Apologies for the delay, I have been revising and relearning much of this in preparation for writing some longer explanations/tutorials of my own. I also plan on writing up some basic projects which you can use for practice. They'll just be very basic stuff to start off with, but I hope to move on to using integrated circuits and micro-controllers fairly quickly (as mentioned in the first post, however, it is better to avoid using the micro-controller until you need it, since much of what I want to focus on right now can be accomplished with more basic components.) If anyone either knows more than I or is unclear on something I have written, I encourage you to let me know and feel free to add your own content et al.
For today, I’m going to focus on the 4 main terms to understand before working on anything else, Voltage, Current, Resistance, and Power.
Voltage ( V ): Voltage is the electrical pressure used to supply power throughout a circuit. For mathematical work, Voltage is measured in Volts and signed by the pronumeral V.
Current ( I ): Is the movement of electrons through a conductor. For mathematical work, current is measured in Amperes (abbrev. Amp) and is signed by the pronumeral I. There are two types of current, Alternating Current (AC) and Direct Current (DC).
Resistance ( R ): Is the reduction of voltage and current in a circuit. For mathematical work, resistance is measured in Ohms and signed by the pronumeral Ω.
Power ( P ): is the output of a circuit. For mathematical work, power is measured in Watts and signed by the pronumeral W.
To explain this by way of example I will use the basic lighting circuit in a house.
Cables run from the mains electricity box (usually located outside the house), through the roof-space and/or walls to a switch, and each switch connects to a light-socket. The switch provides an insulated control for the user so that there is, in theory, no risk to the user when they are switching a light on or off. When in the “Off” position, the circuit is open, so whilst there is voltage applied along the cables to the switch, there is no connection between the active and neutral wires, therefore, there is no current. When the switch is moved to the “On” position, the circuit closes and the voltage from the mains input can push the current along the active and neutral wires up through the light-bulb. A light-bulb filament provides a certain amount of resistance so that when the current and voltage cross it, it outputs power as heat and light.
An analogy of sorts can be drawn from the working of electricity to the common garden hose. When the tap/faucet is closed, the water cannot flow through the hose. When the tap/faucet opens, the pressure (voltage) is able to push the water through the hose and out the end (current). If you were to place a thumb over the end or crimp the hose somewhere (resistance), the pressure inside would increase (Bernoulli’s Theorem stating that a high pressure equates to a low speed and a low pressure to a high speed) which would decrease the rate at which the water flowed. You would also notice an increase in tension on the hose or a pushing against your thumb, this is similar to the power output of a circuit. Whilst there are differences between water and electricity, thinking of things as working in this way is helpful when making distinctions or trying to understand what is happening.
To explain the maths behind these four terms, it helps to have on hand an aid such as this:
In the primary example, you can work out the power output of the light-bulb by substituting P = V * I for whatever voltage and current values are used in your country (for me, 240V * 10A = 2400W (which can be shortened to 2.4kW*)). The value you get here may seem quite high, but this example assumes that there is no circuit resistance. If you want a more accurate reflection of reality, transpose the problem by finding the power dissipation on the light-bulb (it should be written on the component, if not, you can use a multimeter to read the resistance of it) and working backwards: I = P / V. Taking the example of a 60W bulb, I would work it out thus: P / V = I (since I now know my power to be 60W and my voltage to be 240V), 60W / 240V = 0.25A or 25mA.
*A note on the International System of Units (SI) and Scientific Notation: When working with electricity, numbers can get quite large or incredibly small. To make things easier, we express things using the International System of Units (or SI). The standard abbreviations you may see are GV, MV, kV, V, mV, µV, nV, or pV (using Volts (V) as an example). These stand for Gigavolts (1 * 10^9), Megavolts (1 * 10^6), kilovolts (1 * 10^3), volts (1 * 10^0), millivolts (1 * 10^-3), microvolts (1 * 10^-6), nanovolts (1 * 10^-9), and picovolts (1 * 10^-12). More information can easily be procured by searching for any of these terms on Wikipedia.