Higher Energy
Curriculum/Grid Infrastructure
Grid InfrastructureLayer 44 min

Transmission Grid Basics

The U.S. transmission grid spans roughly 160,000 miles of high-voltage lines connecting generators to population centers. Without it, every city would need its own power plant. The grid exists because it is cheaper to move electricity than to generate it locally.

The transmission system uses three voltage tiers. Extra-high voltage (EHV), 345-765 kV, carries bulk power over hundreds of miles between regions. High voltage (HV), 115-230 kV, connects large generators to substations. Sub-transmission, 34.5-69 kV, links transmission substations to local distribution networks. At each transition, transformers step voltage down. The reason for high voltage: for any given power level (P = I x V), raising voltage lowers current. Since resistive losses follow I2R, doubling voltage cuts losses by 75% for the same power delivered.

Quantify losses. A line carries 1,000 MW at 345 kV. Current: I = P/V = 1,000,000 kW / 345 kV = 2,899 A. Losses proportional to I2. Now step up to 765 kV. Current: 1,307 A. Loss ratio: (1,307/2,899)2 = 0.20. Losses drop by 80%.

Why not go even higher? Insulation costs, corona discharge (visible ionization of air around the conductor), and right-of-way width all increase with voltage. Above 765 kV, these costs outweigh the loss reduction benefits.

If higher voltage always reduces losses, why does the grid use different voltage levels instead of one universal high voltage?

Match voltage to function. EHV is economical for long-distance bulk transport. Stepping down is necessary because homes and businesses cannot safely use 765 kV. Each tier matches its voltage to its distance and load requirements.

The transmission grid is the highway system of electricity: high-speed, high-capacity corridors connecting production to consumption.


Question 1 of 2

Doubling transmission voltage while carrying the same power reduces resistive losses by approximately:

Doubling voltage halves current (P = IV). Since losses = I2R, halving current reduces losses by a factor of 4, or 75%.

The answer is D