What is the total capacitance of a circuit containing three capacitors of 0.25 μF, 0.03 μF, and 0.12 μF connected in parallel?

What is the total capacitance of a circuit containing three capacitors of 0.25 μF, 0.03 μF, and 0.12 μF connected in parallel?

• A. 0.4 microfarads
• B. 0.38 microfarads
• C. 0.67 microfarads
• D. 0.2 microfarads

Answer: (A)

Capacitors in parallel simply add their capacitances because the same voltage appears across each one. The total stored charge is the sum of the charges on each capacitor, which adds up exactly when they share the same voltage. Compute the sum: 0.25 μF + 0.03 μF + 0.12 μF = 0.40 μF. So the total capacitance is 0.40 microfarads. This matches the value shown for the option that lists 0.40 μF. Other numbers would come from leaving out a capacitor or misadding, but with all three in parallel, you just add them.

Capacitors in parallel simply add their capacitances because the same voltage appears across each one. The total stored charge is the sum of the charges on each capacitor, which adds up exactly when they share the same voltage.

Compute the sum: 0.25 μF + 0.03 μF + 0.12 μF = 0.40 μF.

So the total capacitance is 0.40 microfarads. This matches the value shown for the option that lists 0.40 μF. Other numbers would come from leaving out a capacitor or misadding, but with all three in parallel, you just add them.