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How do i find the difference in voltage from point A to B

  1. May 17, 2009 #1
    i have the following
    http://lh3.ggpht.com/_H4Iz7SmBrbk/Sg_Mj_v4_LI/AAAAAAAAA_c/Egdgr0F8hfY/s720/Untitled.jpg [Broken]

    the numbers are the resistance of the resistors in ohms

    and have already found the following

    I1=-18/23A
    I2=-10/23
    I3=8/23

    and now need to find the difference in voltage between points A and B,

    what i did was take the loop from point A, going clockwise to point B, and i get

    2I1 + I3 =1 + v

    so i get v=-36/23 +8/23 -1 = -15/23V

    but the correct answer is meant to be -67/23V, and i notice tat if i had said v=-36/23 -8/23 -1 i would have reached the correct answer, but i cannot see why there should be a negatice sign on I3 since I3 is positive 8/23 and i chose I3 clockwise
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 17, 2009 #2

    jambaugh

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    Science Advisor
    Gold Member

    Dell,
    To begin with I don't get the same signs as you on the currents. In fact I get all as positive (as shown in your diagram). But I may have the polarity convention on the batteries reversed in which case all the currents would be negative.

    Intuitively (assuming I haven't reversed battery convention) the three volt battery outweighs the 1 volt battery which would drive I1 positive. Similarly I3 should be positive and given more resistance in the I3 segment more I1 than -I3 would contribute to I2 making it positive. (assuming I have reversed battery convention this all holds but with pos<->neg).

    Remember that the voltages on resistors oppose current flow.

    To set up the system to solve for currents I took the Kirchoff's law (sp?)
    I2+I3=I1
    and the top loop:
    3v - 2I1 -1v - 1I2 = 0
    and the bottom loop:
    1I2 - I3 + 2v - 6I3 = 0
     
  4. May 17, 2009 #3
    okay, getting there, i seem to have a mistake in my equations,,
    i got
    I3 = I1 - I2 which is like what you got
    7I3 - I1 = -2v for the bottom loop, also like you got
    2I1 +I2 = -2v which is opposite sign to what you got


    so lets examine the top loop, i took a clockwise loop and chose both I1,I2 to be clockwise, so they will be positive, now i go into 3v's positive side and into 1v's negative so in the end i get
    I2+3v+2I1-1v=0
    I2+2I1=-2v

    but you someohow got the opposite,
     
  5. May 17, 2009 #4
    never mind, i got it,, thanks
     
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