The Transmission Problem
Part of Power Transmission
Why sending electricity over distance is fundamentally wasteful — and how the math of I²R losses shapes every decision in power system design.
Why This Matters
Building a generator is only half the problem. The generator typically sits near its power source — a river, a ridge with good wind, a slope with a water head. The people who need the power live somewhere else: in the valley, in the workshop cluster, across the field. Getting electricity from generator to load without losing most of it along the way is the central engineering challenge of any electrical system.
The transmission problem is not obvious until you run the numbers. Wire looks like a good conductor — and compared to wood or air, it is. But copper and aluminum have measurable resistance, and that resistance causes power losses that grow rapidly with distance and current. A generator 200 meters from your workshop, connected with wire that seems adequate, might deliver only a fraction of its rated output. The rest heats the wire.
Understanding the transmission problem at a mathematical level — not just intuitively — is what separates functional electrical infrastructure from wasted effort. The formula is simple. Its implications are far-reaching.
The I²R Law
Every conductor resists the flow of current to some degree. When current flows through that resistance, it dissipates power as heat. The formula is:
P_loss = I² × R
P_loss = power lost (watts)
I = current flowing (amps)
R = resistance of the conductor (ohms)
The critical feature of this formula is the square relationship between current and loss. This is not a linear proportionality — it is exponential in practical terms.
| Current (amps) | I² | Relative loss |
|---|---|---|
| 1 | 1 | 1× |
| 2 | 4 | 4× |
| 5 | 25 | 25× |
| 10 | 100 | 100× |
| 20 | 400 | 400× |
Double the current, quadruple the losses. Ten times the current, one hundred times the losses. This is why high current is the enemy of long-distance power transmission.
Wire Resistance Is Real and Significant
Every wire has a resistance that depends on:
- Material: Copper has lower resistivity than aluminum, which is lower than iron
- Length: Resistance is proportional to length
- Cross-section: Resistance is inversely proportional to cross-sectional area (thicker wire = lower resistance)
Typical resistances for copper wire at room temperature:
| Wire Gauge | Resistance per meter (Ω/m) | Resistance per 100m (Ω) |
|---|---|---|
| 18 AWG | 0.0213 | 2.13 |
| 14 AWG | 0.0084 | 0.84 |
| 12 AWG | 0.0053 | 0.53 |
| 10 AWG | 0.0033 | 0.33 |
| 8 AWG | 0.0021 | 0.21 |
| 6 AWG | 0.0013 | 0.13 |
| 4 AWG | 0.00083 | 0.083 |
Remember: a circuit has two conductors — current flows out through one wire and returns through the other. The total resistance is the resistance of both wires combined. A 100-meter run uses 200 meters of wire.
A Worked Example: The Failure Case
Your water wheel generator produces 500 watts at 12V DC. You want to power a workshop 200 meters away.
Step 1 — Calculate current:
I = P / V = 500W / 12V = 41.7A
Step 2 — Calculate wire resistance (using 10 AWG copper, the heaviest wire you have):
Total wire length: 200m out + 200m return = 400m
R = 400m × 0.0033 Ω/m = 1.32Ω
Step 3 — Calculate power loss:
P_loss = I² × R = 41.7² × 1.32 = 1740 × 1.32 = 2,297 watts
You are trying to lose 2,297 watts in wire while only generating 500 watts. This is impossible in the way stated — what actually happens is that the terminal voltage collapses because the wire resistance and the load resistance form a voltage divider. The load receives far less than 12V, and the delivered power approaches zero.
Even with absurdly large wire (2 AWG, about 0.00033 Ω/m):
R = 400m × 0.00033 = 0.132Ω
P_loss = 41.7² × 0.132 = 229 watts
You lose 229 watts — 46% of your generated power — just in copper. Heavier wire improves things but the fundamental problem remains: 41.7 amps is too much current to send 200 meters.
The Same Power at Higher Voltage
Same 500 watts, but using a transformer to step up to 120V before transmission:
Step 1 — Calculate current at 120V:
I = P / V = 500W / 120V = 4.17A
Step 2 — Wire resistance (same 10 AWG, same 400m total):
R = 1.32Ω
Step 3 — Power loss:
P_loss = I² × R = 4.17² × 1.32 = 17.4 × 1.32 = 23 watts
23 watts lost instead of 2,297. That is a 100:1 improvement from a 10:1 voltage increase.
