Norton Equivalent

Discussion in 'Homework Help' started by xz4chx, Oct 23, 2012.

  1. xz4chx

    Thread Starter Member

    Sep 17, 2012
    I need to find Norton Eqivalent by 1 kΩ resistor and then find v_Q
    So I first calculated the short circuit current where the 1kΩ resistor was.
    Since v_Q in the situation would be 0V because of the short dependent source would be gone shorting out the 3.3kΩ resistor and 680Ω resistor. And then I would have
    -12V/4700Ohms + i_{sc} + .005A = 0
    After that I need to calculate R_{th} which is  v_T/i_T
    So I put a 1V test source and i_T in the direction of the voltage rise. The 1V test source is - toward bottom and + towards top of circuit.
    So then this gives me
     -12V/4700Ohms-i_T+1.5v_Q = 0
    Since v_Q is in parallel with my test source but in opposite direction v_Q=-v_T
    So I end up with
     -12V/4700Ohms-i_T+1.5(-1) = 0
    So  i_T = -1.503 [A]
    Divided by  v_T gives me R_{th} which I get to equal .667Ω
    So I set up the Norton Equivalent and then need to solve for v_Q
    which I am not 100% on but I was thinking it would just be the voltage of the Norton Equivalent. So I changed it to a Thevenin Equivalent and got
    v_Q = -1.6317[mV]

    Then answer in the textbook is  v_Q= -8.53[V]

    I want to know where I went wrong as my answer is no where near the right answer.

    I sorry I don't have any pictures of my work, it was hard to take a good picture of it.
  2. Jeffrey Samuel

    New Member

    Oct 25, 2012
    Your Rth calculation is flawed for sure as the resultant is about the 4.7K||3.98K which can never be in the lower end to the best of my knowledge