ok consider the ckt in the attachment...if i label the junctions of the ckt starting from the top left corner as 1,2,3,4 going clockwise...then in this case is the excitation symmetrical?..this ckt surely cannot be folded..but if i rotate the square by 180 degrees then the junctions become 1->3, 2->4, 3->1 & 4->2....under this transformation the ckt remains similar to the original.then can i say this ckt is symmetrical ? is the current distribution shown correct?..if the ckt is symmetrical then current thru the 10 ohm resistors shud be same and current thru 5 ohm resistors shud be same. Is my analysis correct? plz help

You can't use symmetry to fold it as the Electrician has said, but there is a pattern there that allows you to sort of unfold it. See if you can see why circuit A is the same as circuit B and why the currents will be the same if you break the connections as in C, which is much easier to analyse.

 

hey thats a cool trick!! thanks!..i understnd this now...but have they written " by symmetry" in my book i cnt make out!..weirrd..thanks for helping me out though!!

 

btw i checked out that link..in my sum if i do rotate the figure then the "schematic" does not change...i think congruency does hold then!!!

 

You can't use symmetry to fold it as the Electrician has said, but there is a pattern there that allows you to sort of unfold it. See if you can see why circuit A is the same as circuit B and why the currents will be the same if you break the connections as in C, which is much easier to analyse.

There's only one little problem with this.

The end-to-end resistance of circuit A and circuit B is 7Ω, but the end-to-end resistance of circuit C is 11.66667Ω.

The nodes that get connected when you go back from C to B don't have the same voltage, so making the connection changes the resistance of the circuit.

 

There's only one little problem with this.

The end-to-end resistance of circuit A and circuit B is 7Ω, but the end-to-end resistance of circuit C is 11.66667Ω.

The nodes that get connected when you go back from C to B don't have the same voltage, so making the connection changes the resistance of the circuit.

That's embarrassing - you're right! The two voltage dividers are the same, but are connected at different points. It seemed like a great idea!

 

oh sheeesh....anyways thanks guys!