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Simple Power in a Circuit Problem

  1. Jan 27, 2007 #1
    1. The problem statement, all variables and given/known data

    Determine I_0 in the network.

    2. Relevant equations

    P = VI
    Passive Sign Convention Rule

    3. The attempt at a solution

    Here's what I did. I calculated the power absorbed/supplied by the bottom left network.

    Element 3: P = VI = (6V)(I_0 A) = 6I_0 W
    Element 5: P = VI = (6V)(4A) = 24 W
    Element 6: P = VI = (8V)(6A) = 48 W
    V-Dependent Source: P = VI = (4 * 2 A)(-6A) = -48 W (I did this because of the passive sign convention.

    6I_0 + 24 + 48 - 48 = 0
    I_0 = (-24) / 6
    I_0 = -4 A

    It says something is wrong, but I'm not sure why. I'm 99% sure I followed the rules for PSC. I've attached an image of the circuit. I've done it only using this loop, but also, if you check down below, I've done it for every element (I get differing answers).

    Thanks :smile: :smile:

    (Circuit picture taken from WileyPLUS homework, please do not redistribute/copy).


    -------

    If I had to calculate each element, in the WHOLE entire network, would this work be correct:

    E1: 64W
    E2: 60W
    E3: I_0 * 6 W
    Independent 24V Source: -192W
    E4: 32W
    E5: 24W
    E6: 48W
    Dependent 4I_x V Source: -48W

    so... (Im going to leave out the W unit in the equation for now, so it doesn't confuse anyone)

    64 + 60 + 6I_0 + (-192) + 32 + 24 + 48 - 48 = 0

    6I_0 + (-12) = 0
    6I_0 = 12

    I_0 = 2A
     

    Attached Files:

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    Last edited: Jan 27, 2007
  2. jcsd
  3. Jan 28, 2007 #2
    Well using KCL at that node, I'd say the 2nd answer is right.

    What does this tell you about the power around a loop (with a circuit with more than one loop)?
     
  4. Jan 28, 2007 #3
    Power Supplied = Power Absorbed

    So 2Amps would be correct, then right?
     
    Last edited: Jan 28, 2007
  5. Jan 28, 2007 #4
    Look at this example I made:

    [​IMG]
     
  6. Jan 28, 2007 #5
    Ah, I see. Makes sense.
     
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