# Electric charge stored in capacitor – problems and solutions

1. Determine the charge in capacitor C_{5}.

__Known :__

Capacitor 1 (C_{1}) = 6 F

Capacitor 2 (C_{2}) = 6 F

Capacitor 3 (C_{3}) = 3 F

Capacitor 4 (C_{4}) = 12 F

Capacitor 5 (C_{5}) = 6 F

Voltage (V) = 12 Volt

__Wanted :__ Charge in capacitor (C_{5})

__Solution :__

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Capacitor C_{2 }and capacitor C_{3 }are connected in series. The equivalent capacitor :

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

C_{A} = 6/3 = 2 Farad

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

1/C_{B} = 1/C_{4} + 1/C_{5} = 1/12 + 1/6 = 1/12 + 2/12 = 3/12

C_{B} = 12/3 = 4 Farad

Capacitor C_{A} and capacitor C_{B} are connected in parallel. The equivalent capacitor :

C_{C} = C_{A} + C_{B} = 2 + 4 = 6 Farad

Capacitor C_{1} and capacitor C_{C }are connected in series :

1/C = 1/C_{1} + 1/C_{C} = 1/6 F + 1/6 F = 2/6

C = 6/2 = 3 Farad

**Electric charge**

__Electric charge in capacitor ____C :__

q = V C = (12 Volt)(3 Farad) = 36 Coulomb

Capacitor 1 and capacitor C_{C }are connected in series so that electric charge in capacitor C = electric charge in capacitor C_{1} = electric charge in capacitor C_{C} = 36 Coulomb.

Capacitance of capacitor C_{C} is 6 Farad and charge in capacitor C_{C }is 36 Coulomb. The voltage of the capacitor C_{C }is : V = q/C_{C} = 36 Coulomb / 6 Farad = 6 Volt

Capacitor C_{C }is the equivalent capacitor for capacitor C_{A} and capacitor C_{B }connected in parallel. Voltage of the capacitor C_{C} (V_{C}) = voltage in capacitor C_{A }(V_{A}) = voltage in capacitor C_{B} (V_{B}) = 6 Volt.

__Electric charge in capacitor ____C___{B }__:__

q_{B} = V_{B }C_{B }= (6 Volt)(4 Farad) = 24 Coulomb

Capacitor C_{B} is the equivalent capacitor for capacitor C_{4} and C_{5 }connected in series. Connected in series so that the electric charge in capacitor C_{B} (q_{B}) = the electric charge in capacitor C_{4} (q_{4}) = the electric charge in capacitor C_{5} (q_{5}) = 24 Coulomb.

2. Five identical capacitors with a capacitance of 20 µF are connected in series and parallel, as shown in the figure below, with a source of voltage 6 volts. Determine total energy stored in capacitor C_{5}.

__Known :__

Capacitor C_{1} = C_{2} = C_{3} = C_{4} = C_{5} = 20 µF

Voltage (V) = 6 Volt

__Wanted :__ Electric charge stored in capacitor C_{5}

__Solution ____:__

**Capacitor**

Capacitor C_{1} and capacitor C_{2} are connected in series. The equivalent capacitor :

1/C_{A} = 1/C_{1} + 1/C_{2 }= 1/20 + 1/20 = 2/20

C_{A} = 20/2 = 10 µF

Capacitor C_{3} and capacitor C_{4} are connected in series. The equivalent capacitor :

1/C_{B} = 1/C_{3} + 1/C_{4} = 1/20 + 1/20 = 2/20

C_{B} = 20/2 = 10 µF

Capacitor C_{A} and capacitor C_{B} are connected in parallel. The equivalent capacitor :

C_{C} = C_{A} + C_{B} = 10 + 10 = 20 µF

Capacitor C_{C} and capacitor C_{5} are connected in series :

1/C = 1/C_{C} + 1/C_{5} = 1/20 + 1/20 = 2/20

C = 20/2 = 10 µF

**Electric charge**

Electric charge in the equivalent capacitor C :

q = V C = (6 Volt)(10 x 10^{-6 }Farad) = 60 x 10^{-6} Coulomb = 60 µC

Capacitor C_{C }and capacitor 5 are connected in series so that electric charge in the equivalent capacitor C = electric charge in capacitor C_{C }= electric charge in capacitor C_{5} = 60 µC.

3. Five capacitors are connected in series and parallel as shown in figure below. Determine electric charges in capacitor C_{1} (1 µ = 10^{-6})

__Known :__

Capacitor 1 (C_{1}) = capacitor 5 (C_{5}) = 9 µF

Capacitor 2 (C_{2}) = capacitor 3 (C_{3}) = capacitor 4 (C_{4}) = 6 µF

Voltage (V) = 12 Volt

__Wanted :__ Electric charge in capacitor C_{1}

__Solution :__

**Capacitor**

Capacitor C_{2 }and capacitor C_{3} are connected in series. The equivalent capacitor :

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

C_{A} = 6/2 = 3 µF

Capacitor C_{A} and capacitor C_{4 }are connected in parallel. The equivalent capacitor :

C_{B} = C_{A} + C_{4} = 3 + 6 = 9 µF

Capacitor C_{1}, capacitor C_{B }and capacitor C_{5 } are connected in series. The equivalent capacitor :

1/C = 1/C_{1} + 1/C_{B} + 1/C_{5} = 1/9 + 1/9 + 1/9 = 3/9

C = 9/3 = 3 µF = 3 x 10^{-6 }Farad

**Electric charge**

Electric charge in the equivalent capacitor C :

q = V C = (12 Volt)(3 x 10^{-6 }Farad) = 36 x 10^{-6} Coulomb = 36 µC

Capacitor C_{1}, capacitor C_{B }and capacitor C_{5 }are connected in series so that electric charge in the equivalent capacitor C = electric charge in capacitor C_{1 }= electric charge in the equivalent capacitor C_{B }= electric charge in capacitor C_{5} = 36 µC.