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 Resistor wattage?
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 23rd March 2010, 01:51 AM #1 diyAudio Member   Join Date: Dec 2008 Resistor wattage? My project calls for a 24k 1w resistor, if I use 2 12k resistors, do they need to be 1/2 w each to equal 1w? Thanks
 23rd March 2010, 01:56 AM #2 Banned   Join Date: Jan 2008 Blog Entries: 2 Yes w
 23rd March 2010, 04:49 AM #3 diyAudio Member   Join Date: Aug 2008 Location: Broomfield, CO losacco, While two 12k, 1/2 watt resistors in series will give you a resistance of 24k they will only be able to carry 1/2 watt of current. You either need the proper resistor or two 48k, 1/2 watt resistors wired in parallel. That way each resistor carries 1/2 the current & you end up with the equivalent of a 1 watt resistor. Do a google search for "resistors in series" and read a few of the articles. Understanding series and parallel circuits is essential for our hobby.
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Join Date: Oct 2003
Location: Sydney
Quote:
 Originally Posted by ronaldw441 losacco, While two 12k, 1/2 watt resistors in series will give you a resistance of 24k they will only be able to carry 1/2 watt of current. You either need the proper resistor or two 48k, 1/2 watt resistors wired in parallel. That way each resistor carries 1/2 the current & you end up with the equivalent of a 1 watt resistor. Do a google search for "resistors in series" and read a few of the articles. Understanding series and parallel circuits is essential for our hobby.

Last time I checked, current was measured in Amps, not Watts. Watts is a measure of power.

Yes, two 48K resistors in parallel will do the job, as you describe, each resistor carries half the current.

The original question, can two 12k 1/2 watt resistors in series replace a 24K 1 Watt resistor. The answer is YES. Each of the resistors will pass the full current, but as they have half the resistance, they will each drop half as many volts as the original, and therefore each will dissipate half the power.

So, in short, either two 12K 1/2 W in series, or two 48K 1/2 W in parallel will work.

Regards,

Chris

 23rd March 2010, 07:33 AM #5 diyAudio Member   Join Date: Aug 2008 Location: Broomfield, CO Chris, Good point. Thank you, Ron

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