Introduction to Basic Electrical Circuits

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This is my area of interest right here. I really want to make this site because there are very few websites out there that i could find during my time taking this course that could simplify what i needed to know. I am making this website to aide the best i can in understanding these concepts. Let's start with the basics and realize schematic symbols.
These are the symbols you must be confortable with and recognize instantly. Shouldn't be too hard. Resistors are elements in a circuit that do exactly what they say, they resist current flow. The first thing with resistors that must be studied are their color codes. There will be either 3 or 4 stripes on every resistor and each of these stripes mean something different, but they have the same value. The color codes tells you exactly what value of a resistor you are working with. The first band tells you the first number, the second band tells you the second number, the third band tells you the multiplier with respect to 10, and the forth band tells the +/- tolerance range of the resistor. As an example lets say we have a resistor with the color band sequence of "brown, green, orange, gold". This says that the first number is 1, second is 5. Since the third band is orange, that says the resistor has the value of 15 * 103. The forth band is gold so it has a tolerance of +/-5%. Thus it is a 15k-Ohm resistor +/-5%. The color code for resistors follows so now you can know the resistance of any resistor just by looking at the color bands.

black = 0	  blue = 6	  none = 20% tolerance
brown = 1	  violet = 7
red = 2		  grey = 8
orange = 3	  white = 9
yellow = 4	  gold = .1 or 5% tolerance
		    green = 5	      silver = .01 or 10% tolerance

Now that we have the basis for understanding resistors, let's look at some formulas that may prove to be helpful with the use of these resistors. I am gonna list 3 formulas and they will be the only formulas that you will need to memorize. There are more than 3 formulas, but they are derivatives of these so real need to memorize any others.

Ohm's Law 		: V = IR
Kirchhoff's Voltage Law : ΣV in a loop = 0
Kirchhoff's Current Law	: ΣI into a node = 0
To understand these formulas we will need definitions of the variables in the formulas.
V = dw/dq = Joules/Coloumbs = Volts
I = dq/dt = Coloumbs/sec = Current
R = resistance
p = dw/dt = VI = Joules/sec = Watt

Coversions

The first conversion factor that will be shown is the delta-wye and wye delta conversions. Remember that wire connections can be moved around so long as they do not effect the circuit.

To begin with, we will convert delta to wye
a = (B * A) / (A + B + C)
b = (A * C) / (A + B + C)
c = (B * A) / (A + B + C)

Now from Wye to Delta
A = [(a * b) + (b * c) + (c * a)] / a
B = [(a * b) + (b * c) + (c * a)] / b
C = [(a * b) + (b * c) + (c * a)] / c
Just as easy.

This next conversion is very useful. Since we know that it is easier to work with a circuit composed of entirely either current sources or voltage sources, how woul dwe convert on eto another. Good old Ohm;s law allows us to do just that. There is just a couple restriction on the transformation. If you have a voltage source you want to transform into a current source, the resistor must be in series. then you merely obtain the current via Ohm's Law and make that the voltage source. IF starting with a current source, Ohm;s Law a voltage source with the reisistor and make that the voltage source.
Look at the flow of current and how it does not change during this conversion. Also note the open circuits. We don't care what happens beyond this part of the conversion.
On to more fun













Ohm's Law

The first law we must examine is Ohm's Law. This law states the relationship between voltage, resistance and current. Let's examine a simple circuit and see how versatile this law is.
Looking at this circuit we see a DC voltage source in series with a resistor. Let's say we have a resistor with a value of 1kΩ and a voltage source equal to 1V. Ohm's Law states: i = v / r = 1V / 1kΩ = 1mA. So long as you have 2 of the values in Ohm's Law you can solve for the third.



Kirchhoff's Voltage Law

Using that same schematic we can solve for current using KVL. A loop is considered any path in a circuit that begins at a point and terminates at the same point. You don't always need a connection between the path taken, this will be explained further later. For now let's consider the bottom left corner our beginning point and we will travel along the circuit path summing up votages on our way. One thing to keep in mind is the polarity of elements because this will make some negatives in the formula. Resistors are always positive on the side of entrance and negative on the side of exit of current flow. Let's sum up the circuit. The first element we encounter is the DC voltage source which equals +1V. Continuing on, the next element is a resistor. Since the current, or our path, goes form positive to negative, we have a voltage drop here. Using Ohm's law, v = ir, the voltage across the resistor equals -ir. Continuing on we reach the point we started and our loop is complete. Writing it all out yields:
1V - ir = 0 → 1V = ir → i = 1V / 1kΩ = 1mA
The same answer we solved using Ohm's Law. Notice that you use KVL to solve for current.

Kirchhoff's Current Law

This will require a different graph since in order to sum the currents entering a node we need a node. A node is any point that has 3 or more connections. An essential node is a node that connects to one or more nodes. This will be ellaborated on in more depth later.
The node we will be evaluating is 'A'. Taking into account the voltage source is 1V and each resistor equals 1kΩ, we will solve using KCL. This simple example allows me to talk about some aspects of a circuit to simplify later more difficult circuits. If we were taking the voltage of node 'A' to ground, it would equal the voltage of the voltage source since 'A' to ground touches the same points as the positive and negative terminals of the voltage source. Having said that, the voltage at node 'A' is 1V. Now to sum the currents entering the node. By Ohm's Law we know i = v / r. We will just choose reference currents so all of them enter the node, even though common sense says some current will be exiting. The ones that exit will just turn out to be negative, no big deal. Determining the current through R1 entering node 'A' equals (0 - VA) / R1 = -(VA/R1). The zero in the equation is coming from the base node which is equal to ground, or 0. This will become more apparent with a more complicated circuit which I will demonstrate later.

The examples I gave you were pretty basic. They only had one resistor. Lets take into consideration the fact we may run into a situation with more than one resistor or even some capacitors and inductors. We will need to know how to make multiple elements equal 1 element since that is all the formulas will allow for.
Once one introduces multiple elements, there are two possible connections: parallel and series. The following formulas will demonstrate how to reduce multiple elements into one element when the elements in series.

SERIES

PARALLEL The first thing that should be noticed about this schematic is the current through R3 = 0. We know this because the voltage sources V1 and V2 are equal and have positive and negative terminals in relationship to each other. If this isn't obvious to you don't worry we will prove this in a second. Let's just start minimizing this circuit.



  • R14 = R10 + R11 + R7









    We have now got a circuit that we can evaluate for current easily. We could have done KVL or KCL with the original circuit if need be, but i did this to demonstrate element reduction and KVL, KCL at the same time.






    KVL

    KVL uses the loop currents, I1, 2, to determine the currents, i1, 2, 3. This will be done by assuming a current traveling in the direction of the arrows for the coressponding current loops. Let's see how this is done.
    V1 + I1R3 - V2 = 0
    I1 = (V2 - V1) / R3
    I1 = 0
    This is a simplistic version. In reality, most times you will have a resistor where V2 is located. In these cases V2 = R(I2 - I1) since the two current loops overlap one another.

    KCL

    Now we can sum up the currents flowing into node 'A'.
    i1 = 0 as per decided from KVL
    i2 = -I2
    i3 = I2 - I1
    i1 + i2 + i3 = 0
    Now that we have seen the basic electrical system and how to impliment the three basic formulas, Lets move on to the     not so basic circuits.