Single-phase VD: (2 x K x I x D) / CM
4. Three-phase VD: (1.732 x K x I x D) / CM
5. Continuous load multiplier: 1.25 per 210.19(A)(1) and 215.2(A)(1)
6. K values: 12.9 copper, 21.2 aluminum

Ohm's Law isn't academic. It's how you size a circuit, diagnose a bad connection, and justify a conductor upgrade to a GC who wants to cheap out. Run the numbers before you pull wire, not after.