When two capacitors are connected in series, the equivalent capacitance is 1.2 μF

When two capacitors are connected in series, the equivalent capacitance is 1.2 μF, and when parallel, it is 5 μF. Determine the capacity of each capacitor.

Given: Cposl (capacitance in series) = 1.2 μF; Cpair (parallel connection) = 5 μF.

1) Serial connection: Cpl = C1 * C2 / (C1 + C2).

2) Parallel connection: Spar = C1 + C2 and C1 = Spar – C2.

3) Second capacitor: Cposl = (Spar – C2) * C2 / (Spar – C2 + C2).

1.2 = (5 – C2) * C2 / 5.

6 = 5C2 – C2 ^ 2.

C2 ^ 2 – 5C2 + 6 = 0.

Vieta’s theorem: C2.1 = 3 μF; C2.2 = 2 μF.

4) The first capacitor: C1.1 = 5 – 3 = 2 μF; C1.2 = 5 – 2 = 3 μF.

Answer: Capacities are equal to 2 μF and 3 μF.