power and rms

Discussion in 'General Electronics Chat' started by matteo_pixies, Jan 28, 2009.

  1. matteo_pixies

    Thread Starter New Member

    Jan 28, 2009
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    hi guys,
    I've a question about rms values and power. I want to calculate the power at a node and I know that:

    - the current varies sinusoidally from 0 to 500mA with a DC value of 250mA
    - the voltage swing is limited to 6V peak-to-peak

    ...and I know that the power is Vrms*Irms.

    I'm pretty sure that Irms = 3/sqrt(2). but ..what about Vrms? how can I include the DC value into the Vrms calculation?

    thanks
    Matteo
     
  2. mik3

    Senior Member

    Feb 4, 2008
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    Why Irms is 3/sqrt2 ?

    Vrms=Vpeak/sqrt2
     
  3. matteo_pixies

    Thread Starter New Member

    Jan 28, 2009
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    Irms is 3/sqrt2 because the wave varies from -3 to +3 so the Ipeak is 3, right?

    what is Vpeak for the voltage?
     
  4. PRS

    Well-Known Member

    Aug 24, 2008
    989
    35
    RMS power is indeed Vrms*Irms for a single phase circuit, which yours seems to be. Your Irms = Ipeak divided by the square root of two.
    Vrms = Vpeak divided by the square root of 2, hence 3/1.414 = 2.12 volts.

    But when you said, "the current varies sinusoidally from 0 to 500mA with a DC value of 250mA," I'm not sure what you mean. If you mean the peak current is 500 mA then divide by the square root of two, but if you mean peak to peak is 500 mA then divide 250mA by the square root of two.

    Then multiply Vrms times Irms and that is your power in terms of rms.
     
  5. mik3

    Senior Member

    Feb 4, 2008
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    Vpeak=Vpeak-peak/2
     
  6. matteo_pixies

    Thread Starter New Member

    Jan 28, 2009
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    My waves are:

    I(t) = 250mA + 250mA*sin(t)
    V(t) = 3*sin(t)

    my problem is that I dont undertand how to get Irms because of the DC value of I(t). If I solve the integral I get:

    Irms = sqrt( 250mA^2 + 250mA^2/2 )

    what do you think?

    thanks
    matteo
     
  7. mik3

    Senior Member

    Feb 4, 2008
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    So Vrms=3/sqrt2

    To fins Irms integrate this and then find its root:

    [250mA + 250mA*sin(t)]^2=(250^2)+[(250*sin(t))^2]+2*250*250*sin(t)
     
  8. matteo_pixies

    Thread Starter New Member

    Jan 28, 2009
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    I did that and I found out, after the integration:
    Irms = sqrt( 250mA^2 + 250mA^2/2 )

    so you say that it should be correct, right?
     
  9. mik3

    Senior Member

    Feb 4, 2008
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    Yes, it is correct.
     
  10. matteo_pixies

    Thread Starter New Member

    Jan 28, 2009
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    thank you very much ;)
     
  11. Ratch

    New Member

    Mar 20, 2007
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    matteo_pixies,

    Your problem seems to be finding the current and power of a element due to multifrequency sources. In your case, the frequency is zero and one other frequency. Since your exitation is sinusoidal, you don't need to use calculus. The zip file contains a fully worked out example from Irving L. Kosow's book Circuit Analysis. You should find it instructive. Notice that the total power from different frequencies add, and the total current is the square root of the sum of the squares of the current from different frequencies. This is shown in steps (e.) and (f.).

    Ratch
     
  12. PRS

    Well-Known Member

    Aug 24, 2008
    989
    35
    I may be wrong, but I think you just add the rms power of the dc component to the rms power of the ac component to get the overall rms power dissipated in the circuit. I may be wrong and if I am shown to be I will not argue. It's been a long time...
     
  13. matteo_pixies

    Thread Starter New Member

    Jan 28, 2009
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    uhmmm... as I mentioned before I computed the integral of the whole

    I(t) = DC + 250mA*sin(t)

    and I got

    Irms = sqrt( DC^2 + 250mA^2/2 )

    so it doesn't look like it adds directly because of the square root ..do you think I should take the DC value out of the integral? ..I dont undertand, I cant find a consistent definition..

    thanks
    matteo
     
  14. Ratch

    New Member

    Mar 20, 2007
    1,068
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    PRS,

    No, you are not wrong. Isn't that what I said and showed in the attachment I sent?

    Ratch
     
  15. Ratch

    New Member

    Mar 20, 2007
    1,068
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    matteo_pixies,

    Oh yes you can. You should look at the material I sent you.

    Ratch
     
  16. The Electrician

    AAC Fanatic!

    Oct 9, 2007
    2,287
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    You're not adding power here, are you?
     
  17. matteo_pixies

    Thread Starter New Member

    Jan 28, 2009
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    Ratch,
    I already did it and the problem they are dealing with its slightly different from the one I'm talking about. Anyway I understood even better my problem reading your notes ..thank you!
     
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