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Kirchoff Problem

  1. Jul 30, 2008 #1
    1. The problem statement, all variables and given/known data
    Here is the link of the given circuit.

    I am tasked to find the voltage of all the resistors using Kirchhoff Current Law and Voltage Law. TAKE NOTE THAT I AM NOT ALLOWED TO USE MESH ANALYSIS which means I need to assign current for all resistors.

    2. Relevant equations

    KCL at node X

    -I1 + IT -I3=0

    KCL at node Y

    -IT + I2 + I4=0

    KVL at Left Loop

    1I1 - 12 + I2 + 5V = 0

    KVL at Right Loop

    -2I3 + 6V - 2I4 + 12 = 0

    and using both eqns for nodes x and y, I arrived for the fifth equation.

    0=I1 - I2 + I3 - I4

    3. The attempt at a solution

    I arrange all the 5 equations to be easily solved.

    0=IT - I1 + 0I2 - I3 + 0I4 + 0
    0=IT + 0I1 + I2 + I3 + I4 + 0
    0=0IT + I1 + I2 + 0I3 + 0I4 + 5V
    0=0IT + 0I1 + 0I2 - 2I3 - 2I4 + 12V
    0=0IT + I1 - I2 + I3 - I4

    These a lengthy process. Its obvious that I1 = I2 and I3=I4 but the answers I am getting is not the same as these.

    Our teacher told us to use these. We are not allowed the short cut which is mesh analysis. What can I do to simplify the solution?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 30, 2008 #2

    berkeman

    User Avatar

    Staff: Mentor

    You should be able to use KVL and KCL separately to solve for the circuit values. You don't use them together to solve it. You should get the same answers for each method, however.

    Can you show us how to solve the circuit with just one or the other method?
     
  4. Jul 30, 2008 #3
    Your complicating the problem. There are only three unique currents: (1) through the left branch, (2) through the right branch, and (3) through the middle. Another simplifying procedure is to combine the resistances into a single equivalent resistance in each branch.
     
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