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Archived Superposition to solve circuit with dependent source

  1. Feb 24, 2014 #1
    1. The problem statement, all variables and given/known data

    Assume that Vs = 10V , R1 = 1.7Ω , and R2 = 0.50Ω. Find the current I in the figure using the superposition principle.

    Steif.ch03.p69.jpg

    2. Relevant equations

    Superposition principle
    KCL, KVL
    V=IR
    Current and voltage dividers

    3. The attempt at a solution

    I = I' + I''

    First replacing the current source with an open circuit, I have:

    (1.7+0.50)*I' = 10 - 2I' using KVL
    Solving, I' = 2.38 A

    Now replacing the voltage source with a short, I use the current divider to get:

    -I'' = 3*0.5/(1.7+0.5) = 0.682 A
    I'' = -0.682 A

    I = 2.38 - 0.682 = 1.7 A, which is incorrect.

    EDIT: Should I have left the dependent source as 2I and solved for I in the final equation, or am I doing something else wrong?

    EDIT2: Yes, I did have to leave it as 2I. Got it :)
     
    Last edited: Feb 25, 2014
  2. jcsd
  3. Feb 5, 2016 #2

    gneill

    User Avatar

    Staff: Mentor

    A complete solution to wrap things up.

    Using superposition:
    1) Suppressing the current source:

    We are left with a single series circuit (blue dotted components are suppressed/removed):

    upload_2016-2-5_14-46-5.png

    Writing Ohm's Law for the circuit we have
    ## I' = \frac{V_s - 2 I'}{R_1 + R_2} ##

    ## I' = \frac{V_s}{R1 + R2 + 2}~~ ## after solving for I'

    ## I' = 2.381~Amps##

    2) Suppressing the voltage source leaves us with:

    upload_2016-2-5_14-45-39.png

    Let v be the voltage at the junction of the resistors. Then writing a node equation there:

    ##\frac{v}{R_1} - I_s + \frac{v - 2 I''}{R_2} = 0##

    But Ohm's Law also gives us ##v = -I'' R_1##. So substituting for v in the node equation:

    ## -I - I_s + \frac{(-I'' R_1 - 2 I'')}{R_2} = 0##

    ##I'' = -I_s \frac{R_2}{R_1 + R_2 + 2} ##

    ##I'' = -0.357~Amps##

    3) Combining the results:
    ##I = I' + I'' = 2.024~Amps##

    Quick Check:
    ##V_s - I R_1 = (I + I_s) R_2 + 2 I##

    ##I = \frac{V_s - I_s R_2}{R1 + R_2 + 2} = 2.024~ Amps##
     

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