Capacitors in series – problems and solutions

1. Four capacitors, C1 = 2 μF, C2 = 1 μF, C3 = 3 μF, C4 = 4 μF, are connected in series. 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 :

1/C = 1/C1 + 1/C2 + 1/C3 + 1/C4

1/C = 1/2 + 1/1 + 1/3 + 1/4

1/C = 6/12 + 12/12 + 4/12 + 3/12

1/C = 25/12

C = 12/25

C = 0.48

The equivalent capacitance of the entire combination is 0.48 μF.

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2. Determine the charge on capacitor C1 if the potential difference between P and Q is 12 Volt…

Capacitors in series – problems and solutions 1Known :

Capacitor C1 = 10 μF = 10 x 10-6 F

Capacitor C2 = 20 μF = 20 x 10-6 F

Potential difference (V) = 12 Volt

Wanted : the charge on capacitor C1 (Q1)

Solution :

The equivalent capacitance :

1/C = 1/C1 + 1/C2

1/C = 1/10 + 1/20 = 2/20 + 1/20 = 3/20

C = 20/3 μF = (20/3) x 10-6 F

Electric charge on the equivalent capacitor :

Q = (C)(V) = (20/3)(12)(10-6) = 80 x 10-6 C

Q = 80 μC

Capacitors are connected in series so that electric charge on the equivalent capacitors = electric charge on capacitor C1 = electric charge on capacitor C2.

The electric charge on capacitor C1 is 80 μC.

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3. Two capacitors, C1 = 2 μF and C2 = 4 μF, are connected in series. The capacitors are charged. The potential difference on capacitor C1 is 2 Volt. The electric charge on capacitor C2 is…

Known :

Capacitor C1 = 2 μF = 2 x 10-6 F

Capacitor C2 = 4 μF = 4 x 10-6 F

The potential difference on capacitor C1 (V1) = 2 Volt

Wanted : Electric charge on capacitor C2.

Solution :

Electric charge on capacitor C1 :

Q1 = C1 V1 = (2 x 10-6)(2) = 4 x 10-6 C

Q1 = 4 μC

Capacitors are connected series so that electric charge on capacitor C1 = electric charge on capacitor C2.

The charge on capacitor C2 is 4 μC.