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

Question about a 'basic' equivalent resistive circuit problem

  1. Aug 3, 2011 #1
    Hello all, I'm a new member. I've decided banging my forehead until its red will not make me any more able to solve my practice problems. I hope I haven't posted this in the wrong forum but here is the question in concern:

    Select R in the circuit so that VL= 5 V

    (see circuit in attachment below)

    Hopefully you will be able to read this, I wrote it in OneNote- here is my attempt at a solution:

    Attached Files:

    • 2.29.pdf
      File size:
      223.1 KB
  2. jcsd
  3. Aug 3, 2011 #2


    User Avatar

    Staff: Mentor

    Welcome to the PF.

    I would try using the KCL on the original circuit. I usually find that KCL works best for me, and is more intuitive than using source transformations, etc.

    Can you give the KCL approach a try?
  4. Aug 3, 2011 #3
    ay carumba, sorry for not posting the units in that last KVL expression... should've read something like:

    -50/(7+R) V + (17 + 3R)/(7 + R)kOhm * ImA + 5 V = 0
  5. Aug 3, 2011 #4
    absolutely!, I appreciate the suggestion. I'll post back my results
  6. Aug 3, 2011 #5


    User Avatar

    Staff: Mentor

    If you take another look at your second diagram (after the first source transformation) you might notice that there is a clear example of a simple voltage divider. The 1K resistor will not affect the output voltage, as no current will flow through it.


    25V source -- voltage divider -- 5V out. Should be able to solve for R with little pain :smile:

    Attached Files:

  7. Aug 4, 2011 #6
    wow thanks a bunch gneil, I assumed that those open nodes were there to represent some device or external circuit that absorbs 5V and could be even represented by a voltage source of 5V. Thanks a lot, I'd literally been racking my brains over the last day and a half on this problem.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook