Perhaps you are a victim of analogies. The water analogy has problems, as do all analogies, but in the end you'll have to just do the maths. Perhaps you just need the raw, lowest level understanding, not some hand-wavy YouTube video explanation. What follows is an example that I hope will help you understand better the interplay between all the elements of a simple circuit, in a far more formal, yet easy-to-understand way. It's a starting point, not a full answer to your question. It will be up to you to follow the links I provide, and learn those concepts fully, to generalise your skills.

Every two-terminal component has a relationship between current \$I\$ through it, and voltage \$V\$ across it. Two quantities. Here's a 5V zener diode's "\$I\$ vs. \$V\$ curve", from which you can figure out, given a voltage what current is flowing, or given a current what voltage exists across it:

This is telling you that if current is positive, the voltage that will appear across the zener diode will be 5V. If current is negative (in the opposite direction), the zener diode will develop -0.6V across it. The negative sign indicates polarity - which end has the higher potential.

This is the relationship between current and voltage in a 10kΩ resistor:

Unlike the zener diode, the relationship here follows Ohm's law \$V= I \times R\$

$$ V = I \times 10{\rm k\Omega} $$

Lastly, here's what a 12V battery \$I\$ vs. \$V\$ relationship looks like:

This graph is saying that it doesn't matter what amount of current is flowing, or in what direction, the voltage across the battery is 12V, and the polarity doesn't change - one end (labelled positive on the battery) always has the higher potential. Twelve volts higher, in this case.

Putting those three elements into a circuit, to form a single loop around which current can flow, looks like this:

^{simulate this circuit – Schematic created using CircuitLab}

I'll explain my thinking when I look at this schematic. There are a few measurement devices, voltmeters and ammeters, which make it look complicated, but they don't influence the circuit at all, and they don't change the circuit's behavior. They only show me what the simulator thinks is happening.

There's a ground symbol at the bottom. Its meaning is special, and you can learn about it here, in an answer I wrote just before this one. It doesn't change the behaviour of the circuit at all, and can be disregarded for the purposes of this explanation. It is just there to make the simulator work. However, after reading that other answer, the potentials I have marked in red will make more sense, as you proceed through this answer.

There's a single loop around which current can flow. This means that current can't leak out of that loop, or can't be injected from an external source. In other words, current around this loop, regardless of where you measure that current, is the same everywhere. It's value is \$I\$, which we will calculate later. The law that enables me to make this statement is Kirchhoff's Current Law (KCL). You absolutely must learn this law, it's not optional. It is often overlooked in YouTube videos, with the assumption that it's obvious and you know it already. If you don't know it, and understand its implications, then it's no surprise that you are confused.

The 12V battery, from the graph above, produces a potential difference of 12V (duh), and this is unconditional. It can be confirmed by looking at the battery graph above - 12V across it at all times, regardless of current through it, and the top end is more positive than the bottom end, as annotated.

The current produced by that battery flows clockwise. I'm not talking about electron flow, I'm talking about conventional current, which always flows within components from their higher potential end to lower, if they are receiving energy, and in the opposite direction when they are donating energy. This is called "passive sign convention". In this case, the battery is producing energy to be absorbed by D1 and R1, in accordance with this principle. Therefore, current flow is clockwise here.

The other law from Kirchhoff that you must grok is Kirchhoff's Voltage Law (KVL), which describes the relationship between the voltages (potential differences) of the elements around a loop. Taking a walk around that loop, potential rises and falls at each node, and you must always arrive back where you started with the same potential, for a net change of zero. Starting from the ground symbol and going clockwise (you can start anywhere and got in either direction, the equations would be equivalent), I would say that you will encounter a rise in potential as you jump over the battery, and falls in potential as you cross the resistor and diode. Written formally this would be:

$$ V_{B1} - V_{R1} - V_{D1} = 0 $$

The more common "offhand" way of stating this relationship in plain English is that the sum of the voltages across the diode and resistor must equal the voltage across the battery. The battery is fixed in voltage (see its graph), irrespective of current flow, and the other two elements, whose voltages are more flexible, must adopt voltages that sum to match that battery, otherwise the laws of physics are broken, and the universe explodes.

