The effective capacitance between points X and Y shown in fig. assuming C₂ = 10 $\mathrm{\mu }$F and that the other capacitors are all 4.00 $\mathrm{\mu }$F is

The equivalent circuit is shown in fig. If a battery is connected between X and Y, the points P and Q are at the same potential because

${\mathrm{C}}_1={\mathrm{C}}_4={\mathrm{C}}_3={\mathrm{C}}_5=4\mathrm{\mu F}$
Hence no charge flows through C_2. Now C₁ and C₅ are in series. Hence $\frac{1}{{\mathrm{C}}^{\mathrm{\prime }}}=\frac{1}{{\mathrm{C}}_1}+\frac{1}{{\mathrm{C}}_5}$
or     ${\mathrm{C}}^{\mathrm{\prime }}=\frac{{\mathrm{C}}_1{\mathrm{C}}_5}{{\mathrm{C}}_1+{\mathrm{C}}_5}=\frac{4\times 4}{4+4}=2\mathrm{\mu F}$
Similarly, C₄ and C₃ are in series and their equivalent capacitance C" is also 2 $\mathrm{\mu }$F. Further, C' and C" are in parallel. Hence net capacitance between X and $\mathrm{Y}={\mathrm{C}}^{\mathrm{\prime }}+{\mathrm{C}}^{\prime \prime }=2\mathrm{\mu F}+2\mathrm{\mu F}=4\mathrm{\mu F}$