# Series and parallel capacitors circuits – problems and solutions

1. What is the total charges in the capacitor circuits below (1 μF = 10^{-6} F)

__Known :__

Capacitor 1 (C_{1}) = 3

Capacitor 2 (C_{2}) = 3 μF

Capacitor 3 (C_{3}) = 3 μF

Capacitor 4 (C_{4}) = 2 μF

Capacitor 5 (C_{5}) = 3 μF

Voltage (V) = 3 Volt

__Wanted :__ Total charge in capacitor circuits (Q)

__Solution :__

**The equivalent capacitor**

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

1/C_{123} = 1/C_{1} + 1/C_{2} + 1/C_{3 }= 1/3 + 1/3 + 1/3 = 3/3

C_{123} = 3/3 = 1 μF

Capacitor C_{123} and C_{4 }are connected in parallel. The equivalent capacitor :

C_{1234 }= C_{123} + C_{4} = 1 + 2 = 3 μF

Capacitor C_{1234} and C_{5 }are connected in series. The equivalent capacitor :

1/C = 1/C_{1234} + 1/C_{5 }= 1/3 + 1/3 = 2/3

C = 3/2 μF

C = 3/2 x 10^{-6 }F

**The total charges :**

The total charges in the equivalent capacitor = the total charges in capacitor circuits :

Q = V C = (3 Volt)(3/2 x 10^{-6 }Farad) = 9/2 x 10^{-6 }Coulomb

Q = 9/2 microCoulomb = 9/2 μC

Q = 4.5 μC

2. If C_{1} = C_{2} = 2 μF, C_{3} = C_{4} = 1 μF and C_{5} = 4 μF, determine the total charges in the capacitor circuits as shown in figure below (1 μF = 10^{-6} F)

__Known :__

Capacitor 1 (C_{1}) = 2 μF

Capacitor 2 (C_{2}) = 2 μF

Capacitor 3 (C_{3}) = 1 μF

Capacitor 4 (C_{4}) = 1 μF

Capacitor 5 (C_{5}) = 4 μF

Voltage (V) = 1.5 Volt

__Wanted :__ The total charges in circuits (Q)

__Solution :__

**The equivalent capacitor :**

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

C_{34 }= C_{3} + C_{4} = 1 + 1 = 2 μF

Capacitor C_{5}, C_{1}, C_{2} and C_{34 }are connected in series. The equivalent capacitor :

1/C = 1/C_{5} + 1/C_{1 }+ 1/C_{2} + 1/C_{34 }

1/C = 1/4 + 1/2 + 1/2 + 1/2

1/C = 1/4 + 2/4 + 2/4 + 2/4

1/C = 7/4

C = 4/7 μF

C = 4/7 x 10^{-6} F

**The total charges :**

The total charges in the equivalent capacitor = the total charges in capacitor circuits :

Q = V C = (1.5 Volt)(4/7 x 10^{-6 }Farad) = 6/7 x 10^{-6 }Coulomb

Q = 6/7 microCoulomb

Q = 6/7 μC

3. Determine the total charges in the capacitor circuits as shown in figure below.

__Known :__

Capacitor 1 (C_{1}) = 3 μF

Capacitor 2 (C_{2}) = 3 μF

Capacitor 3 (C_{3}) = 4 μF

Capacitor 4 (C_{4}) = 4 μF

Capacitor 5 (C_{5}) = 8 μF

Voltage (V) = 10 Volt

__Wanted :__ The total charge in the circuits (Q)

__Solution :__

**The equivalent capacitor :**

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

C_{12 }= C_{1 }+ C_{2 }= 3 + 3 = 6 μF

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

1/C_{34} = 1/C_{3} + 1/C_{4 }= 1/4 + 1/4 = 2/4

C_{34 }= 4/2 = 2 μF

Capacitor C_{12}, capacitor C_{34} and capacitor C_{5} are connected in parallel. The equivalent capacitor :

C = C_{12} + C_{34 }+ C_{5 }= 6 + 2 + 8 = 16 μF = 16 x 10^{-6 }Farad

**The total electric charges :**

The total charges in the equivalent capacitor = the total charges in capacitor circuits :

Q = V C = (10 Volt)(16 x 10^{-6 }Farad) = 160 x 10^{-6 }Coulomb

Q = 160 microCoulomb = 160 μC