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

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

• A. 0.4 uF
• B. 0.75 uF
• C. 0.28 uF
• D. 0.15 uF

Answer: (A)

Capacitors in parallel share the same voltage, so the total capacitance is the sum of each capacitor’s value. Add them: 0.25 μF + 0.03 μF + 0.12 μF = 0.40 μF. So the circuit’s total capacitance is 0.40 μF (often written as 0.4 μF). This happens because in parallel the capacitances simply accumulate, increasing the overall ability to store charge at the same voltage.

Capacitors in parallel share the same voltage, so the total capacitance is the sum of each capacitor’s value. Add them: 0.25 μF + 0.03 μF + 0.12 μF = 0.40 μF. So the circuit’s total capacitance is 0.40 μF (often written as 0.4 μF). This happens because in parallel the capacitances simply accumulate, increasing the overall ability to store charge at the same voltage.