Using a Resistor to Power Multiple Devices: A Simple Guide

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SUMMARY

This discussion clarifies the use of a 12V, 17.9A external power supply to power a 12V, 10.83A device. It confirms that the power supply's rating indicates the maximum current it can provide, not the constant current output. Users do not need to use a resistor in parallel; the lower current device will only draw the necessary current from the supply. The conversation emphasizes the importance of not operating the power supply at its maximum capacity for optimal performance.

PREREQUISITES
  • Understanding of DC power supply specifications
  • Basic knowledge of electrical current and resistance
  • Familiarity with Ohm's Law
  • Experience with connecting electronic devices to power supplies
NEXT STEPS
  • Research the implications of using power supplies at their maximum ratings
  • Learn about the differences between AC and DC power supplies
  • Study Ohm's Law and its applications in circuit design
  • Explore safe practices for connecting multiple devices to a single power supply
USEFUL FOR

Electronics enthusiasts, hobbyists, and professionals working with DC power supplies, as well as anyone involved in connecting multiple devices to a single power source.

jaydnul
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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:

\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.

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
 
Last edited:
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Jd0g33 said:
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
Jd0g33 said:
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
Jd0g33 said:
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 differencecheers
Dave
 
Oh Ok. Thanks a bunch Dave!
 

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