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

Ohm's Law in one line

V = I × R. Voltage equals current times resistance. Rearrange it two ways and you cover most field problems: I = V / R and R = V / I. Power ties in through P = V × I, which rearranges to P = I²R and P = V² / R.

Memorize the wheel or don't. What matters is recognizing which two values you have and solving for the third. On a service call you rarely measure all three at once, so the law is a bridge between what your meter shows and what you need to know.

This is part 4 of the crash course. Parts 1 through 3 covered the basics, series and parallel behavior, and AC complications. Here we run field examples you actually see in residential, commercial, and light industrial work.

Example 1: Voltage drop on a long branch circuit

You're feeding a 15 A continuous load 180 feet out on a 20 A circuit at 120 V. NEC 210.19(A) Informational Note No. 4 recommends branch circuit voltage drop stay at or under 3%. That's 3.6 V max on a 120 V circuit.

Resistance of #12 copper is roughly 1.93 ohms per 1000 ft. Round trip on 180 ft is 360 ft, so R = 0.695 ohms. V = I × R = 15 × 0.695 = 10.4 V drop. That's 8.7%. Way over. Bump to #10 (1.21 ohms/1000 ft): 15 × 0.436 = 6.5 V. Still over. #8 gets you to 4.1 V, close. Either shorten the run or go larger.

• #14 Cu: 3.07 ohms/1000 ft
• #12 Cu: 1.93 ohms/1000 ft
• #10 Cu: 1.21 ohms/1000 ft
• #8 Cu: 0.764 ohms/1000 ft
• #6 Cu: 0.491 ohms/1000 ft

Example 2: Sizing a resistive heater circuit

Customer wants a 4500 W baseboard heater on 240 V. NEC 424.3(B) says fixed electric space heating is a continuous load, so ampacity and overcurrent are sized at 125%.

I = P / V = 4500 / 240 = 18.75 A. Multiply by 1.25 for continuous: 23.4 A. That puts you on a 30 A circuit, #10 Cu minimum per NEC 240.4(D) and Table 310.16 after adjustments. Resistance of the element itself is R = V / I = 240 / 18.75 = 12.8 ohms, useful if you're meggering or troubleshooting an open element.

If you read infinite resistance across the element terminals with the disconnect off, the element is open. If you read something close to 12.8 ohms on that 4500 W unit, element is intact and your problem is upstream.

Example 3: Why that motor tripped the breaker

A 5 hp single phase 240 V motor draws about 28 A running (Table 430.248). Locked rotor can hit 6 to 8 times that, call it 170 A for a few cycles on start. Ohm's Law tells you why: at standstill the rotor has no back EMF, so effective impedance collapses to winding resistance only. V / R spikes the current.

If the motor is stalling on start, measure winding resistance cold. Compare phase to phase on three phase, or line to line on single phase split capacitor. Big mismatch means a shorted turn or open winding. A reading near zero on the run winding with the cap disconnected means the cap shorted and welded the start circuit.

1. Kill power, lock out per NEC 110.25 and OSHA 1910.147.
2. Discharge any capacitors with an insulated resistor tool.
3. Meter resistance across windings, record values.
4. Compare to nameplate or manufacturer data.
5. Megger to ground at 500 V or 1000 V, expect well over 1 megohm.

Example 4: GFCI nuisance trip math

Class A GFCI trips at 4 to 6 mA of ground fault current per UL 943. NEC 210.8 dictates where they're required. On a long outdoor circuit feeding a pond pump, cumulative capacitive leakage through the cable plus any moisture path can get you there without a real fault.

If a 0.002 microfarad per foot cable runs 250 feet on 120 V 60 Hz, capacitive reactance is Xc = 1 / (2π f C). C = 500 nF total, Xc is about 5300 ohms. Leakage current I = V / Xc = 120 / 5300 = 22 mA worst case line to ground if the shield or conduit is grounded and wet. That alone can trip the GFCI. Shorten the run, use a dedicated GFCI closer to the load, or switch to a device rated for the application.

Example 5: Load calc sanity check

Panel schedule shows 14.4 kW connected at 240 V single phase. I = P / V = 14400 / 240 = 60 A. Cross check against NEC 220 part III demand factors before sizing the service. If you measure 48 A on the main with the house loaded up, that's 11.5 kW actual demand, a demand factor of 0.80. Useful for upgrade conversations with the customer.

Always verify nameplate wattage against measured current before quoting a service upgrade. Half the panels that "need 200 A" are running under 60 A peak.

Quick reference

Keep these rearrangements in your head and you'll solve 90% of field questions without a calculator.

• V = I × R, I = V / R, R = V / I
• P = V × I, P = I²R, P = V² / R
• Continuous load factor: 1.25× per NEC 210.19, 210.20, 215.2, 215.3
• Voltage drop target: 3% branch, 5% total feeder plus branch (NEC Informational Notes)
• Three phase power: P = √3 × V × I × PF

Next up in part 5: reactive power, power factor correction, and why your amp clamp reads more than the kW meter on the pole.