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<blockquote data-quote="Savant" data-source="post: 155974" data-attributes="member: 546108">bringing both pairs to the amp &#039;seperately&#039; leaves it 4 ohm.. you have to connect each pair to eachother to change that.. the + from the amp to 2 4 ohm sets is still 4 ohm..&nbsp; you have to have one of each pair connected to the other to drop/raise the resistence.. that make sense?If you had 50 speakers, all individual.. and ran from the + on the amp to each speaker, and the - on the amp to each, and NONE of the speakers were wired to another speaker, the resistence would be that of the speaker (say 4 ohm) and the output of the amp (say 50 watts for argument) would get split to each speaker.. each speaker would get 1 watt.. that make sense? so in your setup, you have 2 - 4 ohm pieces that aren&#039;t related to each other, each gets 1/2 the output.. since each &#039;set&#039; is 2 speakers, each speaker get 1/4 the total output..//content.invisioncic.com/y282845/emoticons/crazy.gif.c13912c32de98515d3142759a824dae7.gif</blockquote>

[QUOTE="Savant, post: 155974, member: 546108"] bringing both pairs to the amp 'seperately' leaves it 4 ohm.. you have to connect each pair to eachother to change that.. the + from the amp to 2 4 ohm sets is still 4 ohm.. you have to have one of each pair connected to the other to drop/raise the resistence.. that make sense? If you had 50 speakers, all individual.. and ran from the + on the amp to each speaker, and the - on the amp to each, and NONE of the speakers were wired to another speaker, the resistence would be that of the speaker (say 4 ohm) and the output of the amp (say 50 watts for argument) would get split to each speaker.. each speaker would get 1 watt.. that make sense? so in your setup, you have 2 - 4 ohm pieces that aren't related to each other, each gets 1/2 the output.. since each 'set' is 2 speakers, each speaker get 1/4 the total output.. [IMG]//content.invisioncic.com/y282845/emoticons/crazy.gif.c13912c32de98515d3142759a824dae7.gif[/IMG] [/QUOTE]

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