1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Electronics - Determine voltage across current source

  1. Sep 11, 2012 #1
    1. The problem statement, all variables and given/known data
    Determine the voltage across the 2A current source
    2m6lq37.png

    2. Relevant equations
    kirchhoff voltage law (kvl)
    kirchhoff current law (kcl)
    ohm's law

    3. The attempt at a solution

    I'm new to the concept of a current source. I'm not sure how there can be voltage drop across the current source, so maybe the question is asking for the voltage drop across the 10 ohm resistor that's in series with it?
    Are all resistors that are parallel with the power source going to have a voltage drop equal to the source voltage? So the voltage across the 5 ohm resistor and the current source is 50V? I doubt the answer could be that simple though
    I think I'm really missing something to understand how to do this question.

    If I need to do KCL, it would have to be at the first junction the 2A flows into right?
    and for KVL, it would be the loop with the 50V power source? Having both equations is leading me nowhere

    Thanks!
     
  2. jcsd
  3. Sep 11, 2012 #2

    gneill

    User Avatar

    Staff: Mentor

    An ideal current source will produce ANY amount of voltage required to maintain its specified current. So never open circuit an ideal current source :eek:

    An ideal voltage source will produce ANY amount of current required to maintain its specified potential difference. So never short circuit an ideal voltage source :eek:

    You'll need to find the total potential across the branch with the current source in it. Whatever the drop is across the 10Ω resistor, the remainder must be due to the current source.
     
  4. Sep 11, 2012 #3
    Oh thank you! I didn't realize a current source was producing voltage to maintain a specific current.
    (subscripts for resistors using their resistances)
    Using KCL for the junction the 2A flows into:
    2 + i_1 - i_5 = 0
    KCL at junction below the 5 ohm resistor
    i_5 - i_1 - i_10 = 0
    i_10 = i_5 - i_1 = 2A
    so V_10 = i_10 * R = (2)(20) = 20V

    So for 50 V across the current source and 10 ohm resistor, if the voltage drop across the resistor is 20V, the voltage produced by the current source is 30V ?
     
  5. Sep 11, 2012 #4

    gneill

    User Avatar

    Staff: Mentor

    Oooh, so close! Check the polarity of the drop across the 10Ω resistor.
     
  6. Sep 11, 2012 #5
    So it's 70V? woah I couldn't imagine the voltage across anything in this circuit to be greater than the source voltage. But I guess it's possible since the current source produces voltage to keep the 2A constantly flowing?!
     
  7. Sep 11, 2012 #6

    gneill

    User Avatar

    Staff: Mentor

    No, there's something amiss still. Your total potential across the branch doesn't look right.

    If you take the junction at the bottom of the circuit as the reference node, and if you let the voltage at the top junction (node) be Va, then you can rewrite your first KCL node equation as:

    ##\frac{Va - 50V}{4Ω + 1Ω} + \frac{Va}{5Ω} - 2A = 0##

    Solve for Va, which also happens to be the voltage across the 5Ω resistor, as well as the branch with the current source.
     
  8. Sep 12, 2012 #7
    Can you explain how you got the first term? Why are the 4 ohm and 1 ohm resistors summed in the denominator (are you treating them as resistors in series?)
    I just learned about node voltage analysis in class a few hours ago; is this a method similar to that?
     
  9. Sep 12, 2012 #8

    gneill

    User Avatar

    Staff: Mentor

    Yes, they are in series in the branch of the circuit that this term describes. If the lower node is the reference point (assumed 0V), and the node in question is at potential Va, then the current that must flow out of the node at potential Va towards the reference node is given by this term in the equation --- the net potential difference divided by the total resistance in the branch.
    It is exactly that, yes.
     
  10. Sep 14, 2012 #9
    I get it now! I just wonder why the professor assigned this question before node voltage analysis was covered.

    Thank you so much for your help!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Electronics - Determine voltage across current source
Loading...