# Capacitors in parallel – problems and solutions

1. Four capacitors, C1 = 2 μF, C2 = 1 μF, C3 = 3 μF, C4 = 4 μF, are connected in parallel. Determine the capacitance of a single capacitor that will have the same effect as the combination.

Known :

Capacitor C1 = 2 μF

Capacitor C2 = 1 μF

Capacitor C3 = 3 μF

Capacitor C3 = 4 μF

Wanted : The equivalent capacitance

Solution :

The equivalent capacitance :

C = C1 + C2 + C3

C = 4 μF + 2 μF + 3 μF = 9 μF

The equivalent capacitance of the entire combination is 9 μF.

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2. Determine the charge on capacitor C2 if the potential difference between point A and B is 9 Volt… Known :

Capacitor C1 = 20 μF = 20 x 10-6 F

Capacitor C2 = 30 μF = 30 x 10-6 F

Potential difference between point A and B (VAB) = 9 Volt

Wanted : the charge on capacitor C2 (Q2)

Solution :

Potential difference :

Capacitors are connected in parallel so that the potential difference between A and B (VAB) = the potential difference on capacitor C1 (V1) = the potential difference on capacitor C2 (V2) = 9 Volt.

Electric charge on capacitor C2 :

Q2 = C2 V2 = (30 x 10-6)(9) = 270 x 10-6 C

Q2 = 270 μC

The electric charge on capacitor C2 is 270 μC.

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3. Three capacitors, C1 = 4 μF, C2 = 2 μF, C3 = 3 μF, are connected in parallel. The capacitor are charged. The potential difference on capacitor C2 is 4 Volt. Determine

(a) Electric charge on capacitor C1, C2 and C3

(b) Electric charge on the equivalent capacitor of the entire combination

Known :

Capacitor C1 = 4 μF = 4 x 10-6 F

Capacitor C2 = 2 μF = 2 x 10-6 F

Capacitor C3 = 3 μF = 3 x 10-6 F

Potential difference on capacitor C2 (V2) = 4 Volt

Wanted : Electric charge on capacitor C3 (Q3)

Solution :

(a) Electric charge on capacitor C3

The potential difference on capacitor C3 :

Capacitors are connected in parallel so that the potential difference on capacitor C3 (V3) = the potential difference on capacitor C2 (V2) = the potential difference on capacitor C1 (V1) = the potential difference on equivalent capacitor (V) = 4 Volt

Electric charge on capacitor C1 :

Q1 = C1 V1 = (4 x 10-6)(4) = 16 x 10-6 C

Q1 = 16 μC

Electric charge on capacitor C2 :

Q2 = C2 V2 = (2 x 10-6)(4) = 8 x 10-6 C

Q2 = 8 μC

Electric charge on capacitor C3 :

Q3 = C3 V3 = (3 x 10-6)(4) = 12 x 10-6 C

Q3 = 12 μC

(b) Electric charge on equivalent capacitor

Q = Q1 + Q2 + Q3

Q = 16 μC + 8 μC + 12 μC = 36 μC

Alternative solution :

The equivalent capacitance :

C = C1 + C2 + C3

C = 4 μF + 2 μF + 3 μF = 9 μF

C = 9 x 10-6 F

The potential difference on the equivalent capacitor :

V1 = V2 = V3 = V = 4 Volt

The electric charge on the equivalent capacitor :

Q = C V = (9 x 10-6)(4) = 36 x 10-6 C

Q = 36 μC 