In other words, the voltage across the diode, and the voltage across the resistor sum to 12V:

$$ V_{R1} + V_{D1} = 12 $$

The algebraic formula of \$I\$ vs. \$V\$ for a zener diode is hellishly complicated, but its \$I\$ vs. \$V\$ graph is not. You can see from that graph that almost any amount of current flowing through it (in the direction we have established) will result in it having 5V across it. This is shown on voltmeter VM2.

By contrast, the resistor's voltage will depend entirely on the current flowing through it, according to Ohm's law \$V=IR\$, also demonstrated in its graph. However, we know already the voltage across it, because we already know the voltage across the zener diode, and we know that the two values must add up to 12:

$$ \begin{aligned} V_{D1} &= 5 \\ \\ V_{R1} + V_{D1} &= 12 \\ \\ V_{R1} &= 12 - V_{D1} \\ \\ &= 12 - 5 \\ \\ &= 7 \end{aligned} $$

That's the value shown on the simulated voltmeter VM1.

We still don't know current \$I\$, but given that we do have values for \$R_1\$ and its voltage \$V_{R1}\$, that easy to calculate using Ohm's law:

$$ \begin{aligned} R_1 &= \frac{V_{R1}}{R_1} \\ \\ &= \frac{7{\rm V}}{10{\rm k\Omega}} \\ \\ &= 700{\rm \mu A} \end{aligned} $$

Knowing \$I\$, it's worth re-checking on the various graphs that the voltages and current we have predicted here correspond with the components' expected state.

What all of this shows is that if the \$I\$ vs. \$V\$ characteristics of all the components of a system are known, then the state of the system can be determined without any knowledge of how the components work, how they correlate with water systems, or the phase of the moon. In this example, perhaps you already "intuited" that the resistor value doesn't change voltages anywhere - that's decided by the zener diode. Maybe you can see already how useful this can be, given that battery voltage tends to drop as it discharges. If you needed a steady +5V potential, regardless of battery voltage, this would be one way to achieve that.

This process is always building a set of simultaneous equations, describing state, and then solving them. Often it's easy, as the above shows, but most often you just can't simplify the solution as I did here, and you have to actually do the maths.

The principles involved aren't very mysterious or difficult to understand, and analogies might do more harm than good. However, from this semi-formal treatment, those analogies might suddenly make more sense, becoming useful in the future. I believe that maybe the best approach is to first understand the maths, the \$I-V\$ curves, then formulate analogies to build intuition later.

Here are the key principles:

• Ground symbol
• passive sign convention
• Kirchhoff's Current Law
• Kirchhoff's Voltage Law
• Ohm's Law

Once you've got some mastery of this, you can move on to time-varying systems, using capacitors and inductors. The same principles I just listed are still applicable.

Now that you've built a formal understanding of this circuit and its behaviour, maybe we can build an analogy. I don't know if others have ever used the analogy I'm about to propose, which involves forces (tension) and extension (length), but it was fun to think like this.

Imagine that current through a thing is tension in it, like an elastic band having tension as it is stretched. Then imagine that the length of the thing (elastic, perhaps) corresponds to the voltage across it. Using the \$I{\rm -}V\$ graphs, let's try to find materials that would exhibit behaviour analogous to them.

Start with the battery, whose "length" is 12 units, regardless of how hard you compress or stretch it. That sounds like a rigid rod to me. Easy.

For the zener diode we have something that offers no opposition to being stretched, unless you try to stretch it more than 5 units in length, at which point it won't extend any more. That sounds rather like a of piece of string, loose and floppy for all lengths under 5 units, but becoming suddenly straight and taught when you try to extend it beyond 5 units. The "loose and floppy" state of the string requires almost no effort, no tension, to change its length, representing the vertical near-zero current part of the graph, but tension (current) suddenly increases when you try to stretch it beyond 5 units. This is the horizontal section of the graph on the right side.

The resistor behaves like an elastic band. The further you stretch it, the harder it pulls back. Tension (current) increases in proportion to extension (voltage), corresponding to the nice straight line of the graph, on the right side. We'll need to need pretend that the elastic has zero length under zero tension (because the graph passes through point (0A, 0V); I don't think there exists an elastic band with this property in reality, but we'll have to use our imagination, and pretend that it does.

