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 17th January 2008, 09:30 PM #1 Carlp   diyAudio Member   Join Date: May 2006 Location: A New England Plate load resistor question OK, maybe this isn't the right forum for this question, but here goes: I need an odd resistance for a plate load at 10Watts and need to pair resistors to make the right resistance. Can I pair two sustantially different resistances at 5 W and survive, or does the smaller resistor need to be higher rating or something? Thanks, Carl
 17th January 2008, 09:46 PM #2 nhuwar   diyAudio Member     Join Date: Jan 2007 Location: Cypress Texas Do you mean in parallel? The resistor thats smaller will pass more current. You can figure this out with ohms law. You can find the paralleled value of them with this formula r1 x r2 / r1 + r2 =total resistance. If you know the wattage being dissipated you know the voltage across them so use ohms law to find the current through each. Hope this helps. Nick
 17th January 2008, 09:46 PM #3 dsavitsk   diyAudio Member     Join Date: Jan 2005 Location: Hartford The heat dissipated by a resistor will be the voltage it drops multiplied by the current going through it (W = VI). You should double this for a margin of safety. So, if you have a pair of resistors in parallel, you need to figure out how much current each is passing and adjust its wattage rating accordingly. __________________ http://www.ecpaudio.com :: http://diy.ecpaudio.com
 18th January 2008, 02:22 AM #4 Carlp   diyAudio Member   Join Date: May 2006 Location: A New England Thanks for the quick answers. Hmmm, I'd forgotten how Ohm's law handles parallel resistors and it's trickier to figure than I remembered. I'm familiar with the fact that if you double the necessary resistance, the total resistance paralleled is the necessary resistance (e.g. 6.25K ohm 10W resistor can be approximated using two 5W 12.5K resistors paralleled - 12.5 x 12.5/25 = 6.25, and half of current passes through each, so 5W should do the trick). I assume if they are 1% resistors, the current flowing through each won't be much different so dissipation is close enough if the 10W has a margin of safety (which it does). Figuring current and power dissipation is where I get confused. If the above resistance drops 180V or so (I assume that's across both since they are paralleled), each passes 14.4mA (2.6W, correct?). But if i use 9k and 20k to approximate 6.25k, current across the 20k is 9mA (for 1.62W dissipation, right?), but for the 9k it's 20mA (3.6W?). Have I got this right? OK, in the end I remembered that I could simply use a 3k and 3.25k in series to get the same 6.25 resistance. In this case, though, they'd still need to be the full 10W or so, correct? Thanks again for helping someone who used to know this stuff but hasn't used math in a LONG time! Carl
 18th January 2008, 02:48 AM #5 jnb   diyAudio Member   Join Date: Dec 2006 For the series style connection, each resistor will pass the same current. This will be the total voltge divided by the total resistance. The voltage found across each resistor will be proportionate to their resistance. The larger resistor will drop the larger porton of the total voltage. Doing the math, the larger resistance will dissipate the most power.
 19th January 2008, 05:14 PM #6 Carlp   diyAudio Member   Join Date: May 2006 Location: A New England Hmmm. Thanks, jnb. It does indeed make sense. I need to start thinking like Ohm did.
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Quote:
 Originally posted by Carlp Hmmm. Thanks, jnb. It does indeed make sense. I need to start thinking like Ohm did.
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 20th January 2008, 06:06 AM #8 jnb   diyAudio Member   Join Date: Dec 2006 And Larry?... Curly??
 20th January 2008, 07:18 AM #9 ray_moth   diyAudio Moderator Emeritus     Join Date: Jan 2004 Location: Jakarta Ah, you're thinking of Stooge's Law.
 20th January 2008, 09:18 PM #10 Carlp   diyAudio Member   Join Date: May 2006 Location: A New England Oh, I think like a stooge all the time

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