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Homework Help: Linearity and superposition theorem

  1. Apr 24, 2016 #1
    1. The problem statement, all variables and given/known data
    For the network of constant current shown in Figure 4 it is known that R1 = 50 Ω and , R = 10 Ω. When the switch P is
    in the 1-position , current I = 50 mA and Ip = 70 mA known i . When the switch P is in
    the 2-position , current I' = 40 mA and Ip' = 90 mA are known . Determine the current Ip''
    when the switch switches to the third position.
    2. Relevant equations

    3. The attempt at a solution
    By linearity and superposition theorem we have a linear network and if we have our active load network we can write I = xU + y , where current I is the response and voltage U represents excitation , constants x and y are determined by the resistance of the active network ,where x has dimension of A/V and y has dimension of A.

    By solving equation 1 and equation 2 I get that,
    x = 0.2 and y = -0.23,
    for the 3rd equation I get:
    Ip" = xU1 + y = 0.2 U1 - 0.23
    I have two unknowns Ip" ,U1 and only 1 equation.
    What am I doing wrong?

    Attached Files:

  2. jcsd
  3. Apr 24, 2016 #2


    User Avatar
    Homework Helper
    Gold Member

    What if you put the parallel network of R, 3R and 4R inside the active network box? You know the value of R and currents I and I'. Is it possible to find I'' using the same method you used above? I am not sure, it's just a suggestion. I'm not familiar with your method.
  4. Apr 24, 2016 #3
    It would only change constants x and y since the resistance of the active network would be changed , I would have form Ip = xI + y , where x would be non dimensional constant and y would be constant with dimension of amperes , still I would end up with 1 equation with 2 unknowns.
  5. Apr 24, 2016 #4
    Hm if my logic is right

    Answer should be for R I"p = 30 mA ?
  6. Apr 24, 2016 #5
    For example look at those simulations, imagine this network in the black box is the active network and you don't know what is in it ( active since it has active sources, if there were only resistors equation would have y = 0)

    Let the V1 be excertation and the voltage measured by the voltmeter be response V2 .
    So for the first two pictures we have
    V1 = 200 V , V2 = 14.3 V
    V1 = 300 V , V2 = 28.6 V
    V2 = xV1 + y , and solving this equation you get
    x = 0.142 and y = -14

    Now suppose you wanted to find the voltage for V1 = 400 V
    you already know constants so it's
    V2 = V1*0.142 -14 = 400 * 0.142 -14 = 42. 8 V .

    Look at the picture below , we get 42.8 V the same result.
    screen shot on windows
    That is the whole idea about this theorem as I said if there were only resistors in the black box we would have constant y = 0 , so equation would look like V2 = x*V1 ,linear homonogeous equation.
    Last edited by a moderator: Apr 14, 2017
  7. Apr 24, 2016 #6


    User Avatar

    Staff: Mentor

    Try writing Ip in terms of that switched resistance (I'll denote it R* for convenience, i.e.,
    Ip = kR* + y

    If you need to find I'' then determine I as a similar linear relationship to R*.
  8. Apr 25, 2016 #7
    70 m = x30 +y
    90 m = x40 +y

    x = 0.002 , y = 0.01

    Ip" = xR + y = 0.002 * 10 + 0.001 = 30mA , may I ask you how did you come up with this?
    I was thinking it could be since R becomes excitation and by compensation theorem

    if a current or voltage is known on the resistor it can be replaced by ideal voltage/current source and vice versa
  9. Apr 25, 2016 #8


    User Avatar

    Staff: Mentor

    The only independent variable (called "input") in this system is the switch position, i.e., the resistance R. So it seemed that you should be looking for the linear relationship between output Ip and input R*.
  10. Apr 25, 2016 #9
    By the way I think it's called proportionality theorem in english literature.
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