More Fun With Numbers:
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petersonra
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Re: More Fun With Numbers:
[ February 18, 2005, 05:08 PM: Message edited by: petersonra ]
[ February 18, 2005, 05:08 PM: Message edited by: petersonra ]
john m. caloggero
Senior Member
Re: More Fun With Numbers:
This is taking me back to 1956 college days.
For a full-wave rectifier, (2/pi) or
(I average = .637 x I max.)
For a half-wave rectifier,(1/Pi) or
(I average = .318 x Imax)
For full-wave, the ratio of average to rms is:
(Irms = 1/1.11 x Iave)
This is taking me back to 1956 college days.
For a full-wave rectifier, (2/pi) or
(I average = .637 x I max.)
For a half-wave rectifier,(1/Pi) or
(I average = .318 x Imax)
For full-wave, the ratio of average to rms is:
(Irms = 1/1.11 x Iave)
Re: More Fun With Numbers:
Sam, we have to consider the square of the voltage. RMS means:
Square ROOT of the MEAN value of the voltage SQUARED. No one has done that.
Take the square wave for example, it delivers half the power that DC of magnitude Vp would deliver.
1) Vrms^2/R = Vp^2/2R, or
2) Vrms = Vp/1.414
That is, the equation,
3) P = Vrms^2/R must be satisfied
[ February 20, 2005, 09:43 AM: Message edited by: rattus ]
Sam, we have to consider the square of the voltage. RMS means:
Square ROOT of the MEAN value of the voltage SQUARED. No one has done that.
Take the square wave for example, it delivers half the power that DC of magnitude Vp would deliver.
1) Vrms^2/R = Vp^2/2R, or
2) Vrms = Vp/1.414
That is, the equation,
3) P = Vrms^2/R must be satisfied
[ February 20, 2005, 09:43 AM: Message edited by: rattus ]
physis
Senior Member
Re: More Fun With Numbers:
What's going on with power and resistance here?
I thought the question was voltage?
Edit: If the mean voltage (as in the square wave) is .5.
Then
The root of the mean square is:
√.5?=.5
I must still be missing something.
[ February 20, 2005, 01:35 AM: Message edited by: physis ]
What's going on with power and resistance here?
I thought the question was voltage?
Edit: If the mean voltage (as in the square wave) is .5.
Then
The root of the mean square is:
√.5?=.5
I must still be missing something.
[ February 20, 2005, 01:35 AM: Message edited by: physis ]
Ed MacLaren
Senior Member
Re: More Fun With Numbers:
Why wouldn't an AC square wave deliver power equal to a steady DC, assuming the rise time is instantaneous?
Ed
[ February 20, 2005, 08:14 AM: Message edited by: Ed MacLaren ]
I would agree that is correct for a DC square wave.
Why wouldn't an AC square wave deliver power equal to a steady DC, assuming the rise time is instantaneous?
Ed
[ February 20, 2005, 08:14 AM: Message edited by: Ed MacLaren ]
Re: More Fun With Numbers:
Ed, you are correct. This same argument applies to the full-wave rectified AC we discussed a few days ago, or any symmetrical wave for that matter.
Sam, We are concerned with the MEAN value of the voltage SQUARED. You SQUARE the voltage first, then take the MEAN, then take the square ROOT of that.
Ed, you are correct. This same argument applies to the full-wave rectified AC we discussed a few days ago, or any symmetrical wave for that matter.
Sam, We are concerned with the MEAN value of the voltage SQUARED. You SQUARE the voltage first, then take the MEAN, then take the square ROOT of that.
Re: More Fun With Numbers:
You averaged the squared voltage by multiplying by the 50% duty cycle.
Now does 5/1.414 = 3.536?
BTW, you can do this for any duty cycle, but it would no longer be a square wave.
[ February 20, 2005, 12:30 PM: Message edited by: rattus ]
You got it Sam,
You averaged the squared voltage by multiplying by the 50% duty cycle.
Now does 5/1.414 = 3.536?
BTW, you can do this for any duty cycle, but it would no longer be a square wave.
[ February 20, 2005, 12:30 PM: Message edited by: rattus ]
Re: More Fun With Numbers:
For any waveform, we can calculate both a RMS value, and an average value. The numbers posted by Petersonraa are average values. If you follow your intuition Sam, you will probably get average values. It is basically the peak voltage multiplied by the percent of the area under the wave. For example, for a square wave, half the area between 0 and Vp is under the wave, and half is above. So the average value is 1/2 Vp.
Like Rattus said, to get the RMS value, square the origional wave, then find the average, then take the square root.
The reason we need RMS values is to compare the power delivered to a load to DC power. Remember that instantenous power (don't worry, I'm not going there )= V^2/R. So to find out what kind of voltage produces the same power as a DC wave, we have to do the extra squaring and then take the square root after we find the average.
Steve
For any waveform, we can calculate both a RMS value, and an average value. The numbers posted by Petersonraa are average values. If you follow your intuition Sam, you will probably get average values. It is basically the peak voltage multiplied by the percent of the area under the wave. For example, for a square wave, half the area between 0 and Vp is under the wave, and half is above. So the average value is 1/2 Vp.
Like Rattus said, to get the RMS value, square the origional wave, then find the average, then take the square root.
The reason we need RMS values is to compare the power delivered to a load to DC power. Remember that instantenous power (don't worry, I'm not going there
)= V^2/R. So to find out what kind of voltage produces the same power as a DC wave, we have to do the extra squaring and then take the square root after we find the average.Steve
Re: More Fun With Numbers:
Let me make the point that we do not usually compute the RMS values for any waveform other than a sinusoid. We are just pointing out that it can be done for any periodic waveform. This exercise helps one understand the concept of effective (RMS) values.
Let me make the point that we do not usually compute the RMS values for any waveform other than a sinusoid. We are just pointing out that it can be done for any periodic waveform. This exercise helps one understand the concept of effective (RMS) values.
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