Calculate Current in Homework Statement | KCL

[john88]

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.

 

[Defennder]

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.

 

[john88]

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)

 

[Defennder]

First I have this situation

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

You could have just reverse the direction of the battery instead of writing a negative voltage source, you know.

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.

 

[john88]

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