# Capacitors in series – problems and solutions

1. Four capacitors, C_{1} = 2 μF, C_{2} = 1 μF, C_{3} = 3 μF, C_{4 }

__Known :__

Capacitor C_{1} = 2 μF

Capacitor C_{2} = 1 μF

Capacitor C_{3} = 3 μF

Capacitor C_{3} = 4 μF

__Wanted :__ The equivalent capacitance

__Solution :__

The equivalent capacitance :

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

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.

2. Determine the charge on capacitor C_{1} if the potential difference between P and Q is 12 Volt…

__Known :__

Capacitor C_{1} = 10 μF = 10 x 10^{-6} F

Capacitor C_{2} = 20 μF = 20 x 10^{-6} F

Potential difference (V) = 12 Volt

__Wanted :__ the charge on capacitor C_{1} (Q_{1})

__Solution :__

The equivalent capacitance :

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

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

_{1}= electric charge on capacitor C

_{2}.

The electric charge on capacitor C_{1} is 80 μC.

3. Two capacitors, C_{1} = 2 μF and C_{2} = 4 μF, are connected in series. The capacitors are charged. The potential difference on capacitor C_{1 }is 2 Volt. The electric charge on capacitor C_{2} is…

__Known :__

Capacitor C_{1} = 2 μF = 2 x 10^{-6} F

Capacitor C_{2} = 4 μF = 4 x 10^{-6} F

The potential difference on capacitor C_{1} (V_{1}) = 2 Volt

__Wanted ____:__ Electric charge on capacitor C_{2}.

__Solution :__

__Electric charge on capacitor ____C___{1} :

Q_{1} = C_{1} V_{1} = (2 x 10^{-6})(2) = 4 x 10^{-6 }C

Q_{1} = 4 μC

Capacitors are connected series so that electric charge on capacitor C_{1} = electric charge on capacitor C_{2}.

The charge on capacitor C_{2} is 4 μC.