1. Not finding help here? Sign up for a free 30min 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!

How to find Voltages and currents of the following circuit

  1. Feb 8, 2017 #1
    1. The problem statement, all variables and given/known data
    The circuit of the given problem is attached to this post (jpg file). All info is given in the photo.
    Problem: find VA, VB, IA.

    2. Relevant equations
    Ohms law
    KCL equations
    P = VI or V^2/R

    3. The attempt at a solution

    - First, I noted that VB = 9 - VA.
    - next, i set up a KCL equation at VA:
    -.004VB + VA/30 + (2VA -18)/30 = 0
    - i plugged in VB:
    -.004(9-VA) + VA/30 + (2VA -18)/30 = 0

    - Solved for VA = 6.115 V
    - Therefore VB must = 2.885 V (given everything so far is correct

    - by Ohm's law, IA = VB / 15 k ohms
    - IA = 0.1923 mA

    - so at this point I have values for all i need to solve the problem, but to check if i was right I wanted to check consv of power.

    - KCL at VA node = .1923 + 11.54 = IB
    IB = 11.7323 mA

    When I checked the power on all 4 items in this circuit, the power did not check out to zero, so somewhere I messed up but I have no idea where.
     

    Attached Files:

  2. jcsd
  3. Feb 8, 2017 #2

    gneill

    User Avatar

    Staff: Mentor

    I'd look to how you handled the resistances versus the controlled current source magnitude. You've dropped the "k" on the resistors implying that currents will be in mA, but left the controlled source multiplier as is. :wink:
     
  4. Feb 8, 2017 #3

    phyzguy

    User Avatar
    Science Advisor

    I think you have the right answer for VA and VB, but, as gneill says, you have messed up A and mA on the currents, so IB is definitely wrong. Also, make sure you pay attention to the signs on the powers.
     
  5. Feb 8, 2017 #4
    Thanks for the help and time!
    but Ok so seeing VA and VB believed to be correct is reassuring and leads me to believe that somewhere it is just a small error when dealing with powers.

    But ok so in this case either my source calc or IA must be wrong since IB calculations come afterwards...

    so IA = 2.885V / 15Kohm = 0.1923 mA
    source: .004(2.885V) A = 0.01154 A which can be translated to 11.54 mA ??

    I feel like I need new glasses right now lol
     
  6. Feb 8, 2017 #5

    gneill

    User Avatar

    Staff: Mentor

    I see Va being larger, much closer to 9 V. Vb will be pretty small. The currents should be small too, all of them less than a milliamp.
     
  7. Feb 8, 2017 #6
    WAIT !! I think I see it. Is it because in my KCL equation where I solve for Va... By only dividing by 30 those are milliamps whereas the source is Amos so to balance it I need to change the .004(9-Va) to 4(9-Va). Which is the issue you first stated ?
     
  8. Feb 8, 2017 #7

    gneill

    User Avatar

    Staff: Mentor

    It is! It's safer to first write the node equation with the k's still in, then multiply through by 1000 to eliminate them. That catches all the conversions.
     
  9. Feb 8, 2017 #8

    phyzguy

    User Avatar
    Science Advisor

    My bad. I made the same mistake as the OP. Ignore my earlier post and listen to gneill.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted