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: [AIPMT 2003]

A) connecting two in series and one in parallel

B) connecting two in parallel and one in series

C) connecting all of diem in series     

D) connecting all of them in parallel

Correct Answer: A

Solution :

[a] 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 can be shown as

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 the 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.