Hey y'all. As you may or may not know I am an engineering student at the University of Mind Your Own Fucking Buisness. I tutor the other undergrads for fun and profit, but mostly fun. I had to break down a low level Electrical Engineering course into a fucking to do list for a kid today and I figured I would transcribe the list over for you. You may already know all that, that's cool, I did too but I didn't three years ago
SO here goes, without further ado... the first electrical engineering class, in jot note form.
Concepts you need to know
Volts = Voltage = Electromotive Force = Potential Difference between two points
Amps = Current = Amperes
Watts = Power = The Actual Electrical.. POWER
Resistance = What goes in does not necessarily come out, it increases with respect to temperature, and everythig has it including sources and wires
If the Amperes is the size of a pipe, and the voltage is the speed in the pipe (deceasing due to resistance as it goes) then power is the flow rate of the fluid. (this helped as the kid was an industrial student)
Smaller Gauge = Bigger wire, think shot gun.
Conductors .. conduct electricity, insulators stop it from killing you 
Air makes a decent insulator, if people can be reasonably expected not to go near it
Now on to Basic Simplification
Ohm's Law, The First Tool in our Toolbox
V=IR, say it with me. V=IR I=V/R and R= V/I.
All important relationships, all can be remembered by not being a dumbass. If you are a dumbass then make the sound VIR repeatedly and hope you know where the = sign goes.
V = Voltage, I = Amperes (yes I know that is dumb and no i will not explain it), R= Resistance, great. Even if we are going off the first letters you can get 66% of the formula down.
Example problem:
A circuit is hooked up to a 12 V voltage source and is found to draw 3 Amps through a single load R1, what is the resistance of R1, what is the Power? If we are neglecting both source and conductor resistance (we are).
R1 = V/I
R1 = 12V/3A
R1 = 4 Ohms (I cant do the symbol on here, bite me.)
Power = ...
SHIT. I forgot, we're idiots today Here;
P = IV
Power, In Watts, is equal to I times V. No I will not spell those out for you again
So for our example:
P = 3A*12V
P = 36 Watts
Damn son.
Parallel Vs Series; Equivalent Resistance Calculations
You need two formulae for this section, and I say that loosely..
With x resistor:
For Series; Req = R1 + R2 .... + Rx
For Parallel; 1/Req = 1/R1 + 1/R2 .... + 1/Rx
If you can't solve for Req in either one, kill yoself fool.
Now, what IS a series circuit? Versus a paralell one?
Lets pretend that Uncle Greg is a perv. Of course he is. He can either go along banging girls one at a time (series) or lots at a time (parallel). This actually raises some good point, if we consider volts his energy level, and amps his attention to each woman.
In series he gets more an more tired as he does each chick, but his attention stays the same. In Parralel (Say in his threesome between you mom and your childhood crush) he doesnt get as tired, but each girl only get a portion of his attention.
This actually works PERFECT because in a series circuit, Voltage is constantly dropping, and your current is constant. In a parallel circuit each branch receives an equal voltage, but the current is different! We can verify this with V=IR and see why that matters in parallel ones... but later.
Oh yeah and any time you wanna simplify these sons a bitches, start farthest from the source and work towards it. But HTH what if there is multiple sources? Well simmer down young fella we'll get there.
Kirschoff's Voltage and Current Laws
Kirschoff was some asshole who played with electricity, and himself, probably. He ended up making two "Laws".. the self entitled prick. One is about voltage, the other current. They're both moderately helpful, lucky for you they're also simple as piss.
KVL Says: The sum of the directed voltages about any closed loop is equal to 0.
In layman's terms: When you reach the end of your circuit and its about to go back into the source, it should be at 0 volts.
KCL Says: The sum of the directed currents leading a node should equal 0.
In layman's terms; any point where wire's meet is like an intersection of working cars. Different amounts go different places but there shouldn't be any left over in the middle.
This leads us into Loop and Node Analysis, which I'll cover in one more section here.
Current Splitting vs Voltage Splitting
Alright so technically the heading is a lie, we're using current dividers and voltage dividers. Not splitters, same shit different alphabet. Anyway;
A Current Divider is basically this, Ix = (Rt/Rx)It
Remember how I said that the current (Uncle Greg's attention) is split amoungst branches(resistors)/(women) in a Parallel circuit? This is the general formula for finding the Current in a Resistor X, given you know the Total Resistance/Total Current as well as the Resistance of Resistor X.
A Voltage Divider is this: (generally and in lo level classes)
V1 = (R1/RT)*Vi
This is used to calculate the drop of voltage across a given resistor(load/etc)
Example Problems:
Using the Concept of Voltage Division Find the Voltage drop across each individual resistor in the following circuit:
A 6 volt source is hooked up in series to 3 resistors, one is 2 Ohm, One is 3 Ohm, the last is 1 Ohm.
