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Old 9th January 2004, 01:31 PM   #1
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Default Impedence question

How are most loudspeakers wired? I believe two 8 ohm speakers wired in parallel equals 4 ohms, and two 8 ohm speakers wired in series equals 16 ohms. Where I get confused is, what happens when you start wiring 3 or 4 speakers together. How do you keep the impedence in check. I have a Denon 3802 (I think) and don't know what it's rated for.
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Old 9th January 2004, 02:59 PM   #2
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The total resistance of any number of series resistances are just the sum of individual resistances.
The total resistance of equal valued resistances is the resistance value divided by the number of parallel resistances (ie. 4 - 8ohm resistors in parallel = 8/2 = 2ohm)
the total resistance of a number of non-equal resistances in parallel can be calculated using the equation below.
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Old 9th January 2004, 03:05 PM   #3
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Can most amplifiers handle raitings down to 2 ohms?

Why would you ever wire them in series?
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Old 9th January 2004, 03:08 PM   #4
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I have a Onkyo TX-SR600E 6.1 amplifier, but i want two speakers in the rear channel. the speakers i want on the rear channel is 6 ohm, and the resistanse should be between 4-16 ohm...

Is it possible to do this?
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Old 9th January 2004, 03:32 PM   #5
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2 ohm for an amp is not a good idea and will probably end up with heaps of smoke. If you want 4 drivers, wire them in series/parallel which gives you the impedance of the single driver. Have a look at this primer for details.
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