Electric Circuit Troubleshooting: How to Find V6 and I6 | Expert Help

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Homework Help Overview

The discussion revolves around troubleshooting an electric circuit to find the voltages and currents at specific points, specifically V6 and I6. Participants are exploring concepts related to circuit analysis, including Kirchhoff's laws and the relationships between resistors in series and parallel.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants are questioning the original poster's understanding of current flow through resistors and the application of Kirchhoff's laws. There are discussions about potential errors in the original poster's data and the need for detailed attempts to identify mistakes.

Discussion Status

The conversation is ongoing, with some participants providing guidance on the importance of showing work to identify errors. There is a recognition of potential misunderstandings regarding circuit analysis, but no consensus has been reached on the specific issues at hand.

Contextual Notes

Participants note the importance of understanding series and parallel resistor configurations, which may be a source of confusion for the original poster. There are indications that the original poster may have made a typo in their data table, which could affect their calculations.

PhysicsN3rd
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Can someone help me and see if this is right and explain to me how to find V6 and I6? If you can, it would be great if you could explain to me the whole thing. Thank you:blushing:
 

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You definitely took a wrong turn somewhere, because you've got two different currents going through [itex]R_1[/itex] and [itex]R_6[/itex]. They should clearly be the same.

If you can, it would be great if you could explain to me the whole thing.

That would require me to do more work than you, which is not right considering that it this is your assignment. The way it usually works here is you post your attempted solution in all its gory details, and we point out to you where you went wrong.
 
Oh sorry I didn't mean like do it for me, i meant as in explaining how the circuit work, not giving the number and stuff. Where did it go wrong?
 
Where did it go wrong?

For any mistakes you may have made, I couldn't possibly answer that question without seeing your work.

Now that I take a closer look at your data table, I think that you just made a typo. In the line that reads:

Code:
            V          I          R
Total            9         0.89     10.15

Did you mean to put those I and R values in for [itex]R_6[/itex]? If so then that would make more sense, because then the current through [itex]R_6[/itex] would be the same as through [itex]R_1[/itex].
 
I'm not quite sure what you've just said. The last line is the total values for V I and R. What I didn't get was V6 and I6, and I'm not sure if the other answers that I've found are correct or not.
 
I have calculated the values of the total resistance and current, and your answers are correct. What you need to do is check Kirchhoff's current law at every node and Kirchhoff's voltage law around every loop, and make sure both laws are satisfied.

I can tell by inspection that you have at least one problem here. You have [itex]0.89 A[/itex] going up the left branch of the large loop (the one that contains [itex]R_6[/itex]). Then when you add up the currents going through [itex]R_4[/itex] and [itex]R_5[/itex] you only get [itex]0.825 A[/itex]. Something is clearly amiss.

But as I keep saying, I will not be able to find your errors in reasoning without seeing your solution. If you want me to continue reading this thread, then please post your work.
 
PhysicsN3rd said:
I'm not quite sure what you've just said. The last line is the total values for V I and R. What I didn't get was V6 and I6, and I'm not sure if the other answers that I've found are correct or not.

Have you learned what it means for resistors to be in parallel or in series?

For example, the current path is split between R6 and R4.

This circuit is very easy to analyze once you have a firm hold on parallel and series resistance.
 

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