1. Not finding help here? Sign up for a free 30min 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!

Simplifying a circuit in order to find the current

  1. Sep 10, 2012 #1
    1. The problem statement, all variables and given/known data
    I solved my original question so I posted another one.
    http://i.imgur.com/dpevY.png
    Find [itex]R_{eq}[/itex] at terminals (c,d)

    2. Relevant equations

    [itex]R_{eq}[/itex] in series = Ʃ R
    [itex]R_{eq}[/itex] in parallel = [itex]\frac{1}{R_{1}}[/itex] + [itex]\frac{1}{R_{2}}[/itex] + ...+ [itex]\frac{1}{R_{n}}[/itex]

    3. The attempt at a solution

    I don't think any of the resistors are in series or parallel.

    For the equivalent resistor at terminals (a,b) the book says that it's 9Ω. I'm not sure how to get that answer.
     
    Last edited: Sep 10, 2012
  2. jcsd
  3. Sep 10, 2012 #2

    gneill

    User Avatar

    Staff: Mentor

    If that resistor lacking a label at the (b) terminal is also 5 Ohms, then the minimum possible resistance at (a,b) will be 10 Ohms. Is that a 5 Ohm resistor there?
     
  4. Sep 10, 2012 #3
    Oh I drew the circuit wrong. There isn't supposed to be a resistor there. I uploaded the correct version
     
  5. Sep 10, 2012 #4

    gneill

    User Avatar

    Staff: Mentor

    Okay, now it makes sense.

    When you go to solve the problem, keep in mind that any "dangling" components without a closed path for current to flow will not contribute to the circuit and may be ignored (erased from consideration). When you go looking for the resistance between terminals c and d, you can erase the terminal a and its 5 Ohm resistor, as well as the terminal b lead.
     
  6. Sep 10, 2012 #5
    Alright, that makes much more sense. I got 12 ohms for the equivalent resistance.
     
  7. Sep 10, 2012 #6

    gneill

    User Avatar

    Staff: Mentor

    Doesn't look right. Can you elaborate the steps you took?
     
  8. Sep 10, 2012 #7
    I added the 3, 6 and 3 ohm resistors in series into a 12 ohm resistor. I then added the 6 ohm resistor that's in parallel with the 12 ohm resistor to make a 4 ohm resistor. Then I added that with the other two 5 ohm resistors.

    I got 14 ohms now, I think i just added wrong.
     
  9. Sep 10, 2012 #8

    gneill

    User Avatar

    Staff: Mentor

    Yes, that looks better :smile:
     
  10. Sep 11, 2012 #9

    CWatters

    User Avatar
    Science Advisor
    Homework Helper

    I make it 8.6 Ohms.
     
  11. Sep 11, 2012 #10

    gneill

    User Avatar

    Staff: Mentor

    How so? Remember, terminals c-d will be open circuited so those 5Ω resistors at that end won't play any role.
     
  12. Sep 11, 2012 #11

    CWatters

    User Avatar
    Science Advisor
    Homework Helper

    oops I can't add up. Should be 12//6 + 5 = 9.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Simplifying a circuit in order to find the current
Loading...