Using Kirchhoff's laws to find current in a circuit

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Discussion Overview

The discussion revolves around using Kirchhoff's Current Law (KCL) and Kirchhoff's Voltage Law (KVL) to determine the currents I1 to I3 in a given circuit. Participants explore the application of these laws in a homework context, addressing potential discrepancies in their calculations.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant reports obtaining two different answers for I1 using KVL and KCL, expressing uncertainty about the correctness of their approach.
  • Another participant confirms the KVL solution for I1 is correct but notes that the KCL approach lacks sufficient equations to solve for all unknowns.
  • A participant questions whether they can substitute the value of I1 obtained from the KVL equation into the KCL equations, indicating a concern about the requirement to solve separately using both laws.
  • Another participant suggests that the problem likely requires solving the circuit using each technique separately for practice.
  • Participants discuss the possibility of identifying another node to write a third KCL equation to aid in solving the circuit.
  • One participant later indicates they have resolved their issue independently.

Areas of Agreement / Disagreement

Participants generally agree on the correctness of the KVL solution for I1, but there is disagreement regarding the sufficiency of the KCL equations provided. The discussion remains unresolved regarding the best approach to integrate findings from both laws.

Contextual Notes

There is a noted limitation in the number of KCL equations relative to the number of unknowns, which may affect the ability to solve the circuit completely using KCL alone.

november1992
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Homework Statement


Use KCL and KVL to determine the currents I1 to I3 in the circuit below:

aodth.png

Homework Equations



Ʃ[itex]v_{n}[/itex] = 0
Ʃ[itex]i_{n}[/itex] = 0
v=ir

The Attempt at a Solution


I got two different answers for [itex]I_{1}[/itex] using the two different laws. I'm not sure if I'm doing it right.

KVL @ loop 1: -18 + 6 + 8[itex]I_{1}[/itex] = 0
8 [itex]I_{1}[/itex] = 12
[itex]I_{1}[/itex] = 1.5AKCL @ node a: 3 = [itex]I_{1}[/itex] + [itex]I_{2}[/itex] + 1
KCL @ node b: [itex]I_{3}[/itex] = 1 + [itex]I_{2}[/itex]
0 = [itex]I_{1}[/itex] + [itex]I_{2}[/itex] -2
0 = -1 - [itex]I_{2}[/itex] + [itex]I_{3}[/itex]
1 = [itex]I_{3}[/itex]
0 = [itex]I_{2}[/itex]
2 = [itex]I_{1}[/itex]
 
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november1992 said:

Homework Statement


Use KCL and KVL to determine the currents I1 to I3 in the circuit below:

aodth.png

Homework Equations



Ʃ[itex]v_{n}[/itex] = 0
Ʃ[itex]i_{n}[/itex] = 0
v=ir


The Attempt at a Solution


I got two different answers for [itex]I_{1}[/itex] using the two different laws. I'm not sure if I'm doing it right.

KVL @ loop 1: -18 + 6 + 8[itex]I_{1}[/itex] = 0
8 [itex]I_{1}[/itex] = 12
[itex]I_{1}[/itex] = 1.5A


KCL @ node a: 3 = [itex]I_{1}[/itex] + [itex]I_{2}[/itex] + 1
KCL @ node b: [itex]I_{3}[/itex] = 1 + [itex]I_{2}[/itex]



0 = [itex]I_{1}[/itex] + [itex]I_{2}[/itex] -2
0 = -1 - [itex]I_{2}[/itex] + [itex]I_{3}[/itex]
1 = [itex]I_{3}[/itex]
0 = [itex]I_{2}[/itex]
2 = [itex]I_{1}[/itex]

Your KVL solution for I1 is correct. For the KCL solution, you have 3 unknowns but you have only written 2 equations. That isn't sufficient to solve for I1 yet...
 
Could I substitute the value that I got for I1 from the KVL equation into KCL?
I wasn't sure if I could do that because I thought the question was asking me to find the values separately using the two different laws.
 
november1992 said:
Could I substitute the value that I got for I1 from the KVL equation into KCL?
I wasn't sure if I could do that because I thought the question was asking me to find the values separately using the two different laws.

It does sound like they want you to solve the circuit with each technique separately. It's a pretty easy circuit to solve by inspection anyway, so they must want you to go through all the steps of each technique separately for practice.

So can you see another node where you could write a 3rd KCL equation?
 
The one between the 8 ohm and 4 ohm resistor?
 
november1992 said:
The one between the 8 ohm and 4 ohm resistor?

That might work...
 
nvm I figured it out, thanks for the help.
 
Last edited:

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