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In particular, if you were going to try to steal power from HDVC transmission lines you'd need one cable attached to current going in 1 direction and the other attached to current flowing in the opposite direction. So this would create a branch current and if there are meters measuring the current flow, the utility companies could detect an unexpected drop in current and locate where it is coming from.

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    \$\begingroup\$ And why do you think this is not the same with AC transmission? \$\endgroup\$
    – Transistor
    May 15 '19 at 20:49
  • \$\begingroup\$ @Transistor Was just about to say that but you beat me to it. \$\endgroup\$
    – DKNguyen
    May 15 '19 at 20:49
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    \$\begingroup\$ easier to detect the faults created by would-be thieves fried to a crisp \$\endgroup\$ May 15 '19 at 21:02
  • \$\begingroup\$ How much power constitutes theft? I doubt if you had a coil near one of these lines they would be able to see watts missing from a MW line. Anybody stupid enough to connect directly would probably die, you need step down transformers and huge insulators to handle high voltage. \$\endgroup\$
    – Voltage Spike
    May 15 '19 at 21:03
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    \$\begingroup\$ HVDC links operate at hundreds of 1000s of Volts - it's not practical to steal power at those voltages. \$\endgroup\$ May 16 '19 at 1:29
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No, using DC rather than AC doesn't make power loss any easier to detect.

But it does make theft much harder to begin with. With an AC transmission line, most forms of theft rely on the AC magnetic and or electric fields radiated by the lines — i.e., there is no direct connection to the line at all.

With a DC line, there is no radiation of that type, other than any noise generated by the power converters at each end, and the random current variations caused by customers switching loads on and off.

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  • \$\begingroup\$ Also using Ohm's law, if an HDVC line has 800 kV, given the dry human skin resistance of ~100 kOhm, that would give 8 amps of current which would fry em real good! \$\endgroup\$
    – Mr X
    May 16 '19 at 6:10
  • \$\begingroup\$ Again, the same thing is true for AC. \$\endgroup\$
    – Dave Tweed
    May 16 '19 at 11:01

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