# Help on resistance

1. Aug 19, 2004

### marlon

hi guys,

i need to calculate the substitution resistance between the points A and B of the chain in the attached picture. Each rectangle is a resistance R , except the resistance on the right-top of the figure. That has value 2R ; and the solution has to be (13/11)R. I know that we have to use Kirchoff's laws, yet I do not seem to be able to solve it, i Get (9/5)R. Maybe you can.

regards
marlon

Last edited: Feb 9, 2006
2. Aug 19, 2004

### chroot

Staff Emeritus
How are you sure it's (13/11) R? It looks like just R to me -- the middle resistance is inconsequential, since it's between two nodes that are always at the same potential.

- Warren

3. Aug 19, 2004

### marlon

yes, you are right. But i made a little mistake. All resistances are R except the resistance on the right -top. It is 2R not R

Solution is certainly (13/11) R

regards
marlon

4. Aug 20, 2004

### marlon

Well, the only thing I was able to do is to apply the two Kirchoff's laws to the chain. In the first knod I have for the currents : I_0=I_1+I_2 Then the two knods in the middel (say I_3 is the current in the middle) = I_1=I_3+I_4 and below I_2+I_3=I_4

Then the second law , I follow the two closed subchains clockwisely

first chain : -I_1R-I_3R = I_2R
second subchain : -I_4R+I_5R+I_3R = 0

regards
marlon

5. Aug 20, 2004

### humanino

google it. I always hated those resistance exercise, they can be especially difficult. This one is not too difficult though. You could use Millman, Norton, Thevenin... whatever. Here is an elementary solution :
http://engr.calvin.edu/courses/engr204/2000/examples/equRes/irred.ans.html

6. Aug 20, 2004

### Leong

$$R_{eq}$$=$$\frac{v}{i}$$ ........(1)

node 1 :
i + $$\frac{v_{1}-v}{R}$$ + $$\frac{v_{2}-v}{R}$$ = 0

i = $$\frac{-v_{1}-v_{2}+2R}{R}$$ ....... (2)

node 2 :
$$\frac{v-v_{1}}{R}$$ + $$\frac{0-v_{1}}{2R}$$ + $$\frac{v_{2}-v_{1}}{R}$$ = 0

$$v_{2}$$ = $$\frac{-2v + 5v_{1}}{2}$$ ..... (3)

node 3:

$$\frac{v_{1}-v_{2}}{R}$$ + $$\frac{0-v_{2}}{R}$$ + $$\frac{v-v_{2}}{R}$$ = 0

$$v_{1}$$ = -v + $$3v_{2}$$ ....... (4)

Use (3) & (4) and substitute the answers into (1) get :

$$v_{2}$$ = $$\frac{7v}{13}$$
$$v_{1}$$ =$$\frac{8v}{13}$$
$$R_{eq}$$ = $$\frac{13R}{11}$$

Node 1, 2 and 3 use Kirchoff's current law.

Last edited: Aug 20, 2004
7. Aug 20, 2004

### marlon

thanks to all of you for helping me out...

regards
marlon

8. Aug 20, 2004

### humanino

you're welcome Marlon I love electronics :yuck:
:tongue2: