question_answer

                                The effective capacitance between points X and y of figure shown is:                                                    

A)                 \[6\,\mu F\]      

B)                 \[12\,\mu F\]    

C)                 \[18\,\mu F\]                    

D)                 \[24\,\mu F\]

Correct Answer: A

Solution :

                          The given circuit can be redrawn as                                 It is a balanced Wheatstone?s 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 = \[6\,\mu F\]