circuit analysis

Discussion in 'Homework Help' started by fran1942, May 5, 2011.

  1. fran1942

    Thread Starter Member

    Jul 26, 2010
    Hello, I have the attached circuit which I am trying to solve.
    I 'think' I have all the currents correct, however I am struggling to make the voltage drops over the resistors in the right hand loop(2) equate to 0 as per Kirchoff's law of voltages.
    The two resistors on the left hand loop(1) drop the power source to 0, so that loop is fine. It is the big loop on the right that I am having trouble with.

    If someone could please take a look and tell me where I am going wrong that would be much appreciated.

    Thanks kindly.
    Last edited: May 5, 2011
  2. StayatHomeElectronics

    AAC Fanatic!

    Sep 25, 2008
    R4 is not in the loop for the voltage calculation. And, the voltage drop over the final resistors should also give you a (-) sign.

    12 - 4 - 2.67 - 5.33 = 0
    fran1942 likes this.
  3. hgmjr

    Retired Moderator

    Jan 28, 2005
    The reason that your voltage summation fails for the right-hand side of your circuit is that you are using voltage multiple time. Notice that if you had stopped at 12-4-8 = 0 you would have solved the first loop. The 2.66v and 5.333v figures associate with the rightmost loop only.

    fran1942 likes this.
  4. miguel cool

    New Member

    Mar 15, 2010
    For lopp 1:
    12 = I1 (9Ω + 1Ω) so I1 = 12/10Ω = 1.2Amp

    but in the opposite direction that you draw I1

    For the loop 2

    R2 = 2Ω . . . . Req = 12Ω //6Ω gives 4Ω
    so the total resistance in the right side is Rt = 6Ω = 2Ω+4Ω
    and current I2 = 12V/6Ω= 2Amp that flow throug R2