Capacitors in Parallel

If two or more capacitors are hooked together in parallel, which means there are different branches through which the charge can travel, the following equations apply:

 

C_eq = C₁ + C₂ + C₃

Q_tot = Q₁ + Q₂ + Q₃

V_total = V₁ = V₂ = V₃

Q_total = C_eqV

 Note: C_eq = equivalent capacitance; if one capacitor were to replace the two or more capacitors that are in the circuit it would need a larger capacitor

Capacitors in Series

If two or more capacitors are hooked together in series, which means there is only one path through which the charge can travel, the following equations apply:

1/C_eq = 1/C₁ + 1/C₂ + 1/ C₃

Q_tot = Q₁ = Q₂ = Q₃

V_total = V₁ + V₂ + V₃

Q_total = C_eqV_total

 

Note: C_eq = equivalent capacitance; if one capacitor were to replace the two or more capacitors that are in the circuit it would need a smaller capacitor

Energy Stored in a Capacitor

Uc = 1/2 QV = 1/2 CV² = Q²/2C

Like all other types of energy, the energy stored in a capacitor is measured in Joules.

See the problem below:

Sample Capacitor Problem

Consider the above network of capacitors, which were uncharged before the battery was connected. Let V₁ and Q₁ be the voltage and charge on capacitor C_1.Use similar notations for C₂, C₃, and C₄.

1. Using charge conservation, energy conservation and the definition of capacitance to complete the table below.
2. How much energy is stored in capacitor C₁?

The following facts are known:

C₂ has a charge of Q₂ = 30. μC

ε (Battery) = 8.0 V

C₁ = 6.0 μF

V₃ = 2.0 V

C₄ = C₂

Q1
Q2	30 μC
Q3
Q4
V1
V2
V3	2.0 V
V4
C1	6.0 μF
C3

Solution:

To find Q₁, we know that C₁ = 6.0 μF and the voltage across that branch must be the same as the battery. Therefore,

Q₁ = C₁V₁= (6.0 x 10^-6 F)(8.0 V) = 4.8 x 10^-5 F = 48 μC

 Since C₂ = C₄, and their charges are the same because they are in series, the voltages must also be the same, V₂ = V₄.

V₂ + V₃ + V₄= 8.0 V

OR

V₂ + V₃ + V₂= 8.0 V (Since V₂=V₄)

 2V₂ + V₃ = 8.0 V

 2V₂ + 2.0 V = 8.0 V (Given V₃ = 2.0 V)

 V₂ = 3.0 V and V₄ = 3.0 V

 

Since three capacitors are in series, their charges are the same, so …

Q₂ = Q₃ = Q₄= 30 μC.

Since C3 has the same charge as C2 and C4 and the voltage is given….

C₃ = Q₃/V₃= 3.0 x 10^-5 C/2.0 V = 1.5 x 10^-5 F = 15 μF

The completed table looks like….

Q1	48 μC
Q2	30 μC
Q3	30 μC
Q4	30 μC
V1	8.0 V
V2	3.0 V
V3	2.0 V
V4	3.0 V
C1	6.0 μF
C3	15 μF

To find the energy in capacitor 1, use …

U₁ = Q₁ ²/2C₁= (4.8 x 10^-5 C)²/[(2)(6 x 10^-6 F)]

U₁ = 1.92 x 10^-4 J

(source)