Question

Three capacitors each of $4\mu F$ are to be connected in such a way that the effective capacitance is $6\mu F$. This can be done by connecting them:
A. two in parallel and one in series
B. all in parallel
C. all in series
D. two in series and one in parallel

Answer 1

Hint: A capacitor is a two terminal component that stores electrical energy in the form of potential energy, and later discharges them. This property is called the capacitance of the capacitor. The capacitors can be connected in series or in parallel with respect to each other.
Formula used:
$\dfrac{1}{C_{s}}=\dfrac{1}{C_{1}}+\dfrac{1}{C_{2}}$ and $C_{p}=C_{1}+C_{2}$

Complete answer:
A capacitor can store electrical energy, and behaves as a temporary battery. They are used mainly to maintain the power supply while batteries are being changed. It can also store information in the form of binary digits. It is the main component used in full wave and half wave rectifiers. (symbol: F), named after the English physicist Michael Faraday. A 1 farad capacitor, when charged with 1 coulomb of electrical charge, has a potential difference of 1 volt between its plates.
Given that $C_{1}=C_{2}=C_{3}=C=4\mu F$ and $C_{net}=6\mu F$

Clearly there are four possibilities,
1. All the three capacitances are connected in series. Then the resultant capacitance will become $\dfrac{1}{C_{net}}=\dfrac{1}{C_{1}}+\dfrac{1}{C_{2}}+\dfrac{1}{C_{3}}=\dfrac{3}{4}$
$\implies C_{net}=\dfrac{4}{3}\mu F$
Clearly, not the required value. Hence C is incorrect.

2. All the three capacitances are connected in parallel. Then the resultant capacitance will become $C_{net}=C_{1}+C_{2}+C_{3}=3\times 4=12\mu F$
Clearly, not the required value. Hence option B is incorrect

3. Two capacitance in parallel and one in series with the resultant, then we have, $C_{net}=C\prime+C$
$\implies\dfrac{1}{C\prime}=\dfrac{1}{C_{1}}+\dfrac{1}{C_{2}}=\dfrac{2}{4}=\dfrac{1}{2}$
$\implies C\prime=2\mu F$
$\implies C_{net}=2+4=6\mu F$
Clearly, this is the required answer.

4. Two capacitances in series and one in parallel with the resultant.
So, the correct answer is “Option A”.

Additional Information:
We know that the charge $Q$ produced due to capacitance $C$ and potential difference $V$ is given as $Q=CV$. Also, the energy of the capacitor is $E=\dfrac{1}{2}CV^{2}$.
We know that $C=\dfrac{KA\epsilon_{0}}{d}$ where, $A$ is the area of the capacitance and $d$ is the width of the dielectric $K$ and the electric constant $\epsilon_{0}=8.854\times 10^{-12}Fm^{-1}$

Note:
The series of capacitors is the sum of reciprocal of its individual capacitors, whereas in resistance the parallel is the sum of reciprocal of its individual resistors. Also remember that the net capacitance due to series connection is always smaller than the parent capacitance and similarly, the net capacitance due to series connection is always greater than the parent capacitance.