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How much power could be captured per inch by wrapping a wire around a standard 120V power cable for each amp-hour flowing through it (or whatever reference point makes sense if it's not that)?

Assume this is a cable with no special shielding - like what might be used for a consumer device.

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  • \$\begingroup\$ It will decrease power factor and sparks may be created between the transmission line and the wrapped wire. \$\endgroup\$
    – Miss Mulan
    Commented Apr 28, 2022 at 17:48
  • \$\begingroup\$ Is that power cable an electrical extension or a 2-14 AWG? cause if you have both the L and N lines on the same power cable, wrapping around some wiring will not do much other than perhaps capture a spike once in a while. Tell us more about how your circuitry is implemented. 1 wire-2wires- is the wrap a single wire or double ? \$\endgroup\$ Commented Apr 28, 2022 at 17:57

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To extract power from a single conductor current carrying cable, you wrap a magnetic core round the conductor, making it a one turn primary of a transformer, and then a secondary coil round that core.

This is often done in a self-powered clip-on power monitor, to harvest a few mW from the conductor to charge its batteries.

If you have a 'standard power cable' with out and return conductors, this will not work, as the magnetic field from both conductors cancels. You have to do this at the distribution panel where you have single conductor available, or make an extension cable where you can get at both.

The limitation in the amount of power you can pick off is the primary inductance of this transformer. This can be improved by increasing the cross sectional area of the core, or putting several primary turns through the core. The impedance can be further improved by resonating the transformer inductance with a parallel capacitor, on the secondary for practicality.

Simply wrapping a wire round the cable makes a capacitor, coupling to the voltage of the input cable. You will not be able to extract very much power at all via this route. It does have the advantage of working half as well on 'standard power cable' as single conductor cable, compared to the 'not working at all' of the magnetic pickoff.

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What you describe, is a very poor transformer with bad coupling.

The power that you take from the wrapped wire is not for free but will load the mains cable (aka primary) in the same way as in a proper transformer.

The coupling constant is close to 0 though, if you wrap around all wires in the mains cable. Thus, almost no power can be transferred inductively. The coupling can be improved somewhat if you wrap between the L and N wires.

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  • \$\begingroup\$ Furthermore, the amount of power that can be "extracted" depends on how much current is going through the primary wire. I.e., if wrapped around a lightbulb wire, it'll only generate when the bulb is on (when current is flowing through the wire, creating a magnetic field for the wrap to interact with.) Otherwise, it's zero. \$\endgroup\$
    – rdtsc
    Commented Apr 28, 2022 at 18:11
  • \$\begingroup\$ @rdtsc Agreed..But I thought that was implicit as the OP stated "per Ah" in their question. \$\endgroup\$
    – tobalt
    Commented Apr 28, 2022 at 18:16

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