I have a little micro head (Overton Flyweight 200) that puts out 200W @ 4 ohms. I'm running 2 Ampeg SVT210AVs with it, which are 8 ohm cabs. It's a great amp and it's actually loud enough to gig with my 5 piece blues/rock band, if I set it at about 1/2 gain and 1/2 - 3/4 master vol. But just for curiosity, I'm wondering what my final wattage to each speaker is.. My thinking is.. when I run both cabs, my 4 ohm amp then "becomes" 8 ohm and since the cabs are both 8 ohm each, everything is matched perfectly. I know that when you double the ohms, the wattage drops by 1/2, so now the amp is sending 100W@8 ohms. But.. is that to each speaker? Or is each speaker now getting that 100W divided by 2 (50W each)? Sorry if that's it's a dumb question. I'm electrically challenged. I searched and nothing really answered my question.

You're partly right, but the amp doesn't 'become' anything. The numbers simply tell you the maximum current the amp can flow into a given load. Lower Ohms=more current is possible. So yes, your system is well matched because the two 8 Ohm cabs present a 4 Ohm load to the amp. This allows it to produce the maximum possible power (200 watts if it's asked for). Each cab is receiving half of this amp power, which would be a maximum of 100 watts each.

Your two 8 ohm cabs together make a 4 ohm load. Your amp will handle 4 ohms and will provide 200 watts which will be equally divided between the 2 cabs. Each cab gets 100 watts.

Note that the 200w is only a nominal figure for the clean wattage available. Different amps provide extra distorted wattage of varying quality, quantity and sustain. The rating on cabs is thermal. The clean power handling of cabs is usually about half, then they get bit of hair on, until they are farting out at rated power levels. You could double your amp rated power but it wouldn't make nearly the effect you'd think. This is part of how everyone gets all gaga over how loud a 200w micro amp gets. The other part is extra loudness takes exponential increases of power to make. To get twice as loud takes 10x the power, and the speaker to handle it cleanly. This maths works comparing 2w 20w 200w, failing at 500w for just about any cab on the planet. Going by what JimmyM says about your cabs, a 400w amp can give them a beating.

Whipped up two quick drawings. They don't show how power is divided.. but at least shows you how wiring up speakers in a cab, or different cabs effects impedance/resistance. PARALLEL SERIES

Hopefully you will see that in parallel, they share and reduce the load by giving 2 paths to travel.. In series, you need to force your way through a full 8 ohm load, and then another in series.. And those loads add together. 2 and 4 speaker cabinets often do the same thing. 2 4 ohm speakers in series OR two 16Ohm speakers in parallel to give you an 8 ohm 2 speaker cabinet. 4 8 ohm speakers in a 4x10 cab will be wired as a two pairs of series then wired in parallel... to get you back to 8 ohms. OR wire 4 16 ohm speakers in parallel.... 16/4 = 4 Ohm.

Hope you've got it. Series loads are more intuitive- just add them up. A little simple math is needed to get a handle on parallel loads- they only divide cleanly when they're even (as in your case).

Thanks for the drawings Scottkarch! I thought I fully understood it until I read the paragraph at the bottom of your series drawing.. shouldn't that last part read: "In series circuits you multiply the ohms by the number of speakers/resistors, in this case 8 ohm x 2 speakers = 16 ohms" if it in series? I understand your last post. So if that was a typo above, then I do get it, lol!

Actually I think "multiply" is wrong.. you would add the ohm rating of each speaker together (if wired in series) to get you final ohm load, right?

Whoops. I'll fix the series text at the bottom tomorrow. Should show 8 + 8 = 16 ohm. Should have double checked. In series just add things up. In parallel divide... And identical impedance is easier to do.

Think of the impedance as the resistance to the power of the amp. If you have 2 places for the power to go at the same time, the resistance is cut in half. As long as the source can keep up with the demand, the flow will double. However, just as a note, the sound will not be louder because the power doubles. It will be louder because you have double the drivers. As downunder says - it takes 10x the Watts to create 2 x the SPL. Really. Electrical power is not the same as acoustic power.

Ok, I'm glad that was a typo Scott. I was like, "yeah i got it... wait a minute, huh?" Thanks everyone!

Of course, the OP should not be misled into thinking that daisy-chaining the cabs will put them in series. This takes a special cable or circuit box to put two cabs in series. They exist, but are scarce as hen's teeth.

There's some misinformation here. http://physics.bu.edu/py106/notes/Circuits.html For parallel circuits: 1/R (total) = 1/R (1st Cab) +1/R (2nd cab) More here: http://www.sengpielaudio.com/calculator-paralresist.htm

And you didn't ask for it but here it is, the rig you are running will sound louder than if you had a 4 ohm 210 cab instead, because all other things being equal, more speakers gives you more volume.

Yeah. The series option is more Likely what you find inside a multi speaker cabinet. To hook 2 Cabs in series you'd probably need a special long 3 plug cable to connect the amp to both cabs At the same time. It's uncommon. I've never seen it but it should work. The site linked to above is very good. I was just trying to make it visual rather than showing formulas. A lot of people can't link a formula to their amp/cabs easily.