Crash course: Ohm's Law for electricians for apprentices (part 1)

What Ohm's Law actually tells you on the job

Ohm's Law is the relationship between voltage (E), current (I), and resistance (R). The equation you memorized in class is E = I x R. Rearranged: I = E / R, and R = E / I. That is the whole thing. Every load calc, every voltage drop check, every troubleshooting call leans on it.

On a live service, voltage is what the utility gives you. Resistance is what the load and conductors impose. Current is what falls out of the math. If you know two of the three, you know the third. That is why a clamp meter and a voltmeter together are worth more than either alone.

Apprentices get tripped up because the textbook uses clean numbers. The field does not. Conductor resistance changes with temperature, terminations add resistance, and motor loads do not behave like resistors at startup. Start with the law, then learn where the law bends.

The power wheel and why you need it

Ohm's Law pairs with the power formula: P = I x E. Combine them and you get the power wheel, which lets you solve for any variable if you know any two others. Memorize the four core forms and you can size a load, a breaker, or a conductor without reaching for a chart.

The four you actually use in the field:

• E = I x R (find voltage from current and resistance)
• I = P / E (find current from wattage and voltage, the one you use most)
• P = I x E (find wattage from current and voltage)
• R = E / I (find resistance from voltage and current, used in troubleshooting)

The form you will reach for ten times a day is I = P / E. A nameplate gives you watts and voltage. You need amps to size the circuit. That is the whole game on service calls and small commercial work.

Single phase versus three phase: do not mix them up

The formulas above assume single phase. Three phase introduces the square root of 3 (roughly 1.732) into the current calculation. For a balanced three phase load: I = P / (E x 1.732 x PF), where PF is power factor. Forget the 1.732 and you will oversize conductors by almost double, or worse, undersize them.

Power factor matters for motors and any inductive load. For resistive loads like heat strips, PF is 1 and drops out of the equation. For motors, assume 0.8 unless the nameplate says otherwise. NEC 430.6(A)(1) tells you to use the tables in 430.250 for motor FLA, not the nameplate, when sizing conductors and overcurrent protection. The nameplate is for overload sizing.

Field tip: if your calculated motor amps do not match the nameplate, you probably used the nameplate when you should have used Table 430.250. Check the article before you pull wire.

Voltage drop: where Ohm's Law earns its keep

NEC 210.19(A) Informational Note 4 recommends branch circuits be sized so voltage drop does not exceed 3 percent, with a combined feeder and branch drop not exceeding 5 percent. It is not a hard code rule, but inspectors and engineers treat it like one, and your equipment will not last if you ignore it.

The single phase voltage drop formula is VD = (2 x K x I x D) / CM, where K is the resistivity of the conductor (12.9 for copper, 21.2 for aluminum at 75C), I is the load in amps, D is the one way distance in feet, and CM is the circular mils of the conductor from NEC Chapter 9, Table 8. For three phase, replace the 2 with 1.732.

Run the math before you pull long runs. A 20 amp circuit at 120 volts over 150 feet on #12 copper drops about 5.8 volts, which is 4.8 percent. You need #10 to stay under 3 percent. That is a real decision you will make on rough-in, and it only takes thirty seconds with a calculator.

Troubleshooting with R = E / I

When a circuit is acting up, resistance is usually the hidden variable. A loose neutral, a corroded lug, or a backstabbed receptacle adds resistance you cannot see. Measure voltage under load and current at the same point, then calculate resistance and compare it to what you expect.

A good 20 amp circuit with a 15 amp load should drop maybe 1 to 2 volts across the hot conductor from panel to device. If you see 8 volts drop, you have roughly 0.5 ohms of extra resistance somewhere in that run. That is a bad splice or a loose termination, and it is a fire waiting to happen. NEC 110.14(D) requires terminations be torqued to manufacturer specs for exactly this reason.

Field tip: voltage drop under load tells the truth. No-load voltage readings lie. Always load the circuit before you condemn a conductor.

What to commit to memory before part 2

Before you read part 2 on parallel circuits and load calcs, lock in the basics. You should be able to solve for any variable in E = I x R and P = I x E without thinking. You should know when to use 1.732 and when not to. You should know that 3 percent and 5 percent are voltage drop targets, not code minimums.

1. E = I x R, and its rearrangements
2. P = I x E, and I = P / E for quick amp calcs
3. Three phase: multiply by 1.732 and account for power factor
4. Voltage drop: 3 percent branch, 5 percent total, per NEC 210.19(A) IN 4
5. R = E / I for troubleshooting loose terminations and bad splices

Get these into muscle memory. Part 2 builds on all of it when we cover series and parallel resistance, and how the NEC load calculation tables are really just Ohm's Law dressed up.