We wish to obtain a capacitance of 5μF, by using some capacitors, each of 2μF. Then, the minimum number of capacitors required is
A
3
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B
4
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C
5
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D
Not possible
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Solution
The correct option is B4 As we can only use 2 μF capacitors to achieve the total capacitance of 5 μF. So, the best combination that can be made is 5μF = 2×2μF+1μF
For this, we need to connect two capacitors of 2 μF each in parallel and connect another two capacitors of 2μF each in series and further connect both set in parallel to achieve the capacitance of 5μF.
Its initial digramatic representation can be shown as
Upon further connecting all these capacitors in the systematic manner, the final circuit can be shown as below:
To acheive 1 μF we need to connect 2 capacitors of 2 μF each in series. 11μF=12μF+12μF
So, from the diagram above, we can say the minimum number of capacitors required to acheive the capacitance of 5μF is 4.