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!

[Circuits] Calculating the Thevenin Equivalent #1

  1. Jan 21, 2014 #1
    1. The problem statement, all variables and given/known data

    a7zCn7r.png

    2. Relevant equations



    3. The attempt at a solution

    I got ##R_{ th }=4\quad Ω## easily but I am having trouble finding ##V_{th}##. Assume the negative terminal of the 32V voltage source is ground. Here is the nodal analysis at node 1:

    $$\frac { 32-V_{ 1 } }{ 4 } +2=\frac { V_{ 1 } }{ 12 } +V_{ 1 }-V_{ 2 }$$

    But I only have one equation and there are two variables.
     
    Last edited: Jan 21, 2014
  2. jcsd
  3. Jan 21, 2014 #2

    gneill

    User Avatar

    Staff: Mentor

    When you remove the load RL, and after assigning a common reference node, there's only one essential node remaining. So you should have one equation in one unknown.

    Note that with the load removed there will be no current flow through the 1 Ω resistor...
     
  4. Jan 21, 2014 #3
    Opps, I should've added that I'm trying to find the voltage between (a) and (b).

    ##V_1## is the voltage on the node in between the 4 ohm and 12 ohm resistor and ##V_2## is the voltage of (a). (b) is ground.

    If there is no current flow through the 1 ohm resistor, does that mean that the voltage of ##V_1## and ##V_2## is the same?
     
  5. Jan 21, 2014 #4

    gneill

    User Avatar

    Staff: Mentor

    Yup. No current means no potential drop.
     
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: [Circuits] Calculating the Thevenin Equivalent #1
Loading...