# Day 111

## 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:

Ceq = C1 + C2 + C3

Qtot = Q1 + Q2 + Q3

Vtotal = V1 = V2 = V3

Qtotal = CeqV

Note: Ceq = 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/Ceq = 1/C1 + 1/C2 + 1/ C3

Qtot = Q1 = Q2 = Q3

Vtotal = V1 + V2 + V3

Qtotal = CeqVtotal

Note: Ceq = 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 C1.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)