Rank in order, from largest to smallest, the resistance (eq)

[Ali Zain]

Homework Statement

There is a figure, I'll try my best to draw/describe.
1. All three resisters are in parallel
___R____
!___R____!
!___R____ !

2. 2 resisters are parallel and one in series, after the parallel (ignore the dots)
___R___
... _____R___
!___R___ !

3. 2 resisters are in series and one in parallel.

____R____R____

______R_______!

Homework Equations

For resistors in parallel: 1/R(eq)= 1/R1+ 1/R2+1/R3...
For resistors in series: R1+R2+R3...

The Attempt at a Solution

Ok!
So, for the first one the resistance (eq)= 1/R+1/R+1/R= 1/3R
for 2. I first added R in parallel= 1/2R and with added R separately, that gave me, 1.5R
for 3. I first added two resistors in series R+R= 2R, then I added this 2R using parallel resistance equation. 1/2R + 1/1R = 1.5 R.
I know for a fact this is wrong, resistance of #3 should be less than that of #2. Since it is in parallel.
Can someone please explain what I'm doing wrong here?
Thank you
Ali

 

[Ali Zain]

I think I got it. for #3. 1/R(eq)= 3/2R. R(eq) must be equal to 2/3R.

 

[epenguin]

I am beginning to think that conductance is an under-used concept often more helpful than resistance. Also that ppl would do well to use more physical less formulaic thinking for these electrical problems - note here the question was only qualitative!

So I'd say the first case 1 has 3 conducting each the least resistance possible, so that must have the highest conductance, lowest resistance.
Comparing 2 and 3, well notice that part of the paths are the same. You could say in both cases at top a current is going through
__ R__R__ so that's the same for both. Then at the bottom in 2 current goes through __ R__R__ , in 3 it only goes through __R__ ...

Well may be you did think that, since you did somehow know your calculated result was wrong .
Intuition and calculation should each be used as a check and correction of the other!