question_answer

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

A) \[6\,\mu F\]

B) \[12\,\mu F\]

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

D) \[24\,\mu F\]

Correct Answer: A

Solution :

[a] 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

\[C=C'+C''=3+3=6\,\mu F\]