Question about resistances please

[carlos125]

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

 

[lewando]

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

 

[carlos125]

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

. but i need to know why . if someone knows please explain

 

[lewando]

Thanks for adjusting the problem statement . 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.

 

[carlos125]

Thanks for adjusting the problem statement . 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 !

 

[lewando]

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!

 

[carlos125]

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.

 

[lewando]

Well something was applied to inject a current, I, into a and outfrom b. Again, resulting in branches, therefore no series resistance.

 

[SammyS]

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.

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 .