Equivalent Resistance: How to Rank Resistors from Highest to Lowest

[MitsuShai]

The question was to rank from highest to lowest equivalent resistance. This is what I came up with, but it was counted wrong and I don't understand why.

http://i324.photobucket.com/albums/k327/ProtoGirlEXE/100_0687-1.jpg

The equivalent resistance of resistors in series are added while in parallel the reciprocal is added.


Each resistor as the same R, so I used R=2

Here are my numbers, in the order that I ranked, wronging:

1. (1/2) + .5= 1 + 2= 3

2. 2

3. .5+ .5+ .5= 1.5

4. .5 + .5= 1

5. 2+2 = 4
(1/4)+ .5 = .75

 

[kuruman]

It would be easier to troubleshoot your answers if you presented them in the order of the figures left to right. However, just by looking at what you have, I see no figure that shows two resistors in series corresponding to 2+2=4.

 

[cupid.callin]

3. .5+ .5+ .5= 1.5

5. 2+2 = 4
(1/4)+ .5 = .75

How have you done these? they seems to be wrong

 

[cupid.callin]

I see no figure that shows two resistors in series corresponding to 2+2=4.

Its for upper part of last figure which he has done correctly

_________________

I guess you have solving parallel connections

 

[MitsuShai]

How have you done these? they seems to be wrong[/QUOTE


Since R is equal in every resistor, I chose R=2


so for parallel the R is summed by the reciprocal (1/2)=.5
in series they are summed

 

[cupid.callin]

for parallel,

[tex]\frac{1}{R_{net}} = \frac{1}{R_1} + \frac{1}{R_2} \ ...[/tex]

so your answer is 1/(1.5)

 

[MitsuShai]

for parallel,

[tex]\frac{1}{R_{net}} = \frac{1}{R_1} + \frac{1}{R_2} \ ...[/tex]

so your answer is 1/(1.5)

Thank you so much for catching that error. I've looked over this problem many times to see if I was making an error, but I find it very hard to catch my own mistakes. :(