In the circuit shown, the equivalent capacitance between the points A and B is
Answer & explanation
Correct answer: option 4
Rearranging the circuit, the points E and D are at the same potentital (by symmetry). Then the capacity between E and D can be removed.
$\frac{1}{C'}=\frac{1}{C}+\frac{1}{C}$
∴ C' = C/2
$\frac{C}{2}$ and $\frac{C}{2}$ are in parallel.
Hence $C_{eq}=\frac{C}{2}+\frac{C}{2}$ = C
∴ OPTION 4 is correct