# 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 $12 \text{V}$ dropped across the device and $2 \text{A}$ through it. I put the device in series with an $8 \Omega$ equivalent resistor (a series of two $3 \Omega$ resistors and two pairs of three $3 \Omega$ resistors in parallel).

May 19, 2017

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

Using Ohm's law
$V = I R$
we get
$28 - 12 = 2 \times R$
$\implies R = \frac{16}{2} = 8 \Omega$

To make a $8 \Omega$ resistor with $3 \Omega$ resistors

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