Capacitors in series and parallel – problems and solutions

1. Three capacitors, C1 = 2 μF, C2 = 4 μF, C3 = 4 μF, are connected in series and parallel. Determine the capacitance of a single capacitor that will have the same effect as the combination.

Capacitors in series and parallel – problems and solutions 1Known :

Capacitor C1 = 2 μF

Capacitor C2 = 4 μF

Capacitor C3 = 4 μF

Wanted : The equivalent capacitance (C)

Solution :

Capacitor C2 and C3 connected in parallel. The equivalent capacitance :

CP = C2 + C3 = 4 + 4 = 8 μF

Capacitor C1 and Cp connected in series. The equivalent capacitance :

1/C = 1/C1 + 1/CP = 1/2 + 1/8 = 4/8 + 1/8 = 5/8

C = 8/5 μF

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2. Five capacitors, C1 = 2 μF, C2 = 4 μF, C3 = 6 μF, C4 = 5 μF, C5 = 10 μF, are connected in series and parallel. Determine the capacitance of a single capacitor that will have the same effect as the combination.

Known :

Capacitors in series and parallel – problems and solutions 2Capacitor C1 = 2 μF

Capacitor C2 = 4 μF

Capacitor C3 = 6 μF

Capacitor C4 = 5 μF

Capacitor C5 = 10 μF

Wanted : The equivalent capacitance (C)

Solution :

Capacitor C2 and C3 are connected in parallel. The equivalent capacitance :

CP = C2 + C3

CP = 4 + 6

CP = 10 μF

Capacitor C1, CP, C4 and C5 are connected in series. The equivalent capacitance :

1/C = 1/C1 + 1/CP + 1/C4 + 1/C5

1/C = 1/2 + 1/10 + 1/5 + 1/10

1/C = 5/10 + 1/10 + 2/10 + 1/10

1/C = 9/10

C = 10/9 μF

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3. C1 = 3 μF, C2 = 4 μF and C3 = 3 μF, are connected in series and parallel. Determine the electric energy on the circuits.

Known :

Capacitor C1 = 3 μFCapacitors in series and parallel – problems and solutions 3

Capacitor C2 = 4 μF

Capacitor C3 = 3 μF

Wanted : The equivalent capacitance (C)

Solution :

Capacitor C2 and C3 are connected in parallel. The equivalent capacitance :

CP = C2 + C3

CP = 4 + 3

CP = 7 μF

Capacitor C1 and CP are connected in series. The equivalent capacitance :

1/C = 1/C1 + 1/CP

1/C = 1/3 + 1/7

1/C = 7/21 + 3/21

1/C = 10/21

C = 21/10

C = 2.1 μF

C = 2.1 x 10-6 F

The electric energy on the circuits :

E = ½ C V2

E = ½ (2.1 x 10-6)(122)

E = ½ (2.1 x 10-6)(144)

E = (2.1 x 10-6)(72)

E = 151.2 x 10-6 Joule

E = 1.5 x 10-4 Joule