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danssoslow
12-01-2007, 12:54 PM
I used a parallel notch filter calculator to determine the values for L,C,& R for a trap atennuating 12db between 12k and 14khz. This will be used with the Dayton RS52's I purchased for my Blazer.
The values given were:

C- 30.03uf
L- .75mh
R- 2.65ohm

I found the cap and inductor on PE's website, but the resistor is confusing me. I'm under the impression the resistor has to be rated for the wattage being used with te speaker, right? If this is so, I'm having a hard time finding a resistor of such little resistance @ 50 watts or better.

At this point, I figure I'm near over my head. I need sme help here! :(

BTW, how have you guys been? :)

Dan

95Honda
12-01-2007, 10:39 PM
You don't need to have a resistor that is equal to the wattage you are using, just the wattage it will be required to dissipate. If the resistor is very low in ohms, it won't have much voltage drop across it and won't be required to dissipate much, like only a small fraction of the total power. If the resistor itself was connected to the load, with nothing else, it would be required to dissipate all of the power it would pull from the amplifier. But in your case, the speaker is going to dissipate most of the power in the circuit, as with all properly designed filters. In fact, the only time the resistor will do anything is really between 12-14K like you set your parameters, and even then it may be a few watts at most. So, you will be totally fine using a 10 watt resistor. You could use that same resitor in a circuit like that with a 100-200 watt amp and still be fine....

danssoslow
12-01-2007, 11:17 PM
Makes absolute sense. You're the man!

Thanks,
Dan