The law of resistance states that the electrical resistance of a uniform conductor is directly proportional to its length and inversely proportional to its cross-sectional area. In a real circuit or installation, this physical law dictates exactly how much voltage your load will actually receive at the end of a wire run and how much parasitic heat the conductor will generate along the way. If you ignore it, your breakers might hold, but your equipment will starve for voltage and your walls might harbor hidden heat.
The Core Formula and a Bench-Tested Numeric Example
In physics textbooks, the law of resistance is written as R = ρ(L/A), where R is resistance, ρ (rho) is the material's resistivity, L is length, and A is cross-sectional area. But on the jobsite or at the workbench, we use a more practical version based on American Wire Gauge (AWG) and circular mils:
R = (K × L) / A
K = Resistivity constant (12.9 for copper, 21.2 for aluminum at 20°C)
L = One-way length of the wire in feet
A = Cross-sectional area in circular mils (cmil)
Let’s run a numeric example using a standard 15A branch circuit. You are running 100 feet of 12 AWG copper wire to a receptacle.
- Find the Area: According to standard wire tables, 12 AWG wire has a cross-sectional area of 6,530 cmil.
- Calculate One-Way Resistance: R = (12.9 × 100) / 6,530 = 0.197 Ω.
- Calculate the Circuit Loop: Current must travel out and back, so the total wire length is 200 feet. Total loop resistance = 0.197 × 2 = 0.394 Ω.
- Calculate Voltage Drop: Using Ohm’s Law (V = I × R), a 15A load will experience a drop of 15 × 0.394 = 5.91V.
On a 120V nominal circuit, a 5.91V drop is a 4.9% loss. The NEC strongly recommends keeping voltage drop under 3% for branch circuits. The breaker won't trip, but a sensitive motor or switching power supply at the end of that run will run hot and inefficient.
Where You Meet the Law of Resistance in Practice
You interact with this law every time you size a wire for anything beyond a short jumper. Here is where it dictates your design choices:
- Solar Panel Strings: To keep L (distance from roof to inverter) manageable without buying massively thick wire, we wire panels in series. This increases voltage and drops current, minimizing the I²R heat losses dictated by the wire's resistance.
- Subpanel Feeders: When running 240V aluminum feeder to a detached garage 150 feet away, the length variable (L) is huge. To keep resistance low, you must artificially increase the area (A) by upsizing from 2 AWG to 1/0 AWG aluminum, even if the breaker size only strictly requires the smaller wire.
- 12V Automotive and Marine Systems: Because system voltage is so low, even a 0.5V drop is catastrophic. Since you can't easily shorten the wire runs in a vehicle, you must use massive cross-sectional areas (like 2/0 AWG for winches) to force the resistance down.
Real-World Scenario Walkthrough: The Melted 30A RV Receptacle
Abstract formulas are fine, but here is what happens when the law of resistance is ignored in a real installation.
The Numbers:
- 10 AWG copper area = 10,380 cmil.
- One-way resistance = (12.9 × 100) / 10,380 = 0.124 Ω.
- Total loop resistance = 0.248 Ω.
- The RV's compressors and power supplies pull a continuous 24A.
- Voltage drop = 24A × 0.248 Ω = 5.95V.
- Power dissipated as heat in the wire = I²R = (24²) × 0.248 = 142 Watts.
The Outcome:
142 Watts of heat is roughly equivalent to leaving a 140W incandescent lightbulb turned on inside your wall cavity, distributed along 200 feet of wire. The NM-B cable gets warm. Worse, because the voltage at the RV drops to ~114V, the RV's modern switching power supplies and inverter-driven AC compressors pull more current to meet their wattage requirements. This pushes the amperage closer to 28A, increasing the heat exponentially.
What Went Wrong:
The homeowner sized the wire for the breaker (10 AWG is legally allowed on a 30A breaker per NEC 240.4), but they completely ignored the length variable in the law of resistance. The terminal lugs at the receptacle eventually overheated from the compounded thermal load, melting the plastic faceplate. The correct fix was upsizing to 6 AWG copper to drop the loop resistance to 0.080 Ω, cutting the heat dissipation down to a safe 46 Watts.
Common Confusions: Resistance vs. Resistivity and Impedance
People frequently mix up three related but distinct concepts when discussing wire behavior. Here is how to keep them straight:
1. Resistance (Ω) vs. Resistivity (Ω·m):
Think of a plumbing system. Resistivity is the inherent friction of the pipe material itself—rough cast iron has high resistivity, while smooth PVC has low resistivity. Resistance is the actual friction you experience pushing water through a specific 50-foot length of that pipe. Resistivity is a property of the material (copper vs. aluminum); resistance is a property of the specific object (this exact spool of 12 AWG wire). For a deep dive into material properties, the Engineering Toolbox resistivity tables are an excellent bench reference.
2. DC Resistance vs. AC Impedance (Z):
The law of resistance strictly calculates the DC resistive component. In AC circuits, wire also exhibits inductive reactance (which opposes changes in current). The combination of resistance and reactance is called impedance. However, for standard 60Hz residential wiring in sizes smaller than 1/0 AWG, the inductive reactance is negligible. For hobbyist and residential DIY math, treating AC impedance as equal to DC resistance yields an error of less than 2%, which is perfectly acceptable for voltage drop calculations.
Quick-Reference: Copper Resistance by AWG
When calculating the law of resistance, you need the cross-sectional area. Below is a reference chart for solid copper wire at a baseline of 20°C (68°F). Note that as wire temperature increases under load, resistance increases by roughly 0.4% per degree Celsius.
| AWG Size | Area (cmil) | Area (mm²) | Resistance per 1,000 ft (Ω at 20°C) |
|---|---|---|---|
| 14 AWG | 4,110 | 2.08 | 3.14 |
| 12 AWG | 6,530 | 3.31 | 1.98 |
| 10 AWG | 10,380 | 5.26 | 1.24 |
| 8 AWG | 16,510 | 8.37 | 0.78 |
| 6 AWG | 26,240 | 13.30 | 0.49 |
| 4 AWG | 41,740 | 21.15 | 0.31 |
| 2 AWG | 66,360 | 33.62 | 0.19 |
Source data aligned with standard Georgia State University HyperPhysics conductor models and NEC Chapter 9, Table 8 baseline metrics.
FAQ: Troubleshooting Resistance in the Field
Q: Can I just use my multimeter's continuity mode to measure the law of resistance on a long wire?
A: No. Standard multimeters inject less than 1mA during a continuity/resistance test. At that microscopic current, the resistance of your test leads and the contact resistance of the probes will completely skew the reading on a low-resistance wire. To measure real-world resistance, apply the actual load, then measure the voltage at the source and the voltage at the load. The difference is your voltage drop; divide that drop by the measured amperage to find the true operational resistance (R = V_drop / I).
Q: Does the law of resistance apply to aluminum wire the same way?
A: The law applies universally, but the K constant changes. Aluminum has a higher resistivity than copper. When using the practical formula (R = K × L / A), you must change the K value from 12.9 to 21.2. This is why aluminum feeder wires must be sized two AWG steps larger than copper to achieve the exact same resistance over the same distance.
Q: Why does my wire get hot if it's sized correctly for the breaker?
A: Breakers only protect against short circuits and massive overloads. They do not protect against the heat generated by normal current flowing through the inherent resistance of a long wire. If your wire is hot to the touch on a 20A circuit, your run is too long for that AWG size, and the I²R power loss is dissipating as physical heat. Upsize the wire immediately.
