Final answer:
The total capacitance of the circuit with three capacitors is 0.17 μF.
Step-by-step explanation:
The total capacitance of a circuit containing three capacitors with capacitances of 0.02 microfarad, 0.05 microfarad, and 0.10 microfarad can be found using the equation for capacitance in parallel.
Cp = C₁ + C₂ + C₃
Substituting the values of the capacitances, we get Cp = 0.02 µF + 0.05 µF + 0.10 µF = 0.17 µF.
Therefore, the total capacitance of the circuit is 0.17 μF.