question_answer

 In the given network capacitance, \[{{C}_{1}}=10\mu \,F,\,{{C}_{2}}=5\mu \,F\] and \[{{C}_{3}}=4\mu \,F\]. What is the resultant capacitance between A and B                                                         [Pb. PMT 1999]

A)            2.2\[\mu \,F\]

B)            3.2\[\mu \,F\]

C)            1.2\[\mu \,F\]

D)            4.7\[\mu \,F\]

Correct Answer: B

Solution :

                   \[C=\frac{({{C}_{1}}+{{C}_{2}})\times {{C}_{3}}}{({{C}_{1}}+{{C}_{2}})+{{C}_{3}}}\]\[=\frac{(5+10)\times 4}{5+10+4}=\frac{60}{19}=3.2\,\mu F\]