Simplifying a circuit in order to find the current

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  • #1
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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.
 
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Answers and Replies

  • #2
gneill
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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?
 
  • #3
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Oh I drew the circuit wrong. There isn't supposed to be a resistor there. I uploaded the correct version
 
  • #4
gneill
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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.
 
  • #5
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Alright, that makes much more sense. I got 12 ohms for the equivalent resistance.
 
  • #6
gneill
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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?
 
  • #7
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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.
 
  • #8
gneill
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Yes, that looks better :smile:
 
  • #9
CWatters
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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
gneill
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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
CWatters
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oops I can't add up. Should be 12//6 + 5 = 9.
 

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