Question

What is the total capacitance of a certain circuit containing three capacitors with capacitances of .25 micro-farad, .03 microfarad, and .12 microfarad, respectively?

(Note: CT = C1 + C2 + C3)

A. 4 μF.
B. 04 pF.
C. 04 μF.

Answer 1

The total capacitance of three capacitors of 0.25 µF, 0.03 µF, and 0.12 µF connected in parallel is 0.40 µF.

Step-by-step explanation:

The total capacitance of a circuit containing three capacitors connected in parallel is obtained by the simple addition of their capacitances. The given capacitances are 0.25 microfarad, 0.03 microfarad, and 0.12 microfarad. We can calculate the total capacitance (CT) using the formula CT = C1 + C2 + C3.

Therefore, CT = 0.25 µF + 0.03 µF + 0.12 µF = 0.40 µF. This makes option C (0.40 µF) the correct answer.