# Doubts in circuit analysis [closed]

I am really confused here

how is the potential at O found? Resistance is equal at all resistors this is an illustration in my book, they have not explained why voltage at O is 50V, just stated that its symmetry, i will accept this reason but the RP resistor and SQ resistor makes it hard to justify the potential at O as 50V it seems that i can't see the symmetry here, can you help me ? [I am not asking for the answer, i know how to find the answer what i need to know is the reason how symmetry helps in calculation]

• Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.
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Commented Jun 30 at 9:38
• As explained in the answer below, if you swap left and right you get exactly the same circuit with current going in the opposite direction (because all resistors have equal resistances) so the circuit is left/right symmetric. Commented Jul 1 at 16:48
• i have found the answer, cut the circuit into half, along the imaginary line (perpendicular to AB)equidistant from the terminals now split the circuit, the second part is just the reverse of the first part, since resistors can be connected either way, we can reverse the second part so we get the same resistance twice in parallel now it becomes easy. the explanation is voltage drop from either side untill the imaginary axis is same so the points coinciding with the imaginary axis are equipotential. This simplifies and explains my doubt. Commented Jul 23 at 17:24

Maybe I am loosing some detail here, but assuming that all resistances are equal, and looking at the symmetry of the circuit, all dpps and currents should be the same in module if you flip the circuit horizontally, i.e. if you invert the polarity of the battery. This can only be the case if $$i_{AO}=i_{OB}$$, hence $$dpp_{AO}=dpp_{AB}/2$$.