Can I safely play a 16 ohm and a 4 ohm cab at once? My amp math says (16+4)/4 equals 5 ohms, so set the transformer switch at 4 ohms. The amps involved have two speaker jacks, can I assume they split the load evenly? Any reason to not do this? Kiwi

George My math says 3.2 ohms... for the special case of two speakers the math collapses to R1 x R2 ------------ = 64/20 = 3.2 ohms. R1 + R2 Still the 4 ohm tap, but closer to No-Go-sville, IMO. and I seem to remember that the 4 ohm speaker will draw the lion's share of the current. I don't remember that math, so I'll stop here. Slick51

No reason not to do it. The total impedance is 3.2 ohms, yes. That's completely safe from a 4-ohm output on any tube amp - the safe range is about half to double, ie 2 ohms being the lower limit... 3.2 is well above that (it's the proportion that matters, not the absolute value). The power is divided in reverse proportion to the impedances, if the cabs are in parallel (with almost all amps they are) - so in this case 4/5 of the power goes to the 4-ohm cab and 1/5 to the 16. This is quite a big difference, but it probably won't be as noticeable as you'd think... depending on what, and how many, speakers are in each cab. (You can actually use this unequal power distribution for attenuation, if you have the right combination of loads - a 16-ohm cab and a 4-ohm dummy load in parallel works well, and reduces the volume quite a lot - about 7dB.)

Thanks, John...I've alwas operated on the 'never under 4 ohms' wife's tale; it's nice to hear the truth. I knew the math, but not the reasoning. Ditto on the current draw...it's a straight ratio as well. That's why I hang around here...someone will always come by with the answer! Slick51