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Thevenin / Superposition

  1. Mar 13, 2008 #1
    Never mind, did superposition incorrectly.

    1. The problem statement, all variables and given/known data


    Find Vload using superposition and thevenin's equivalent

    2. Relevant equations

    3. The attempt at a solution

    Well for network on the left of Rload, I found Vth = 5 V and Rth = 15.1 kohms
    For the network on the right of Rload, I found Vth = 7.5 V and Rth = 8150 ohms.

    So using superposition I would do:
    1) 5 V * 10k / (10k+15.1k) = 1.99 V on Rload
    2) 7.5 V * 10k / (10k + 8150) = 4.13 V
    Adding those I get 6.12 V.

    However, when I simulate this in PSPICE I get 4.33 V.
    I also checked each side and I get 1.99 V on the left (when I disconnect the right) and I get 4.13 V on the right (when I disconnect the left).

    What am I doing wrong, I can't even remotely see where that 4.33 V comes from.
    Last edited: Mar 13, 2008
  2. jcsd
  3. Mar 14, 2008 #2
    When you calculate the voltage on the load due to one voltage source, you can't just remove the others, but need to replace them with their internal resistance.

    for 1) you'd get 5 v * (10K // 8150) / ((10K // 8150) + 15.1k)
    (A // B) is the value of the resistance of A parallel with B, or AB/(A+B)
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