Circuit Question

  • Thread starter ianb
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  • #1
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Homework Statement



http://img514.imageshack.us/img514/1223/questionju1.jpg [Broken]

Homework Equations



Parallell: [tex]\frac{1}{R_{eq}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}[/tex]

Series: [tex]R_{eq} = R_{1} + R_{2}[/tex]

The Attempt at a Solution



Part 1 - When the switch is not closed:

[tex]R_{eq_2} = R + ((90+10)^{-1} + (90+10)^{-1})^{-1}[/tex]
[tex]R_{eq_1} = R + 50[/tex]

Part 2 - When the switch is closed:

[tex]R_{eq_2} = R + (90^{-1} + 10^{-1})^{-1} + (90^{-1} + 10^{-1})^{-1}[/tex]
[tex]R_{eq_2} = R + 18[/tex]

Since [tex]R_{eq_1} = 2R_{eq_2}[/tex], then

[tex]R + 50 = 2(R + 18)[/tex]
[tex]R + 50 = 2R + 36[/tex]
[tex]R = 14[/tex]

The answer in the back, though, is 24. What have I done wrong?
 
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Answers and Replies

  • #2
1,860
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Good question, it looks to me like you did it right. Are you sure you didn't make a numerical error (this part looks okay too, though I didn't actually calculate)? How trustworthy are the answers in your book?
 
  • #3
17
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Thanks, mindscrape. I suppose the book could have misprinted the answer. One more thing, though, if you don't mind:

When the switch is closed, I calculated two parallel circuits. Are these two circuits, in turn, parallel to each other, or in series to each other (like I did above)?
 
Last edited:
  • #4
23
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When the switch closes, it creates a node between the four resistors. The resulting parallel connections will be in series with each other - the way you calculated it. :smile:
 

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