What is the total capacitance of a circuit containing three capacitors in parallel with capcitances of .02 microfarad, .05 microfarad, and .10 microfarad, respectively?

What is the total capacitance of a circuit containing three capacitors in parallel with capcitances of .02 microfarad, .05 microfarad, and .10 microfarad, respectively?

• A. .170 uF
• B. .0125 uF
• C. .025 uF
• D. .20 uF

Answer: (C)

In a parallel arrangement, the total capacitance is the sum of all individual capacitances because each capacitor shares the same voltage and contributes its own store of charge. Add them: 0.02 µF + 0.05 µF + 0.10 µF = 0.17 µF. So the total is 0.17 µF, which matches the value listed as 0.170 µF. The other numbers don’t fit this simple sum rule; they would not equal the combined capacitance of all three in parallel.

In a parallel arrangement, the total capacitance is the sum of all individual capacitances because each capacitor shares the same voltage and contributes its own store of charge. Add them: 0.02 µF + 0.05 µF + 0.10 µF = 0.17 µF. So the total is 0.17 µF, which matches the value listed as 0.170 µF. The other numbers don’t fit this simple sum rule; they would not equal the combined capacitance of all three in parallel.