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Electrical engineering circuit analysis

  1. Sep 18, 2010 #1
    1. The problem statement, all variables and given/known data

    Question 1: http://img690.imageshack.us/i/79806671.jpg/

    Question 2: http://img833.imageshack.us/i/48920688.jpg/

    2. Relevant equations

    V=IR, P=I^2R

    3. The attempt at a solution

    Answer to Question 1: http://img831.imageshack.us/i/imgmp.jpg/

    Answers to Question 2 Part 1 :http://img545.imageshack.us/i/img0001e.jpg/

    Answers to Question 2 Part 2 :http://img94.imageshack.us/i/img0002te.jpg/

    Here are my answers for Question 2, if the solution is too long for you to bother with =D
    a) Vth = 13V, Isc = 13A, Rth = 1Ohm
    b) Vx = 12V
    c) Vx = 12V
    d) 8 watts, 2 watts, and 12 watts
    e) Independent sources 16V and 10V absorb power, dependent source 1.2Vx delivers power

    Feel free to give me any inputs! Your help is really appreciated!
  2. jcsd
  3. Sep 18, 2010 #2


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    At t=3, Q=5. How do you get a negative current?
  4. Sep 18, 2010 #3
    Since Q/T = I

    We can say that taking the derivative of Q with respect to T will yield I, which is actually the gradient of Q and T, or of the graph.

    From t=2 till t=4, gradient is negative, hence yielding a negative value. Are there anymore errors that you have spotted in the solution?
  5. Sep 18, 2010 #4


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    In the future, please post only one problem per thread (different questions related to the same problem are okay, but just don't post completely different problems). It gets really confusing if people are trying to help you with different problems all in the same thread.

    So for "Question 1" (actually, Problem# 1), your solutions to parts A and B are correct.

    Part C, the rms current is incorrect.
    For RMS,
    (1) First take the square of the function
    (2) Find the "mean" of that squared function, i.e. the average of the squared function. This generally done by integrating the squared function (i.e. finding the area under the curve of the squared function) and then dividing by the time.
    (3) Take the square root of whole result.

    In your particular problem, you didn't find the mean correctly. (The mean of a and b is not a/2. The mean is (a + b)/2).

    More generally, rms for a continuous function is:

    [tex] f_{rms} = \sqrt{\frac{1}{T_2-T_1}\int_{T_1} ^{T_2}[f(t)]^2 dt} [/tex]
    Last edited: Sep 18, 2010
  6. Sep 18, 2010 #5
    I see, the answer would be 5 then? Thank you for your input, try to have a look at question 2 too, thank you!
    Last edited: Sep 18, 2010
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