hello, I am making a project that is a voltage divider, except one of the resistor will be replaced with a pot. It's a 5ohm resistor and 5ohm pot, my question is what power rating should the resistors be at? The vcc is a 9 volts battery so,
I know that when the pot is at zero resistance the second resistor will have 9 volts going through it so the current is 1.8 A . So the power equation tells us it needs to be 16.2 watts. But I'm not sure where to find a 16.2 watt resistor. Am I miss calculating? Since the voltage through the pot is 4.5, the current will be 0.9 and the power will be at 4.04 watts. So since both these power ratings are greeting than 1watt should I be looking for a 16.2 w resistor and a 5 watt pot? Or am i miss calculating?

 

Most pots are 0.5-2W. If you give us a better idea of what the divider is being used for, we may be able to offer other ideas.

 

More importantly what are you trying to do with this setup?

Reason being pulling 1.8 amps off a 9 volt battery will drain it in minutes.

 

Sorry I've been away from this forum for awhile, so what this voltage divider was for was to adjust the brightness of the LEDs in a light chaser circuit. However, last night I tried using a adjustable voltage regulator in my bread board circuit and everything seem a lot happier(Mainly no smoke or heating up). Here's a picture of the schamtic of the circuit with a LM317 where the voltage divider would have been.