I think my math is right here but just want to make sure.

I have a Markbass 3x10 at 6 ohms and I’m thinking of getting the matching 2x10 at 4 ohms for smaller gigs.

If I ever decide to run them together the setting on my amp would need to be on 2 ohms, right?

6 ohms x 4 ohms 24

————————— —

6 ohms + 4 ohms 10 = 2.4 ohms required

Fortunately my Carvin BX700 has a switch between 2 and 4 ohms so it can be done. I doubt I’ll do it often but before I get the 2x10 I just wanted to confirm it can be done.

Thanks!

If the amp is rated 2 ohms minimum, you need to make sure the load (2.4 ohms) is > 2 Ohms (It is

).

The other thing that would be useful is to figure out is how the two dissimilar cabs share power.

Arrange the impedance values into a fraction; assuming 4 ohm and 6 ohm cabs.

4/6

Simplify the fraction

2/3

Add the numerator and denominator

2 + 3 = 5. This is the number of shares the power is broken into.

Divide the available power by the number of shares. Let's assume the amp makes ~583W at 2.4 ohms.

583/5 = 116.6W per share

*This estimated is based on the assumption the amp makes 700W at 2 ohms and produces the same voltages into a 2.4 ohm load. I'll post the Ohm's/Watt's Law formula wheel and estimate at the end.*
The available power is distributed inversely to the impedances. So the 4 ohm cab gets 3 shares and the 6 ohm cab gets 2 shares.

4 ohm cab power = 3 x 116.6 =

**349.8W** out of 583W total

6 ohm cab power = 2 x 140 =

**233.2W** out of 583W total

Check results

Total Power = 349.8 + 233.2 = 583W

Make sure each cab can safely handle the expected power.

Formula Wheel

For the power estimate, I assumed the amp is rated 700W at 2 ohms and it produces the same voltage at 2.4 ohms

From the formula wheel:

V = sq rt of PZ

Let's rearrange this to

V^2 = PZ

Insert known values (700W and 2 ohms)

V^2 = 700 x 2 = 1400

The next formula we will use is

P = V^2/Z

We just calculated V^2 = 1400. Solve for P with Z = 2.4 ohms

P = 1400/2.4 = ~583.333333...(W)