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### GR0877 #68

Posted: Sat Sep 17, 2011 1:17 pm
Hello all,
I recently took the 08 test as part of my preparation for the test in Oct and I was using tashwoods solutions to go over the test but I do not understand #68. Here is the question The problem I am having is how to find the total resistance. Tashwoods solution says the resistor grid is composed of 3 2R resistors in parallel to give R_tot = 2R/3 and therefore the current through the battery is I = 3V/2R but why is the total resistance 2R/3? Do you ignore the two horizontal resistors for some reason? I would be eternally grateful if someone could explain this to me or provide a link (or book reference) to somewhere that goes over resistors in grids. I've googled it but have only found info on how resistors add in parallel and in series.

Thanks,
Jim

### Re: GR0877 #68

Posted: Sat Sep 17, 2011 2:00 pm
Electrons, like all things, are lazy.

### Re: GR0877 #68

Posted: Sat Sep 17, 2011 3:16 pm
try applying kirchhoffs laws? also i just realized that kirchhoff is spelled with 2 h's. crazy

### Re: GR0877 #68

Posted: Sat Sep 17, 2011 5:14 pm
So I think I figured it out a quick way to solve it but I'd like someone to confirm if they can. Taking Nergu's suggestion I broke down the resistor grid into 4 separate loops as shown and got the following: Loop 1 - $$V_1=V_2+V_4$$
Loop 2 - $$V_2=V_3+V_5$$
Loop 3 - $$V_7=V_6+V_4$$
Loop 4 - $$V_8=V_7+V_5$$
Rearranging a few things gives
$$V_5=V_2-V_3=V_8-V_7$$
$$V_4=V_1-V_2=V_7-V_6$$
But since components in parallel have the same voltage and the resistance is the same everywhere then we can assume
$$V_2=V_3=V_1$$
Resulting in
$$V_5=V_4=0$$
Thus the resistance in each leg is
$$R_{1,6}=R_{2,7}=R_{3,8}=2R$$
And total resistance is given by
$$\frac{1}{R_{tot}}=\frac{3}{2R}\Rightarrow R_{tot} = \frac{2R}{3}$$

### Re: GR0877 #68

Posted: Sat Sep 17, 2011 5:21 pm
On second thought, referring to the image above if you do the loop of 1,2,7,6 then you find
$$V_2+V_7-V_6-V_1=0$$
But this is exactly what is given for $$V_4$$ above when rearranged so
$$V_4=0$$ and by symmetry (or just doing the same thing with loop 3,8,7,2) then $$V_5=0$$

I feel dumb sometimes when I figure these problems out or read the solution and see how easily they can be solved. ### Re: GR0877 #68

Posted: Mon Sep 26, 2011 10:53 am
There is a simple way to solve it. The circuit is symmetric for a given resistor. For example, lets take the down left resistor, It has the same value as the upper right resistor. which means that the same current must flow through each of them. if both have the same current then none is going through the horizontal resistors. (and if so, one does not need to take it into account when calculating the total resistance because the voltage difference is zero)

### Re: GR0877 #68

Posted: Thu Oct 06, 2011 4:05 am
First, solve the problem without the horizontal resistors. What you find is that in the middle, it is equipotential. Nothing happens when you connect two equipotentials with resistors, or any wires, for that matter. You can short it if you like, and the answer would be the same.

### Re: GR0877 #68

Posted: Wed Oct 19, 2011 6:49 pm
Please tell me there is some actual point of having horizontal resistors that do nothing

### Re: GR0877 #68

Posted: Sun Dec 18, 2011 1:10 pm
wavicle wrote:Please tell me there is some actual point of having horizontal resistors that do nothing

Of course....not. The correct answer to these sorts of questions are "What sort of dumbass experimental set-up are you creating?" But that's not an available answer, so go with (D).

### Re: GR0877 #68

Posted: Sun Dec 18, 2011 9:56 pm
wavicle wrote:Please tell me there is some actual point of having horizontal resistors that do nothing
It's to find out if you can recognize obvious errors in an electronic circuit.

### Re: GR0877 #68

Posted: Mon Dec 19, 2011 12:54 am
CarlBrannen wrote:
wavicle wrote:Please tell me there is some actual point of having horizontal resistors that do nothing
It's to find out if you can recognize obvious errors in an electronic circuit.
As I said, electrons are lazy.

### Re: GR0877 #68

Posted: Sat Mar 17, 2012 1:24 pm
scasplte2 wrote:So I think I figured it out a quick way to solve it but I'd like someone to confirm if they can. Taking Nergu's suggestion I broke down the resistor grid into 4 separate loops as shown and got the following: Loop 1 - $$V_1=V_2+V_4$$
Loop 2 - $$V_2=V_3+V_5$$
Loop 3 - $$V_7=V_6+V_4$$
Loop 4 - $$V_8=V_7+V_5$$
Rearranging a few things gives
$$V_5=V_2-V_3=V_8-V_7$$
$$V_4=V_1-V_2=V_7-V_6$$
But since components in parallel have the same voltage and the resistance is the same everywhere then we can assume
$$V_2=V_3=V_1$$
Resulting in
$$V_5=V_4=0$$
Thus the resistance in each leg is
$$R_{1,6}=R_{2,7}=R_{3,8}=2R$$
And total resistance is given by
$$\frac{1}{R_{tot}}=\frac{3}{2R}\Rightarrow R_{tot} = \frac{2R}{3}$$
best method is to solve it by symmetry along the vertical mid line. so fold it along this line. now use balanced wheatstone bridge,,,, ask if u still have problem

### Re: GR0877 #68

Posted: Mon Oct 08, 2012 12:35 pm
I solved this using the assumption that current follows the path(s) of least resistance. Is this generally a valid assumption to make for non-trivial resistor assemblies (i.e. not of the form $$R_s=\sum R_i$$ or $$\tfrac{1}{R_p}=\sum\tfrac{1}{R_i}$$)?

### Re: GR0877 #68

Posted: Mon Oct 08, 2012 2:39 pm
Minovsky wrote:I solved this using the assumption that current follows the path(s) of least resistance. Is this generally a valid assumption to make for non-trivial resistor assemblies (i.e. not of the form $$R_s=\sum R_i$$ or $$\tfrac{1}{R_p}=\sum\tfrac{1}{R_i}$$)?
It's never true that all current goes through the path of least resistance; the resistance and current are related by Ohm's law. However, if you simply mean that the paths electrons take through the above circuit ignore the horizontal resistors, that's true due to the symmetry of the resistor set-up. If it weren't symmetric, there could be cross-talk between the three vertical wires through those resistors. However, since they have equal resistance, there's no reason for current to move from one vertical wire to another; in other words, there is equal current going right-left and left-right through the horizontal resistors, so the net current through them is 0, and there is no voltage drop due to these resistors.