Question about resistances please

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Homework Statement



just a questiion aren't the 1-ohm resistance(upper left) and the 3-ohm resistance in series? . The definition for resitance in series is "Two or more elements are in series if they are cascaded or connected sequentially " ,aren't they sharing the same wire? so why aren't they in series?the wire i'm talking about doesn't separate in two branches too , i'm confused
look:View attachment 54025
thanksssss
 

Answers and Replies

  • #2
lewando
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Yes, they are in series. The 1Ω and 5Ω connected at "b" are also in series.
 
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Yes, they are in series. The 1Ω and 5Ω connected at "b" are also in series.

Hi tahnks for your reply well i don't think they are in series because in the solution they don't use the rules of resistances in seires or parallel to solve the problem .check it out
sdfgf.jpg
. but i need to know why . if someone knows please explain
 
  • #4
lewando
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Thanks for adjusting the problem statement :smile:. When you apply a voltage across a and b or a current through a and b, then there is a branch at a and at b, so they are not in series.
 
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Thanks for adjusting the problem statement :smile:. When you apply a voltage across a and b or a current through a and b, then there is a branch at a and at b, so they are not in series.

actually the original figure doesn't show that the branch at point "a" bifurcates. it only appears in the solution , but i guess i finally get it , i think it's because black dots (nodes) mean that there is always a bifurcation . now everything makes sense , thanks !!!!!
 
  • #6
lewando
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I read your original post literally. In my mind, a dot is a dot. To be more clear, show the attached voltage/current source. Then the branches will stand out. Glad you are clear!
 
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I read your original post literally. In my mind, a dot is a dot. To be more clear, show the attached voltage/current source. Then the branches will stand out. Glad you are clear!

haha there is no a voltage source in this problem, the problem just say that we have to prove that equivalent resistance between "a" and "b" is 27/17 ohm.That's it. :smile:
 
  • #8
lewando
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Well something was applied to inject a current, I, into a and outfrom b. Again, resulting in branches, therefore no series resistance.
 
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  • #9
SammyS
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haha there is no a voltage source in this problem, the problem just say that we have to prove that equivalent resistance between "a" and "b" is 27/17 ohm.That's it. :smile:
If you consider what is meant by "equivalent resistance", then it is implied that you are finding what resistor could be placed between a & b and provide the same resistance to current flow as the given set of resistors. In other words, it's implied that there is a potential difference applied across a & b and an equivalent resistor would allow the same current to flow as given set of resistors.

This set of resistors cannot be analyzed on the basis of series/parallel analysis, unless you first do Y-Delta transformation.

The other common way to solve this is with Kirchhoff's circuit laws .
 
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