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Urgent help! Circuit problem for Physics.

  1. Mar 30, 2015 #1
    I need to solve for the equivalent resistance of the circuit below:https://s.yimg.com/hd/answers/i/7936cef74d9a45cfb7c9f051c19041dd_A.jpeg?a=answers&mr=0&x=1427780127&s=19cab3f26be71f84b6bb2d7d291a5217 [Broken]
    Voltage is 100volts. R1=6ohm, R2=8 ohm, R3= 4 ohm, R4=8 ohm, R5= 4 ohm,R6= 6ohm, R7=ohm R8=10 ohm, R9=6 ohm, R10= 2 ohm. Thanks :)
    2. Relevant equations
    V=IR
    V=IP
    P=I^2R

    SERIES:
    VTOT=V1+V2
    ITOT=I1=I2
    RTOT=R1+R2
    PARALLEL:
    VTOT=V1=V2
    ITOT=I1+I2
    REQ=(1/R+1/R2)^-1


    3. The attempt at a solution
    I believe R1 and R2 are in a series. So adding them up would mean R1+2=14ohm We know the voltage is 100volts so I did 100/14=I I=7.14 amps then I believe we could find the voltage of R1 and R2. Voltage of R1= 6x7.14=42.85V Voltage of R2=8x7.14=57.12V Knowing that VTOT=V1+V2, 57.12+42.85=pretty close to 100V.
    Then the triangle circuits just make me go what... I believe they're in a series? Not sure.
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Mar 30, 2015 #2

    SammyS

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    R1 and R2 are not in series. Not all the current that passes through R1 passes through R2.

    Start at the far end. Look at R8 and R10 . Then toss in R9 . ...
     
    Last edited by a moderator: May 7, 2017
  4. Mar 30, 2015 #3
    Ok, so I believe R8 and R10 are in a series. Or am I supposed to do R8, R10, and R9 together? Anyways, I did R8 and R10 first making it become 12 ohm and after did 100/12=8.3 amps. Now finding the voltage of each I did Voltage of r8=83.3(8.3x10) and then r10=16.67ohm. Not sure what you mean by tossing in R9. Idk if you could do it singularly but I guess you do V=IR Current is same throughout in series so 8.3x6=50V? Is this right or not?
     
  5. Mar 30, 2015 #4

    SammyS

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    A step at a time. It's way too soon to assign voltage & current.

    Yes, R8 & R10 are in series.

    The voltage drop (still unknown) across them is the same as across R9 so How is R9 connected to them ?
     
  6. Mar 30, 2015 #5
    So is R9 parallel to them? Since the voltages are same in a parallel circuit?
     
  7. Mar 31, 2015 #6

    SammyS

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    Correct. So that's 6Ω in parallel with 12Ω.

    That combination is in series with R7 .

    ... etc.
     
  8. Mar 31, 2015 #7
    Ok. Btw the directions say I need to include voltage and current for each resistor so thats why I am finding those too. So I'm now guessing that you mean do r8 and r10 as a series then make it parallel to r9. So assuming that, R9's voltage is 100V? Since the voltage across R8 and R10 is 100V. But you said its unknown still so not sure. REQ=4ohms=(1/6+1/12)^-1 So then we have to find current which is V/r=I 100/4=25amps I am stopping here until you confirm what I am doing is correct. Actually, I'll just keep going. R6 and R7 (1/14)+(1/4) = 3.1 ohms Then becomes parallel with R3 and R4. (1/12+1/3.1)^-1=2.46 ohms becomes parallel with r2 and r5 (1/2.46+1/12)^-1=2.04 Then only R1 is left and I think becomes a series? So its 8.04=REQ?
     
    Last edited: Mar 31, 2015
  9. Mar 31, 2015 #8

    SammyS

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    You are correct regarding the equivalent resistance of R9 combined with R8 and R10; it's 4Ω. The rest is incorrect.

    So replace (at least mentally) the combination of those three with a 4Ω resistor. Any current through that combination must also pass through what resistor.

    Again, it's premature to assign a voltage or current to any of these resistors. There will be far less than 100 volts available to the resistors at the far right end of this circuit. *

    What sort of circuit analysis have you done so far in your course? There are other methods for solving this problem, but they're more sophisticated either from a physics point of view or from an algebra point of view. You seemed to be familiar with series/parallel analysis. That is a valid method here.

    * (For Example)
    Suppose the equivalent resistance for all ten resistors is 8Ω . The current supplied to the circuit is then 12.5A, which is also the current through R1 and also the current through that big combination of resistors R2 through R10. That gives a voltage drop of 75 Volts across R1 and 25 Volts across the combination of the other nine resistors.

    R2 being in parallel with the combination of the other eight R's has a voltage drop of 25 Volts across it thus a current of 25/8 Amps through it, the rest of the current (9.375 Amps) passing through the remaining eight resistors. ... etc.​
     
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