# how to calculate resistance needed for voltage drop

by taeagan
Tags: resistance, voltage
 P: 3 Hi. I'm hoping someone can help me out with a pretty simple physics question. My electrical circuit skills are pretty rusty, so please bear with me. I'm using a 19V DC laptop adapter to power a 12V DC pump motor. The pump is rated for 3.5A. I have an on/off switch between the power supply and the motor. I'm trying to figure out what kind of circuit I need to put in between to drop the voltage from 19V out of the battery to 12V at the motor. By my (likely overly simplified) calculations I think that I need a 2ohm resistor rated for 25W+. Is it as simple as just putting a resistor in between the + side of the battery and the motor? Do I need anything more complicated than that? I have to think that I'm wrong here. Thanks!
 Sci Advisor PF Gold P: 2,604 Does you 19V DC adapter indicate at what current it can supply at that voltage or what its power rating is? Most of these wall warts output voltage can sag as you draw power from them. Welcome to Physics Forums.
 P: 3 The power supply reads that it is rated for DC output 120W or 6.32A.
PF Gold
P: 2,604

## how to calculate resistance needed for voltage drop

Sounds like it's good enough for what you are doing. So if you want the voltage at your pump to be 12 volt, you'll need a drop of 7 volts. At the rated pump current of 3.5 amp you would need a resistance V/I=7volts/3.5amp=2Ω. The power dissipation would need to be I2R=(3.5amp)2(2Ω)=24.5watt. So your calculations are correct.

One thing to consider is if the pump load varies (depending on what's being pumped), the pump current draw will vary, hence the voltage you supply the pump with will also vary. One could consider a regulator circuit if this is a problem.
 P: 4 Cant we use a ZENER DIODE ???