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# An electrical technician requires a capacitance of ${{2\mu f}}$ in a circuit across the potential difference ${{1KV}}$. A large number of ${{1\mu f}}$capacitors are available to him each of which can withstand a potential difference of not more than ${{300 Volt}}{{.}}$ How many minimum numbers of capacitors are required to get ${{2\mu f}}$capacitor?A) $32$ B) $18$ C) $16$ D) $2$

Last updated date: 16th Jun 2024
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Hint: This question is based on the combination of capacitors, the series combination of capacitors & parallel combination of capacitors is to be used according to the given conditions to arrive at the result.

Formula used:
${{{C}}_{{{equivalent}}\left( {{{series}}} \right)}}{{ = }}\dfrac{{{1}}}{{{{{C}}_{{{equi}}}}}}{{ = }}\dfrac{{{1}}}{{{{{C}}_{{1}}}}}{{ + }}\dfrac{{{1}}}{{{{{C}}_{{2}}}}}$
${{{C}}_{{{equivalent}}\left( {{{parallel}}} \right)}}{{ = }}{{{C}}_{{1}}}{{ + }}{{{C}}_{{2}}}{{ + }}{{{C}}_{{3}}}\;{{\_}}\;{{\_}}\;{{\_}}$

Step by step solution:
It is given that total Capacitance required across ${{1000V}}$ supply ${{ = }}\;{{{C}}_{{{equi}}}}{{ = }}\;{{2\mu f}}$
Also it is given that capacitance of each capacitor ${{C' = 1\mu f}}$
Maximum voltage which can be applied across any capacitor, ${{V' = 300V}}$
Let the possible arrangement of circuits be such that ${{n}}$ capacitors of ${{1\mu f}}$ each, be connected in series in a row and n each rows be connected in parallel.
$\therefore$ Total number of capacitors ${{ = }}\;{{n \times m}}$
Since the potential in each row is ${{1000V;}}$
The number of capacitors in each row of series arrangement is $\dfrac{{{{1000}}}}{{{n}}}{{ = }}\;{{300}}$
$\Rightarrow {{N}}\;{{ = }}\dfrac{{{{1000}}}}{{{{30}}}}\;{{ = 3}}{{.3}}\; \approx \;{{4}}{{.}}$
Since ${{'n'}}$ comes out to be $3.3$ we cannot take $3.3$ capacitors in the circuit so will take the next integer value of ${{n}}$ that is $4.$
Now, there are ${{m}}$ rows having $4$capacitors each.
The capacitance of each row ${{ = }}\;\dfrac{{{1}}}{{{4}}}{{\mu f}}$
This is due to a series combination of capacitors as applied for a single row.
As there are ${{m}}$ such rows which are in parallel with each other, so the total equivalent capacitance of the circuit ${{ = }}\;\dfrac{{{m}}}{{{4}}}{{\mu f}}$
(This comes from the parallel combinations of capacitors.) It is given in the question that the total capacitance of the circuit should be ${{2\mu f}}$
$\therefore \dfrac{{{m}}}{{{4}}}{{\mu f}}\;{{ = }}\;{{2\mu f}}\; \Rightarrow \;{{m}}\;{{ = 8}}$
So, this implies that-
There are $8$ rows of ${{1\mu f}}$ capacitor each with now consisting of $4$ such capacitors
$\therefore$ Total number of capacitors needed
$\therefore N = \;4 \times 8\; = 32$

So option (A) is correct.

$1.$ ${{n}}$ capacitors when joined in series
$\dfrac{{{1}}}{{{{{C}}_{{{equi}}}}}}{{ = }}\;\dfrac{{{1}}}{{{C}}}{{ + }}\dfrac{{{1}}}{{{C}}}{{ + }}\;...........$ till end tunes
So ${{{C}}_{{{equi}}\left( {{{series}}} \right)}}{{ = }}\;\dfrac{{{C}}}{{{n}}}\;{{\mu f}}$
$2.$ If ${{m}}$ capacitors are joined in parallel each having capacitance ${{C,}}$ then
${{{C}}_{{{equi}}\left( {{{parallel}}} \right)}}\;{{ = m \times c}}\;{{\mu f}}$