How Dangerous Is It?

Electricity and Magnetism Level 3

A daredevil with questionable judgment bounces off of a trampoline and grabs onto a single phase conductor (phase A) of a high-voltage power line with both hands. As he hangs from the power line with his feet off of the ground, his hands are $1$ meter apart.

What is the potential difference between his two hands (in volts RMS)?

Details and Assumptions:

The line is $100\text{ km}$ long.
The nominal voltage level is $230\text{ kV}$ RMS phase-to-phase. This corresponds to $\frac{230}{\sqrt{3}} \text{ kV}$ phase-to-neutral.
Both ends of the line are at nominal voltage level, and there is a $15^\circ$ angular difference between the two ends.
For simplicity, assume that the voltage gradient is uniform along the line's length. This is a reasonably good assumption for a line of this length.
Please do not actually try this!

The answer is 0.34665.

1 solution

Steven Chase
Jan 15, 2018

Use the phase-to-neutral voltage to calculate the total voltage difference between the two line ends:

$\Delta V = \frac{230,000}{\sqrt{3}} (1 \angle 0^\circ - 1 \angle 15^\circ)$

Then divide the total voltage difference by the line length (in meters) to find the voltage difference between the daredevil's two hands:

$\Delta V_{1m} = \frac{230,000}{\sqrt{3} \times 100,000} (1 \angle 0^\circ - 1 \angle 15^\circ) \\ |\Delta V_{1m}| = 0.34665$

While this would most likely not cause any harm, do not try this stunt!

Post-script: Here's an expansion of the complex math, for calculation purposes

$1 \angle 0^\circ - 1 \angle 15^\circ = 1 - cos(15^\circ) - j \, sin(15^\circ) \\ |1 \angle 0^\circ - 1 \angle 15^\circ| = \sqrt{2 - 2 \, cos(15^\circ)} \\ |\Delta V_{1m}| = \frac{230,000}{\sqrt{3} \times 100,000} \sqrt{2 - 2 \, cos(15^\circ)} \approx 0.34665$

I think there is a typo in the final number, I got 0.34765V. Also, here is a question: How can be the line voltage the same, except for the phase? My guess would be that even if there is no load on the transmission line, the voltage at the receiving end will be always smaller than the source voltage, because of the capacitive currents to the ground and the ohmic losses in the wire.

Laszlo Mihaly - 3 years, 4 months ago

Greetings. I did the math again and got the same result that is posted. Regarding the question, in practice, the two ends will usually not be at the same voltage level, but they will be within about 5 percent of nominal. In this problem, we're essentially assuming that both ends of the line are connected to "infinite buses".