Since Req = Rt = R1 + R2 + R3, we Know that total Resistance is 6 Ohm , convenient.
A this point if you're not an inbred you can tell me the voltage drop across each but just in case your mother is in fact also your sister...
V1 = (R1/RT)*Vin
V1 = (3Ohm/6Ohm)*6 Volt
V1 = 3 V
V2 = (r2/rt)*vin
V2 = (2ohm/3ohm)* 3 Volt (Note new totals)
V2 = 2 V
V3 = Oh shiiit this is the last resistor and since we know that voltage must go to 0 in our idealized world, we can just say V3 = the remainder, aka, 1 Volt.
What we learned: The Voltage Drop across a resistor in series is dirctly proportional to the resistance of said resistor.
Current Divider Example:
A circuit has two resistors connected in parallel, one is 4 Ohm, the other is receiving 3 Amps out of the total 9 Amps entering the parallel branches. Solve for all variables.
Hold up wait a second.
We know that Ix = (Rt/Rx)It and we also know that 1/Rt = 1/R1 + 1/R2
We have: R1 I2 and It.... but we need more! we also know that thanks to KCL (look above) that the current in the first branch + the current in the second branch must equal the current entering, so with simple math we deduce I1 = 6 Amps , if we had a voltage at this point we could solve for the Resistance with ohms law but I didn't give it to myself.
R1 = 4 Ohms
I1 = 6 Amps
R2 = ?
I2 = 3 Amps
So NOW in our formula solving for the Resistance of the Second should be easy, Ill tell you that we should solve for Rt and work back to solve R2
6 Amps = (Rt/4Ohms)10Amps
4(6/10)=Rt
Rt = 2.4 Ohms
Now we go (using the formula because im teaching a first year this)
1/2.4 Ohms = 1/4 Ohms + 1/ x Ohms
Multiple it all to get a common denominator:
10/24 - 6/24 = 1/x
4/24 = 1/x
1/6 = 1/x
x= 6 Ohms
All that just to show you two things:
Always look at all your basic formula, that was long winded but easy. and
The Current in a branch is inversely related with the resistance.
A note on Maximum Power Transfer
The maximum power is achieved when the source resistance is equal to the load resistance.
That's it. All he needed to know, all y'all need to know.
Thevenin's Theorum
Any electrical circuit with only voltage/current sources and resistors can be replaced by a voltage source in series with an equivalent resistance.
Once again, just something he needs to memorize. Good for you all to know too, maybe sound intelligent at a party or something... Idk... I dont go to cool parties much any more ;'(
Node and Loop Analysis
and
Theorum of Super Position
I decided to do these together because they all kinda work together (and i only want to need to describe one circuit lmao) For each of these concepts we will be using this bastardized circuit:
---------------~~~----------------------------------------~~~-------------
| 2 Ohm | 3 Ohm |
| + $ $
--- $ 5 Ohm $ 8 Ohm
6V $ $
-------------------------------------------------------------------------------
The only difference being that for super positions we will deal with this one:
---------------~~~---------------------------------------~~~----------------
| 2 Ohm | 3 Ohm | +
| + $ --
--- $ 5 Ohm 4V
6V $ --
-----------------------------------------------------------------------------|
Whattt, yes. Two sources.
Please bear with me on the shitty diagrams btw.
Loop Analysis:
This one works off the basis of KVL, where every closed loop can be seen to have a 0 voltage by the end. We can take apart more complex circuits and look at them as simply a system of loops. So in our example. How many loops do you think we should look at? Well the correct answer is; it depends.
So I'll make this a questions so I can explain as we go...
Find the voltage drop across the 8 Ohm resistor.
Now we could just find the equivalent resistance to find the amperage.. then work back out to the full circuit. But once we get decent with loop (mesh) analysis... it is much faster and scales a bit better.
First we write down a few of the loops as equations, about the voltages, knowing that V=IR and that all the resistors have a voltage drop making each closed loop = 0...
6 Volts - 2 ohm(I1) - 5 Ohm (I1-I2) = 0 //The left hand loop, notice the shared branch has partial current
- 5 Ohm (I2-I1) - 3Ohm (I2) - 8 Ohm(I2) = 0 // You can see that the voltage drop across one branch of a parallel circuit is equal ot the voltage drop across the other... moderately important.
Simplifying we get:
6v - 7 I1 + 5I2 = 0
-16 I2 + 5 I1 = 0
We can take the second loop and use substitution to solve for a current;
5 I1 = 16 I2
I1 = 16/5 I2
6v -7(16/5 I2) + 5I2 = 0
6v = 87/5 I2
I2 = 30/87A (I didn't show units, bite me)
I2 = 0.345 A
and
-16 I2 + 5 I1 = 0
so I1 = (16(0.345A))/5
or 1.104A
**at this point i want to say I have not verified any of these numbers and its very possible i fucked something up, I was just trying to show him the basic idea and to be frank if i had a math error he wouldnt have noticed :p**
We already said that V=IR and the V=8Ohm*0.335A or 2.68 Volts
Next we will use the node method to double check our answer.
