the impedance formula. sorry.

[darthjujuu]

yea, i'm gonna be THAT guy, and bring this up again.

this is more for lulz than anything. this kid i know (DEFINITELY NOT ME) thinks it's a good idea to run his jcm900 into two 8ohm cabs ANDDD an ampeg 8x10. i told him his amp would explode. i know the math for multiple cabs of the same impedance, but i don't know how it adds up with 3 cabs of differing impedance. so if anyone would like to crunch the numbers and see just how quickly his JCM will burst into flames, i'd be much obliged.

 

[Wolfeman]

well running two 8 ohm cabs will put a 4 ohm load on the trans.

And I know for a fact that going any lower that 4 on a JCM900 transformer will eventually stress it too much.

So adding anything else to the mix will not work at all. Let him do it though, then he will be forced to buy a better amp

 

[Wolfeman]

Oh just to add, if that ampeg is 4ohms(pretty sure it is) the total load will be 2ohms on the amp.

I don't think it would last too long like that, it would either fry some windings in the transformer or destroy the power tubes first.

 

[darthjujuu]

just what i needed.

 

[tarnationsauce2]

An 8 ohm cab and a 4ohm cab in parallel are not quite 2 ohms.

The formula is (R1*R2)/(R1+R2)
So (4*8=32) / (4+8=12) = 2.667ohms

Either way he will be pushing the amp too hard. The transformer will eventually fail and the power tubes life will diminish quickly.

 

[egan.]

If you built the cables to run them series/parallel you could get to 8ohms.

[(8ohm marshall1 x 8ohm marshall2)/(8ohm marshall1 + 8ohm marshall2)] + 4omh ampeg = 8ohms.

Of course I don't really get why you would want to do that.

 

[Wolfeman]

An 8 ohm cab and a 4ohm cab in parallel are not quite 2 ohms.

The formula is (R1*R2)/(R1+R2)
So (4*8=32) / (4+8=12) = 2.667ohms

Either way he will be pushing the amp too hard. The transformer will eventually fail and the power tubes life will diminish quickly.

True, but he said an 8ohm an 8ohm and a (most likely)4ohm.... Assuming all are being run parallel, it's a 2 ohm load.

 

[ahjteam]

jcm900 into two 8ohm cabs ANDDD an ampeg 8x10

I checked from Ampeg's website that the ampeg 8x10 is 4 ohms in mono, so two 8 ohm speakers and one 4 ohm speaker results in 2 ohms load. Most likely first you will blow your fuse and ruin your valves.

http://www.sadowsky.com/audio/impedance.html

Depends on the amp too what happens, for example I have a Marshall Mode Four amp that has a SS poweramp, and it reads in its manual on page 8 that some sort of protection circuit activates if you feed it less than 8 ohm load

As I was finding out info on the subject, I found these cool photos:

So if he could change the impedance on the 8ohm cabinets to 16 ohms and 8ohms on the 8x10, he would get 8 ohms from 8+16+16. But that would still be most likely INSANELY loud.

 

[zackpennington]

So if he could change the impedance on the 8ohm cabinets to 16 ohms and 8ohms on the 8x10, he would get 8 ohms from 8+16+16. But that would still be most likely INSANELY loud.


He couldn't do that. He could only change the ohmage with multiples of the number of speakers.

 

[ScootBaloo]

So if he could change the impedance on the 8ohm cabinets to 16 ohms and 8ohms on the 8x10, he would get 8 ohms from 8+16+16. But that would still be most likely INSANELY loud.

I didnt understand one thing about that comment.

Note that loudness is soley dependent on the impedance regardless
of the speaker configuration.

Power = V*V / R

 

[tarnationsauce2]

True, but he said an 8ohm an 8ohm and a (most likely)4ohm.... Assuming all are being run parallel, it's a 2 ohm load.

Ahhh I overlooked the part where he said two 8-ohm cabs.

 

[John_C]

Running in parallel with a low impedance cab tends to give you a low total impedance as the power is far happier to travel across the 4 ohm load than the larger load.

e.g.
2 8ohm cabs in series, connected in parallel to a 4 ohm cab = 1 16ohm cab connected in parallel to a 4ohm cab = 1/((1/16)+(1/4)) ohms = 1/(5/16) ohms = 16/5 ohms = 3.2 ohms

When working in parallel, the lowest impedance cab is always dominant in determining the overall impedance, even a 64ohm and a 4ohm resistor in parallel still only gives you around 3.8ohms total resistance (apologies for swapping around impedance and resistance)