potentiometers

Thread Starter

jhsmith87

Joined Oct 4, 2012
4
i am looking to get a 1k ohm resistor, although my local store only the lowest one is a 5k ohm. and this is going on a circuit that is running on 12v 3a dc power supply. it will be going through a coil first so that may reduce some power, but after the coil i need the potentiometer even if i get the 5k ohm one that they have the specs on it say wattage ratting .5w tolerance 20%.

so wouldnt that fry the potentiometer? or is the coil that is wound 500 or more times of 26g wire be enough resistance to keep it from frying?
 

Kermit2

Joined Feb 5, 2010
4,159
12/1k = .0012 amps

12 x .0012 = .0144 watts

Seems you would be ok, UNLESS you try to reduce the resistance of the pot down below 100 ohms. At that point you would be in the 'let the smoke out' zone.

Amps = volts div. by resistance.
Watts = volts times amps

I leave it to you to figure out how low you can take the resistance before things start to glow red and smoke. :)

A schematic of HOW you plan on connecting power and ground and how you plan to wire in the pot. would be extremely helpful to those who would help you.

(using the pot to control the base of a power transistor connected in shunt mode would be more reasonable - depending again on how you plan to interconnect things.)
 

ScottWang

Joined Aug 23, 2012
7,509
If your VR 5K only used 2 pins, then you can do like this:
1K + 250Ω = 1.25K
1.25K parallel VR5K : 1 .25K // 5K = 1K.
So you got one VR5K.
 

Ron H

Joined Apr 14, 2005
7,063
12/1k = .0012 amps

12 x .0012 = .0144 watts

Seems you would be ok, UNLESS you try to reduce the resistance of the pot down below 100 ohms. At that point you would be in the 'let the smoke out' zone.

Amps = volts div. by resistance.
Watts = volts times amps

I leave it to you to figure out how low you can take the resistance before things start to glow red and smoke. :)

A schematic of HOW you plan on connecting power and ground and how you plan to wire in the pot. would be extremely helpful to those who would help you.

(using the pot to control the base of a power transistor connected in shunt mode would be more reasonable - depending again on how you plan to interconnect things.)
You slipped a decimal point. 12V/1k=12mA.

Power=V^2/R=12*12/1k=144mW.
 
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