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Power through current sources

Discussion in 'Electronics Homework Help' started by Constant, Feb 16, 2015.

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  1. Constant


    Feb 15, 2015
    So the voltage across the two upper branches is 8V.

    Working through say the top branch, the voltage difference across the current source must be 8V-6V=2V.

    Hence P=IV = 2 X 4mA = 8mW.

    Why is my answer incorrect?

    Attached Files:

  2. Harald Kapp

    Harald Kapp Moderator Moderator

    Nov 17, 2011
    I say your answer isn't incorrect. I arrive at the same value both by a back of the envelope calculation and by simulation. In my view the answer provided in the task description is wrong.
  3. Ratch


    Mar 10, 2013
    Yes, the answer key is wrong. The 4 ma source absorbs 8 mw and the 2 ma source supplies 20 mw.

  4. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

    Nov 28, 2011
    Just trying to clarify...

    I guess the original circuit can be simplified down to this:

    epoint 272717.png

    Do you agree with that diagram?

    Then the power dissipated in each current source can be calculated using the Power law, taking into account the direction of current flow.

    So you guys are assuming that the arrow in the current source represents the direction of conventional current flow? AFAIK that's true; I've never seen a current source where the arrow direction represented electron flow.
  5. Ratch


    Mar 10, 2013
    I don't know how the rest of this group interprets the diagram, but I go by what a voltmeter and ammeter would indicate.

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