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Homework Help: The amount of power supplied/received by the dependent current source

  1. Sep 5, 2012 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    Ʃ [itex]V_{n}[/itex] = 0
    Ʃ [itex]I_{n}[/itex] = 0
    v = ir

    3. The attempt at a solution

    I combined the two 2Ω resistors into a single 4 Ω resistor because they're in series.

    Then I tried to use KCL but I got 2 unknowns in one equation.

    0.2A + [itex]\frac{Vi}{4}[/itex] = I
    Last edited: Sep 5, 2012
  2. jcsd
  3. Sep 5, 2012 #2
    Hi November,

    Are you familiar with Mesh Current analysis? I think you are getting ahead of yourself when you add the two resistors together. Try leaving them separate and applying Mesh Current analysis. That should give you two equations and two unknowns.

    Here is a short lesson on Mesh Current if you are unfamiliar with it or need a refresher:
    http://www.allaboutcircuits.com/vol_1/chpt_10/3.html (The first problem is very similar to yours.)
  4. Sep 5, 2012 #3
    I don't know the voltage of any of the devices in the circuit, so how can I use KVL?
    Last edited: Sep 5, 2012
  5. Sep 5, 2012 #4
    Hmm, I may have given a more complicated method than necessary. If you were to use Mesh Current, you would draw the 2 current loops. When you have a current source in a loop, then you can just define that loop current as the value of the current source (i.e. I1 = 0.2 A, I2 = V1/4). I have noticed there is an easier way, using what you already have done.

    If you only look at the top 2Ω resistor and write out Ohm's law in regards to it, that will give you a second equation that you can use with your first one in order to solve the problem.
  6. Sep 5, 2012 #5
    Vi = 2I

    0.2A + [itex]\frac{Vi}{4}[/itex] = I

    0.2A = I - [itex]\frac{I}{2}[/itex]

    0.4A = I

    Did I do this right?
  7. Sep 5, 2012 #6
    That's what I got!

    You can always do an extra check by calculating the power produced (by the two current sources) vs. power consumed (by the two resistors). These should always be the same.
  8. Sep 5, 2012 #7
    The formula for power is P = IV, P= [itex]I^{2}[/itex]R
    So how can I check the power output of a current source?
  9. Sep 5, 2012 #8
    P = IV still applies.

    You know the current through the two resistors (0.4 A), and the equivalent resistance of the two series resistors is 4Ω. So to find the voltage from the top node to the bottom node, it's just:

    V = IR
    V = 0.4*4
    V = 1.6 volts

    Then P = IV!

    Edit: And maybe I should be clear that 1.6 volts is also the voltage across each current source.
  10. Sep 5, 2012 #9
    Do you mean the top and bottom nodes of the resistor or the current source? If you mean the resistor, then wouldn't the voltage drop be split between the two paths at the bottom node?
  11. Sep 5, 2012 #10
    But it's in parallel so they would have the same voltage right?
  12. Sep 5, 2012 #11
    Yep! I was just making a few doodles on the picture to illustrate that, so I'm going to post it anyways, even though you already got it. :smile:

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