Question

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).

Answer 1

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.