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Thevenin Resistance

  1. Oct 18, 2007 #1
    1. The problem statement, all variables and given/known data


    • Basically asking me to find the Thevenin/Norton's equivalent across a-b.
    • Then figure out the maximum power transferred
    • Then figure out max power delivered

    2. Relevant equations
    • Maximum power is transfered when [tex]R_{l} = R_{th}[/tex] .
    • Max power delivered is [tex]\frac{V_2}{4R_{th}}[/tex]

    3. The attempt at a solution
    I honestly don't know where to start this. I know Thevenin equivalent requires me to break this down into a voltage source, one resistor in series, but I really have no clue where to start. I tried using nodal analysis but this left me with nothing.

    --My poor attempt at nodal analysis--

    [tex]-IR = 4[/tex]

    [tex]I = \frac{-4}{R}[/tex]

    [tex]5I_1[/tex] = -3[tex]I_1[/tex]

    --sub [tex]\frac{-4}{R}[/tex] for [tex]I_1[/tex]--

    [tex]5\frac{-4}{R} = -3\frac{-4}{R}[/tex]

    [tex]\frac{-20}{R}= \frac{12}{R}[/tex]

    [tex]-20 = 12[/tex]

    A push in the right direction would be greatly appreciated.
    Last edited: Oct 18, 2007
  2. jcsd
  3. Oct 18, 2007 #2
    You will need to do nodal analysis twice. Once with the independent sources on, once with them off (the later to find the Thevenin/Norton equivalent resistance).

    Set the bottom to the ground node and the top is [tex]V_{th}[/tex]. This will yield one equation with one unknown- a straightforward solution.

    To find [tex]R_{th}[/tex] turn off the independent source (the 4A current source), connect a test source between A and B and find the equivalent resistance from there.
  4. Oct 18, 2007 #3
    Thank you very much! I will try this tomorrow morning and post my results. Again, thanks for the response.
  5. Oct 18, 2007 #4
    No problem, it was good review for me as I have an exam on it monday :P
  6. Oct 19, 2007 #5
    Okay. I think I got it.

    Now, using KCL at the node above the dependent current source I have:

    [tex]0 = 4 - 3_i_1 - i_1[/tex]
    [tex]4 = 4_i_1[/tex]
    [tex]\ldots i_1 = 1[/tex] amp

    [tex]V_{oc} = 5 V[/tex] [tex]\ldots[/tex] [tex]V_T = 5 V [/tex]
    [tex]I_{sc} = 4 A [/tex]

    [tex] R_{th} = \frac{V_{oc}}{I_{sc}}[/tex]

    [tex]R_{th} = \frac{5}{4} \Omega[/tex]

    So, a voltage source of 5V in series with a resistor of [tex]\frac{5}{4} \Omega[/tex]. Now to figure out the max power delivered and max power transferred.
    Last edited: Oct 19, 2007
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