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4ohms head


tkelly150
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2 8 Ohm cabinets connected in parallel (which is usually how the speaker outputs on amps are connected) equals 4 Ohm, so you should be good.

 

2 8 Ohm cabinets in series though equals 16 Ohm, but it is with almost 100% certainty that the speaker outputs of your amp will be connected in parallel.

 

Edited by Baloney Balderdash
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7 minutes ago, 0175westwood29 said:

100% dont mean to be harsh to the OP we all have to learn but that a simple google search or its actually in modern amps in the manual.

Yes, how about we just close Bass Chat and in its place just leaving an image saying "Use Google!".

 

 

Edited by Baloney Balderdash
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1 minute ago, Baloney Balderdash said:

Yes, how about we just close Bass Chat and in its place just leaving an image saying "Use Google!".

 

 

thats not waht i meant but a pinned thread for impedance was more what i meant so if people are here they can look at that thread and very quickly get the answer

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1 hour ago, fretmeister said:

There's a link to an impedance calculator in my signature.

 

Everybody should save it.

 

 

 

 

It is a great ready reckoner. Is there a possible enhancement with the minimum load rating for the amp taken into consideration? Does that make a lot of difference?

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Calculating parallel impedances is so simple as just about every one in the world has a calculator built into their phone. Formula is R1 times R2 divided by R1 plus R2. Using 8Ω and 4Ω: 4Ω times 8Ω=32, 8Ω plus 4Ω = 12. 32 divided by 12 = 2.67Ω. QED :)

 

Edited by BassmanPaul
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6 hours ago, asingardenof said:

It is a great ready reckoner. Is there a possible enhancement with the minimum load rating for the amp taken into consideration? Does that make a lot of difference?

 

 

It's not my site. I just put it in my Signature because I thought it would be useful to anyone with an amp!

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1 hour ago, BassmanPaul said:

Calculating parallel impedances is so simple as just about every one in the world has a calculator built into their phone. Formula is R1 times R2 divided by R1 plus R2. Using 8Ω and 4Ω: 4Ω times 8Ω=32, 8Ω plus 4Ω = 12. 32 divided by 12 = 2.67Ω. QED :)

 

 

Yeah... but that requires remembering the formula.

 

It's actually easier just to learn the common combinations by rote.

 

8&8 = 4 ohms

8&4 = 2.67 ohms

4&4 = 2 ohms.

 

There's hardly any 16 ohm bass cabs so no need to remember those.

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20 hours ago, fretmeister said:

 

Yeah... but that requires remembering the formula.

 

It's actually easier just to learn the common combinations by rote.

 

8&8 = 4 ohms

8&4 = 2.67 ohms

4&4 = 2 ohms.

 

There's hardly any 16 ohm bass cabs so no need to remember those.

The Barefaced 12 ohm option on the Two10S makes things interesting

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28 minutes ago, asingardenof said:

The Barefaced 12 ohm option on the Two10S makes things interesting

 

If someone is using a pair they can have 6 ohms and they really won't hear a difference in the real world. Impedance is not a static figure anyway - it varies in use.

 

Or if they have a 2 ohm capable amp they can run a pair of them in 4 ohm mode.

 

Even if I had a 2 ohm capable amp I'd run a pair of them at 6 ohms. Less of a chance for an amp problem and just turn it up a tiny tiny bit more.

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2 hours ago, fretmeister said:

Even if I had a 2 ohm capable amp I'd run a pair of them at 6 ohms. Less of a chance for an amp problem and just turn it up a tiny tiny bit more.

If an amplifier has been correctly designed to operate, as all the ones I own are, into a 2Ω load they will run just as well at 2Ω as they will at 8Ω.

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