Now since the loop method worked off of KVL you probably guessed that the Node Analysis method would work off of KCL...
Node Analysis:
We know that the current entering and leaving a node must equal out to 0, (no cars left in the intersection) and we also know that current is equal to V/R.
One thing I can't stress enough is that when you are using Node analysis.. you MUST Specify a reference point!
THis decides the signs in our equation and without it... well you're not getting the correct answers.
FOr our I will put it on the source side of the bottom part of our circuit, denoted by a *.
As with in loop analysis we must first write equations for each node, luckily this circuit only has one, it s where circuits meet :p Between the 2 and the 3 ohm resistors on the top length. The equations are as follows (in the format V/r +/- V/R ..... = 0 Because the CURRENT (V/R) entering/leaving he node must equal 0.
So for our example:
Va = THe voltage at out node, called A.
((Va - 6v )/ 2 Ohm) + Va/5 Ohm + Va/11 Ohm = 0
Youll see that for any given Va under 6v which it has to be... the branch with the higher resistance has less amps (current dividing) and that elecricity wont flow towards the positve end of a battery (we get a negative there, implying that the current flows into the node, not out of it on that wire)
Simplifying we see that:
110* all of it: 55(Va-6v) + 22Va +10Va = 0
55 Va - 330V + 22Va + 10Va = 0
87Va = 330V
Va = 330/87 // Notice how the Voltage is 11 times the current in this part?
Va = 3.79
Va*8/11=V(R8ohms) //Voltage divider for the one we want
2.68 Volts
***Serious Disclaimer, I spent the past 20 minutes finding out that chopping off the last few digits in one of my answers made onew wrong by a few hundreths of a volt.
One is 0.07 V off based on my numbers here but the process is correct.
Finally. Super Position. I'm gonna rush through this one, as I have... real work. Lol.
Basically super position says that we can seperate a problem into layers, then take the directional sum at any point to get the real answer. That's as best as I can explain it. If there is two sources, you have two layers. One with the first source, one with the second. If the other source is a voltage source you replace it with wire, if it is a current source you leave it open.
So for our examples we end up with two layers. One that has a 4v source, and one that has a 6v source. We are going to say... find the voltage drop across the 3 ohm resistor
It's worth mentioning you can use any method to solve with multiple sources (in fact nodal analysis excels at it) but for certain problems superpositioning is much better so you must know it
So the loop equations for the first layer (6v) are as follows:
6 Volts - 2 ohm(I1) - 5 Ohm (I1-I2) = 0
- 5 Ohm (I2-I1) - 3Ohm (I2) = 0
Which we simplify into
-8 Ohm(I2) + 5Ohm(I1) = 0
I1 = 8/5(I2)
I2=5/8(I1)
6V -2(I1)-5(I1) + (5(5/8)*I1) = 0
6V -7I1 + 25/8I1 =0
*8: -56(I1) +25(I1) = -48V
//Once again sorry im not writing ohm in to all these
-31(I1) = -48V
I1 = 48/31A
I2= 5/8(48/31)A
Now that you have both Amperages you can find the voltage across that resistor, but im going to do it at the end to show you something
for with the 4 volt source:
4 Volts - 3 ohm(I2) - 5 Ohm (I2-I1) = 0
- 5 Ohm (I1-I2) - 2Ohm (I1) = 0
Lalala simplify...
-7(I1)+5(I2)=0
I1=5/7(I2)
4V-8(I2)+25/7(I2)=0
4v = -25/7(I2)+8(I2)
28V = 31(I2)
I2=28/31A
I1= 5/7(28/31)A
Did you keep track of which loop is which? good. cause i didn't.
Anyway:
6V:
I1 = 48/31A
I2= 5/8(48/31)A
4V
I1= 5/7(28/31)A
I2=28/31A
Since the 3 Ohm Resistor is NOT shared (its a bit different if it is) I can just say
V = I*R+I*R = R(Ia+Ib)
V=3Ohm(0.9677+0.9032) OR CAN I... DUN DUN DUN.... I can't. Because remember how I said directional? yeah, the four volt is going opposite the 6 volt, so the drop is actually the difference between the two.
V=3Ohm(0.9677-0.9032)
V=0.1935 Volts
ANNNND... We're outta here. Good bye. Next time Im tutoring someone Ill try and do this again. Maybe it'll be a cooler class... or maybe it'll be calculus. Bleh.
