# Electrical energy in capacitor circuits – problems and solutions

1. Determine the electrical energy in the circuit shown in the figure below (1 µF = 10^{-6} F)

__Known :__

Capacitor 1 (C_{1}) = 3 µF

Capacitor 2 (C_{2}) = 1 µF

Capacitor 3 (C_{3}) = 2 µF

Capacitor 4 (C_{4}) = 6 µF

Capacitor 5 (C_{5}) = 4 µF

Electric voltage (V) = 5 volt

__Wanted :__ Electrical energy in circuits

__Solution :__

__The equivalent capacitor :__

Capacitor 2 and capacitor 3 are connected in parallel. The equivalent capacitor :

C_{A }= C_{2 }+ C_{3} = 1 + 2 = 3 µF

Capacitor 1, capacitor A and capacitor 4 are connected in series. The equivalent capacitor :

1/C_{B} = 1/C_{1} + 1/C_{A} + 1/C_{4} = 1/3 + 1/3 + 1/6 = 2/6 + 2/6 + 1/6 = 5/6

C_{B} = 6/5 µF

Capacitor B and capacitor 5 are connected in parallel. The equivalent capacitor :

C = C_{B} + C_{5} = 6/5 + 4 = 6/5 + 16/4 = 24/20 + 80/20 = 104/20 = 5.2 µF

C = 5.2 x 10^{-6} Farad

__Electrical energy in circuit :__

E = ½ C V^{2} = ½ (5.2 x 10^{-6})(5^{2}) = (2.6 x 10^{-6})(25)

E = 65 x 10^{-6} Joule

2. Determine the electrical energy in capacitor circuits shown in figure below (1 µF = 10^{-6} F)

__Known :__

Capacitor 1 (C_{1}) = 4 µF

Capacitor 2 (C_{2}) = 6 µF

Capacitor 3 (C_{3}) = 12 µF

Capacitor 4 (C_{4}) = 2 µF

Capacitor 5 (C_{5}) = 2 µF

Electric voltage (V) = 40 volt

__Wanted :__ Electrical energy in circuits

__Solution :__

__The equivalent capacitor :__

Capacitor 1, capacitor 2 and capacitor 3 are connected in series. The equivalent capacitor :

1/C_{A} = 1/C_{1} + 1/C_{2} + 1/C_{3} = 1/4 + 1/6 + 1/12 = 3/12 + 2/12 + 1/12 = 6/12

C_{A} = 12/6 = 2 µF

Capacitor 4 and capacitor 5 are connected in series. The equivalent capacitor :

1/C_{B} = 1/C_{4 }+ 1/C_{5 }= 1/2 + 1/2 = 2/2

C_{B }= 2/2 = 1 µF

Capacitor A and capacitor B in parallel. The equivalent capacitor :

C = C_{A} + C_{B} = 2 + 1 = 3 µF

C = 3 x 10^{-6} Farad

__Electrical energy in circuits :__

E = ½ C V^{2} = ½ (3 x 10^{-6})(40^{2}) = (1.5 x 10^{-6})(1600)

E = 2400 x 10^{-6} = 2.4 x 10^{-3} Joule

3. Determine the energy stored in the electrical circuit shown in the figure below.

__Known :__

Capacitor 1 (C_{1}) = 4 F

Capacitor 2 (C_{2}) = 4 F

Capacitor 3 (C_{3}) = 4 F

Capacitor 4 (C_{4}) = 4 F

Capacitor 5 (C_{5}) = 2 F

Electric voltage (V) = 12 volt

__Wanted :__ Electrical energy in circuits

__Solution :__

__The equivalent capacitor :__

Capacitor 1, capacitor 2 and capacitor 3 are connected in parallel. The equivalent capacitor :

C_{A} = C_{1 }+ C_{2 }+ C_{3} = 4 + 4 + 4 = 12 F

Capacitor 4 and capacitor 5 are connected in parallel. The equivalent capacitor :

C_{B} = C_{4 }+ C_{5} = 4 + 2 = 6 F

Capacitor A and capacitor B are connected in series. The equivalent capacitor :

1/C = 1/C_{A} + 1/C_{B} = 1/12 + 1/6 = 1/12 + 2/12 = 3/12

C = 12/3 = 4 Farad

__Electrical energy in circuits :__

E = ½ C V^{2} = ½ (4)(12^{2}) = (2)(144)

E = 288 Joule