Current connecting two parts of a circuit

[RX-78-2]

in the diagram below Io is zero. But i can't figure out why. Can anyone explain this to me.

 

[RX-78-2]

ouch, my mistake in the title it should say conductor connecting two parts of a circuit.

 

[LeBrad]

There is no element connecting the two parts, just a wire, and since the wire is assumed to be a perfect conductor, there is no voltage drop across any segment of the wire. Current flows between two ends of a device with a potential difference across it because it wants to equalize the potential at both ends. If there is no potential difference, there will be no current flow to equalize the difference.

 

[berkeman]

Suppose Io were non-zero. Where is the return current? Or are you building up static charge...

 

[RX-78-2]

There is no element connecting the two parts, just a wire, and since the wire is assumed to be a perfect conductor, there is no voltage drop across any segment of the wire. Current flows between two ends of a device with a potential difference across it because it wants to equalize the potential at both ends. If there is no potential difference, there will be no current flow to equalize the difference.

so a potential difference causes current to flow. like water flowing because of pressure pump right?

what do you mean by equalize potential though? i don't know what that means.

 

[RX-78-2]

Suppose Io were non-zero. Where is the return current? Or are you building up static charge...

yeah, i was confused. I assumed everytime a current met two forks in the circuit path like a parallel circuit for instance it would just split up the current flow. So i assumed that Io had come from the current in the right-hand circuit after splitting up.

Because i remember learning that a current if given several different branches to flow through the most current would flow through the branch with the least resistance. So I assumed currents from the two parts of those circuits would flow through that empty branch. which was absolutely wrong...

 

[Ouabache]

You diagram indicates a circuit representation with a dependent source.
You may learn more about these kinds of representations by searching previous posts here.
For example, here is one. My reply on that thread gives insight, directly related to your question.
hint: where is ground?

 

[RX-78-2]

You diagram indicates a circuit representation with a dependent source.
You may learn more about these kinds of representations by searching previous posts here.
For example, here is one. My reply on that thread gives insight, directly related to your question.
hint: where is ground?

I see so that's the ground where the potential is taken from. So I'm guessing the circuit would still be the same if the model illustrated that single branch to be just a point node...

thank you for the link and thanks for the responses.

 

[RX-78-2]

I modified the circuit now and i put in resistor R3 behind the negative terminal of the independent voltage source.

I assumed the voltage drop for this new resistor to the ground would be the reverse of the voltage R1 to the ground. but the only way for that to be possible is if the currents running through each Resistor to be different. but Io is zero so the currents through both resistors should be the same. I don't get it.

I forgot to add an image i'll add it below this post. sorry for the spam.

 

[RX-78-2]

here is the image

 

[doodle]

Suppose Io were non-zero. Where is the return current? Or are you building up static charge...

berkeman is spot on.

The current Io will always be zero because current goes around in a loop. To put it loosely, if current Io goes left in your circuit, then somewhere in the same circuit, it has to go back right, too. If you must make Io nonzero, then place a wire connecting the right terminal of R1 to the left terminal of R2, and then you'd see (a nonzero) Io going right along this wire.