The effective capacitance between points x and y of figure shown is: [AIPMT 1999]

[a] The given circuit can be redrawn as It is a balanced Wheatstones bridge \[\left( as\frac{{{C}_{AB}}}{{{C}_{BD}}}=\frac{{{C}_{AC}}}{{{C}_{CD}}}==\frac{6}{6} \right)\] So, POTENTIAL of B and C are equal and \[20\,\mu F\] capacitor is ineffective. The SIMPLIFIED circuit is shown as: Capacitors of \[6\,\mu F\] and \[6\,\mu F\] in upper ARMS are in series order, so \[C'=\frac{6\times 6}{6+6}=\frac{36}{12}=3\,\mu F\] SIMILARLY, \[6\,\mu F\] and \[6\,\mu F\] in lower arms are in series order, so \[C''=\frac{6\times 6}{6+6}=3\mu F\] Now, C and C are in parallel order, hence \[C=C'+C''=3+3=6\,\mu F\]