Question

To form a composite $16\,\mu F,1000\,V$ capacitor from a supply of identical capacitors marked $8\,\mu F,250\,V$ we require a minimum number of capacitors
A. $40$
B. $32$
C. $8$
D. $2$

Answer 1

Hint: In order to solve this question we need to understand the definition of capacitor which states that a capacitor is a device that is used to store energy in the form of charges. Capacitance depends on the geometry of plates used and these capacitors can be connected in circuit by two means one known as series combination and other known as parallel combination. In series combination charge stored across each capacitor is same and in parallel combination voltage developed across each capacitor is same.

Formula used:
Equivalent capacitance of series combination:
If capacitor ${C_1}$ and ${C_2}$ are connected in series then Equivalent capacitance in series is $\dfrac{1}{C} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}}$
And voltage developed across first capacitor is given by ${V_1} = \dfrac{{{V_t}{C_2}}}{{{C_1} + {C_2}}}$ , and across second capacitor is ${V_2} = \dfrac{{{V_t}{C_1}}}{{{C_1} + {C_2}}}$ where ${V_t}$ is total voltage of circuit.
Equivalent capacitance of Parallel combination:
If capacitor ${C_1}$ and ${C_2}$ are connected in Parallel the Equivalent capacitance in parallel is \[C = {C_1} + {C_2}\]
And the charge develop across each capacitor is given by ${q_1} = \dfrac{{Q{C_1}}}{{{C_1} + {C_2}}}$ and across second capacitor is ${q_2} = \dfrac{{Q{C_2}}}{{{C_1} + {C_2}}}$ where $Q$ is total charge developed.

Complete step by step answer:
Now, according to the given question each capacitor value is $8\,\mu F,250\,V$. So if we connect $4$ such capacitor in a series then make $8$ parallel combination of such series combination we can achieve our goal of $16\,\mu F,1000\,V$ Equivalent series capacitance is given by
$\dfrac{1}{{{C_1}}} = \dfrac{1}{8} + \dfrac{1}{8} + \dfrac{1}{8} + \dfrac{1}{8}$
$\Rightarrow {C_1} = \dfrac{8}{4} \\
\Rightarrow {C_1}= 2\mu F$

Similarly equivalent capacitance of each such series combination will be equal and can be written as, ${C_2} = {C_3} = {C_4} = {C_5} = {C_6} = {C_7} = {C_8} = 2\mu F$
Hence net equivalent capacitance is,
$C = {C_1} + {C_2} + {C_3} + {C_4} + {C_5} + {C_6} + {C_7} + {C_8}$
$\Rightarrow C = 8 \times (2\mu F) \\
\Rightarrow C= 16\mu F$
Also voltage developed across each capacitor in series is $V = \dfrac{{1000}}{4} = 250V$. As all capacitance values are equal. So in total we need $32$ capacitors for our desired result

Hence, the correct option is B.

Note: It should be remembered that for sake of simplicity here parallel plate capacitors are used but cylindrical plate capacitors could also be used still the result would be the same. Also here capacitors are considered as ideal that’s why no energy loss during charging but practically energy would be lost due to heating of plates.