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Electric Circuit Analysis: Calculating power absorbed/supplied by each element

  1. Jan 8, 2013 #1
    Once again, my textbook seems to do a poor job of explaining another concept just like I commented earlier here. Anyways, here is the relevant part of the textbook: http://www.mediafire.com/view/?cw5t2wr9vqqprnv

    To calculate p4 in example 1.7, voltage of 8V is multiplied by a current of -0.2I. Firstly, why is the current negative? Secondly, why do they substitute a value of 5A for I. Where did they get this value?

    I tried to do practice problem 1.7, but I don't seem to get the right answer for p3 when I use a voltage of 3V and a current of 0.6(3)A.
  2. jcsd
  3. Jan 8, 2013 #2
    Because the dependent current source is supplying the power. So current flows from its positive to its negative. But when you calculate power of a branch you take the current value (which is to be multiplied) same as what will be entering the branch from positive.

    They defined the dependent power that way. What ever will be the I (which is going through p1 brance) will be multiplied by 0.2 and that will be the current coming out of dependent source.
  4. Jan 9, 2013 #3
    Where did they get 5 from?
  5. Jan 10, 2013 #4
    Well it says on your diagram I = 5 amps.

    Incidentally, I cannot acces your link on my ofice computer, so others may well have the same problem.

    It is better to post the doc or image here at PF, and also to properly reference the book as others may have it.

    You will also likely receive more attention that way.
  6. Jan 10, 2013 #5
    I don't have time to do that at the moment as I'm heading to school right now, but isn't it 5 amps for only p2? For p3, it's 6 amps.
  7. Jan 10, 2013 #6
    Read "en.wikipedia.org/wiki/Dependent_source" [Broken].

    The value of the second current source is not constant, but changes when the current through p2 branch changes (and the proportionality constant is 5).

    The I indicated in the figure defines the current through p2, not p3. Whatever is the value of current through p3 it will not necessarily be same as p2.
    Last edited by a moderator: May 6, 2017
  8. Jan 13, 2013 #7
    Ohh! I get it now! I feel so stupid. For some reason, I thought of I as the current flowing through p4 instead of thinking of I as a variable such as X that is clearly defined as 5A in the top left of the diagram. I'm so stupid. :blushing:
    Last edited by a moderator: May 6, 2017
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