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5 resistor circuit

  1. Jun 5, 2012 #1
    1. The problem statement, all variables and given/known data
    Five resistors with resistance R1 = 4.00 Ω, R2 = 6.00 Ω, R3 = 2.00 Ω, R4 = 7.00 Ω, R5 = 4.00 Ω, are connected to a battery with emf = 8.00 Volts as shown in the figure. Find the equivalent resistance of the circuit. Give you answer with a unit of "Ohm".


    2. Relevant equations
    parallel: Req = 1/((1/R1)+(1/R2)+....)
    series: Req = R1 + R2 + R3 +....


    3. The attempt at a solution
    well im thinking that R1 and R3 is in series and R2 and R4 is in series but from there im not sure what to do. the R5 gets me confused
     

    Attached Files:

  2. jcsd
  3. Jun 5, 2012 #2

    ehild

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    Neither R1 and R3 nor R2 and R4 are in series. Do you know what series connection means?

    The resultant resistance seen by the battery is the voltage of the battery divided by the current through the battery. To find the current, apply Kirchhoff's laws.

    ehild
     
    Last edited: Jun 5, 2012
  4. Jun 5, 2012 #3
    Use wheat stone bridge idea to solve this.
     
  5. Jun 5, 2012 #4
    i dont really know it ..

    but uhhh oh .. im not sure if i can apply them here it seems too complicated my mind just gets all frazzled. can you give me one eqn to work off?
     
  6. Jun 5, 2012 #5

    ehild

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    Do you need the equivalent resistance between points P and P'?

    ehild
     
    Last edited: Jun 5, 2012
  7. Jun 5, 2012 #6
    uhhh i guess? idk! im confused already. isn't that just R5 though?
     
  8. Jun 5, 2012 #7

    ehild

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    No, the other resistances are also connected to those points. Do you really need the equivalent resistance between P and P'? Show the original text of the problem, please.

    ehild
     
  9. Jun 5, 2012 #8
    oh yea true .. well then i guess not .. but that is the original text, all i have in the original post is the whole problem
     
  10. Jun 5, 2012 #9

    ehild

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    Equivalent resistance refers to two points. If you do not know those points I can not help.
    Have you learnt something about Wheatstone Bridge? Look after in your textbook. Is there anything about it?

    ehild
     
  11. Jun 5, 2012 #10
    nothing in my textbook about it .. professor briefly went over it today in class. theres like 4 formulas that he got.
    I1R1=Va-Vb
    I1R1=Vb-Vc
    I2R2=Va-Vd
    I2R=Vd-Vc

    where c is the left corner of the 'diamond' , a is the right corner, b is P in our case, d is P' in our case .. R is our R3 and R3 in his example is my R4


    can't you just do from the left corner of the diamond that we'll call c like he did all the way to the right corner which is a?
     
  12. Jun 5, 2012 #11
    Those can't be right. Your professor might have done something like the following,
    but you seem to have made some errors in copying them.

    Assuming that the equivalent resistance is wanted between the points where the battery is connected.

    There are 2 unknown voltages, in the points P and P'. The points connected to the battery are at 8V and 0V.

    If you know these unknown voltages, you can compute all the currents in the 5 resistances, The current between two points is the potential difference between
    those two points divided by the resistance between them.
    for example, the point connected to the battery is at 8V and the point P is
    at V_P, so the potential differenct between them is (8-V_P) and the current
    through it: I_1 = (8 - V_P)/R_1

    When you've computed all 5 currents as a funtion of V_P and V_P', you can than use Kirchhoffs current law in the points P and P' to get 2 simultaneous equations for V_P and V_P', wich you can solve.
     
  13. Jun 5, 2012 #12
    ok well I1 is easy... i assume I3 is the same thing? I_3=(8-V_P')/R3 right?
    so when you want to do R2 and R4 how do you do it .. is it like I_2=(V_P-8)/R2?
    and then the same thing for R4 .. I_4=(V_P'-8)/R4 ?
    if thats the case then I_5=(V_P-V_P')/R5 ?


    and then the current laws:

    for P: I1-I2-I5=0
    for P': I3+I5-I4=0

    this is all right?
     
  14. Jun 6, 2012 #13

    ehild

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    Everything is right except the equation for I4. Check the PD across R4.

    ehild
     
  15. Jun 6, 2012 #14
    ummm I_4=(V_P'-V_P+8)/R4? since the potential difference at the right corner just past R2 and R4 is V_P-8? no i dont think thats right .. why isn't it what I had? what's wrong with the 8?
     
    Last edited: Jun 6, 2012
  16. Jun 6, 2012 #15
    i don't understand why I4 is not right. Isn't it similar to I2?
     
  17. Jun 6, 2012 #16
    I2 passes 2 resistors
    I4 passes 4 resistors
     
  18. Jun 6, 2012 #17
    3? well i dont get what the equation would be still with that information :cry:
    I_4=(V_P-V_P'-8)/R4 so is this wrong then too?
     
    Last edited: Jun 6, 2012
  19. Jun 6, 2012 #18
    First equation/circuit remove R3, R4 and R5. You just got R1 and R2. Noted the I11 and I2.
    Second equation/circuit, remove R2. Noted the I12 I3,I4 &I5

    Combine them and you get the current and voltage on each resistor.
     
  20. Jun 6, 2012 #19
    what youre saying makes no sense. well i got the first part, but the second part made absolutely no sense to me? i don't understand how to get the I4 equation i dont understand the potential difference across R4
     
  21. Jun 6, 2012 #20
    would the potential difference across R4 be -V_P?
    if across R3 it is (8-V_P) and across R5 it is (V_P-V_P') then [(8-V_P)-(V_P-V_P')] is the 'initial' and the final is 8. so [(8-V_P)-(V_P-V_P')]-8 is just -V_P?
    or is it I4=(V_P'-2*V_P)/R4
     
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