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The capacitor two capacitors in series give a total capacitance of 0.03 μF. If they are connected in parallel, the total capacitance is 0.16 μF. Find the values of both capacitors.
1. 0.02 μF 0.14 μF
2. 0.08 μF 0.08 μF
3. 0.04 μF 0.12 μF
4. 0.06 μF 0.10 μF

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Correct Answer - Option 3 : 0.04 μF 0.12 μF

Concept:

When capacitors are connected in parallel, the total capacitance is the sum of the individual capacitors' capacitances.

\(C_{{eq}}(parallel) = {C_1} + {C_2} + {C_3} + \ldots + {C_4}\)

When capacitors are connected in series, the total capacitance is less than the least capacitance connected in series.

\(\frac{1}{{{C_{eq}(series)}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}} + \ldots + \frac{1}{{{C_n}}}\)

Calculation:

The capacitor two capacitors in series give a total capacitance of 0.03 μF

\( \Rightarrow \frac{{{C_1}{C_2}}}{{{C_1} + {C_2}}} = 0.03\)

If they are connected in parallel, the total capacitance is 0.16 μF.

C1 + C2 = 0.16

C1C2 = 0.16 × 0.03 = 4.8 × 10-3

C1 = 0.04 μF and C2 = 0.12 μF

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