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!

4 resistors in series and parallel

  1. Apr 18, 2009 #1
    1. The problem statement, all variables and given/known data

    I have 4 resistors R1=R2=R3=R4=40[tex]\Omega[/tex]

    I need to find the Equivalent Resistance when they are connected in series and when they are connected in parallel.

    The problem is my answer for the parallel differs from the answer given in the book.

    2. Relevant equations

    Series: R=R1+R2+R3+R4
    Parallel: R=[tex]\frac{R_{1}*R_{2}*R_{3}*R_{4}}{R_{1}+R_{2}+R_{3}+R_{4}}[/tex]


    3. The attempt at a solution

    The first is easy: 4*40=160

    Parallel: R=[tex]\frac{40^{4}}{40*4}[/tex]=16000[tex]\Omega[/tex]

    in the book the answer is: 10[tex]\Omega[/tex]
     
  2. jcsd
  3. Apr 18, 2009 #2
    You're almost there, however the second equation is wrong i'm afraid, get that checked up on (the parallel one)
     
  4. Apr 18, 2009 #3

    Doc Al

    User Avatar

    Staff: Mentor

    OK.
    :bugeye: Where did you get this equation? (Look up resistors in parallel.)
     
  5. Apr 18, 2009 #4
    The parallel resistor formula I took from the same lesson of the same book. Very weird. Anyway, thanx guys :)
     
  6. Apr 18, 2009 #5

    Cyosis

    User Avatar
    Homework Helper

    I take it the formula from that lesson only had two parallel resistors? In that case [tex]\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2} \Rightarrow R=\frac{R_1R_2}{R_1+R_2}[/tex]. Now add a third resistor to the first equation and see that the generalization you made is not correct.
     
  7. Apr 18, 2009 #6
    I saw the wrongness of my ways now. Thank you
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: 4 resistors in series and parallel
Loading...