1.34 ohm load

[cleansoundzz 10+ year member]

I have a Crossfire VR 1000D rated at 1332 watts @ 1 ohm @ 14.4 volts. Does anyone know how many watts I would get at 1.34 ohms? The amp itself has a 120a fuse on it. It would be used for 3 10's.

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[kovemaster559 5,000+ posts]

No

 

[528hz 10+ year member]

Slightly less than what it is rated at @ 1 ohm, but you would have to clamp it to find out definitive numbers.

 

[kmanian 5,000+ posts]

1000 +/-

 

[Louisiana_CRX 5,000+ posts]

1000 +/-

x2 plus the amp should run a little cooler

 

[Rashaddd 5,000+ posts]

1332/1.33 gives you your answer

right about 1k, probably a little bit more as amps are more efficient at higher loads, you might see 1050-1100

 

[bumpin buick 5,000+ posts]

1332/1.33 gives you your answer

Really is that how you figure that out ?//content.invisioncic.com/y282845/emoticons/wow.gif.23d729408e9177caa2a0ed6a2ba6588e.gif

 

[grimreper912003 5,000+ posts]

1332/1.33 gives you your answer
right about 1k, probably a little bit more as amps are more efficient at higher loads, you might see 1050-1100

does that always work?

 

[cleansoundzz 10+ year member]

The guy at Crossfire told me 1100 watts. I get 1001when I divide 1332/1.33.

 

[CrossFired 10+ year member]

The guy at Crossfire told me 1100 watts. I get 1001when I divide 1332/1.33.

I get 1063 watts @14 volts

 

[squeak9798 5,000+ posts]

does that always work?

No. It "just so happens" to work out (theoretically) when the power you are dividing by is a 1ohm power figure. Just plugging the desired impedance into the denominator will be inaccurate for most any other impedance.

You could apply basic ohms law. It won't be perfect (well, it would be in a perfect world //content.invisioncic.com/y282845/emoticons/biggrin.gif.d71a5d36fcbab170f2364c9f2e3946cb.gif ), but it'll give you decent approximation of what to expect.

To show how the "formula" you quoted would be wrong if it were at any other impedance, let's say the amplifier did 1332w @ .5ohm and we were wondering what it would do @ 1.34ohm.

If we were just to divide the given power by the impedance we were wanting the predicted power figure for, it would give us;

1332/1.34 = 994w

Using ohms law;

sqrt(1332*.5) = 25.81V

25.81^2/1.34 = 497w

If you wanted to use the actual formula Rashaddd posted, it would be;

Given power/(desired impedance/impedance of given power)

For example;

1332/(1.34/.5) = 497w

You can see why this works eloquently when we are given power @ 1ohm, because any number divided by 1 is itself....so it cuts out the actual math of the denominator. But I don't like that formula because it doesn't show you the mechanics of ohms law that are at work.....

This later formula is probably the one Rashaddd used, but without showing you how the denominator was actually arrived at.

 

[cleansoundzz 10+ year member]

So the accurate number would be 994 watts. Give or take a few watts.

 

[helotaxi 5,000+ posts]

But the amp will never do either of those numbers in the real world so why worry about it.

The simple way to calculate is to divide the rated power by the rated impedance and multiply that by the desired impedance. The square of the voltage is a constant. The problem with that is that most every amp is capable of holding a higher rail voltage at a higher impedance...