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: Direct Current circuits

  1. Mar 9, 2010 #1
    1. The problem statement, all variables and given/known data
    What is the equivalent resistance of the combination of identical resistors between points a and b in the figure below?


    2. Relevant equations
    Series= Req= R1 + R2 + R3 etc
    Parallel = Req= (1/R1 + 1/R2 + 1/R3 etc)^-1


    3. The attempt at a solution
    Req= R1 + (1/3R) + R3

    I dont know what to do from here.
     

    Attached Files:

    Last edited: Mar 9, 2010
  2. jcsd
  3. Mar 9, 2010 #2

    rl.bhat

    User Avatar
    Homework Helper

    Parallel= Req= R1 + R2 + R3 etc
    Series= Req= (1/R1 + 1/R2 + 1/R3 etc)^-1

    These equations are wrong. The correct ones are
    series= Req= R1 + R2 + R3 etc
    Parallel = Req= (1/R1 + 1/R2 + 1/R3 etc)^-1
     
  4. Mar 9, 2010 #3
    Thank you is this correct I only have one more submission
     
  5. Mar 9, 2010 #4

    rl.bhat

    User Avatar
    Homework Helper

    Req= R1 + (1/3R) + R3
    Check the second term in the equation.
     
  6. Mar 9, 2010 #5
    I tried that answer already and it says "Check the syntax of your response".
     
  7. Mar 9, 2010 #6

    ideasrule

    User Avatar
    Homework Helper

    The second term is right if he means 1/3 R and not 1/(3R). However, what is R1 and R3 in terms of R? Can you simplify the equation?
     
  8. Mar 9, 2010 #7
    There are no numerical values for R. I asked my friend and the correct answer is 2 + (3)^-1.

    Not sure how she got it.
     
  9. Mar 9, 2010 #8

    ideasrule

    User Avatar
    Homework Helper

    I know there are no numerical values for R, which is why I asked how R1 and R3 relate to R. Look at the diagram and figure out what R1 and R3 are in terms of R.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook