Help Calculate Resistance Between Points A & B - Kirchoff's Law

  • Thread starter Thread starter marlon
  • Start date Start date
  • Tags Tags
    Resistance
AI Thread Summary
The discussion revolves around calculating the equivalent resistance between points A and B in a circuit using Kirchhoff's laws. Marlon initially miscalculated the resistance as (9/5)R but later confirmed the expected solution is (13/11)R, with the only differing resistance being 2R. Contributors suggested applying Kirchhoff's current law at various nodes to derive the correct equations. The final calculations showed that the equivalent resistance indeed equals (13/11)R, confirming Marlon's assertion. The thread highlights the complexity of resistance calculations and the utility of Kirchhoff's laws in solving such problems.
marlon
Messages
3,779
Reaction score
11
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:
Physics news on Phys.org
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
 
chroot said:
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

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

regards
marlon
 
you're welcome Marlon :wink: I love electronics
:biggrin: :-p
 
Back
Top