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!

Very, very simple DC circuit

  1. Mar 2, 2014 #1
    Hi everyone, first post.
    Just wanted to say I've been using parts of this forum for a long time for help in understanding all kinds of concepts, and have found it immensely helpful over the years. Thank you for all of your contributions!
    So, here goes with my hopefully very simple question:

    1. The problem statement, all variables and given/known data

    From a practice exam with answer key.
    Written: 'What is the value of the voltage U?'
    The drawing: a simple closed loop with 3 circuit elements in series (from top to bottom on one side): an ideal voltage source U_0 (positive terminal 'on top'), a resistor R, and an ideal current source I_0 (current emerges 'on top').
    The voltage in question U appears across two terminal legs placed above and below the ideal current source, and has the same orientation as the voltage source (i.e. positive terminal 'on top').
    I've attached a picture as well.


    2. Relevant equations

    I used KVL in the main loop (clockwise):
    -U_0 + R*I_0 -U = 0


    3. The attempt at a solution

    The above equation gave me my solution, or so I thought. U = R*I_0 - U_0.
    However, the listed solution is U = R*I_0 + U_0.

    I would assume that R*I_0 and U_0, written symbolically as they are in this problem, could never have the same sign (following the KVL loop and polarities).

    But I don't feel that I understand how to treat voltages across ideal current sources anyway. So perhaps my problem is there. I would appreciate any explanation of how to approach that in general as well.

    I'm still a beginner and make dumb mistakes in circuit analysis all the time and root most of them out eventually, but for whatever reason I'm not getting it here. I think I've overthought it. Thanks in advance!
     

    Attached Files:

  2. jcsd
  3. Mar 2, 2014 #2
    Try to draw the circuit in this way
     

    Attached Files:

  4. Mar 2, 2014 #3
    Thanks for your reply, Jony130 :-) The drawing is helpful to me, didn't even think of redrawing it. Now I have a new technique to use when I get stuck.

    But even after trying that, I (think I) still have the same question.

    Let me write out how I'm thinking about it from your drawing:
    For KVL loop in main loop, I make my loop clockwise starting between the resistor and voltage source:
    Voltage across voltage source: -U_0
    Voltage across terminals (voltage across current source): -U
    Voltage across resistor: +R*I_0
    Total KVL equation: -U_0 + R*I_0 -U = 0
    Add U to both sides, I get my answer: U = R*I_0 MINUS U_0

    The solution in the answer key: U = R*I_0 PLUS U_0
    Both R*I_0 and U_0 in the final equation are positive and I don't see why yet.

    (sorry for the bolding, I don't want to be rude but I just wanted to make my question clearer)
    Thanks
     
  5. Mar 2, 2014 #4
    I mark the component polarities

    attachment.php?attachmentid=67167&stc=1&d=1393770251.png

    So we have Uo - U + Io*R = 0


    Or

    - Uo + U - Io*R = 0
     

    Attached Files:

    Last edited: Mar 2, 2014
  6. Mar 2, 2014 #5
    Thank you again :-)
    I'm in Germany and here, as far as I know, the arrow on a voltage source is actually opposite polarity to what you have drawn (tail at + end and tip at - end)... so we still have the same problem.
     
  7. Mar 2, 2014 #6
    You mark the direction of a current source in clockwise direction. And this means that the voltage at point A must be higher then the voltage at point B (Io*R higher ).
    attachment.php?attachmentid=67173&stc=1&d=1393774657.png
    So from this we can say that the voltage at point A is equal to:
    Va = Vb + (Io*R) = Uo + VR. And the voltage at current source is
    Vac = Va - Vb = (Uo + VR) - 0V = Uo + VR because we treat Vc as our reference point (GND).

    But if we reverse the direction of a current source we have this situation
    attachment.php?attachmentid=67172&stc=1&d=1393773853.png
    And now the voltage at point A is Io*R lower then the voltage at point B.
    So Vac = Vo - Io*R

    In my drawings the arrow tip points to the most positive end (highest potential) of the component.
    OK so for your polarity we have this situation
    attachment.php?attachmentid=67174&stc=1&d=1393774670.png

    We treat point C as our GND (reference point).
    The voltage at point B is equal to -Vo and the voltage at point A is Io*R higher the the voltage at point b.
    Vac = Vb + Io*R = -Vo + Io*R
     

    Attached Files:

    Last edited: Mar 2, 2014
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Very, very simple DC circuit
  1. Very simple RC-circuit (Replies: 11)

Loading...