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Circuit analysis - current dissipated and pot. difference

  1. Nov 14, 2015 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations


    3. The attempt at a solution

    1. upload_2015-11-14_11-50-11.png

    2. i) Use mesh analysis to solve for circulating currents I1 and I2.
    ii) What is the current dissipated by resistor R1?
    iii) What is the potential difference across resistor R3?

    3

    i)
    . upload_2015-11-14_11-53-32.png

    Loop ABEFA

    E1-I1RI-(I1-I2)R3=0
    10-10I1-40(I1-I2)=0
    -50I1+40I2=-10
    50I1-40I2=10 .... EQUATION 1

    upload_2015-11-14_11-57-59.png

    Loop BCDEB

    -I2R3-I1R3-I2R2-E2=0
    -40I2-40I1-20I2-I2=0
    -40I1-60I2=12 ... EQUATION 2

    EQ 1 X 4 = 200I1-160I2=40
    EQ 2 X 5 = -200I1-300I2=60 +

    = -460I2=100
    I2=-0.217A

    Substitute I2 into equation 1: 50I1-40(-.217)=10
    I1=0.0264A



    ii) I=V/R
    = 10/10
    =1A

    iii) I3=I1+I2 (KCL)
    =(0.0264)+(-0.217)
    = -0.1906A


    VR3 = I3xR3
    = -0.1906x40
    = -7.624V



     
  2. jcsd
  3. Nov 14, 2015 #2

    NascentOxygen

    User Avatar

    Staff: Mentor

    In their passage through R3, currents I1 and I2 are in opposite directions, so partly cancel. This means their contributions to the R3 voltage will have opposite polarities, one will have a + sign, the other a - sign.

    Before using your I1 and I2 values in parts (ii) and (iii) you should have first checked that they correctly satisfy Kirchoff's Laws and your circuit. Do they?
     
  4. Nov 14, 2015 #3
    Firstly - do we take exit or entrance polarities? By this I mean if we have are working in a clockwise direction would the effect of I2 on R3 be +I2R3 or -I2R3? The examples in my study guide suggest that we use exit polarities. If so then I would do the following:

    Loop BCDEB (clockwise)

    +I2R3-I1R3+I2R2+E2=0
    40I2-40I1+20I2+I2=0
    -40I1+60I2=12 ... EQUATION 2


     
  5. Nov 14, 2015 #4
    Okay I think I have it now:

    Loop BCDEB

    +12V+40ohm(I2-I1)+2ohmI2=0
    40ohmI1-60ohmI2=12V

    Eq 1. x 4 200ohmI1-160ohmI2=40V
    Eq. 2 x 5 200ohmI1-300ohmI2=60V -
    140ohmI2=-20V
    I2=-0.14A

    Subst I2 into equation 1: 50ohmI1-40ohm(-.014)=10V
    50ohmI1=4.4V
    I1=0.088A

    I3=I1-I2
    = 0.088-(-0.14)
    = 0.288A

     
  6. Nov 14, 2015 #5

    NascentOxygen

    User Avatar

    Staff: Mentor

    It makes no difference whether you consider clockwise to be positive or negative, just so long as you stay consistent.

    That's right, so far.
     
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