Calculate Current in Homework Statement | KCL

  • Thread starter Thread starter john88
  • Start date Start date
  • Tags Tags
    Current
Click For Summary

Discussion Overview

The discussion revolves around calculating the current in a circuit using Kirchhoff's Current Law (KCL). Participants explore different methods for analyzing the circuit, including superposition and current division, while addressing potential issues with their approaches.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant questions whether current can flow when the potential difference is zero, suggesting that a current source can still produce current under these conditions.
  • Another participant asserts that current will flow through an ideal wire despite a zero potential difference, proposing the use of the superposition principle to solve the problem.
  • A participant describes their application of the superposition method, detailing the steps taken to analyze the circuit, including replacing the current source with an open circuit and the voltage source with a short circuit.
  • There is uncertainty expressed about the correctness of replacing the voltage source with a short circuit and whether the removal of certain resistors is appropriate.
  • Another participant points out that a resistor was incorrectly removed in the analysis but notes that they arrived at the same answer using a different method, indicating a potential inconsistency in the approach.

Areas of Agreement / Disagreement

Participants express differing views on the validity of their circuit analysis steps, particularly regarding the treatment of voltage and current sources. There is no consensus on the correctness of the methods used or the final answers derived.

Contextual Notes

Participants highlight potential errors in their circuit analysis, such as the incorrect removal of resistors and the implications of using the superposition principle. These points remain unresolved, contributing to the uncertainty in the discussion.

john88
Messages
16
Reaction score
0

Homework Statement





Homework Equations



KCL


The Attempt at a Solution



I have some question for this circuit.

http://img510.imageshack.us/my.php?image=questpt3.jpg

I want to calculate the current I, my first thought was that it can't go any current because the potential difference is zero. But then I thought that the current source can produce a current even if there is zero potential, so there is a current I anyway. I am correct?

I would appreciate some help.
 
Physics news on Phys.org
There will be a current through the wire. The potential difference across any ideal wire is 0, but still there will be current flow if the other circuit element are present. You can make use of superposition principle to do this problem.
 
ok so if I use the superposition method I get this:

First I have this situation

http://img387.imageshack.us/my.php?image=firstum9.jpg

First I replace the current source whit an open circuit and I end up whit this:

http://img66.imageshack.us/my.php?image=second2ud4.jpg

So I = -(2.71/R9)


Next I replace the voltage source whit a short-circuit

http://img399.imageshack.us/my.php?image=second1br1.jpg

(im not sure if that step is right, I replaced the voltage source whit a short-circuit. So the resitance R10 got short circuted and I took it away correct?)

Then I = I1 (current source) not sure of that either...

which means that the total current I is equal to I1 -(2.71/R9)
 
john88 said:
You could have just reverse the direction of the battery instead of writing a negative voltage source, you know.

john88 said:
First I replace the current source whit an open circuit and I end up whit this:

http://img66.imageshack.us/my.php?image=second2ud4.jpg

So I = -(2.71/R9)
What happened to the 2k resistor?

Next I replace the voltage source whit a short-circuit

http://img399.imageshack.us/my.php?image=second1br1.jpg

(im not sure if that step is right, I replaced the voltage source whit a short-circuit. So the resitance R10 got short circuted and I took it away correct?)
Yes, and so is the R9 resistor, so you have I = I1 here as well.
 
Well ye I took away the 2k resistor, and that's not correct but I still end up whit the same answer when I use the current divider, but its not negative. I = 2.72/6500

So now I_tot = I + I1

This must be the right answer
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
2
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 9 ·
Replies
9
Views
6K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 16 ·
Replies
16
Views
3K