The circuit is rather like connecting these three elements end-to-end in a loop. The analogy permits you to imagine what happens to the rather abstract voltage and current in more tangible terms, length and tension, with which you are familiar. Imagine that the rod-battery is 12 units long. This would stretch the zener-string to its maximum 5 units, and the remaining 7 units difference must be made up by stretching the elastic-resistor to a length of 7.

The analogy remains correct as we increase battery voltage. As you increase the rod's length, the diode-string remains at 5 units long, but the elastic resistor will stretch to bridge the gap between the ends of the diode and battery.

What happens when the rod-battery is shorter than 5 units? The string-diode becomes loose, the elastic-resistor length drops to zero, and there's no tension in the system, zero current. There will be zero voltage across the elastic-resistor, and whatever's left over is the distance between the two ends of the string-diode. In other words, rod-battery length (voltage) equals string-diode length, and elastic-resistor length is zero. Now you can picture the graph of current as voltage goes between 0 and 12V, and you'll find it will correspond to a simulation of this same "sweep" of battery voltage:

^{simulate this circuit}

This shows current stuck at zero (no tension) until rod-battery length exceeds 5 units, at which point the string-diode becomes taught and ceases to extend, and the elastic-resistor begins to stretch, with tension (current) increasing with the elastic's extension.

The analogy has flaws. What happens for negative battery voltage? What does a negative length even mean? Still, it works for this simple example. I don't know how useful the analogy will be for general use, because I haven't tested it in other scenarios, I just made it up.

The important thing is that the system settles into a condition in which every element's state sits at some position on its \$I-V\$ curve, simultaneously satisfying all the equations we derived earlier:

$$ \begin{aligned} V_{B1} &= 12{\rm V} \\ \\ V_{D1} &= 5{\rm V} \\ \\ V_{R1} &= V_{B1} - V_{D1} \\ \\ I &= \frac{V_{R1}}{R_1} \\ \\ \end{aligned} $$

Any analogy is just a necessarily flawed scenario which tries to mimic these rather abstract equations. The analogy can fail (zero and negative length elastic doesn't exist or make sense, negative tension can't be represented, and so on), but the equations don't have such constraints, and always lead to the correct answer. Trust the maths, every time, but using analogies whenever you can will often permit you to predict behaviour intuitively, without having to trawl through and solve dozens of horrendous simultaneous equations.

For example, this rod-battery analogy works well to visulalise the effect of connecting in parallel two batteries of differing voltage (length). Imagine trying to connect their two ends - you'd have to compress one rod while stretching the other, and the force (tension, current) you would have to exert to achieve this would be enormous. That's exactly what would happen - the resulting current would be huge, and it's very likely that one of the two rods would fracture.

Those \$I-V\$ curves are right at the heart, as are KCL and KVL, so learn them. There's no short-cut here, do the maths, but analogies are sometimes helpful.

Some general advice, in no particular order:

Start by learning the fundamentals. Understand what voltage is, what current is, and the equations that define each component. Start with passives: resistors, capacitors, and inductors. Then do diodes. Then transistors.

Play with a simulator to build intuition; I suggest starting with circuitjs and graduating to LTspice.

Make sure you know at least differential calculus, though integral calculus and differential equations will help a lot too.

A good textbook will help. Horowitz and Hill's The Art of Electronics is a good one for beginners.

Trying to jump straight to working with microcontrollers is, in my opinion, the wrong thing to do. Don't complicate things by getting an arduino involved, not until you understand the basics.

People learn in different ways. I don’t know about you, but I find that being able to visualize and interact with circuits is very helpful.

To that end, I haven’t found a better electronics learning aid than the Falstad circuit simulator, a JavaScript-based tool that lets you observe circuit voltages and currents interactively. It’s very easy to use and comes with example circuits for you to try out. I use it here quite a bit: roughly half my nearly 2000 answers here on Stackexchange have a Falstad sim included.

Speaking of which, another aid is right here in front of you: yup, this Stackexchange, peopled with knowledgeable folks who understand both electronics and the learning process.

How to understand circuits

I understand capacitors, diodes, resistors, transistors on a simple level, but for the life of me I cant figure out how or why they or what they do in a circuit.

