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

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

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

Q_{total} = C_{eq}V

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/C_{1} + 1/C_{2} + 1/ C_{3}

Q_{tot} = Q_{1} = Q_{2} = Q_{3}

V_{total} = V_{1} + V_{2} + V_{3}

Q_{total} = C_{eq}V_{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 CV^{2} = Q^{2}/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 V_{1} and Q_{1} be the voltage and charge on capacitor C_{1.}Use similar notations for C_{2}, C_{3}, and C_{4}.

- Using charge conservation, energy conservation and the definition of capacitance to complete the table below.
- How much energy is stored in capacitor C
_{1}?

The following facts are known:

C_{2} has a charge of Q_{2} = 30. μC

ε (Battery) = 8.0 V

C_{1} = 6.0 μF

V_{3} = 2.0 V

C_{4} = C_{2}

Q1 | |

Q2 | 30 μC |

Q3 | |

Q4 | |

V1 | |

V2 | |

V3 | 2.0 V |

V4 | |

C1 | 6.0 μF |

C3 |

Solution:

To find Q_{1}, we know that C_{1} = 6.0 μF and the voltage across that branch must be the same as the battery. Therefore,

Q_{1} = C_{1}V_{1}= (6.0 x 10^{-6} F)(8.0 V) = 4.8 x 10^{-5} F = 48 μC

Since C_{2} = C_{4}, and their charges are the same because they are in series, the voltages must also be the same, V_{2} = V_{4}.

V_{2} + V_{3} + V_{4}= 8.0 V

OR

V_{2} + V_{3} + V_{2}= 8.0 V (Since V_{2}=V_{4})

2V_{2} + V_{3} = 8.0 V

2V_{2} + 2.0 V = 8.0 V (Given V_{3} = 2.0 V)

V_{2} = 3.0 V and V_{4} = 3.0 V

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

Q_{2} = Q_{3} = Q_{4}= 30 μC.

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

C_{3} = Q_{3}/V_{3}= 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 …

U_{1} = Q_{1} ^{2}/2C_{1}= (4.8 x 10^{-5} C)^{2}/[(2)(6 x 10^{-6} F)]

U_{1} = 1.92 x 10^{-4} J

(source)