Engineering Simplifying a circuit in order to find the current

AI Thread Summary
The discussion focuses on finding the equivalent resistance (R_eq) at terminals (c,d) of a circuit. Participants clarify that certain components can be ignored if they do not create a closed path for current. Initial calculations yield varying results, with one participant arriving at 12 ohms and another at 14 ohms after reevaluating their steps. The correct approach involves recognizing that the 5 ohm resistors at the open terminals do not contribute to the resistance calculation. Ultimately, the equivalent resistance at terminals (a,b) is confirmed to be 9 ohms, aligning with the textbook answer.
november1992
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Homework Statement


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

Homework Equations



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

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.
 
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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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