Let me clarify right from the start: If it is about genuine understanding, and not just memorization, then it is a difficult and thankless job. This explains why people avoid over-analyzing things. But since you show a desire for this and express it emotionally, I will share my philosophy of understanding, illustrating it with a concrete example.

What does understanding mean

Every circuit, no matter how simple it is, is based on some basic idea. To understand the circuit, we need to understand what that idea is.

At this stage, the details only hinder. The circuit should be cleared of them and only the most essential things should remain.

Basic ideas are revealed through "ideal" components, for example, transistors with zero base-emitter voltage, op-amps with infinite gain, zero voltage between the inputs, zero input currents, infinite input resistance, zero output resistance, and so on.

Looking for functional idea of components

I just want a list of every component on earth that states in plain english "if this, then this...

At this initial stage of qualitative understanding, you do not need to delve too deeply into the internal structure of the components; you need a functional idea of the component. For example:

• a resistor dissipates energy by converting, according to Ohm's law, voltage into current (I = V/R) and vice versa (V = I*R),

• a capacitor stores energy by slowly changing its voltage,

• a diode dynamically changes its "resistance" in the forward direction when the applied voltage or current changes,

• a transistor changes its "resistance" under the control of the input voltagee, and so on...

Discerning sub-circuits

Simple circuits are composed of components, and more complex circuits are composed of simpler ones. We understand circuits by recognizing familiar sub-circuits within them.

Applying the idea

Once we have extracted the valuable basic idea, we can use it to understand other circuits, from which new basic ideas can be extracted. Thus, we can accumulate a collection of basic ideas.

An example of understanding

I just want to take small baby steps and learn something small everyday then put it into practice with a breadboard so i can fully grasp it, instead of assembling circuits I dont understand and accomplishing nothing because i dont understand it!!!

Since you mention the capacitor first, I will illustrate a possible step-by-step scenario of understanding based on this element. This is the famous op-amp circuit of the inverting integrator.

Initial full circuit

Here is the classic circuit diagram of this device. Let's apply a 1 V step input voltage and observe in a graphical form how the voltages change through time.

^{simulate this circuit – Schematic created using CircuitLab}

We see that the circuit is "ideal" and produces a linearly increasing voltage at its output. But what is this due to? This is what we must understand...

Resistor as V-to-I converter

It strikes us that the voltage V- at the inverting input is constantly 0 V. It turns out that the resistor R is grounded on the right (the so-called virtual ground) and the entire input voltage is applied across it. According to Ohm's law, the output current is Iout = Vin/R. Therefore, we conclude, the resistor plays the role of a passive voltage-to-current converter.

^{simulate this circuit}

Capacitor as I-to-V integrator

This current flows through the capacitor C, which we can initially assume is grounded on the right. Thus, it plays the role of an integrator with a current input and a voltage output, and the resistor turns it into an integrator with a voltage input. This is the well-known RC integrating circuit.

^{simulate this circuit}

Disadvantages: However, a problem arises: the capacitor's voltage slows its rate of change and does not change linearly, as we expect, but exponentially. Aha! We realize why - the capacitor's voltage is subtracted from the input voltage, and the current gradually decreases, I = (Vin - Vc)/R.

Compensated Vc

It seems that there is some trick here related to the op-amp. Aha! This is why the right end of the capacitor is not grounded, but connected to the op-amp output! We can imagine the op-amp as a following negative voltage source -Vc...

^{simulate this circuit}

... that sets its voltage (tan) equal to the voltage across the capacitor (gold) and adds it in series to the input voltage. As a result, the voltage V- (blue) in the middle point is zero.

So, the basic idea upon which an op-amp inverting amplifier is built is the following:

The op-amp copies the capacitor's voltage and adds it to the input voltage. As a result, the voltage drop across the capacitor is compensated, and the current I = Vin/R is determined only by Vin and R.

We can call this idea voltage compensation.

Equivalent circuit

So, the total voltage across the network of C and -Vc in series is zero, and we can think of it as "piece of wire" or "virtual short".

^{simulate this circuit}

Practical circuit

Real operational amplifiers have a limited supply voltage, usually bipolar (+10 V and -10 V); so, let's replace the ideal op-amp with a real one having supply terminals.

