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tags: [physics, electricity]
---

# Voltage

## Difference of potential and the transfer of energy

We noted in the discussion of [current](Current.md) that current flows when
there is a difference of potential between two points with negatively charged
atoms at one point and positively charged atoms at the other.

_Difference of potential_ is the same thing as voltage. Voltage is essential to
current because it is the force that enables the current to flow.

Without voltage there can be no current because in their natural state, the
electrons in an atom are **in random motion with no direction. To produce a
current, energy must be imparted to the electrons so that they all flow in the
same direction.**

Voltage is the application of this energy. Any
[form of energy](Voltage_sources.md) that dislodges electrons from atoms can be
used to produce current. Thus:

> Voltage is the work required per coulomb to move a charge from one point to
> another.

### Voltage exists even without current

Given that voltage is the force that generates current, it would be natural to
think that voltage only exists when a voltage source (such as a
[battery](Cells_and_batteries.md)) is connected to a circuit. This however is
not the case. Even if a 9V battery isn't connected to anything it still has a
difference of potential of 9-volts accross its terminals. Remember voltage is
_potential energy_ not just the actualisation of that energy.

## Voltage rise and voltage drops

In circuits there are actually two types of voltage:

1. Voltage rise
2. Voltage drop

### Voltage rise

When we introduce potential energy into a ciruit in the form of voltage, this is
a voltage rise. The current flows from the negative terminal of the voltage
source and returns to the positive terminal of the voltage source.

A 12V battery connected to a circuit gives it a voltage rise of 12 volts.

### Voltage drop

Voltage drop is the corrolary to voltage rise. It is the loss of energy that the
electrons of the circuit current experience as a result of encountering
resistance.

As they move through the circuit the electrons encounter a **load** which is
what we call resistance to the flow of current. As they run into this, they give
up their energy. The relinquishing of energy happens in the form of a conversion
of electrical energy to heat (hence why computers heat up). The amount lost is
equal to the amount of energy imparted by the voltage rise.

> The voltage drop in a circuit equals the the voltage rise of the circuit
> because energy cannot be created or destroyed, only changed to another form.
> When a voltage rise is converted to a voltage drop we say that **the energy
> has been _consumed_ by the circuit**.

#### Examples

- If a 12V source is connected to a 12V lamp, the source supplies a 12V voltage
  rise and the lamp produces a 12V voltage drop.
- If two identical 6V lamps are connected in series to the same 12V source, each
  lamp produces a 6V drop for a total of 12 volts
- If two different lamps are connected in series to to a 12V source: a 3V and a
  9V lamp, the 9V lamp produces a drop of 9V and the 3V lamp produces a drop of
  3V. The sum of the voltage drops equals the voltage rise of 12 volts.

These examples demonstrate that the voltage rise: voltage drop ratio always
evens out.

### Kirchoff's Voltage Law

The relationship between voltage rise and voltage drop is expressed in
Kirchoff's Voltage Law:

> The sum of the voltages around any closed loop in a circuit must be zero.

The application of the Law is illustrated in the following diagram:

![](/img/voltage-drop.png)

The explanation for the voltage drop at the positions $V^{A}$ and $V^{D}$ are
obvious enough: they are at the beginning and end of the loop so are equal to
the maximal voltage rise and minimal voltage drop, respectively.

We can work out the voltage of the remaining voltage points by inverting
[Ohm's Law](Ohms_Law.md): $V = I \times R$:

For the voltage at $V^{B}$:

$$
  0.5\textsf{mA} \times \textsf{4}k\Omega = 2 \textsf{V}
$$

For the voltage at $V^{C}$:

$$
  0.5\textsf{mA} \times \textsf{6}k\Omega = 3 \textsf{V}
$$

The total drops (2V, 3V, 5V) are equal to the initial volatage rise (10V)

## Distinguishing _voltage_ from _electric field_

It can be confusing that two different symbols often seemed to be used
interchangeably when talking about voltage: $V$ and $E$. However, while they
broadly point to the same phenomenon there is a difference in emphasis.

- $V$ stands for volts or voltage conceived purely in terms of the difference in
  potential between two points: the positive and negative terminals
- $E$ stands for electric field. This is the field that surround each electric
  charge and exerts force on all other charges in the field, attracting or
  repelling them. So it is more the physical process that the volatage
  represents rather than the numerical representation of the potential between
  the terminals.
