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Homework Help: Equivalent resistance & current

  1. Mar 22, 2004 #1
    ok, so the problem states: the current in a loop circuit that has a resistance of R1 is 3A. The current is reduced to 1.6A when an additional resistor R2=2ohms is added in series with R1. What is the value of R1?

    I drew two diagrams. My logic was that the voltage would be the same, b/c the voltage is supplied by the battery which does not change. Thus you have V=I1R1 for the first scenario; and you have V=I2(R1+R2). Since V is the same, you can set the two equations equal to each other, and solve for R1.

    Is the logic right so far? If so, maybe I just have an algebratic error...

    (3A)R1=(1.6A)(R1+2ohms)
    (3/1.6)=(R+2)/R
    1.875=1+R/2
    .875=R/2
    R=1.75 ohms???
     
  2. jcsd
  3. Mar 22, 2004 #2
    That is almost correct, you made a mistake here:
    [tex]\frac{3}{1.6} = \frac{R+2}{R}[/tex]
    Needs to become:
    [tex]1.875 = 1 + \frac{2}{R}[/tex]
     
  4. Mar 22, 2004 #3

    ShawnD

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    Science Advisor

    I get R1 = 2.285 from breaking that down 2 different ways.

    [tex](3)(R_1) = (1.6)(R_1 + 2)[/tex]

    [tex]3R_1 = 1.6R_1 + 3.2[/tex]

    [tex]1.4R_1 = 3.2[/tex]

    [tex]R_1 = 2.285[/tex]



    [tex](3)(R_1) = (1.6)(R_1 + 2)[/tex]

    [tex]\frac{3}{1.6} = \frac{R_1 + 2}{R_1}[/tex]

    [tex]1.875 = 1 + \frac{2}{R_1}[/tex]

    [tex]0.875 = \frac{2}{R_1}[/tex]

    [tex]R_1 = \frac{2}{0.875}[/tex]

    [tex]R_1 = 2.285[/tex]
     
  5. Mar 22, 2004 #4
    thank you! thank you! thank you!
     
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