Question

Three capacitors each of 12farad are available The minimum and maximum capacitance which may be obtained are

Answer 1

Answer:

Max= 36 farad(parallel)

Min= 4 farad(series)

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Answer 2

Given,

3 capacitor each of capacitance, C = 12 farad

To Find,

The minimum and maximum capacitance.

Solution,

The minimum capacitance can be obtained by connecting all capacitors in series. Therefore,

$\frac{1}{C}$ = $\frac{1}{12} +\frac{1}{12} +\frac{1}{12}$=$\frac{1}{4}$

C= 4F

The maximum capacitance can be obtained by connecting all capacitors in parallel. Therefore,

c = 12+12+12=36F

∴ Hence the minimum capacitance is 4F and the maximum capacitance obtained is 36F.