OK, I got myself a new amp , a Peavey head and a big ol' speaker to go with it. Now, if I take a look on the backside of the top there is two slots for speakers. One is of course connected to my speaker, and one is open. Now, if I want to get another speaker (since there is an open slot), will my volume increase ? if so, how much? depends on size? My head puts out 220W. Hope you understand my question, /lovebown

Need more info. What are the impedances of the cabs? Model numbers of the head and cab or any info? Exact wording of the labeling around the jacks (not slots ).... It'll say something like 220W at 4 ohms, or jacks in parallel, or something.... Chris

yeah, basicly, your amp probly puts out 220w @ 4ohms, because companys like to act like they have ton's of wattage, so they say the bigger number (the lower the impedince goes, the higher the output.) If your cab is 8ohms, then your head isnt putting out 220w, rather probly 140w(?). Now, when you add another 8ohm cab, it will put up the impedince to 4 ohms. (two 8ohm cabs add up to 4ohms, one 4 ohm cab, and one 8 ohm cab adds up to about 2.6ohms. ), it you go below the allowed impedince, you will adventuly blow out your poweramp. SO... If your cab is a 4ohm cab, you CANNOT add another cab, because it will damage your head IF it cannot run at 2 ohms (two 4ohm cabs = 2ohms, 4 8ohm cabs = 2ohms). But if your cab is 8ohms, then it can handle without a doubt another cab unless your head is only rated at 8ohms (i've never heard of such a thing). Also, just because you have another speaker output, doesnt mean you have to use it. YOu could use a daisy chain... Look up the topic, it should be back a page or two. My best recomendation is just read all the tech info you can find on this page. It helps a ton to know what you're doing. I learned everything i know off this page, and you've got some VERY smart guys on this board.

Err ok, at the back of the head there are two slots, jacks, whatever, and above them it says 210 W @ 2 ohms , however, what greatly troubles me is that it says 700 W / 50hz at one place ??? Either way, at the back of the cab I can't see any info on it.. all I know is its big fat and theres only one speaker. /lovebown

Now we're getting somewhere. Don't worry about the 700 Watts thing, that is how much power it sucks from the outlet when putting out pull power. The important rating is the 210W at 2 ohms. So if your cab is 4 ohms, then you can safely hook up another 4 ohm cab which would make your amp see its minimum 2 ohm load. You'll need to find out the impedance of your cab. Know anyone with a multimeter so you can measure resistance of the speaker? If you measure about 6, it's an 8 ohm cab, and if you measure about 3, it's a 4 ohm cab. It'd be nice to know, then we could give you an exact answer. I doubt that it is 2 ohms, it's probably 4, so you could add another 4 ohm cab and get the full 210 Watts out. Chris

Ok thanks, all I really care about is if i can get another cab without ruining the head. Im guessing the speaker i have now is @4ohms but... thx, /lovebown

...and with regard to how much volume that equals... it depends a lot (a whole helluva lot) on the cab itself. Look up threads on efficency to see how much. The simple answer is that you will have more speakers and more power, hence more volume.

Wait a minute...P=IV 700W=I x 120V 700W/120V=I I=5.833333333 Amps That's just under 6 amps of current going into the head! Sorry, I just took my Electronics GCSE, and most stuff I work with is under 0.5 Amps.

What's GCSE? Yeah, that power consumption rating is probably fudged high a little bit. For UL listing purposes, that number has to be higher than what is actually measured on an amp, but it can only be like 10% off or something. It's for use in determining power loads, etc. When I was designing treadmills, we would measure the actual power on a unit and round up to the next round number for a little headroom for production variances. Plus, since it's electronics with a capacitor input filter after the rectifier, the current waveform is non-linear and has nasty current surges on the peaks of the sine wave. Since P=I^2*R, those spikes explain the high power shown. At idle, that thing is probably drawing less than 0.5A. Chris

50hz, I'd assume the voltage is 220v or 240v, no? If so, then I=P/V: with a 220v = 3.18 Amps with a 240v = 2.916 Amps Still seems pretty small compared to the 8 Amp fuse my amp uses. Or does that formula even have anything to do with what size/rating fuse my amp requires?

Well, he's in Sweden, so I don't know how much voltage comes through the sockets over there. I assumed he was on 120V, but without confirmation I'm not sure. And 3A stills seems very high.... GCSE is General Certificate of Secondary Education, it's the last exam you take in high school before going to college(I live in the UK). I have an A* in it, and by god I'm not afraid to use it!

Yep, you can compare that to the fuse, the fuse should be about 125% of that, I'd guess. (Don't go changing your fuse based on that number I pulled out of my a$$. I really am just guessing.) Chris

Yep, probably is 220V. I didn't catch that. Rock on with your GCSE! I say, become an electrical engineer. I envision British college life as what I used to watch on "The Young Ones" Brit TV show from the 80's. If you've never seen the show, see it. Motorhead plays in their kitchen!! They have senseless violence and crude humor. What more could you ask for? Chris

Okay - one thing people should know about numbers is the concept of "significant figures". The 700 watts and 120 volts are rather rough figures. You'll notice no decimal places, and even then they are not precise, because they could be 5%-10% off. So, in this case, the precision is maybe 2 figures. This would mean that the "70_" part of 700 and the "12_" are meaningful. This means that the wattage draw could be anywhere from 650-749 watts, in theory. They use 700 as a round estimate. Likewise the voltage could be 115-124 volts. However, as someone else said, the ratings tend to be conservative to avoid safety problems, so may be reflective of the maximum. So what's my point? If your basis for a calculation is, at best, only 2 significant figures (SFs) of precision, then your result should also show the same level of precision. This means that the current, instead of 5.833333333 amps, should be expressed as 5.8 amps. If you instead had been given 699 watts (3 SFs) and 119 volts (3 SFs), then it would be correct to type 5.87 amps (but not 5.87394958...!). Same with impedance. 8 ohms in parallel with 4 ohms (to one SF) would be 3 ohms (to one SF) or, at best, 2.7 ohms - but NOT 2.666666667. This is because the 8 was not expressed as 8.000000000 nor the 4 as 4.000000000. There isn't enough precision. End of number theory lesson #821. - Mike (the numbers nerd)