I have a device that needs 12"V" and 2"A" to run. I have a 28"V" source. How do I make the device work with one or more 3Omega resistors?

This was a question on a test and I assume it implies there needs to be exactly 12"V" dropped across the device and 2"A" through it. I put the device in series with an 8Omega equivalent resistor (a series of two 3Omega resistors and two pairs of three 3Omega resistors in parallel).

1 Answer
A08
May 19, 2017

The device is rated to run at 12V, 2A, and supply voltage of source is 28V.
The excess voltage must drop, while it draws 2A current, across the series resistor.

Using Ohm's law
V=IR
we get
28-12=2xxR
=>R=16/2=8Omega

To make a 8Omega resistor with 3Omega resistors

  1. Two numbers of 3Omega resistors connected in series =6Omega resistance.
  2. Three numbers of 3Omega resistors connected in parallel =1Omega resistance
    Hence, 8Omega resistance =Series connection of Two 3Omega resistors and two sets of three 3Omega resistors connected in parallel.
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