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Old 26th March 2008, 05:23 PM   #1
Clipped is offline Clipped  Thailand
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Default accuracy of ohms law

how do you know which one to use when calculating power in watts?

P=I*E
P=E2/R
P=I2*R

i just converted a 225 hcca to 34.5 volt rails....so if i followed the second way of figuring power thatd be:

(imagine full rail voltage at outputs)

P=34.5*34.5/4ohms = 297 watts per channel

if it was in mono it would be:

P = 69 * 69 /4 ohms = 1190 watts mono

what am i missing? i dont think im getting 1190 watts from this amp.

im guessing the only truly accurate way is the first way?
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Old 26th March 2008, 05:34 PM   #2
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I've seen a lot of strange and outrageous claims before on this forum, but doubting Ohm's law is a new one

Those equations you posted apply to RMS voltages, not peak, so you need to figure out what is the maximum capable peak-to-peak voltage swing across the speaker, the use Vrms = Vpp/(2*sqrt(2)), then use ohm's law.
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Old 26th March 2008, 05:47 PM   #3
Clipped is offline Clipped  Thailand
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not doubting ohms law...i am a law abiding citizen.

just figured i was missing something, bad subject title

everytime i look for the way to calculate power i find like 2-3 different ways and dont know which one to trust...

your formula is too much for this mathematic defficient brain to hande...care to simplfy what each variable stands for?


....i thought output voltage cannot exceed rail voltage???
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Old 26th March 2008, 06:10 PM   #4
mag is offline mag  Switzerland
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Hi Clipped,
if your rail voltages are +/- 34.5V and the amplifier has no voltage drop under load you will have a sinus on the load of 34.5Vpeak.

1st method: P=U^2/R

34.5Vpeak are 34.5Vpeak/1.41=24.5Vrms

Your RMS power will be (24.5Vrms)^2/4ohm=149Wrms

2nd method: P=R*I^2

With 34.5Vpeak on a 4ohm load your current will be Ipeak=34.5Vpeak/4ohm=8.62Apeak.
Your RMS current will be 8.62Apeak/1.41=6.11Arms,
thus the RMS power is (6.11Arms)^2*4ohm=149Wrms

3rd method: P=U*I

The RMS power will be Prms=Urms*Irms=24.5Vrms*6.11Arms=149Wrms

As you can see the three methods have the same solution, it is the ohm's law...

Now, if you put the amp in mono (brigded mode) your peak voltage on the load is doubled, so that you will have 69Vpeak.

69Vpeak is 69/1.41=48.9Vrms and then
Prms=(48.9Vrms)^2/4=598W

Putting an amp in brigde will double the voltage on the load and thus multiplies the power by 4. This is truth only in ideal case, normally the output voltage is not doubled because of the increased current demand increase the losses in the output stage and also the power supply may loose some volts. A good amp, designed to work in brigde mode normally multiplies the power be 3 or something like that

ciao
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Old 26th March 2008, 06:16 PM   #5
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34.5V rail, in THEORY, produces 69V peak to peak. You need to work in RMS quantities, so for a sine wave, Vrms = Vp-p / 2.828 = 24.4Vrms.

Into a 4 ohm load (theory again, speakers are not resistors), you get 24.4/4 = 6.1A.

Do the watt calculations:

P = V^2/R = 24.4^2/4 = 149
P = I*V = 6.1 * 24.4 = 149
P = I^2*R = 6.1^2*4 = 149

Ohms law is one essentially V=IR. P=IV is an offshoot of that. You can derive all the above equations for power based on these two equations.
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Old 26th March 2008, 06:57 PM   #6
sreten is offline sreten  United Kingdom
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Default Re: accuracy of ohms law

Quote:
Originally posted by Clipped
how do you know which one to use when calculating power in watts?

P=I*E
P=E2/R
P=I2*R

Hi,

They are all the same thing.

P=I*V, but I=V/R so P also = V*V/R, and V=I*R so P also = I*I/R.

/sreten.
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Old 26th March 2008, 07:28 PM   #7
Nordic is offline Nordic  South Africa
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They are the same but differ in that they calculate the same thing useing 2 diffirent known values...

Current and voltage
Voltage and resistance
Current and resistance

The method you use will depend on which of these two values are known or measurable...
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Old 26th March 2008, 08:52 PM   #8
sreten is offline sreten  United Kingdom
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typo :

P=V*I, but I=V/R so P also = V*V/R, and V=I*R so P also = I*I*R

/sreten.
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Old 26th March 2008, 10:21 PM   #9
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Why not use a clamp meter, volt meter, and an oscilloscope to find real output??
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Old 27th March 2008, 03:05 AM   #10
Clipped is offline Clipped  Thailand
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so basicly multiply the rail voltage by 0.707 , square it by 2 then divide by R.

so, 34.5 * 0.707= 24.3915 (using calculator in mobile phone)

24.3915 * 24.3915 =594.94527 / 4 = 148.73632

ok thats what i thought, but i read somewhere that wasnt the right thing to do....i guess that site was wrong.

thanks for all the input, this goes to back it up.
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