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Homework Help: Open Voltage and Short Circuit

  1. Aug 24, 2011 #1
    1. The problem statement, all variables and given/known data

    So the first part is to find Voc in this circuit:

    So I did this:
    -10 + 1200*i1 + 3300*(i1 - i2) - 12 = 0
    since i2 = 0 A,
    -22 + 4500*i1 = 0
    i1 = 22/4500 A = 0.0049 A

    Then, for Voc:
    3300*(i2 - i1) + Voc + 12 = 0
    since i2 = 0 A,
    3300*(−22/4500) + Voc + 12 = 0
    Voc = 62/15 V = 4.13 V

    Then the next part is to find the short circuit current Isc in the same circuit:

    Assuming that my calculations for i1 were right, let i1 = 0.0049
    Isc = i2 so then do mesh current analysis for the second mesh, which will get:

    3300(i2 - i1) + 12 = 0
    3300(i2 - 0.0049) + 12 = 0
    3300i2 - 16.137 + 12 = 0
    3300i2 = 4.137
    i2 = 0.001254 A
    .:. Isc = 0.00125 A

    Would this be correct? Thanks in advance for any help given. :)
  2. jcsd
  3. Aug 24, 2011 #2


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    Staff: Mentor

    Since you know that terminals a and b are open circuited, you only have one loop. Writing in i2 as a variable is unnecessary.

    You've solved for the current in the loop, and the result is okay. You might consider taking advantage of unit prefixes. Thus i1 = 4.9 mA.

    Now, regarding Voc, note the polarity of the 12V supply: starting at terminal b and proceeding through the 12V supply there is a DROP of 12V. Proceeding on through R2, since the current i1 flows from top to bottom of R2, there will be a potential rise as we pass through. So Voc = -12V + i1*R2.

    Nope! You can't do that since the circuit changes when the open circuit at a-b becomes a short circuit; You now have two loops instead of one, and they will interact via their common components (R2 in particular).

    You'll have to write the two KVL mesh equations and solve for the currents, or use another method as you see fit (Source transformations (Norton), superposition, etc.). Superposition would be particularly simple for this problem.
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