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Thevenin Problem

  1. Sep 5, 2011 #1
    1. The problem statement, all variables and given/known data

    This one has a few parts, so let's start with part a. See attachment for circuit. I've got two drawings in the attachment. I'll label them A and B. Image A is the original circuit.

    2. Relevant equations

    None.

    3. The attempt at a solution

    My first step is to find set RL to open. I can use a voltage divider equation and find the voltage to be 16V.

    The next (which is where I am stuck) is to find RL as a short. Shown by image B. My professor wasn't the best in showing this as he used some trick I have no clue how solved this way.

    I can find ITOT using...

    [itex]I=\frac{V}{R} \Rightarrow \frac{24}{3+6//4}[/itex]

    Solving for ITOT gives me 4.4A. My professor used this value and somehow came up with the current over the 4Ω resistor.

    Then plugged it into..

    [itex]R_{th} = \frac{V_th}{I_{short}}[/itex]

    To find the Thevenin equivalent.

    How does one find the current when RL is a short?
     

    Attached Files:

  2. jcsd
  3. Sep 5, 2011 #2

    Zryn

    User Avatar
    Gold Member

    Did you mean for the parallel resistors in Diagram B to be 6R and 4R? They are labeled 6R and 3R currently.

    Knowing the circuit series current allows you to determine how the current will split when it enters a parallel connection. The general way to do this is using the current divider formula:

    [itex]I_{n}[/itex] = [itex]I_{s}[/itex] * [itex]\frac{R_{other}}{(R_{other} + R_{n})}[/itex]

    Where:

    [itex]R_{other}[/itex] is the total resistance of the other resistors, [itex]R_{n}[/itex] is the resistance which you wish to find the current through, [itex]I_{s}[/itex] is the current being divided and [itex]I_{n}[/itex] is the current you want to find.
     
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