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how does the ohms thing work??

yeah I recently bought an amp and it has a maximum output of 300w @ 4 ohms, the instructions say I can handle a 4 ohm cab or two 8 ohm cabs or one 2 ohm cab and two 16 ohm cabs, I have a 2x10 4 ohm cab, it is only 4 ohms I cant switch it to 8 (I don't know how this works so please excuse me if this sounds stupid), well my question is, can I add another cab even if my actual cab is 4 ohms? I’m planning on getting a 1x15 and if I can, how many ohms should this one be? I hope some of you more experienced guys reading this can throw me a line, after you're done laughing that is hehehe, thanks, please reply.
 
The Ohm is a measurement of resistance. This is the resistance of your speaker. If you hook up your cabinets in parallel electrically, the ohmage is halved. So the two 8 ohm cabinets would now equal 4 ohms. If you have two 16 ohm cabinets and a two ohm cabinet it still equals four. So, if you want to buy the 1x15 you can, but you can't use the other 4 ohm cabinet in conjunction with the 15. If you want a 15 and 2x10s you have to buy two 8 ohm cabinets.
 
Don't feel to bad. This is a problem I have always had and can only partially answer. If you use two 8 ohm speakers
with your amp you will get the full 300w at 4ohms.
Unfortunately two 4ohm speakers will need 2ohms of amplification. And I truly cannot tell you what a 4ohm and 8 ohm speaker will need to draw to be efficient. The bottom line is that your system needs to match up to be efficient. I am finally getting my rig together with an Acoustic image Focus2R
which has 600 watts into 4ohms and 300 into 8ohms. I match them with two different cabinets and EU 1x10x5x3 with 8ohms and 250watts and an eden 1x10 at 8ohms and 250 watts. With the two cabinets I draw 4ohms and have the 600watts at my fingertips. This system is really efficient so I can get the most out of my speakers. I have experienced "headroom" for the first time in years since the SVT days I left way behind me. Te only problem is you could overdrive the speakers and tear a cone.