Question

Given a number of capacitors labelled as \[(8\mu F,250V).\] Find the minimum number of capacitors needed to get an arrangement equivalent to $(16\mu F,1000V):$
$
(A){\text{ 4}} \\
{\text{(B) 16}} \\
(C){\text{ 32}} \\
(D){\text{ 64}} \\
$

Answer 1

Hint: Capacitors are an electronic component which stores energy with the charge and potential difference across the terminals. When one capacitor is attached back to back with others they are said to be in series, whereas when both the terminals of the capacitor are attached with both the terminals of the other capacitor, they are said to be in parallel capacitors.

Complete step by step answer:
Given that the capacitors are rated as $250$volts and these capacitors are needed to be equivalent to $1000$ volts.
Therefore, the minimum number of the capacitors in each of the row is $ = \dfrac{{1000}}{{250}} = 4$
Therefore, the number of capacitors to be connected is $ = 4$ ...... (a)
Now, when these four capacitors are connected in series the effective capacitance is –
$\dfrac{1}{{{C_{effective}}}} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}} + \dfrac{1}{{{C_3}}} + \dfrac{1}{{{C_4}}}$
Place the value of capacitor, $C = 8\mu F$
$\dfrac{1}{{{C_{effective}}}} = \dfrac{1}{8} + \dfrac{1}{8} + \dfrac{1}{8} + \dfrac{1}{8}$
Simplify the above equation-
$
\dfrac{1}{{{C_{effective}}}} = \dfrac{4}{8} \\
\Rightarrow\dfrac{1}{{{C_{effective}}}} = \dfrac{1}{2} \\
\therefore{C_{effective}} = 2\mu F \\
$
To get the capacitance of$16\mu F$,
The number of effective combinations is $ = \dfrac{{16}}{2} = 8$ ........ (b)
From the equations (a) and (b), we get –
The minimum number of capacitors required to get an arrangement is the product of the number of capacitors for each series combination and number of such combinations.
$\therefore 4 \times 8 = 32{\text{ capacitors}}$
The required answer is – The minimum number of capacitors needed to get an arrangement equivalent to $(16\mu F,1000V)$ is $32$ capacitors.
Hence, from the given multiple choices – the option C is the correct answer.

Note: The above problem can be solved by using other method as below –
Let “m” be the capacitors joined in the series and “n” be the capacitors joined in parallel.
Therefore, the capacitor $C = \dfrac{8}{m}$ and
The equivalent capacitance is ${C_{equi.}} = n \times \dfrac{8}{m} = 16$
$ \Rightarrow n = 2m$
Now, the potential for the given arrangement is $mV = 1000 \Rightarrow m = \dfrac{{1000}}{{250}} = 4$
Now place the value of “m” to get the value of “n”
$n = 2m \Rightarrow n = 2(4) = 8$
Hence, the total number of capacitors required$ = mn = 8 \times 4 = 32$