my current rig consists of a behringer 410 cab at 4 ohms and a behringer 210 cab at 8 ohms. i recently purchased a gk 1001 mark ll head. now if i understand correctly, 8 ohms plus 4 ohms produces 2.6 ohms. can the gk head handle that kind of load or would i be better off eliminating the 210 cab altogther?

The amp shows the correct loads on the rear and in the manual. No, it won't work with this load if you connect them both to the Main output. If you were bi-amping, you could do this but you would still need to make sure the 8 Ohm cabinet is connected to the High Pass output.

It can only handle a 4 ohm load, so you would be eliminating the 2X10. It's not really a question of better off . It's having a working amp or a blown amp.

I checked the amp's manual online: http://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&ved=0CDMQFjAA&url=http%3A%2F%2Fwww.gallien-krueger.com%2Fmanuals%2F1001RB-II_700RB-II.pdf&ei=3WYFUdPLC4nm0QHR5oCABA&usg=AFQjCNF_5xKAET4xsXhSAVOPJzUZI1XuoQ&bvm=bv.41524429,d.dmQ and it seems that there is a 4 ohm minimum load, so no, you can't run a 2.67 ohm load. I'd go with just the 410.

You can't run them both. The min ohm load for the amp is 4. You have less than that so you'll make bad things happen. Lets say that your have an amp that can run down to 2 ohms. You could run both cabs BUT the cabs will sound very different. The ohm mismatch causes this, there's nothing you can do about it and it will drive you crazy. So I don't mismatch ohm ratings even when I used a Crown that was good to 1ohm.

right on, i appreciate all your help guys. i was advised about that potential problem but wanted to seek some extra advice/tips as well. im looking to upgrade to a gk 412 sometime. thank yall again.

The 1001 can drive a minimum speaker load of 4 ohms, so your 4x10 alone is the only possibility. The 1001 is rated at 700W into a 4 ohm load. How much power can your 4x10 handle?

Wrong............. *IF* the posters amp could handle a 2 ohm load, his setup would be the correct way to run both cabinets to ensure proper power distribution. Each individual speaker would be getting the same amount of power.

The actual calculation that you may want to store someplace is: R1 x R2 / R1 + R2 R1 times R2 divided by R1 plus R2 Where R1 is the impedance (ohms) of the first cabinet Where R2 is the impedance (ohms) of the second cabinet So if you have two 8 ohm cabinets, the formula looks like: 8 X 8 / 8 + 8 = 64 / 16 = 4 So two 8 ohm cabinets = 4 ohms load If you have two 4 ohm cabinets, the formula looks like: 4 X 4 / 4 + 4 = 16 / 8 = 2 So two 4 ohm cabinets = 2 ohms load If you mix an 8 ohm cabinet and a 4 ohm cabinet, the formula looks like: 8 X 4 / 8 + 4 = 32 / 12 = 2.67 So an 8 ohm cabinet and a 4 ohm cabinet = 2.67 ohms. Hope this helps to better understand how the total load numbers are calculated.

The formula gets mangled like that for two cabs. The actual formula is the combined resistance equals the reciprocal of the sum of the reciprocals. 1/R = 1/R[SUB]1[/SUB] + 1/R[SUB]2[/SUB] + etc

ugh, the math always makes me cross-eyed! an easy way to ballpark it in your head with two cabs is to think of it as "half the average"; the average between 2 8Ω cabs is 8, so half that is 4. the average, or "halfway point", between an 8Ω cab and a 4Ω cab is like 6Ω. half that is 3Ω. (the real answer is 2.6Ω, which is pretty close).

Either reciprocal or product over sum works fine for only two loads. So there is no "actual formula" for that scenario being that the TWO formulas are interchangeable. But I think the OP got the point several posts ago anyway.

Huh? That is the formula, some butcher it with the convenience that two cabs work out to the same answer even though the maths carries no logic with it.

I am in school for electronics. We don't study speaker cabs on a daily basis. However, we DO STUDY resistance every single day. For calculating combined resistance (actually impedance but that's neither here nor there) in parallel circuits BOTH METHODS are STILL currently taught with regard to TWO LOADS. They are called reciprocal and PRODUCT OVER SUM. The latter can ONLY be used for TWO LOADS. So, if you want to call the ENTIRE HIGHER EDUCATION SYSTEM BUTCHERS, please go right ahead. That will most certainly make you right, and them wrong...... you logical and them silly. Oy. I can't believe I let you drag me into a debate even though the OP got the point many posts ago.

He already showed a practical number for the impedance, but didn't carry it to the third significant digit. At this point, the .07 Ohms is strictly academic. The first digit is the most important, followed by the second. The last is really insignificant.