POWER = I2R or V2/R ? - AVS Forum
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post #1 of 9 Old 11-16-2004, 06:15 AM - Thread Starter
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We all know any decent power amp will provide more power to a 4 ohm speaker load than a 8 ohm speaker load.

So, I would assume V(squared)/R is correct? But why are there 2 formulas for Power, where in the 1st one (I2R) , increasing the Resistance INCREASES Power and in the 2nd (V2/R), increasing Resistance DECREASES Power?
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post #2 of 9 Old 11-16-2004, 06:25 AM
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Some output-transformerless amplifiers actually produce less power into lower impedances. Futterman OTL-1's are an example. Their rated power is into 16 ohms I believe and pretty much halves as you drop down.

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post #3 of 9 Old 11-16-2004, 06:33 AM
 
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"where in one, increasing the Resistance lowers Power and in the other, decreasing Resistance increases Power?
"

Both equations produce the same result, one uses resistance and voltage, the other, resistance and current. They're easy to prove.
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post #4 of 9 Old 11-16-2004, 07:52 AM
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Quote:
Originally posted by Peter_Klim
We all know any decent power amp will provide more power to a 4 ohm speaker load than a 8 ohm speaker load.

So, I would assume V(squared)/R is correct? But why are there 2 formulas for Power, where in one, increasing the Resistance lowers Power and in the other, decreasing Resistance increases Power?
If you have constant power, voltage will increase as resistance increases, and current will decrease. As resistance is lowered, voltage will lower and current will increase. You can use these relationships to show that power will really remain the same as resistance is varied if your power supply can maintain a constant output.

Both are derived from combining Ohm's Law (V=I*R) and Watt's Law (P=V*I).
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post #5 of 9 Old 11-16-2004, 08:09 AM - Thread Starter
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Quote:
Originally posted by TonyBDA
"where in one, increasing the Resistance lowers Power and in the other, decreasing Resistance increases Power?
"

Both equations produce the same result, one uses resistance and voltage, the other, resistance and current. They're easy to prove.

Sorry guys, I updated my quote. What I meant to say is that in one equation, an increase in Power comes with an increase in Resistance, but in the other an increase in Power comes with a decrease in Resistance.

So which one is applied to audio amps? Or are they just not applicable to audio amps?

And yes, I understand that even though good amps do supply more power with a decrease in R, there is a point that all amps will start producing less power when the R is too low....so...maybe I just answered my question - neither applies to audio?
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post #6 of 9 Old 11-16-2004, 08:14 AM
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The equations are two ways of stating the same thing depending on which terms are known. They are identical, so they both apply to everything.

Changing resistance will have an effect on voltage and current depending on what type of source you have, so you can't look at the equation as changing resistance and nothing else.
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post #7 of 9 Old 11-16-2004, 08:45 AM
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Solid state amps in general act like voltage sources, not current sources. Power goes up as resistance decreases until the amp hits the limit of current it can supply.
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post #8 of 9 Old 11-16-2004, 09:45 AM - Thread Starter
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Thanks guys, now I understand.

rynberg,

"Solid state amps in general act like voltage sources, not current sources" - this statement to me put all the pieces together, thanks.

Just one last question - then why are they called "amplifiers" instead of "Voltafliers"? ;)
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post #9 of 9 Old 11-16-2004, 09:58 AM
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I think it's because the english word amplify and the idea of a device as an amplifier (of anything) existed long before stereo amplifiers were invented.

http://www.etymonline.com/index.php?...earchmode=none
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