At 240V:
I = 500/240 = 2.08A
P_loss = 2.08² × 1.32 = 4.33 × 1.32 = 5.7 watts
5.7 watts lost out of 500 generated — 98.9% efficiency on the transmission line.
The pattern: Every time you double the transmission voltage, losses fall by a factor of four (because current halves, and I² means the loss quarters). This is the entire justification for high-voltage transmission lines.
Percentage Loss: The Useful Metric
When evaluating a transmission line design, calculate percentage loss rather than absolute watts:
% loss = (P_loss / P_generated) × 100
Design targets depend on system scale:
- Homestead-scale (single building, short runs): accept up to 10% loss
- Community grid (multiple buildings, longer runs): aim for under 5% loss
- Established infrastructure: under 3% loss is the standard goal
For the 500W system at 120V, 200 meters, 10 AWG:
% loss = (23 / 500) × 100 = 4.6%
Acceptable for a community grid. The step-up transformer and distribution transformer eat some efficiency (typically 2-5% each), but the total system loss remains manageable.
Distance Multiplies the Problem
The transmission problem becomes more severe as distance increases because wire resistance is directly proportional to length.
Same 500W at 120V, but now 2,000 meters (2 km) instead of 200 meters:
Total wire: 4,000m at 0.0033 Ω/m = 13.2Ω
P_loss = 4.17² × 13.2 = 229 watts
% loss = (229/500) × 100 = 45.8%
At 10× the distance, losses multiply by 10. You need to step up to a higher voltage, use heavier wire, or both.
At 1,200V transmission voltage for the same 2 km run:
I = 500/1200 = 0.417A
P_loss = 0.417² × 13.2 = 0.174 × 13.2 = 2.3 watts
% loss = 0.46%
This is why long-distance transmission lines operate at tens of thousands of volts. The physics demanded it. Edison tried to build a DC power system at 110V and found he could not serve customers more than a kilometer from his generators without enormous copper costs. Tesla and Westinghouse’s AC system, which could use transformers to step up to high voltage for transmission and back down for use, won because the transmission problem is mathematically insurmountable at low voltages.
Practical Design Process
When designing a transmission line, follow this sequence:
1. Determine the load. What total wattage must the line carry? Add all loads with a 25-50% growth margin.
2. Choose the transmission voltage. Higher is better for efficiency, but more dangerous to build and maintain. For a small community grid: 240V is practical and relatively safe. For runs over 500 meters at significant wattage, consider 480V or higher.
3. Calculate the required current.
I = P / V
4. Calculate the total wire resistance. Total wire length = 2 × one-way distance. Look up resistance per meter for candidate wire gauges.
5. Calculate power loss and percentage.
P_loss = I² × R
% loss = P_loss / P_total × 100
6. Adjust until acceptable. If losses are too high: increase transmission voltage, increase wire gauge, or both. In practice, voltage increase is almost always more cost-effective than wire gauge increase, because wire cost and weight grow rapidly with gauge while transformer construction is a one-time investment.
7. Verify voltage at the far end.
Voltage drop = I × R
Received voltage = Transmitted voltage - Voltage drop
For a 120V line with 5% loss: 6V drop; received voltage is 114V. Most loads tolerate ±10% voltage variation, so this is fine.
The Limits of DC at Low Voltage
Direct current (DC) at low voltage faces the transmission problem with no escape. You cannot step up DC voltage without a rotating machine (a motor-generator set) or a DC-DC converter, neither of which existed in early electrical history. This is why Edison’s Pearl Street Station, using 110V DC, required a generating station every kilometer throughout Manhattan.
Alternating current, by contrast, uses transformers — simple, static devices with no moving parts, extremely reliable, and capable of stepping voltage up or down with very high efficiency. The transformer is the solution to the transmission problem. Its invention (or rather, its practical development by Gaulard, Gibbs, Stanley, and others in the 1880s) made continent-spanning power grids possible.
For a post-collapse builder, this means: if you are building any electrical system that transmits power more than a few tens of meters, design it around AC from the start. Build a transformer before you run your distribution wire. The efficiency gain is not marginal — it is the difference between a functional grid and a system that wastes most of its output heating wire.