On this page
Electricity often comes from a power station hundreds of kilometres away. On a short wire you can treat the line as a simple resistance, but on a long line that shortcut fails, and in the 1920s the correct calculation took engineers a long time. Edith Clarke built a tool that did it in minutes. This post covers who she was, what her calculator did, why long lines are different, a worked example on transmission losses, common mistakes, and what you can take from her way of working.
Who was Edith Clarke?
The National Inventors Hall of Fame, which inducted her in 2015, gives her life as 10 February 1883 to 29 October 1959. She was born in Howard County, Maryland, one of nine children, and used her inheritance to study mathematics at Vassar College, graduating in 1908. She taught mathematics and physics, then in 1911 began civil engineering at the University of Wisconsin.
The Handbook of Texas says she left Wisconsin to work for AT&T as a computing specialist, a "computor" doing calculations on transmission lines for engineers, and from 1915 to 1918 supervised a group of women doing such calculations. In 1918 she went to MIT, and according to the Hall of Fame, in 1919 she became the first woman to earn a master's degree in electrical engineering there.
Even with that degree she could not find work as an engineer, so General Electric hired her to supervise computers in its Turbine Engineering Department. The Hall of Fame says that after a year teaching physics in Constantinople, she returned in 1922 to a salaried electrical engineer's post at GE, becoming the first woman professionally employed as an electrical engineer in the US. In 1926 she was the first woman to present a paper at the annual meeting of the American Institute of Electrical Engineers, and between 1923 and 1945 she published 18 technical papers. She retired from GE in 1945, and in 1947 joined the University of Texas at Austin as the first woman professor of electrical engineering in the country, teaching until 1957.
What did the Clarke calculator do?
Her US patent No. 1,552,113, filed on 8 June 1921 and granted on 1 September 1925, states the problem. For short lines, engineers usually assumed the line's capacitance and inductance were concentrated at a few points. For a line a few hundred miles long, she wrote, that assumption is not justifiable, and "even for such distances as 200 miles, errors would result". Accurate work had to treat inductance and capacitance as spread evenly along the line, and the formulas for that took a great deal of work.
Her graphical calculator let an engineer start from the voltage, power and power factor wanted at the receiving end and find the current and voltage at the generator end "in the short space of a few minutes". She claimed an error of only a small fraction of one percent for lines of about 500 miles. The Hall of Fame says it solved the line equations, which involve hyperbolic functions, ten times faster than earlier methods.
Why are long lines different from short ones?
Alternating current is a wave travelling along the line. For a rough estimate, take its speed as close to the speed of light, 3 × 10⁸ m/s. At 50 Hz, one wavelength is λ = c/f = 3 × 10⁸ / 50 = 6.0 × 10⁶ m, or 6,000 km.
A 500-mile line is 805 km, about 13% of a wavelength. The voltage at the far end is then about 360° × 0.134 = 48° out of step with the near end. A 200-mile line (322 km) still gives about 19°. Over such distances you cannot treat voltage and current as the same all along the wire, which is the point of Clarke's patent.
Worked example: why lines run at high voltage
Sample numbers: a line must deliver 10 MW, and its total resistance is 10 Ω. Take power factor 1 first.
At 11 kV. I = P/V = 10 × 10⁶ / 11 × 10³ = 909 A. Loss = I²R = 909² × 10 = 8.26 MW. Most of the power is lost, so this is impossible.
At 110 kV. I = 90.9 A. Loss = 90.9² × 10 = 82.6 kW, or 0.83%.
At 220 kV. I = 45.5 A. Loss = 20.7 kW, or 0.21%. Doubling the voltage cuts the loss to a quarter.
Power factor 0.8 at 110 kV. I = P/(V cos φ) = 10 × 10⁶ / (110 × 10³ × 0.8) = 114 A. Loss = 113.6² × 10 = 129 kW, more than at unity power factor.
This is why transformers step the voltage up for transmission and down for use. Our post on transformers and AC generators explains how.
What mistakes do students make with power transmission?
Using supply voltage in V²/R for the loss. Find the current first, then use I²R.
Ignoring power factor. For the same real power, a lower power factor means more current and more loss.
Mixing peak and rms values. Power formulas use rms. See our post on AC power and rms values.
The JEE (Main) 2026 syllabus lists "power in AC circuits", "wattless current" and "transformer" in Unit 14 and electromagnetic waves in Unit 15. Long-line theory is beyond the syllabus. Check your own year's syllabus.
If power factor problems keep going wrong at the same step, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
What can a student take from Clarke's way of working?
Turn repeated work into a tool. She had done these calculations by hand, so she knew exactly which steps to simplify.
Keep going when the door is shut. Turned down as an engineer, she took the job she could get and kept producing work until GE hired her as one.
Pass it on. She ended her career teaching engineering students.
Next step: find the line loss for the 10 MW example at 66 kV with power factor 0.9.
