I need an education here.

800 watt head, 2 ohm stable

2x12, 4 ohm 800 watt cab

1x15, 8 ohm 500 watt cab

Anyone see any problems here? I’m not clear on how the power gets divided up. Would the 500 watt 1x15 be at risk?

I should say, I’m talking about both cabs hooked up at the same time. The amp in question is 400 watts at 8 ohms. So, I should be good with either cab solo.

Thnx

As

@agedhorse said, you should be fine.

When the cabs have the same impedance, you simply divide the impedance by the number of cabs. So two 4 ohm cabs will total 4/2 = 2 ohms and two 8 ohm cabs will total 8/2 = 4 ohms.

When the impedance does not match, the math is a bit more complicated. If you connect two cabs with different impedance in parallel, use this formula where Z is impedance:

For example if you have an 8 ohm cabinet and a 4 ohm cabinet:

The amp should be fine with any load that is 2 ohms or higher. The amount of power the amp can make is related to the impedance presented by the speakers. So as

@gtirard mentioned, the power at 2.67 ohms will likely be less than 800W.

It's common for amp specs to include power ratings for 2 ohms, 4 ohms, and 8 ohms. Unfortunately this means you are often left to guess in regards to how much power an amp can make at other impedances. For example, 800W Mesa Subway amps are rated 400W at 8 ohms, 800W at 4 ohms, and 800W at 2 Ohms. AFAIK they make approximately 600W at 2.67 ohms

If you have a 4 ohm cab and an 8 ohm cab in parallel, the 4 ohm cab will receive twice as much power as the 8 ohm cab. The easy way to figure this is to divide available power by 3; then give the 4 ohm cab two shares and the 8 ohm cab one share.

For example if the available power is 600W: \

- 600/3 = 200W per share
- 4 ohm cab gets 2 x 200 = 400W
- 8 ohm cab gets 1 x 200 = 200W

Total power is 400 + 200 = 600W.