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EE - Effective Resistance

  1. Oct 20, 2009 #1
    EE -- Effective Resistance

    1. The problem statement, all variables and given/known data
    I don't know how to post schematics so I hope this makes sense. There are four resistors in the circuit. An 18 Ohm resistor is the first resistor out of the 5V voltage source followed by three resistors hooked up in parallel. They have a resistance of 100ohms 25ohms and 30ohms. Find the equalization resistance.



    2. Relevant equations
    Hey guys this is my first post to bear with me while I figure things out. And thanks for any help in advance. So I'm having trouble calculating the effective resistance across circuits and I can't figure out why! The math seems to easy but I seem to be getting them wrong. Could someone explain to me what I am doing wrong.

    for 2 resistors in parallel... Req= R1*R2/(R1+R2) (will yield a resistance less than the lower resistor)
    for more than 2 resistors in parallel... 1/Req = 1/R1 + 1/R2 + 1/R3... (will yield a resistance less than the lower resistor)
    for resistors in series... Req = R1+R2 (will yield a resistance greater than the largest resistors)

    3. The attempt at a solution
    So I started by find the Req for the parallel resistors. 1/Req = 1/r1 + 1/r2 + 1/r3 = (1/100) + (1/25) + 1/(30) which I got meant Req = 12ohms. I then noticed the first resistor and the combined resistors are in series. So I added R1 + R2 = Req = 30ohms. The back of the book says the answer is 6ohms =/ is there something I need to be doing with the voltage? I'm just so confused, simple math and I'm coming up short every time. If you guys need a diagram please let me know how and I will post one. Thanks again
     
    Last edited: Oct 20, 2009
  2. jcsd
  3. Oct 20, 2009 #2

    The Electrician

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    Re: EE -- Effective Resistance

    Take a picture of the schematic diagram and post it here. Use the "manage attachments" below the reply window.
     
  4. Oct 20, 2009 #3
    Re: EE -- Effective Resistance

    Hope this works
     

    Attached Files:

  5. Oct 20, 2009 #4
    Re: EE -- Effective Resistance

    Guess I should explain how I simplified the whole circuit now.

    I combined the 10ohm and 40ohm (in parallel) ... (10)(40)/50 = 8ohm
    I combined the 8ohm and the 22ohm (in series)... 8+22 = 30ohm
    I combined the 100ohm the 25ohm and the 30ohm (in parallel)... (1/100)+(1/25)+(1/30)=12ohms
    I combined the 12ohm and the 18ohm(in series)... 12+18 = 30ohm
     
  6. Oct 20, 2009 #5

    The Electrician

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    Re: EE -- Effective Resistance

    You'll notice the image says "pending approval". It can't be viewed until one of the moderators notices it and approves it.

    If you upload an image to one of the free image hosting sites, and post the link here, you can get a quicker response.
     
  7. Oct 20, 2009 #6
    Re: EE -- Effective Resistance

    http://img30.imageshack.us/img30/3494/circuir.jpg [Broken] there that should work I think, thanks for the advice
     
    Last edited by a moderator: May 4, 2017
  8. Oct 20, 2009 #7

    The Electrician

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    Re: EE -- Effective Resistance

    Your calculations are correct. Textbooks are known to make mistakes.

    You will note that if the source is 5 volts, loaded by 30 ohms, the current will be 1/6 amps.

    That's the only way I see to come with a "6" in this problem.
     
  9. Oct 20, 2009 #8
    Re: EE -- Effective Resistance

    The problem stated excatly is "Find the equivalent resistance seen by the source in the circuit." Does everything still appear to be the same. A kid in my class said that he was able to match the answers out of the book so I still think I'm missing something =/
     
  10. Oct 20, 2009 #9

    berkeman

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    Re: EE -- Effective Resistance

    Please post your attachment in PDF format next time. Fewer worries about macros and such. You can get a PDF writer for free from PrimoPDF if you don't have one already.
     
  11. Oct 20, 2009 #10

    The Electrician

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    Re: EE -- Effective Resistance

    Look at the circuit. There's an 18 ohm resistor in series with the source, first thing. There's no way you can do anything after that point (right end of the 18 ohm resistor) to create less than 18 ohms as seen by the source, if you can only add various combinations of real resistors.

    You could do so if you allowed negative resistors, but I don't see that happening here.

    Could you be looking at the wrong schematic?

    You better ask the other guy for some details of what he did.

    Please come back here and tell us what the problem was when you find out.
     
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