Simple DC circuit with battery and resistors

Discussion in 'Homework Help' started by epsilonjon, May 30, 2011.

  1. epsilonjon

    Thread Starter Member

    Feb 15, 2011
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    Question:

    "The resistance R3 in the diagram is the equivalent resistance of a pressure transducer. This resistance is specified to be 200Ω ± 5%. The voltage source is a 12V ± 1% source capable of supplying 5W. Design this circuit, using 5%, 1/8W resistors for R1 and R2, so that the voltage across R3 is Vo=4V±10%."

    [​IMG]

    Attempt:

    I'm ignoring the error margins for the moment and just working with the specified values.

    The voltage drop accross R2 must be 8V, therefore I*R2=8. Each resistor can only dissipate 1/8W, so

    \frac{8^{2}}{R_{2}}\leq\frac{1}{8}

    R_{2}\geq512.

    The equivalent resistance of the circuit is

    R_{T}=R_{2}+\frac{R_{1}R_{3}}{R_{1}+R_{3}} = R_{2} + \frac{200R_{1}}{R_{1}+200}.

    Therefore

    12=IR_{T}

    12=\frac{8}{R_{2}}(R_{2}+\frac{200R_{1}}{R_{1}+200})

    12R_{2}=8R_{2}+\frac{1600R_{1}}{R_{1}+200}

    R_{2}=\frac{400R_{1}}{R_{1}+200}

    R_{1}+200=\frac{400R_{1}}{R_{2}}.

    So if R2>512Ω (required by the power restrictions), you are bound to require a negative value for R1. Am I doing something silly here, or does the question not work?

    Thanks for any help! :)

    Jon.
     
    Last edited: May 30, 2011
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Your equations looks good.
     
  3. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    For R3= 210Ω and Vo = 3.6V and 4.4V for no load
    We have Ith = 3.6V/210Ω = 17.14mA
    and Rth = 0.8V/17.14mA = 47Ω
    Rth = R1||R2 = (R1*R2) / (R1+R2) = 47Ω
    Vth = 12V * R1/(R1+R2) = 4.4V
    R2/R1 = 1.72
    R2 = 128Ω and R1 = 74Ω
    But then the power is to big.

    So i really don't know how we can meet all those requirements.
     
  4. epsilonjon

    Thread Starter Member

    Feb 15, 2011
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    Thanks for the swift reply. :)

    I'm glad it's not me misunderstanding things, but rather the question which is poorly conceived. I spent £50 on this textbook too! :mad:
     
  5. Jony130

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    Feb 17, 2009
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    What is the name of the text book?
     
  6. epsilonjon

    Thread Starter Member

    Feb 15, 2011
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    Introduction to Electric Circuits by Richard C. Dorf and James A. Svoboda.
     
  7. jegues

    Well-Known Member

    Sep 13, 2010
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    That text has alot of goofy problems in it.

    The answers to the problems are usually horrid. (Really retarded numbers for what could be straightforward and simple computation)
     
  8. TBayBoy

    Member

    May 25, 2011
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    you should be able to look up how the answer was derived here

    <SNIP>

    just put in the chapter and problem number.
     
    Last edited by a moderator: Jun 1, 2011
  9. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    The answer to this problem is to using two 200Ω resistors connected in series instead of R2. And R1 treat as a open circuit.
     
  10. epsilonjon

    Thread Starter Member

    Feb 15, 2011
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    Unfortunately you have to sign up, and it costs money per month :-(
     
    Last edited by a moderator: Jun 1, 2011
  11. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    I saw the solution. Read my previous post.
     
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  12. epsilonjon

    Thread Starter Member

    Feb 15, 2011
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    Ah I see now, thanks. Kind of a strange solution, but I guess it is trying to get you to think 'outside the box' or whatever. Unfortunately I just saw it as 'the question is wrong' :p.

    Thanks!
     
  13. Kingsparks

    Member

    May 17, 2011
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    TBayBoy.

    In reference to your signature; "Have your Tribble spayled or neutered."

    I don't have to worry about it, I think the Bandersnach got it under the yum yum tree. (Or was it the tum tum tree.) English Lit was a loooong time ago. :rolleyes:
     
    TBayBoy likes this.
  14. dcarson7

    New Member

    Apr 28, 2011
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    http://www.pearsonhighered.com/floyd/

    Some good textbooks there.

    Step by step explanations, questions and answers at the back of the book.

    Still using my earlier versions, some 8 years on.
     
  15. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    I may be mistaken, but isn't 1/8 of a watt 0.125 and NOT 0.16?

    The constraint was using E24 one-eight watt resistors to meet the criteria of 4 Volts output +/- 10%

    It's still a BS question.
     
    Last edited: Jun 4, 2011
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