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: Solving for current between 2 nodes

  1. Feb 24, 2014 #1
    1. The problem statement, all variables and given/known data

    Assume that I = 22mA , V = 6.0V , and R = 350Ω. Determine the current between B and D using the superposition principle.


    2. Relevant equations

    Superposition Principle
    Voltage and Current dividers
    KCL, KVL

    3. The attempt at a solution

    I_BD = I_BD' + I_BD''

    Shorting the voltage source first, I have:

    R_eq of resistors now in parallel = 100*300/(100+300) = 75

    Applying the current divider to the remaining circuit, I have:

    I_BD'' = -0.022 * 300/825 = -0.008

    (I also solved this using mesh analysis and got the same answer)

    Now replacing the current source with an open circuit, I have:

    R_eq from A to C to D to B = 750


    Combining that with the 300 Ohm resistor it is now in parallel with, then with the 100 Ohm resistor in series, I have:

    R_eq = 314.3

    Solving for total current with V=IR, I_total = 0.019 A

    Applying the current divider:

    I_BD' = 0.019*300/1050 = 0.0054 A

    Combining I_BD' and I_BD'', I_BD = -0.0026 A which is incorrect. Not exactly sure what I'm doing wrong here.
  2. jcsd
  3. Feb 24, 2014 #2


    User Avatar

    Staff: Mentor

    Presumably the current IBD is meant to refer to the current from B to D. If so, check the assumptions you're making about the direction of current flow for each case.
  4. Feb 24, 2014 #3
    I think I see my mistake, but I just want to be sure before putting in an answer because I'm on my last attempt. I believe I mixed up the sign with the current source removed, so it should be -0.0054 A and the total I_BD should be -0.0134. Does this sound correct?
  5. Feb 24, 2014 #4


    User Avatar

    Staff: Mentor

    Your new current for when the current source is suppressed looks good. But also check the sign for the case where the voltage supply is suppressed. What direction will the current BD flow then?
  6. Feb 24, 2014 #5
    So I guess both my signs were mixed up then, which sounds right but my problem with that is it gives me 0.0026 A for I_BD, which I had put in earlier and it said I was wrong.
  7. Feb 24, 2014 #6


    User Avatar

    Staff: Mentor

    So you've got:
    When I source is suppressed: -5.45 mA
    When V source is suppressed: 8.00 mA

    making the sum IBD = 2.54 mA

    Your system may require that your answer have a certain number of significant digits and particular units. How exactly do they want the input to be given?
  8. Feb 24, 2014 #7
    It specifically asks for 2 significant figures. I could try 0.0025 A but I'm not sure if it'll make a difference, usually it accepts the answer if you're only a tad off.


    I'll probably end up talking to my professor about it tomorrow, because this seems like the correct answer.

    EDIT: I took the risk and put in 0.0025 A, which is indeed correct. Thanks for your help.
  9. Feb 24, 2014 #8


    User Avatar

    Staff: Mentor

    Well, 0.0026 is not the quite the same thing as 0.0025. 2.54 mA doesn't round to 2.6 mA.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted