Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Equivalent Resistance Question

  1. Feb 25, 2006 #1
    Find the equivalent resistance between points A and B for the group of resistors shown in Figure 21-29, where R1 = 93 and R2 = 40 .

    [​IMG]

    ok i know that [tex]R_{1}[/tex] and 35[tex]\Omega[/tex] are in series and [tex]R_{2}[/tex] is in parallel to the top 2 resistors but I am obviously making errors either in how i set up the problem or in my calculations. I have tried (1/[tex]R_{2}[/tex]) + (1/2R) = 3/2R (as shown in my book) and then invert to show 2/3(R) and the answer i'm getting is wrong.. please help :confused:
     
  2. jcsd
  3. Feb 25, 2006 #2

    Physics Monkey

    User Avatar
    Science Advisor
    Homework Helper

    You should first combine the two resistors in series, and then combine the result with the other resistor in parallel. What are the laws of combination for resistors in series? What about parallel?
     
  4. Feb 25, 2006 #3
    for series: [tex]R_{eq} = R_{1}+R_{2}+R_{3}...[/tex]
    for parallel: [tex]1/R_{eq}= 1/R_{1}+1/R_{2}+1/R_{3}...[/tex]

    so you're saying: 93[tex]\Omega[/tex]+35[tex]\Omega[/tex]+(1/40[tex]\Omega[/tex]) OR (1/93 +35) + (1/40) ???
     
    Last edited: Feb 25, 2006
  5. Feb 25, 2006 #4
    nevermind i'm an idiot.. i was forgetting to take the inverse of the answer :redface: :redface: :redface: :redface: :redface: :redface:
     
  6. Feb 25, 2006 #5

    Physics Monkey

    User Avatar
    Science Advisor
    Homework Helper

    Clearly the first statement can't be right since you're adding things that have different units. The second looks ok as long as you mean 1/(93+35). Like I said, first combine the two resistors in series. Take that result and combine it with the other resistor in parallel.
     
  7. Feb 25, 2006 #6
    thanks for the help
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Similar Discussions: Equivalent Resistance Question
Loading...