question_answer

Three capacitors each of capacity\[4\mu F\]are to be connected in such a way that the effective capacitance is\[6\mu F\]. This can be done by

A)  connecting two in series and one in parallel

B)  connecting two in parallel and one in series

C)  connecting all of them in series

D)  connecting all of them in parallel

Correct Answer: A

Solution :

                 Key Idea: In series order, the net capacitance is, \[\frac{1}{C}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{2}}}+\frac{1}{{{C}_{3}}}+.......\] In parallel order, the net capacitance is, \[C={{C}_{1}}+{{C}_{2}}+{{C}_{3}}+.....\] We have given, \[{{C}_{1}}={{C}_{2}}={{C}_{3}}=4\mu F\]  The network of three capacitors is shown. Here,\[{{C}_{1}}\]and\[{{C}_{2}}\]are in series and the combination of two is in parallel with\[{{C}_{3}}\]. \[{{C}_{net}}=\frac{{{C}_{1}}{{C}_{2}}}{{{C}_{1}}+{{C}_{2}}}+{{C}_{3}}\] \[=\left( \frac{4\times 4}{4+4} \right)+4\] \[=2+4=6\mu F\]               The corresponding network is shown. Here,\[{{C}_{1}}\]and\[{{C}_{2}}\]are in parallel and this combination is in series with\[{{C}_{3}}\]. So,          \[{{C}_{net}}=\frac{({{C}_{1}}+{{C}_{2}})\times {{C}_{3}}}{({{C}_{1}}+{{C}_{2}})+{{C}_{3}}}\]                      \[=\frac{(4+4)\times 4}{(4+4)+4}=\frac{32}{12}\]                 \[=\frac{8}{3}\mu F\]  The corresponding network is shown. All of three are in series. So,       \[\frac{1}{{{C}_{net}}}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{2}}}+\frac{1}{{{C}_{3}}}\]                 \[=\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{3}{4}\] \[\therefore \]  \[C=\frac{4}{3}\mu F\]  The corresponding network is shown. All of them are in parallel. So           \[{{C}_{net}}={{C}_{1}}+{{C}_{2}}+{{C}_{3}}\] \[=4+4+4=12\mu F\] Hence, only choice  is correct.