Condition for the use of perpendicular symmetry in solving resistance problems

It is possible to use parallel symmetry even when there is more than one battery, but what about perpendicular symmetry?

I tried to solve some examples using KVL but it didn't work out as I expected. If it is not possible to use it in all the cases then in which cases is it applicable? Also I didn't quite understand how we just separate the wires from a junction if we have established perpendicular symmetry; kindly provide an intuitive proof.

Finally, this all came in my mind because of the following problem: (The question was too long to solve by using KVL as there would be 7 loops)

NOTE: The resistance of the middle resistance is not specified; it is R (not 2 ohm).

• Try using The Nodal Analysis method. May 15 '17 at 12:49
• I don't believe I've heard the phrase "perpendicular symmetry" before; do you mean connected in series? May 15 '17 at 16:33
• @HalHollis, by perpendicular symmetry, I mean that the circuit is symmetrical along the line perpendicular to the main branch. For example, youtube.com/watch?v=kygPc0TIAP8. May 16 '17 at 12:30

This is a homework problem, so I'm not going to completely solve it for you but I'll give some hints that might apply to other problems.

First imagine you did use all Kirchoff's laws, and solved for $I$ the current in the middle branch. Kirchoff's laws are linear, so in terms of $V_{bottom}=100V$ and $V_{top}=50V$, your solution will have the form $$I=A V_{bottom}+B V_{top}$$ for some coefficients $A,B$ which we don't know until we solve the problem.

So what this means is that if we solve a new circuit where we set the bottom voltage to zero (i.e. replace the battery with a plain wire) the new current in the middle branch $I_{top}$ is $$I_{top}=B V_{top}.$$ And similarly if we replace the top battery and solve for the middle branch $$I_{bottom}=A V_{bottom}.$$

But if we take out the top battery, it is easy to see by symmetry or by thinking about potential drops that there will be no current in the middle branch. $$I_{bottom}=A V_{bottom}=0.$$

So $I=I_{top}$. We can pretend the bottom battery is not there at all as far as the middle branch is concerned!

Then by symmetry we can not only replace the bottom battery by a wire, we can completely take out that branch since the potentials it is connecting would be equal anyway. This then lets you use series and parallel resistor rules to simplify the circuit even more, and by this point the circuit has been simplified enough to solve in a couple lines with Kirchoff's laws.

• and what about the condition for perpendicular symmetry. In this case, you solved the problem by breaking it up into smaller sub - problems but if there are 4 cells then what? Also, could this question have been simply approached by using perpendicular symmetry? May 16 '17 at 12:35
• one thing which I don't understand is that why is the value independent of R (not specified) ? May 16 '17 at 15:16
• I hadn't heard the term 'perpendicular symmetry' either, but I watched the short video and don't think you should apply the principle (directly) when there are extra batteries in the circuit. If the top battery was not there yes you could apply this idea to solve the circuit, and in a sense you can say I'm using perpendicular symmetry when I conclude $I_{bottom}=0$. But since the top battery is there, current is flowing inwards to the center in the top two resistors, not directly across from right to left like is assumed in the video. May 17 '17 at 6:56
• The value isn't independent of R. There is nonzero current flowing through the middle branch and it will depend on the values of the resistors. May 17 '17 at 6:59