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This question has been bothering me since my university days but I never got a clear answer for it. At the university I was told that if a device need 50Watts DC then it will need fairly less Watts in AC(if we design a AC to DC converter for it).

For example I have a device which needs 50 W DC i.e 
Power = 20 Volts (DC) x 2.5 Ampere (DC) = 50 Watts DC

If I create an AC to DC Converter of normally attainable efficiency how much AC watts will this device need to operate.

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    \$\begingroup\$ Basic numeracy: \$20\cdot 5 = 100\$. \$\endgroup\$ – jonk Aug 29 '16 at 5:10
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    \$\begingroup\$ The DC watts divided by the converter efficiency. \$\endgroup\$ – Ignacio Vazquez-Abrams Aug 29 '16 at 5:11
  • \$\begingroup\$ I suspect you are leaving out some portion of the conversation that occurred in your University days. But it is obviously impossible for the output power of an AC-DC converter to be higher than its input power. \$\endgroup\$ – mkeith Aug 29 '16 at 6:23
  • \$\begingroup\$ @jonk sorry I typed the ampere wrong. it is 2.5A \$\endgroup\$ – Naeem Ul Wahhab Aug 29 '16 at 7:50
  • \$\begingroup\$ 50W plus change. If your AC/DC converter is 80% efficient, that would be 62.5W. \$\endgroup\$ – Brian Drummond Aug 29 '16 at 10:55
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If a device needs 50 W, it needs 50 W. If the input waveform is shaped like an ice cream cone, it still needs 50 W. Think of the simplest device: A heating element. A simple resistance that needs 50 W to operate.

No matter what you do, you can't dream up a device that operates this using less power. If so, you could extract the power, feed a fraction back, use the rest, and now you have a perpetual motion machine.

It is likely that you or the person who told you this confused current with power. It is generally true that a device normally operated on DC will draw much less current from an AC outlet using a suitable adapter.

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  • \$\begingroup\$ This is to do with voltage conversion more so than AC vs DC. An efficient voltage converter can supply more output current than input current provided that the output current is at a lower voltage than the input. \$\endgroup\$ – mkeith Aug 29 '16 at 6:25
  • \$\begingroup\$ @pipe You mean the device needing 50 Watts DC will also need 50 watts AC if we are using and adapter or converter? \$\endgroup\$ – Naeem Ul Wahhab Aug 29 '16 at 7:48
  • \$\begingroup\$ @TheNoble-Coder If you use an adapter or converter, it will need at least 50 Watts, because you will inevitably have losses in the converter as well. \$\endgroup\$ – pipe Aug 29 '16 at 10:41
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If your load device needs 50 watts than your supply needs to make supply which shoud be able to deliver 50 watts.

Like for AC to DC converter of 20 Volts and 5Amp DC

Power (P) = 20*5 = 100 watts

If your Supply voltage is 230V than current will be

Current (I) = P/V = 100/230 = 0.4347 Amp.

Considering Ideal (zero loss) converter.

Hope this helps

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