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two ohm load. I have my subs wired to 2.67 load and im wonderin how to hook this wiring to the amp. Does it automatically adjust its self or is there a way to bridge it. The channels go ++-- without a diagram showing which way is to bridge.

2007-03-20 16:11:25 · 4 answers · asked by Anonymous in Cars & Transportation Car Audio

4 answers

i'm assuming that you have 3 subs, each 8 ohm svc or 4ohm dvc (each coil in series on each sub) in parallel. that's the only way i could calculate from 2.67 ohms. i think you set it up the lowest you can safely go, so just go from one (+) to one (-). inside the amp, both (+) and (-) terminals are hard wired together so it doesn't matter which of the two you connect your two wires to (that being a mono amp). the pair of each terminals are offered only for flexibility and ease of installing multiple speakers because amps of that rating usually benefit from larger speaker wire than normal.

2007-03-21 12:00:24 · answer #1 · answered by dali_lama_2k 3 · 0 0

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I have a Kenwood KAC-9102d and I want to know how to wire this in a?
two ohm load. I have my subs wired to 2.67 load and im wonderin how to hook this wiring to the amp. Does it automatically adjust its self or is there a way to bridge it. The channels go ++-- without a diagram showing which way is to bridge.

2015-08-10 05:25:51 · answer #2 · answered by Jervis 1 · 0 0

It's a mono amp; there is no "bridging" it. When you bridge an amplifier you're combining two channels into one channel; that amp has only one channel to start with.

As far as wiring it to a 2-ohm load: the speakers are the load, and the only way to change the load at the amplifier is to change the way the speakers are wired. Impedance (at least the type you're talking about) is entirely determined by the speakers, not the amp. We'd have to know what kind and how many speakers you have in order to tell you how to wire it for 2 ohms.

2007-03-20 16:20:10 · answer #3 · answered by KaeZoo 7 · 0 0

Kenwood 9102d

2016-12-10 14:04:58 · answer #4 · answered by ? 4 · 0 0

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