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Homework Help: HOw much power is drawn from a one volt battery?

  1. Sep 12, 2010 #1
    1. The problem statement, all variables and given/known data

    each bulb in the circuit has a resistence of one ohm. how much power is drawn from the one volt battery?

    to see the circuit diagram, click on this link:

    it is problem 16 on page 123

    To see the circuit diagram, go here:
    http://www.lightandmatter.com/bk4c.pdf [Broken]


    GO to the link, scroll to page 123, and the diagram is problem 16.


    2. Relevant equations

    Ohms Law
    Loop rule
    Junction rule


    3. The attempt at a solution

    I have
    D=B+E
    A+B=C

    and using Loop rule three times:
    Va+Vc=1
    Vd+Ve=1
    Va=Vd+Vb

    but i can't go on to solve the the five equatios cause i get lost. Can you help me?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 12, 2010 #2

    Redbelly98

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    Since you know the resistances, you can rewrite these in terms of the voltages.

    For example, D = Vd/Rd, etc.

    Shouldn't that middle equation be
    Vd + Ve + Vf = 1​
    ?
     
  4. Sep 12, 2010 #3
    regarding the middle equation, i assumed e anf F to be one single bulb with a resistance of 2 ohms.

    I'm not sure how to re wrie each in relation to volatge like you say? Can you help?
     
  5. Sep 12, 2010 #4
    P=V^2/R for each.
     
  6. Sep 12, 2010 #5

    Redbelly98

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    Okay.
    Sure.
    From Ohm's law, we know the current through bulb D is
    D = Vd / Rd​
    So we can replace D in the equations with Vd/Rd instead.
    I.e., instead of
    D = B + E​
    we get
    Vd/Rd = B + E​
    Then do the same for B and E.
     
  7. Sep 12, 2010 #6
    Ok, so my new equations are:
    Vd/Rd=B+E
    Which becomes
    Vd/Rd=Bd/Rb+Ve/Re
     
  8. Sep 12, 2010 #7

    Redbelly98

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    Yes, good. Here are two more things to do:

    1. Since you know what Rd, Rb, and Re are, you might as well put those numbers into the equation.

    2. Do the same thing for the A+B=C equation.

    After doing those, try to solve for the 5 unknowns Va, Vb, etc.

    I'm logging off for the night, so good luck.
     
  9. Sep 13, 2010 #8
    Substituting Rd, Rb and Re = 1 into equation gives:
    Vd/1=Bd/1+Ve/1


    A+B=C
    becomes ?

    sorry, I'm trying!
     
  10. Sep 13, 2010 #9

    Redbelly98

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    Okay, let's work some more with that equation before moving on.

    Doesn't Re=2, since you have combined e and f into one single bulb with a resistance of 2 ohms?

    What is "Bd"? I think you meant to write Vb there.

    Also, a simpler way to write "Vd/1" would be ____?
     
  11. Sep 13, 2010 #10
    Vd^-1
     
  12. Sep 13, 2010 #11

    Redbelly98

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    What? No. Look at these examples:

    [tex]\frac{1}{1} = 1[/tex]

    [tex]\frac{2}{1} = 2[/tex]

    [tex]\frac{3}{1} = 3[/tex]​

    You're just dividing each number by 1, so you get the same number.

    [tex]\frac{Vd}{1} \ = \ ?[/tex]​
     
  13. Sep 13, 2010 #12
    sorry Vd/1 is the same as Vd
     
  14. Sep 13, 2010 #13
    Vd=Vb+Ve
     
  15. Sep 13, 2010 #14

    Redbelly98

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    Almost. Just one little error to address:
     
  16. Sep 13, 2010 #15
    Vd=Vb+2Ve
     
  17. Sep 13, 2010 #16

    Redbelly98

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    No, try again. Look at what you wrote in post #6.

    I'm logging off for the night but will be back online tomorrow.
     
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