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Finding the magnitude of a circuit

  1. Oct 13, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the magnitude of the current in each branch of the circuit shown below, in which B1 = 1.96 V. Specify the direction of each current.

    diagram: http://img183.imageshack.us/img183/6951/physics.gif [Broken]

    2. Relevant equations
    I1+I2-I3=0
    I1+I2=I3


    3. The attempt at a solution
    22I1+56(I1+I2)-5=0
    78I1+56I2=5 EQUATION 1

    -56(I1+I2)-75I2+B1
    -56I1-131I2=-1.96
    56I1+131I2=1.96 EQUATON 2

    75I1+56I2=5
    I1=(5-56I2)78
    PLUGGED THAT INTO EQUATION 2 AND GOT: -.0179.<---- thats wrong.

    i understand how to solve the equations for an unknown variable. what i want to know is why is my answer coming out wrong? Please help. Once i get this answer I will be able to solve the rest. Thanks.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Oct 13, 2009 #2
    The potential across the 5V voltage source has the wrong sign.
     
  4. Oct 13, 2009 #3
    so would the equation instead be 22I1+56(I1+I2)+5=0 instead of 22I1+56(I1+I2)-5=0
     
  5. Oct 14, 2009 #4
    Yes. You have a current I_1 going through the right through the 22 ohm resistor. This means the left side of this resistor must be more positive than the right side. This potential difference has the same direction as the potential difference across the 5V resistor.

    Personally I almost never use the Voltage law. I assign an unknown potential of V to the left
    side of the circuit and 0 to the right side. You can then compute all currents, and finally use the current law on them.
    Only one equation to solve and less problems with the direction of the currents. (you still need to assign a direction to all currents and stick with it)
     
  6. Oct 14, 2009 #5
    so far for one of my answers i got I2=-.0893
    i just checked and webassign said it's wrong. What am i doing wrong so that it is coming out wrong.
    help a bit more its due in about an hour and half.
     
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