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

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

• A. .0125 uF
• B. .170 uF
• C. .075 uF
• D. .010 uF

Answer: (B)

Capacitors in parallel simply add their capacitances. So add the three values: 0.02 μF + 0.05 μF + 0.10 μF = 0.17 μF. This is the total capacitance if the capacitors are connected in parallel. If they were in series, the total would be found from the reciprocal sum: 1/Ct = 1/0.02 + 1/0.05 + 1/0.10 = 50 + 20 + 10 = 80, giving Ct = 0.0125 μF. The given total of 0.17 μF corresponds to a parallel connection.

Capacitors in parallel simply add their capacitances. So add the three values: 0.02 μF + 0.05 μF + 0.10 μF = 0.17 μF. This is the total capacitance if the capacitors are connected in parallel.

If they were in series, the total would be found from the reciprocal sum: 1/Ct = 1/0.02 + 1/0.05 + 1/0.10 = 50 + 20 + 10 = 80, giving Ct = 0.0125 μF. The given total of 0.17 μF corresponds to a parallel connection.