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Solving a linear circuit #2

  1. Jan 19, 2017 #1
    1. The problem statement, all variables and given/known data
    This is the second problem from our test preparation cycle and i partially solved and understood it.
    IMG_2330.JPG
    The circuit in the picture is in the state of switch being open (STATE 1: SWITCH OPEN). Calculate ##E1## so that the change in the electrical energy of the capacitor is ##ΔW=4uJ##.
    2. Relevant equations
    3. The attempt at a solution

    ##ΔW=1/2ΔU^2_{1/2}C=4uJ##
    ##ΔU^2_{1/2}=\pm 2V## [there are two values and we cant determine which is correct so we work both]
    ##ΔU=ΔU_{ab}## [that voltage is the same for points A and B]
    IMG_2331.JPG
    Starting to simplify the circuit and added current generator to represent the change in that branch of the circuit:
    ##Δ=R1+R2+R3## [starting to transfer a triangle into a star]
    ##RA=\frac{R1R2}{Δ}##;##RB=\frac{R1R3}{Δ}##;##RC=\frac{R2R3}{Δ}## [finished with that]
    ##ΔU_{ab}=a*ΔIGK## [figured its a superposition principle where we find Uab if only IGK works]
    ##ΔIGK=\pm 2/55 mA## [immediate jump to the result, didnt quite get that, how did they get that? im gonna try to derive it myself] [dont get #1]
    TRY:
    ##I_{branch}=ΔIGK*\frac{RB+R5}{RA+RB+R4+R5}## [tried the current divider to get the current in left part]
    ##U_{ab}=I_{branch}*(R4+RA)+ΔIGK*RC## [didnt check the results but i will, does this seem good though?]
    Continuation of the solution:
    IMG_2332.JPG
    ##ET=E\mp IR=\mp 2V## [so the theorem of compensation, putting the voltage source instead of the whole branch]
    ##I=ΔIGK## [the change that happens is that the current flows through the branch]
    IMG_2333.JPG
    ##ET'=E1/8## [now this all of a sudden, how did they get to this?] [dont get #2]
    ##RT'=55## [the equivalent thevenin resistance i get this but the upper part i cant get still]
    ##ET'-ET+RT'I=0## [why +RT'I and not -RT'I?] [dont get #3]
    ##ET'=0 \Rightarrow E1=0##
    ##ET'0=4 \Rightarrow E1=32##
     
  2. jcsd
  3. Jan 20, 2017 #2

    haruspex

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    I'm not happy with your star resistance diagram in the second image.
    If I label the left side of R2 as X and the right side of R3 as Y, for your equations to be right, RA connects to point A, RB connects to point Y, and RC connects to point X. The way you have drawn it suggests RA and RC swapped around.
     
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