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 = \[6\,\mu F\]You need to login to perform this action.
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