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Electronics/Voltage.md
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@ -40,7 +40,7 @@ A 12V battery connected to a circuit gives it a voltage rise of 12 volts.
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 electrons. 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. The amount lost is equal to the amount of energy imparted by the voltage rise.
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. 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**.
@ -52,6 +52,34 @@ As they move through the circuit the electrons encounter a **load** which is wha
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](/Electronics/Physics_of_electricity/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.