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