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SnooKiwis6943 t1_j6267x9 wrote

Reply to comment by thursdayfern in I can charge my Apple IPhone from 0-100% in 2.35sec by StrictProfessional82

Actually, in theory yes it does work. Current is equal to voltage divided by resistance (I=V/R). The higher the current, the faster the charge. The longer the charging cable, the more resistance that cable has and the lower the current is, resulting in a slower charge. Fun fact the wireless charging (mag safe) that Apple makes for the iPhone only comes with a built in cable that is of relatively short length and only available in that short length for this very reason. The mag safe uses induction charging which is very inefficient and requires a lot more amps to charge your phone. If the MagSafe charger had a longer wire, you would need more than the 20 watt charging block to power the charger due to the resistance in the longer wire.

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thursdayfern t1_j627tet wrote

I had a feeling someone would bring up Ohm’s law, and that’s why I said “in practice” and not “in theory”, because you are correct, in theory a shorter cable means less resistance means faster charging.

In practice, a 1m usb cable has less than 1 ohm resistance; I would imagine the difference in charge time between a 1cm cable and a 2m cable would be measured in seconds.

Also yeah, MagSafe is relatively inefficient for charging, but it is one of the most efficient methods of wireless charging, if not the most efficient (because alignment magnets).

A longer cable wouldn’t change much at all in this equation. You can test this yourself with a usb c extension cable (assuming it supports the necessary PD standards).

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ElephantRattle t1_j6316nc wrote

I’ve done a lot of meditation in the implication of Ohm’s Law.

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ZappySnap t1_j62dvpf wrote

The longer the cable, the more voltage drop, not current drop. And yes, if the device pulls the same current that lower voltage will reduce the wattage slightly. (And charge speed is not current. It’s power: IxV) However, voltage drop in a cable that is of normal length isn’t going to be an issue, and putting a higher wattage brick behind it isn’t going to do squat.

If a 20W charger is used, the puck would still attempt to draw 15W of power because that’s the rated charging load. The brick isn’t going to shove more power into the device to make up for voltage drop. (And if it did, it would be as a result of higher current, which would also increase voltage drop further).

You decrease the effects of voltage drop by either increasing the conductor size or by raising the source voltage (and thereby decreasing current for the same load).

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