The electrical capacity of two parallel-connected capacitors is 5.0 nF, and those connected in series are 1.2 nF.

The electrical capacity of two parallel-connected capacitors is 5.0 nF, and those connected in series are 1.2 nF. Determine the capacitances of the capacitors included in the connections.

Data: Cpair (electrical capacity of parallel-connected capacitors) = 5 nF; Sposl (capacitance of series-connected capacitors) = 1.2 nF.

Parallel connection: Spar = C1 + C2 = 5 and C1 = 5 – C2.

Serial connection: 1 / Sposl = 1 / C1 + 1 / C2 = (C1 + C2) / (C1 * C2) = 1 / 1.2; 1.2 = C1 * C2 / 5 and C1 * C2 = 6; C1 = 6 / C2.

6 / C2 = 5 – C2.

C2 ^ 2 – 5C ^ 2 + 6 = 0.

By Vieta’s theorem: C2 = 2 nF and C1 = 3 nF.

Answer: The capacitances of the capacitors included in the connection are equal to 2 nF and 3 nF.