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HELP! Resistors Parallel/series

  1. Mar 5, 2005 #1
    Ok the question is:

    A resistor (resistance=R) is connected first in parallel and then in series with a 2.00 ohm resistor. A battery delivers five times as much current to the parallel combination as it does the series combination. Determine the two possible values for R.

    So let's see: I can use I=V/R

    For series it's I=V/(R+2)

    For parallel it's 5I=V/(1/R+1/2ohm) Ok, now I equate the Currents (I)

    and I get V/(R+2)=V/5(1/R+1/2). I can cancel out the V's and I'll have:

    Now I solve for R?

    Am I on the right track? :confused:
     
  2. jcsd
  3. Mar 5, 2005 #2
    Your parallel equation is wrong:

    [tex]R_{eq}=(\frac{1}{R_1}+\frac{1}{R_2}+\cdots+\frac{1}{R_n})^{-1}[/tex]

    So if you say [itex]I=\frac{V}{R_{eq}}[/itex] you end up with:

    [tex]I=\frac{V}{(\frac{1}{R_1}+\frac{1}{R_2}+\cdots+\frac{1}{R_n})^{-1}}[/tex]

    or more simply:

    [tex]I={V}{(\frac{1}{R_1}+\frac{1}{R_2}+\cdots+\frac{1}{R_n})}[/tex]

    Your thought process is correct though.

    Good luck.
     
  4. Mar 5, 2005 #3
    Your equation for equivalent resistnace in a parallel circuit is wrong. The correct equation is:

    [tex]\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}[/tex]

    Otherwise you seem to be on the right track.
     
  5. Mar 5, 2005 #4
    Ok this is weird. Ok so jumping to the equating of the two equation step...

    I get :

    V/(R+2)=V/5 x (1/R+1/2). Now, I cancel the Vs and try to solve for R

    1/(R+2)=1/5 x (1/R+1/2). Now I bring the 5 over on the otherside and now I will solve and get a quadratic equation:

    R^2-6R+4=0. Solving for R's I get R=5.236 and R=.7639 ohm.

    It looks right? Thanks.
     
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