What kind of symmetry I can use to get the result? I think I have to redraw a circuit, but how?
34.7k 34 34 gold badges 146 146 silver badges 269 269 bronze badges asked Mar 27 at 13:19 weycdicdib weycdicdib 29 2 2 bronze badges\$\begingroup\$ As drawn here, you could argue that nothing is connected to the terminals other than the 5 Ω and 2 Ω resistors, and all the others are just floating, because there are no junction dots drawn. That would give you an equivalent resistance of 7 Ω. I don't think your teacher would be satisfied with that answer, though. \$\endgroup\$
Commented Mar 27 at 14:14What kink of symmetry I can use to get the result? I think I have to redrew a circuit but how?
The top array of 3 resistors is identical to the lower array of 3 resistors AND they connect to three identical nodes. This means you can parallel them and reduce the resistor count by three: -
Then it's down to using your preferred circuit theory. Mine is source transformation but, yours may be different. But you can do a bit more when exploring the symmetry idea.
For instance, can you see that because there are the same values of resistance in opposite corners, if you split the vertical middle resistor into two separate equal resistors, the mid-point would be exactly half the applied voltage (1 volt applied in this example): -
So, you can reduce this circuit further and get the same upper node voltage as per my simulation: -
Of course I'm using the simulator as an aid (because it is trustworthy) but, do you see that we have lost a further two resistors (but only gained one voltage source).
It's a simpler case to solve for the currents in R1, R2 and R5A and, of course, the same currents will flow in their opposing corner counterparts.
