Solve Kirchhoff Problem: Find Voltages of Resistors

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AI Thread Summary
The discussion focuses on solving a circuit problem using Kirchhoff's Current Law (KCL) and Kirchhoff's Voltage Law (KVL) without mesh analysis. The user has derived five equations based on KCL and KVL but is struggling with the complexity of the solution. They note that their current assignments lead to discrepancies in the answers, despite the teacher's instructions to avoid shortcuts. Suggestions include simplifying the circuit by combining resistances into equivalent resistances and focusing on the three unique currents present. The conversation emphasizes the importance of using KCL and KVL separately to achieve consistent results.
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Homework Statement


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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.

Homework 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

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?
 
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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?
 
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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