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!

Simple circuit.

  1. Dec 18, 2011 #1
    1. The problem statement, all variables and given/known data
    practiceprob.jpg


    2. Relevant equations
    I am trying to do some self study here before I take my circuits course. I am stuck on this problem for some reason.. The equation I am coming up with for the node is : 6 - Vo/2 - Io/4 - Vo/8 = 0.. After getting at equation I cant seem to come to the answer. Do I need to do another equation here? Thanks


    3. The attempt at a solution
     
  2. jcsd
  3. Dec 18, 2011 #2

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Welcome to PF Would!

    Ohm's law for the 2 ohm resistor tells you that i0 = v0/(2 Ω). That's your second equation. Now you have everything in terms of only one unknown, and you can therefore solve the system.
     
  4. Dec 19, 2011 #3
    Thanks a lot for the help! I must be doing some simple math wrong here.. but im getting : 6 - Vo/2 - Vo/4 - Vo/8 = 0 then multiplying all of that by 8 to get 48- 4Vo - 2Vo - Vo = 0 then ultimately getting Vo = 6.85V but that is wrong... What am I doing wrong there? Should I be assuming that the current in across the 8 ohm resistor is moving upward and therefore making the equation 48 - 4Vo - 2Vo + Vo = 0? It works out that way but if it is like that could you tell me why? Since it is not marked i assumed it was a voltage drop across the 8 ohm resistor..
     
  5. Dec 19, 2011 #4

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    The third term in your equation is simply wrong. In your original KCL equation, this term was i0/4, which was correct. Then we determined that i0 = v0/2. Hence, i0/4 = (v0/2)/4 = v0/8. The third term (in red) should therefore be v0/8.
     
  6. Dec 19, 2011 #5
    Wow I feel stupid after looking back at that. Thank you for the help.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Simple circuit.
  1. Simple Circuit (Replies: 4)

Loading...