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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