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Electrodynamics problem (circuit with two sources)

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

    In the circuit (see attachment) we have 2 sources. If we decrease the EMF of the first source (ε1) by 1.5V, the current changes in every branch of the circuit. How does the EMF of the source ε2 has to be changed, so that the current through the ε1 source would be the same as previously?

    2. Relevant equations

    Kirchhoff's laws

    3. The attempt at a solution

    I guess what we need to do here is to find the dependency of ε1 on ε2. Is using the Kirchhoff's equations the best way here? I wrote down such eq:
    [itex]
    I_3R + I_1r_1 + I_5R = \epsilon_1
    [/itex]
    [itex]
    3I_4R + I_6R + I_1r_1 = \epsilon_1
    [/itex]
    [itex]
    I_2r_2 + I_6R - I_5R = \epsilon_2
    [/itex]
    [itex]
    I_3 + I_4 = I_1
    [/itex]
    [itex]
    I_5+I_6 = I_1
    [/itex]
    [itex]
    I_3 = I_5 + I_2
    [/itex]
    [itex]
    I_4 + I_2 = I_6
    [/itex]
    Here r1, r2 - internal resistances of the sources (since there is current flowing through the sources, they must have internal resistances, right?)
    I3 - top left resistor
    I5 - down left resistor
    I4 - top right resistor
    I6 - down right resistor
    I1 - ε1 current
    I2 - ε2 current
    But couldn't solve the system :/
    Maybe there is a better way to do this? It seems to me there should be something simpler than system of 8 equations?
    Any help appreciated!
     

    Attached Files:

  2. jcsd
  3. Jun 1, 2014 #2
    Assume that r1 and r2 are negligible compared to R unless otherwise stated and can be removed from the equations.
     
  4. Jun 1, 2014 #3
    But then I1 and I2 are removed, too, and (I made a mistake in the first post here) it is not possible to find the dependency of I1 on the other parameters (like ε1, ε2)? Or am I wrong and we are looking for something else?
     
  5. Jun 1, 2014 #4
    No, I1 and I2 are not removed. They show up in the junction equations.
     
    Last edited: Jun 1, 2014
  6. Jun 2, 2014 #5
    Yes, thanks a lot, that was it!
     
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