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On the adapter it says Input ~230V 1.5A, Output 29V 1.5 A.
If the adapter draws 1.5 A at 230 V and transfers them at 29V wouldn't that mean a power consumption of (230-29)V *1.5A = 301.5W in the adapter? That can't be right, so how does it actually work?

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    \$\begingroup\$ The "230V 1.5A" on the label is wrong. Is the secondary accurate or does it measure wrong? I would not use an adapter that has a wrong label like that. What else could be wrong with it? \$\endgroup\$
    – Audioguru
    Apr 16, 2023 at 13:50

2 Answers 2

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Ignoring efficiency losses:

  • If your load draws 1 A at 29 V the power supply will have to supply \$ P = VI = 29 \times 1 = 29 \text W \$.
  • It will have to draw the same power from the mains at 230 V. We can work out the current drawn from the mains by rearranging the same equation. \$ I = \frac PV = \frac {29}{230} = 0.13 \text A \$.
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  • \$\begingroup\$ How come there can be more input current than output current? I would've thought it has to be the same because of Kirchhoffs circuit laws. \$\endgroup\$
    – klingeron
    Apr 16, 2023 at 14:41
  • \$\begingroup\$ Found an example explanation here: electronics.stackexchange.com/a/541752/337651 So the extra current comes from the ground? \$\endgroup\$
    – klingeron
    Apr 16, 2023 at 14:48
  • \$\begingroup\$ Kirchhoff's circuit laws apply to a given circuit. The circuit into the power adapter is a different circuit from the circuit out of the power adapter. Therefore, Kirchhoff's law doesn't apply. Also, the output current is higher, not lower than the input current, due to a different law, which does apply: conservation of power. As output voltage is lower, output current must be higher. And, no, the extra current doesn't come from the ground. \$\endgroup\$ Apr 16, 2023 at 15:14
  • \$\begingroup\$ That answer leaves the open question where the current would be coming from instead. Or would it not be a contradiction to have more electrons flow out than electrons are flowing in? \$\endgroup\$
    – klingeron
    Apr 16, 2023 at 16:40
  • \$\begingroup\$ Ignore electrons. Think of current and voltage. \$ P_{OUT} = P_{IN} - P_{LOSSES} \$. Therefore, since \$ P = VI \$ we can say, \$ V_{IN}I_{IN} - losses= V_{OUT}I_{OUT} \$. For a given power, stepping the voltage down steps the current up. Have a read on how ordinary transformers work to get the basic theory on this. \$\endgroup\$
    – Transistor
    Apr 16, 2023 at 17:08
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On the adapter it says Input ~230V 1.5A

That's "up to 1.5 A". That is, from ~0 A to 1.5 A, depending on what's connected to the power adapter's output.

wouldn't that mean a power consumption of (230-29)V *1.5A = 301.5W in the adapter?

That's "up to 50 W" (29 x 1.5 = 43.5 W, input power a bit higher due to inefficiencies = ~ 50 W). That is, the input power into the adapter is from ~3 W to ~50 W, depending on what's connected to the power adapter's output.

how does it actually work?

The load determines the power. No load = almost no power into the adapter. Mid-power load = mid-power into the adapter. Full load = ~50 W into the adapter.

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    \$\begingroup\$ The 300W is probably accounting for a few microseconds of inrush current, with some healthy safety factor built in. \$\endgroup\$ Apr 16, 2023 at 14:02
  • \$\begingroup\$ The adapter is from an electric recliner chair, so maybe because the load on the chair can change a lot suddenly while reclining there can be large spikes sometimes \$\endgroup\$
    – klingeron
    Apr 16, 2023 at 14:54

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