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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.

Additional Information:
$1.$ ${{n}}$ 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
${{{C}}_{{{equi}}\left( {{{parallel}}} \right)}}\;{{ = m \times c}}\;{{\mu f}}$

Note: The equivalent capacitance is defined as the capacitance of one single capacitor which should replace the given set of capacitors such that the charge and the voltage across the system remains the same.