Question

minimum no. of capacitor of 2uf each required to obtain a capacitance of 5uf

Answer 1

ans : 4

if two capacitors will be in series and two capacitors will be in parallel as shown in figure then, equivalent capacitance will 5 microfarad.

Let see AB and BC capacitors are in series
so, equivalent capacitance {X} of it = 2microfarad/2 = 1 microfarad { in series 1/Ceq = 1/C1 + 1/C2}

DE and GF capacitors are in parallel ,
so, equivalent capacitance { Y} of it = 2 + 2 = 4 microfarad { in parallel, Ceq = C1 + C2 }

now, X and Y capacitors are in parallel then,
Ceq = 4 + 1 = 5 microfarad.

hence, minimum number of capacitors = 4

Answer 2

"The minimum number of capacitors required are four.

Solution:

We know that the total capacitance of capacitors connected in series is

$\frac{1}{C_{e q}}=\sum \frac{1}{C_{n}}$

The total capacitance of capacitors connected in parallel is

$C_{e q}=\sum C_{n}$

Thus, in order to obtain$C_{e q}=5 \mu F$, a combination of series and parallel capacitors are required.

The minimum $C_{e q}$ that can be obtained in parallel combination is $4 \mu \mathrm{F}$, that is when two capacitors are connected in parallel.

Thus, the remaining $C_{eq} which is 1 \mu \mathrm{F}$ has to be obtained from the series combination which is connected parallelly to the parallel combination.

n=2

$\frac{1}{C_{e q}}=\sum \frac{1}{\frac{1}{2}+\frac{1}{2}}$

$\frac{1}{C_{e q}}=1 \mu \mathrm{F}$

Two capacitors have to be connected in series.

So, that the total $C_{eq}$ is a simple algebraic sum of $4 \mu \mathrm{F}+1 \mu \mathrm{F}=5 \mu \mathrm{F}$"