How Do You Find the Equivalent Resistance in This Circuit?

[James889]

Hi,

I have the following circuit:
[PLAIN]http://img689.imageshack.us/img689/9694/parallel.png

I need to find the equivalent resistance between the points A and B.

I have tried several ways, but no matter how i try to combine them i always end up with the wrong answer.

/James

 

[Filip Larsen]

You can step by step reduce the network by combining two resistors at a time into one replacement resistance using either a serial or parallel connection. As a start you can for instance reduce the two top right resistances with one using a parallel connection. After three more reductions you are left with only one resistance between A and B, which is the equivalent resistance of the network.

 

[James889]

Are the top two resistors then parallel with the bottom two resistors?

 

[Filip Larsen]

Indirectly, yes. If you have trouble visualizing the structure of the network, then try make a sketch of the same network but with A to the left and B to the right (or top and bottom). It should hopefully then be more obvious that you initially can make two independent parallel reductions, and after those a serial reduction and then finally a parallel reduction.

 

[James889]

Hi,

So i tried
\frac{35\cdot42}{35+42} = 19.09

\frac{70*30}{70+30} = 21

\frac{21*19.09}{21+19.09} = 9.999

And finally \frac{9.999*7}{9.999+7} = 4.11

But the correct answer is supposed to be 11.635 ohm

Any ideas?

 

[The Electrician]

You need 19.09 in series with 7, then that combination in parallel with 21

 

[James889]

You need 19.09 in series with 7, then that combination in parallel with 21

ok, to me it looked like the 7 ohm resistor was in parallel with the rest.

 

[vela]

When two elements are in parallel, one element is connected to the same two nodes as the other element. In your circuit, the 35-ohm and 42-ohm resistors are both connected to node A, and their other ends are connected together at the top. Therefore, they are in parallel. Similarly, the 70-ohm and 30-ohm resistors are both connected to node A on one end and node B on the other, so they are also in parallel.

This is not the case for the resulting 21-ohm and 19.09-ohm resistors, however. While both are connected to node A, one is connected to the top of the 7-ohm resistor, and the other, to the bottom. They're not connected to the same two nodes, so they are not in parallel.

 

[Tommo1]

I think that the key to this question is "art" over "maths": you must redraw the circuit in a manner that is less confusing.

[PLAIN]http://www.xphysics.co.uk/x/R1

Then redraw...

[PLAIN]http://www.xphysics.co.uk/x/R2

and again...

[PLAIN]http://www.xphysics.co.uk/x/R3

Simples. :)