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Calculating Resistance In Circuit

  1. Mar 17, 2008 #1
    1. The problem statement, all variables and given/known data

    A loop circuit has a resistance of R and a current of 1.8A. The current is reduced to 1.3A when an additional 2.7 ohm resistor is added in series with R.

    What is the value of R?

    2. Relevant equations

    I is constant throughout a series circuit
    V is shared in a series
    Total resistance=the sum of the resistors in series

    V=I x R

    3. The attempt at a solution
    I substituted R (in the equation V=IR) with 4.7+R because this would represent the total resistance in the series. I knew that I was 1.3A. I calculated the Voltage on the 4.7 resistor (using V=IR) and got 6V. So then I substitued V total with 6 + V.

    I ended up with the following equation. 6+V=1.3(4.7+R). I then tried substituting V (of the unknown resistor) with 1.3R, but then I realized that couldn't possibly work, because the R's would cancel. Then I tried 1.8R instead so I would end up with the equation .5R=.11, R=.22 but that doesn't sound right.
    Last edited: Mar 17, 2008
  2. jcsd
  3. Mar 17, 2008 #2
    You don't have to figure out V. In the first case, you have [tex]\frac{V}{R}=1.8[/tex]

    And in the second case, you have [tex]\frac{V}{2.7+R}=1.3[/tex]

    Two unknowns and two equations.
    Last edited: Mar 17, 2008
  4. Mar 17, 2008 #3
    Okay. I got an answer of 7.02 using those equations, but it was incorrect.
  5. Mar 17, 2008 #4
    Nevermind, I see where the mistake was. I used 2.7, it's supposed to be 4.7
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