Simple power supply question

  • Thread starter jaydnul
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Main Question or Discussion Point

I have a power brick (transformer) that sends 12V, 17.9A to a device. I have another device that requires 12V, 10.83A. If I put a 1.7 ohm resistor in parallel to draw some of the current:

[itex]\frac{12V}{17.90A-10.83A}=1.697312[/itex]

Could I use it to power the 10.83A device? I'm just not sure if it is as simple as that.

Also, when it says 17.9A on the transformer, does that just mean that is the MAX current that can be drawn, or is that what it is always sending?

Edit: Sorry I should add that it is an external power supply, so it is outputting DC not AC.

Thanks
 
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Answers and Replies

  • #2
davenn
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I have a power brick (transformer) that sends 12V, 17.9A to a device. I have another device that requires 12V, 10.83A. If I put a 1.7 ohm resistor in parallel to draw some of the current:

12V17.90A−10.83A =1.697312 \frac{12V}{17.90A-10.83A}=1.697312

Could I use it to power the 10.83A device? I'm just not sure if it is as simple as that.
no you don't need all that

Just power the 12V 10.83 A device off the 12V 17.9A supply .... the lower current device will only draw what it needs from the 12V supply


Also, when it says 17.9A on the transformer, does that just mean that is the MAX current that can be drawn, or is that what it is always sending?
That's the max it can supply, and preferably you never want to push a PSU to its max


Edit: Sorry I should add that it is an external power supply, so it is outputting DC not AC.
yup I guessed that, tho my earlier answer wouldn't have made any difference


cheers
Dave
 
  • #3
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Oh Ok. Thanks a bunch Dave!
 

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