Question

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 them in series
B) Connecting them in parallel
C) Connecting two in series and one in parallel
D) Connecting two in parallel and one in series

Answer 1

Use the formula of equivalent capacitance of both series and parallel connection to check whether they are connected in series or parallel or two in series, one in parallel or two in parallel, one in series. Equivalent capacitance, of which case is similar to the given effective capacitance, is the required answer.

Complete step by step solution:
Let, the three capacitors are ${{C}_{1}}$ , ${{C}_{2}}$ and ${{C}_{3}}$ .
By the given problem, ${{C}_{1}}={{C}_{2}}={{C}_{3}}=4\mu F$ .
Case-1: let the given three capacitors are connected in series. Then, the total capacitance will be,
${{C}_{tot}}=\dfrac{1}{\dfrac{1}{{{C}_{1}}}+\dfrac{1}{{{C}_{2}}}+\dfrac{1}{{{C}_{3}}}}$ where ${{C}_{1}}$ , ${{C}_{2}}$ , ${{C}_{3}}$ are the three given capacitors.
Therefore, using the values of the three capacitors, we get,
$\Rightarrow {{C}_{tot}}=\dfrac{1}{\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}}\mu F$
$\Rightarrow \dfrac{1}{\dfrac{(1+1+1)}{4}}\mu F$
$\Rightarrow \dfrac{4}{3}\mu F$
Case-2: let the given three capacitors are connected in parallel. Then, the total capacitance will be,
${{C}_{tot}}={{C}_{1}}+{{C}_{2}}+{{C}_{3}}$
Now, using the values of the given three capacitors, we get,
$\Rightarrow {{C}_{tot}}=(4+4+4)\mu F$
$\Rightarrow 12\mu F$
Case-3: let the given three capacitors are connected as two in series and one in parallel. Then, the total capacitance of the first two capacitors connected in series will be,
${C'}=\dfrac{1}{\dfrac{1}{{{C}_{1}}}+\dfrac{1}{{{C}_{2}}}}$
Therefore, ${C'}=\dfrac{1}{\dfrac{(1+1)}{4}}\mu F$
$\Rightarrow \dfrac{4}{2}\mu F$
$\Rightarrow 2\mu F$
Now, if the third capacitor is connected in parallel with the first two, the total effective capacitance will be,
${{C}_{tot}}={C'}+{{C}_{3}}$
Now, as ${C'}=2\mu F$ and ${{C}_{3}}=4\mu F$ we get,
$\Rightarrow {{C}_{tot}}=(2+4)\mu F$
$\Rightarrow 6\mu F$
Therefore, the total capacitance in this case is $6\mu F$.
Case-4: let the given three capacitors are connected as two in parallel and one in series. Then, the total capacitance of the first two capacitors connected in parallel will be,
${C'}={{C}_{1}}+{{C}_{2}}$
Therefore, ${C'}=(4+4)\mu F$
$\Rightarrow 8\mu F$
Now, if the third capacitor is connected in series with the first two, the total effective capacitance will be,
${{C}_{tot}}=\dfrac{1}{\dfrac{1}{{C'}}+\dfrac{1}{{{C}_{3}}}}\mu F$
$\Rightarrow \dfrac{1}{\dfrac{1}{8}+\dfrac{1}{4}}\mu F$
$\Rightarrow \dfrac{8}{3}\mu F$
As it is given, that the effective capacitance is $6\mu F$,

So, the correct answer is (C), Connecting two in series and one in parallel.

Note: In series connection, capacitances diminish, while capacitances add-in parallel connection. The total capacitance in case of series connection is less than the capacitance of any one individual capacitor’s capacitance.