What is the Net Resistance in a Circuit with Multiple 2.8Kohm Resistors?

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Homework Help Overview

The discussion revolves around determining the net resistance in a circuit containing multiple 2.8Kohm resistors. Participants are analyzing the configuration of the resistors and how they are connected, exploring series and parallel combinations.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss breaking down the circuit into simpler components and question the validity of their interpretations of series and parallel connections. There are attempts to clarify the arrangement of resistors and whether certain resistors can be combined.

Discussion Status

The discussion is ongoing, with participants providing feedback on each other's reasoning and circuit breakdowns. Some guidance has been offered regarding the correct identification of series and parallel connections, but there is no explicit consensus on the final configuration.

Contextual Notes

Participants are working with a visual representation of the circuit, which may lead to differing interpretations of the connections. There are indications of confusion regarding the definitions of series and parallel arrangements, as well as the implications of combining resistors.

thermocleanse
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Homework Statement



Each R = 2.8Kohms. Find net resistance.

Please see attached picture.

Homework Equations





The Attempt at a Solution

 

Attachments

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To find the net resistance, you need not calculate the current. You can do it by using series and parallel combination formula.
 
OK, so, now that I've broken down the circuit to what is pictured, can i break it down anymore? the last picture of the first attachment shows a circuit with 3 resistors, but i don't know at this point how to find the net resistance...
please help, thanks.
 
I think you went wrong at the first step. You made a resistor disappear that is NOT parallel to anything. It does not matter in what direction resistances are drawn.

If R1 and R2 are parellel, one side of R1 is connected to one side of R2, and the other side of R1 is connected to the other side of R2.
In the original circuit there are no resistances that are parallel (but there will be once you start replacing series resistances)
 
so, from the original circuit, do you agree that, when all combined, there are 2 resistors in parallel and one resistor that runs diagonal within the circuit?
 
thermocleanse said:
so, from the original circuit, do you agree that, when all combined, there are 2 resistors in parallel and one resistor that runs diagonal within the circuit?

No. If you combine everything that can be combined, you will have only one resistor left.
 
i've attached what i got. please check it out. i have all my process written out, with each circuit breakdown and whether that part is parallel or series. please check it out and let me know if/where i went wrong. thank you.
 

Attachments

Your last step to combine R2 and R4 is invalid. These are not parallel.
One side of R2 is connected to one side of R4, the other side of R2 is NOT connected to the other side of R4, so they are not parallel
There's a pair of series resistances in your next to last circuit.
 
please check this attachment out. i think that's it...
if it is, thank you. if it isn't, thank you, anyway...
 

Attachments

  • #10
You combined R2 and R6 which are not in series. Two resistors are in series if they are connected to each other AND if nothing else is connected to the wire that connects the resistors. This last condition isn't valid.
There are 2 parallel resistances you can combine at the point where you made that mistake.
 
  • #11
ok...check this out...i think i got it...
 

Attachments

  • #12
Your answer is correct.
 

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