# Capacitors in parallel – problems and solutions

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

__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 :

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

C = 4 μF + 2 μF + 3 μF = 9 μF

The equivalent capacitance of the entire combination is 9 μF.

2. Determine the charge on capacitor C_{2 }if the potential difference between point A and B is 9 Volt…

__Known :__

Capacitor C_{1} = 20 μF = 20 x 10^{-6} F

Capacitor C_{2} = 30 μF = 30 x 10^{-6} F

Potential difference between point A and B (V_{AB}) = 9 Volt

__Wanted ____:__ the charge on capacitor C_{2 }(Q_{2})

__Solution :__

**Potential difference :**

Capacitors are connected in parallel so that the potential difference between A and B (V_{AB}) = the potential difference on capacitor C_{1 }(V_{1}) = the potential difference on capacitor C_{2} (V_{2}) = 9 Volt.

__Electric charge on capacitor ____C___{2} :

Q_{2} = C_{2} V_{2 }= (30 x 10^{-6})(9) = 270 x 10^{-6} C

Q_{2} = 270 μC

The electric charge on capacitor C_{2} is 270 μC.

3. Three capacitors, C_{1} = 4 μF, C_{2} = 2 μF, C_{3} = 3 μF, are connected in parallel. The capacitor are charged. The potential difference on capacitor C_{2} is 4 Volt. Determine

(a) Electric charge on capacitor C_{1}, C_{2} and C_{3}

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

__Known :__

Capacitor C_{1} = 4 μF = 4 x 10^{-6} F

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

Capacitor C_{3} = 3 μF = 3 x 10^{-6} F

Potential difference on capacitor C_{2} (V_{2}) = 4 Volt

__Wanted ____:__ Electric charge on capacitor C_{3 }(Q_{3})

__Solution :__

**(a) ****Electric charge on capacitor ****C**_{3}** **

__The potential difference on capacitor ____C___{3} :

Capacitors are connected in parallel so that the potential difference on capacitor C_{3} (V_{3}) = the potential difference on capacitor C_{2} (V_{2}) = the potential difference on capacitor C_{1} (V_{1}) = the potential difference on equivalent capacitor (V) = 4 Volt

__Electric charge on capacitor ____C___{1} :

Q_{1} = C_{1} V_{1 }= (4 x 10^{-6})(4) = 16 x 10^{-6 }C

Q_{1} = 16 μC

__Electric charge on capacitor ____C___{2} :

Q_{2} = C_{2} V_{2 }= (2 x 10^{-6})(4) = 8 x 10^{-6 }C

Q_{2} = 8 μC

__Electric charge on capacitor ____C___{3} :

Q_{3} = C_{3} V_{3 }= (3 x 10^{-6})(4) = 12 x 10^{-6 }C

Q_{3} = 12 μC

**(b) ****Electric charge on equivalent capacitor**

Q = Q_{1} + Q_{2} + Q_{3}

Q = 16 μC + 8 μC + 12 μC = 36 μC

Alternative solution :

__The equivalent capacitance :__

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

C = 4 μF + 2 μF + 3 μF = 9 μF

C = 9 x 10^{-6} F

__The potential difference on the equivalent capacitor :__

V_{1} = V_{2} = V_{3} = 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