I²R Losses

Part of Power Transmission

Understanding resistive power loss in conductors — the fundamental problem that high-voltage transmission solves.

Why This Matters

I²R losses are the most important concept in power transmission engineering. Every conducting wire has resistance. Current flowing through resistance dissipates power as heat. The formula P = I²R tells you exactly how much power is wasted in any conductor — and the squared relationship to current means that reducing current is far more effective than reducing resistance when trying to minimize losses.

This is not an abstract engineering concern. In a rebuilding scenario, you may be transmitting power hundreds of meters from a water-wheel generator to a community workshop. If you do not understand I²R losses, you will wire your system at too-low voltage, put too much current through too-thin wire, and find that your generator produces far less useful power at the load than you expect. In the worst case, the wire overheats, the insulation burns, and you have a fire.

Understanding I²R losses also explains every major decision in power system design: why high voltage is used for transmission, why wire gauge is critically important, why good connections matter (bad connections have high resistance), and why the power factor of inductive loads increases your wire losses without delivering additional useful work.

The Physics

Ohm’s Law and Resistance

When current flows through a conductor, electrons collide with the atomic lattice of the metal. Each collision transfers kinetic energy from the electron to the lattice, which vibrates more — which is heat. The rate of this energy transfer is the resistive power loss.

For a uniform conductor:

V = I × R     (Ohm's Law)

Where:
  V = voltage across the conductor (volts)
  I = current through the conductor (amperes)
  R = resistance of the conductor (ohms)

Power Loss Formula

Power is voltage times current:

P = V × I

Substituting V = I × R:
P = I × R × I = I²R

Or, substituting I = V/R:
P = (V/R) × V = V²/R

For a transmission conductor, both forms are useful:

• P = I²R is used when you know the current flowing in the wire
• P = V²/R is used when you know the voltage drop across the wire

The Squared Relationship

The most important insight is that loss scales with the square of current. This is not linear:

Current (fraction of rated)	Loss (fraction of full-rated loss)
10% of full current	1% of full losses
50% of full current	25% of full losses
100% of full current	100% of full losses
150% of full current	225% of full losses
200% of full current	400% of full losses

Running a wire at twice its rated current produces four times the heat. This is why overloaded wires overheat so rapidly — the relationship is not gradual.

Calculating Losses in Real Circuits

Example 1: Short-Distance Low-Voltage System

A solar system on a house roof charges batteries 15m away through two 10 AWG copper conductors.

System voltage: 48V DC
Maximum current: 25A
Wire: 10 AWG copper (3.28 mΩ/m per conductor)
Total conductor length: 2 × 15m = 30m
Total resistance: 0.00328 × 30 = 0.0984 Ω

Voltage drop: V = I × R = 25 × 0.0984 = 2.46V
% drop: 2.46/48 = 5.1%
Power loss: P = I² × R = 625 × 0.0984 = 61.5W
% of delivered power: 61.5/(48×25) = 5.1%

This 5% loss is acceptable for a 48V system. The wire delivers 94.9% of generated power to the battery.

Example 2: Medium-Distance Higher Power

A water wheel generates 2,000W at 240V AC. The mill it powers is 300m away.

Current: I = 2,000/240 = 8.33A
Wire: 12 AWG copper (5.21 mΩ/m per conductor)
Total conductor length: 2 × 300m = 600m
Total resistance: 0.00521 × 600 = 3.126 Ω

Voltage drop: V = 8.33 × 3.126 = 26.0V
% drop: 26.0/240 = 10.8%
Power loss: P = 8.33² × 3.126 = 216.8W
Efficiency: (2000 - 217) / 2000 = 89.2%

10.8% loss is significant but may be acceptable. The mill receives 89.2% of generated power. To reduce to 5%, upgrade to 8 AWG wire (2.06 mΩ/m):

Resistance: 0.00206 × 600 = 1.236 Ω
Power loss: 8.33² × 1.236 = 85.7W
% drop: 4.3%

Better. Or use higher voltage transmission — see High Voltage Solution.

Example 3: Long-Distance Transmission

A community generator produces 20,000W. The settlement it serves is 1km away. Available wire: 4 AWG aluminum (1.3 × 2.82/1.72 = 2.13 mΩ/m).

At 240V AC:

Current: I = 20,000/240 = 83.3A
Resistance: 0.00213 × 2,000m = 4.26 Ω
Voltage drop: 83.3 × 4.26 = 354.8V

Impossibly high — the wire resistance requires more voltage than you have.

At 4,800V AC:

Current: I = 20,000/4,800 = 4.17A
Resistance: 4.26 Ω (same wire)
Voltage drop: 4.17 × 4.26 = 17.8V
% drop: 17.8/4,800 = 0.37%
Power loss: 4.17² × 4.26 = 74.0W
Efficiency: (20,000 - 74)/20,000 = 99.6%

The 20× voltage increase (240V to 4,800V) reduces losses by 400× (from impossible to 0.37%).

Connection Resistance and Its Consequences

Every joint, terminal, and connector in a circuit adds resistance. In a well-made joint, this resistance is negligible — a few milliohms. In a bad joint, it can be several ohms.

A loose or corroded connection carrying 20A:

Extra resistance: 2 ohms (typical poor connection)
Extra power loss: 20² × 2 = 800W at that single joint

This 800W of heat at a single point will burn through wire insulation,
ignite wood, and start fires. This is the most common cause of electrical fires.

Testing connections: Measure voltage drop across any joint while current is flowing. A good joint shows millivolts. A bad joint shows multiple volts — a direct measurement of wasted power.

Maintaining connections:

• Tighten all screw terminals to rated torque after initial installation and after each thermal cycle
• Use antioxidant compound on all aluminum connections
• Apply electrical-grade corrosion inhibitor (NO-OX-ID or equivalent) to outdoor connections
• Physically inspect all visible connections annually for discoloration, which indicates excess heating

I²R in Transformers and Motors

Transformer windings have resistance. The current through the secondary winding produces I²R losses in the copper wire of the secondary — this is “copper loss” in transformer engineering.

For a transformer with 0.5 ohm secondary resistance carrying 20A secondary current:
Copper loss = 20² × 0.5 = 200W — this is why transformers run warm

For a transformer on full load vs half load:
Full load copper loss: I² × R
Half load copper loss: (I/2)² × R = I²R/4 — copper losses drop to 25%

This is why lightly loaded transformers are much more efficient than heavily loaded ones, and why transformers should be sized close to their expected load rather than dramatically oversized.

Motor windings follow the same relationship. A motor running at full load current has much higher I²R heating than a motor running at half load. A motor that stalls (maximum current at zero speed) has maximum I²R heating — this is why stalled motors overheat and burn out rapidly.