Simplifying a circuit in order to find the current

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Discussion Overview

The discussion revolves around finding the equivalent resistance (R_eq) in a circuit at specified terminals (c,d) and (a,b). Participants explore different methods of simplifying the circuit and calculating the resistance based on the configuration of resistors.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant initially states uncertainty about whether any resistors are in series or parallel.
  • Another participant suggests that if a resistor at terminal (b) is also 5 Ohms, the minimum resistance at (a,b) would be 10 Ohms, questioning the presence of a resistor there.
  • A participant acknowledges drawing the circuit incorrectly and mentions that "dangling" components can be ignored when calculating resistance.
  • One participant calculates the equivalent resistance as 12 Ohms but is challenged to elaborate on their steps.
  • Another participant describes their method of combining resistors in series and parallel, arriving at a different value of 14 Ohms, indicating a potential calculation error.
  • One participant proposes an equivalent resistance of 8.6 Ohms, prompting a question about the role of certain resistors when terminals are open circuited.
  • A later reply corrects their calculation to arrive at 9 Ohms for the equivalent resistance at terminals (a,b).

Areas of Agreement / Disagreement

Participants express differing views on the equivalent resistance values, with no consensus reached on the correct approach or final answer. Multiple competing calculations and methods are presented.

Contextual Notes

Participants mention the importance of considering circuit configurations and the impact of open circuits on resistor contributions, indicating potential limitations in their calculations.

november1992
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Homework Statement


I solved my original question so I posted another one.
http://i.imgur.com/dpevY.png
Find [itex]R_{eq}[/itex] at terminals (c,d)

Homework Equations



[itex]R_{eq}[/itex] in series = Ʃ R
[itex]R_{eq}[/itex] in parallel = [itex]\frac{1}{R_{1}}[/itex] + [itex]\frac{1}{R_{2}}[/itex] + ...+ [itex]\frac{1}{R_{n}}[/itex]

The Attempt at a Solution



I don't think any of the resistors are in series or parallel.

For the equivalent resistor at terminals (a,b) the book says that it's 9Ω. I'm not sure how to get that answer.
 
Last edited:
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november1992 said:
For the equivalent resistor at terminals (a,b) the book says that it's 9Ω. I'm not sure how to get that answer.

If that resistor lacking a label at the (b) terminal is also 5 Ohms, then the minimum possible resistance at (a,b) will be 10 Ohms. Is that a 5 Ohm resistor there?
 
Oh I drew the circuit wrong. There isn't supposed to be a resistor there. I uploaded the correct version
 
november1992 said:
Oh I drew the circuit wrong. There isn't supposed to be a resistor there. I uploaded the correct version

Okay, now it makes sense.

When you go to solve the problem, keep in mind that any "dangling" components without a closed path for current to flow will not contribute to the circuit and may be ignored (erased from consideration). When you go looking for the resistance between terminals c and d, you can erase the terminal a and its 5 Ohm resistor, as well as the terminal b lead.
 
Alright, that makes much more sense. I got 12 ohms for the equivalent resistance.
 
november1992 said:
Alright, that makes much more sense. I got 12 ohms for the equivalent resistance.

Doesn't look right. Can you elaborate the steps you took?
 
I added the 3, 6 and 3 ohm resistors in series into a 12 ohm resistor. I then added the 6 ohm resistor that's in parallel with the 12 ohm resistor to make a 4 ohm resistor. Then I added that with the other two 5 ohm resistors.

I got 14 ohms now, I think i just added wrong.
 
Yes, that looks better :smile:
 
For the equivalent resistor at terminals (a,b) the book says that it's 9Ω. I'm not sure how to get that answer.

I make it 8.6 Ohms.
 
  • #10
CWatters said:
I make it 8.6 Ohms.

How so? Remember, terminals c-d will be open circuited so those 5Ω resistors at that end won't play any role.
 
  • #11
oops I can't add up. Should be 12//6 + 5 = 9.
 

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