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: Network of resistors

  1. Jul 6, 2009 #1
    1. The problem statement, all variables and given/known data

    This is the diagram:
    https://wug-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?cc/DuPage/Phys1202/summer/homework/Ch-21-DC-Circuits/resistor_network/ex1s95p3.gif [Broken]

    R1 = 3W, R2 = 6W, R3 = 11W, R4 =8W, V=6V

    A) What is the current through the resistor, R1 in the above circuit? -2A

    B) What single, equivalent resistor could replace all of the resistors in this circuit?

    C) What is the current supplied by the battery?

    D) What is the current through the resistor, R2 ?

    2. Relevant equations

    1/R (parallel) = 1/R1 + 1/R2 + 1/R3 + 1/R4
    R (series) = R1 + R2 + R3 + R4
    E = IR

    3. The attempt at a solution

    I solved part a. For part be, it seemed to me that R1 and R4 were connected in series, so I combined those two resistors by adding the resistances. Then I added that resistance to R2 and R3 using the equation for parallel resistors.

    So, I had: 1/R = (1/11W) + (1/6W) + (1/11W) = 2.87W. This answer is wrong, and I am not sure what to do. Any help would be greatly appreciated :)
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jul 6, 2009 #2
    Okay, so I solved parts b and c also. Now I am working on part d.

    I thought that I2 would = V/R2 = 6/6 = 1, but this is wrong. Any ideas?

  4. Jul 6, 2009 #3
    You need the correct potential drop across the resistor. I'll redraw the circuit for you, convince yourself that the one I have drawn is equivalent. I hope it will then be clearer to you.

    Redrawing the circuit into this form is often a good idea when solving such problems. It's now very clear which parts are in parallel, and which are in series.

    Attached Files:

  5. Jul 6, 2009 #4


    User Avatar
    Homework Helper

    The voltage across R2 is necessarily different from the 6V. Because the voltage at that node between R2||R4 and R3 is determined by 6v * R3 / (R3 + (R2||R4)). Subtract that from 6v to give the drop across R2. With that voltage drop divided by R2 you should have I through R2.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook