power calculation : Doubts

Discussion in 'Homework Help' started by nishu_r, Jun 24, 2012.

  1. nishu_r

    Thread Starter Member

    Jun 2, 2012
    31
    0
    In order to calculate P, the real power absorbed by an impedance, we are to use P=|I_{rms}|^2*Z, and why the modulus of I_{rms} (the magnitude) used instead of I_{rms} directly, if used what would it yield.

    This is a sincere question as i am not able to comprehend why, and i think the text book i use, is giving only a vague insight into the power calculations from a problem solving point of view.
     
  2. cork_ie

    Member

    Oct 8, 2011
    348
    58
    I am not sure I quite understand your question.
    Are you enquiring why P is proportional to the square of I RMS rather than just being directly proportional to I RMS ?

    If so:
    P= VRMS x I RMS ( assuming a resistive circuit where I & V are both in phase)

    VRMS = I RMS x z , hence P = [ I RMS ]sq. x Z
     
  3. jimmy101

    New Member

    Jun 23, 2012
    9
    2
    I believe he means why have the absolute value function at all since
    (|x|)^2=(x)^2
    for all x. So the absolute value function does nothing.
     
  4. WBahn

    Moderator

    Mar 31, 2012
    17,715
    4,788
    Are you sure that you have copied that from the textbook correctly? Because real power is a real quantity that the expression, irrespective of the form of I, is multiplied by Z which, in general, is a complex quantity.

    As for the absolute value. In general, the phasor for the following time-domain current:

    <br />
i(t) \ = \ I_o \cos(\omega t + \phi)  <br />

    would be written

    <br />
I(j\omega) \ = \ I_o e^{(j \phi)} \ = \ I_R + jI_I<br />

    Now, these I's are amplitudes of sinusoids. But, we know that the RMS value of a sinusoid can be obtained by just dividing it be the square root of two. So we could right:

    <br />
I_{RMS}(j\omega) \ = \ I_{o_{RMS}} e^{(j \phi)} \ = \ I_{R_{RMS}} + jI_{I_{RMS}}<br />

    The point being that the RMS phasor current is still a complex quantity. What you need for the power calculations is the magnitude of the RMS phasor, not the square of it (the square of a complex number is a complex number with twice the phase angle). That is why you take the "absolute value" (which, for complex numbers, means the magnitude) and square that.

    But to get real power, you would need to multiply by the resistive part of the impedance, not the entire impedance.
     
    nishu_r likes this.
  5. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    WBahn I think you are correct in questioning the OP's transcription or interpretation of the equation in post #1.

    Perhaps the text would follow something along the lines ....

    {|I_{rms}|}^2Z={|I_{rms}|}^2(R+jX)\\={|I_{rms}|}^2(|Z|cos\theta +j|Z|sin\theta)\\={|I_{rms}|}^2|Z|cos\theta+j{|I_{rms}|}^2|Z|sin\theta\\=P+jQ

    The relationship if fact gives the complex or rather phasor/vector power - not the real power as the OP has stated.
     
  6. nishu_r

    Thread Starter Member

    Jun 2, 2012
    31
    0
    Yes it does give the complex power, S .And, I am terribly sorry for the flaw in the question.

    Now i understand clearly. thanks.
     
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