Five capacitors 10 $\mathrm{\mu }$F capacity each are connected to a D.C. potential of 100 volts as shown in fig. The equivalent capacitance between the points A and B will be equal to
 

The given circuit is equivalent to Wheat stone's bridge. when the bridge is balanced, the upper two condensers between A and B will be in series. Their resultant capacitance would be
$\frac{1}{{\mathrm{C}}^{\mathrm{\prime }}}=\frac{1}{10}+\frac{1}{10}=\frac{2}{10}=\frac{1}{5}$
or     ${\mathrm{C}}^{\mathrm{\prime }}=5\mathrm{\mu F}$
Similarly, the lower two condensers between A and B are also in series. Their resultant capacitance would be 
$\frac{1}{{\mathrm{C}}^{\prime \prime }}=\frac{1}{10}+\frac{1}{10}=\frac{2}{10}=\frac{1}{5}$
or    ${\mathrm{C}}^{\prime \prime }=5\mathrm{\mu F}$
The capacitor C' and C'' will then be in parallel. Hence the effective capacitance between A and B will be $\mathrm{C}={\mathrm{C}}^{\mathrm{\prime }}+{\mathrm{C}}^{\prime \prime }=5+5=10\mathrm{\mu F}$