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Proof by Circuit analysis

  1. Oct 18, 2015 #1
    1. The problem statement, all variables and given/known data
    I have to prove that RI = RSRP/(RS - RP) by circuit analysis.
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    3. The attempt at a solution

    So I tried to create a system of equations to do this:

    For the first circuit using kirchoffs rules:
    V-IA(RS + RI) = 0 [1]

    For the second circuit:

    V-IBRS - IPRP = 0 [2]

    V-IBRS - (IA)/2 * RI = 0 (It's IA/2 because of kirchoff's rules regarding the junction therefore it's equivalent to the first circuit's I divided by 2) [3]

    IB = IP + IA/2 [4]

    First I solved for V for first 3 equations. Then for equation 4, i solved for IA/2 to make it easier for me, then I substituted it and tried to eliminate as much variables as I can by equating [1] and [2]. then whatever result I get for that, I then equate it to [3]. After doing this, I get
    IB(2Rs + RI) = IP(2Rs + RI -RP)

    Solving this using my head, I know I wont be able to prove the Ri equation. Any tips or other methods to do this?
     
  2. jcsd
  3. Oct 18, 2015 #2

    gneill

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    Staff: Mentor

    Are there any particular circuit conditions under which this expression is meant to hold? The problem statement doesn't mention anything, but to me these are just two completely independent circuits with variable resistors. Why variable?
     
  4. Oct 18, 2015 #3
    I'm not really sure what you mean but this is from a lab and for the series circuit, full scale deflection of the meter is required. The parallel circuit, one-half scale deflection of the meter was required, which is why the parallel resistor was added.I originally used the parallel circuit to try to prove the Ri equation but I ended up with:
    RI=RSRP/(RP - RS) which is wrong.
    So I asked my professor, and this [the attempted solution: using both circuits] is what he hinted me to do and I just can't seem to solve it.
     
  5. Oct 18, 2015 #4

    gneill

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    Staff: Mentor

    Well, that's context that is important! So somehow you have to incorporate the meter conditions into the analysis. What does half deflection versus full deflection imply? (think: current)

    EDIT: Or think voltage across the meter. Either way works.
     
    Last edited: Oct 18, 2015
  6. Oct 19, 2015 #5
    I 'm sort of getting it now. Since the first circuit is in series, lowering the resistance increased the deflection. Using Ohm's law, I =V/R, the meter indicated more current was able to flow through. For the 2nd circuit, using kirchoff's law, increasing the resistance by a little bit, decreased the deflection by signifcantly since it was in parallel. therefore the current in the meter decreased.

    I still can't think of a way to incorporate that into the analysis...Right now my mind is heading towards something about manipulating the current but I don't know how to do it in terms of the equations
     
  7. Oct 19, 2015 #6

    gneill

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    Staff: Mentor

    Start by writing expressions for the current through the meter for both circuits.
     
  8. Oct 20, 2015 #7

    rude man

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    Gold Member

    The only thing making sense to me is:
    1. using circuit 1, record Rs1 and il.
    2.Using circuit 2, adjust Rs2 and Rp for same iL as for ckt. 1. Record Rs2 and Rp.
    Then you can express Rl in terms of Rs1, Rs2 and Rp. Rs2 will be < Rs1.
     
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