Question

10 μF, and a 18 μF, and a capacitor are connected in series. How much capacitance would a single capacitor need to have to replace the three capacitors?

Answer 1

6.429 μ F

Step-by-step explanation:

C1 = 10 μ F

C2 = 18 μ F

Let C be the equivalent capacitance of the circuit.

They are connected in series combination

1 / C = 1 / C1 + 1 / C2

1 / C = 1 / 10 + 1 / 18

1 / C = (18 + 10) / 180

1 / C = 28 / 180

C = 180 / 28 = 6.429 μ F