1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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...

    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


    User Avatar
    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!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook