Calculate Total Resistance: 5R/3

[aviv87]

Hey, I am trying to calculate the total resistance for this but I always come up with the wrong answer.
A picture is available
http://img109.exs.cx/img109/5889/physics.jpg
In this picture every blue line is supposed to be a resistor with resistance R (yeah, I know it's ugly )
The answer should be $$R_{T} = \frac {5R}{3}$$.

Thanks ahead!

 

[Tide]

I got 5R/3 also.

This might help: If you number the resistors consecutively - top to bottom and left to right then resistor number 4 and its counterpart near the bottom contribute nothing to the resistance of the array. Then each of the remaining triangles comprised of a two resistor plus a single resistor branch are equivalent to (2/3)R and the rest is arithmetic.

 

[aviv87]

First of all thanks for answering =)
I understood that one of the resistors in A's nearest triangle and in B's doesn't affect the total resistance, but I don't understand why the resistance of the other triangle is (2/3)R.
Can you please explain to me how you got that?

 

[Tide]

In the triangles two of the resistors are in series with each other (giving a combined resistance of 2R) and they are in parallel with the third so the trio gives

[tex]\frac {1}{R_{triangle}} = \frac {1}{R} + \frac {1}{2R}[/itex]<br /> <br /> so that <br /> <br /> $$R_{triangle} = \frac {2}{3} R$$[/tex]

 

[ceptimus]

As the diagram is mirror symmetric about the horizontal centre line you only have to calculate the resistance for (say) the top half.

As the diagram is mirror symmetric about the VERTICAL centre line, the horizontal resistors on the centreline will have the same potential on both sides, so no current will flow in them, and they can be removed.

This leaves you with two arms starting from the node A which are the same as each other, so you only have to calculate one of them.

Taking the upper left quadrant, we can see it is just:

    A
    +
    |
    R
    |
+-R-+--(this resistor has no current flowing)--
|   |
R   R
|   |
+-+-+
  |

So we have 2 Rs in series with one in parallel, and then another R in series with that whole bunch.
So the total resistance of the above diagram is

Rt = R((2 * 1)/(2 + 1) + 1) = 5R/3

The upper half of your original diagram has two of my diagrams in parallel, giving half the resistance of one, but then the same arrangement repeats in the lower half, doubling the resistance again. So the whole of your mesh has the same resistance as mine does.

Note: when you have two resistors A and B in parallel, the equivalent resistance is (A * B)/(A + B)
That is where the (2 * 1)/(2 + 1) comes from.

 

[Tide]

Nice reduction, Ceptimus!

 

[aviv87]

Thank you both very much! I got it now =)