# Capacitors in series and parallel – problems and solutions

1. Three capacitors, C_{1} = 2 μF, C_{2} = 4 μF, C_{3} = 4 μF, are connected in series and parallel. Determine the capacitance of a single capacitor that will have the same effect as the combination.

__Known :__

Capacitor C_{1} = 2 μF

Capacitor C_{2} = 4 μF

Capacitor C_{3} = 4 μF

__Wanted :__ The equivalent capacitance (C)

__Solution :__

Capacitor C_{2} and C_{3} connected in parallel. The equivalent capacitance :

C_{P }= C_{2} + C_{3 }= 4 + 4 = 8 μF

Capacitor C_{1} and C_{p }connected in series. The equivalent capacitance :

1/C = 1/C_{1} + 1/C_{P } = 1/2 + 1/8 = 4/8 + 1/8 = 5/8

C = 8/5 μF

2. Five capacitors, C_{1} = 2 μF, C_{2} = 4 μF, C_{3} = 6 μF, C_{4} = 5 μF, C_{5} = 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 :__

Capacitor C_{1} = 2 μF

Capacitor C_{2} = 4 μF

Capacitor C_{3} = 6 μF

Capacitor C_{4} = 5 μF

Capacitor C_{5} = 10 μF

__Wanted :__ The equivalent capacitance (C)

__Solution :__

Capacitor C_{2} and C_{3} are connected in parallel. The equivalent capacitance :

C_{P }= C_{2} + C_{3 }

C_{P} = 4 + 6

C_{P }= 10 μF

Capacitor C_{1}, C_{P}, C_{4} and C_{5} are connected in series. The equivalent capacitance :

1/C = 1/C_{1} + 1/C_{P }+ 1/C_{4 }+ 1/C_{5}

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

3. C_{1 }= 3 μF, C_{2} = 4 μF and C_{3 }= 3 μF, are connected in series and parallel. Determine the electric energy on the circuits.

__Known :__

Capacitor C_{1} = 3 μF

Capacitor C_{2} = 4 μF

Capacitor C_{3} = 3 μF

__Wanted :__ The equivalent capacitance (C)

__Solution :__

Capacitor C_{2} and C_{3} are connected in parallel. The equivalent capacitance :

C_{P }= C_{2} + C_{3}

C_{P} = 4 + 3

C_{P }= 7 μF

Capacitor C_{1 }and C_{P }are connected in series. The equivalent capacitance :

1/C = 1/C_{1} + 1/C_{P}

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 V^{2}

E = ½ (2.1 x 10^{-6})(12^{2})

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