Now it is interesting to see where current flows (green arrows). It starts from Vin, passes through R and C, enters the op-amp output, exits the negative supply terminal, flows through the negative supply voltage -V, and returns to Vin.

^{simulate this circuit}

When the output voltage approaches either of the supply voltages, it stops changing (the op-amp "saturates"). The magic of this trick ceases: V- and Vc continue to change but exponentially.

More applications of the same idea

After we arrived at this powerful idea, we can try to see it in other similar circuits, and thus explain them. Here are a few of them:

Differentiator

^{simulate this circuit}

Inverting amplifier

^{simulate this circuit}

Log converter

^{simulate this circuit}

Antilog converter

^{simulate this circuit}

Generalization

If we continue, we will see this idea in all op-amp inverting circuits with negative feedback. Therefore, we can generalize it as follows:

The op-amp copies the voltage across the element connected between the op-amp output and the inverting input, and adds the copy to the input voltage source. As a result, the undesired voltage drop across the element is neutralized.

Each of these op-amp circuits consists of an imperfect passive circuit helped by an op-amp. So, we can conclude that op-amp inverting circuits are improved passive circuits.

Op-amp circuit builder

I propose you explore multiple such circuits in a more attractive way, using Op-amp circuit builder. I created it more than 20 years ago with the help of Macromedia Flash. Adobe discontinued Flash Player in 2020, but thanks to Ruffle the movie continues to live.

Here is a brief product description:

The introductory part of the tutorial reveals the basic idea. This is followed by the circuit builder where, in the center of the screen, is the circuit diagram of the device being built (voltages are visualized with red voltage bars, and currents with green loops). There are identical libraries of elements on both sides. When you click on an element from the left library, it flies over and lands in the place of Element 1 (between the input source and the inverting input of the op-amp); when you click on an element from the right library, it lands in the place of Element 2 (between the inverting input and the op-amp output). With the arrows inside the diagram, you can explore the circuit step-by-step, and with the arrows in the bottom right corner of the screen, you can navigate between the four construction stages.

What you only need to know are:

1. Kirchoff's current law
2. Kirchoff's voltage law
3. The relationship between current and voltage of a component

1. Kirchoff's current law says that all current entering a node of a circuit, must also be leaving from that node.

^{simulate this circuit – Schematic created using CircuitLab}

\$I_{in}=I_{out}\$

2. Kirchoff's voltage law says that around a closed loop the voltage drop is 0.

^{simulate this circuit}

\$V_{1}-V_{box}=0\rightarrow V_{1}=V_{box}\$

3. The relationship between current and voltage of a device.

Assume you have this circuit:

^{simulate this circuit}

The relationship between current I and voltage V for a resistor is: \$ I_{R}=\frac{V_{R}}{R} \$ which in our case is \$ 0.01V_{R}\$.

The relationship between current I and voltage V for a capacitor is: \$ I_{C}=C\frac{dV_{C}}{dt} \$ which in our case is \$ \frac{dV_{C}}{dt}\$

The relationship between current I and voltage V for a diode is: \$I_{D} = I_{S}(e^{\frac{V_{D}}{V_{T}}}-1) \$ which in our case is \$ 0.0025(e^{\frac{V_{D}}{V_{T}}}-1) \$

Now let's analyze the circuit. Remember Kirchoff's current law:

Iin at any node must be equal to Iout so take the node with the dot: \$Iin=Iout\$ but what is Iin? Isn't it the current through the resistor which makes it \$0.01Vr\$?

And what is Iout? Isn't it the sum of the current of the diode + the current of the capacitor? But isn't \$Ic=dVc/dt\$ and \$Id=0.0025(e^{Vd/Vt}-1)\$. So since \$Iin\$ must be equal to \$Iout\$ and \$Iin=0.01Vr\$ and \$Iout=dVc/dt+0.0025(e^{Vd/Vt}-1)\$ so write the equation \$0.01Vr-(dVc/dt+0.0025(e^{Vd/Vt}-1))=0\$.

Remember Kirchoff's voltage rule: \$V_{1}-Vr-Vc=0\$. But why didn't we include the voltage drop on the diode? Because \$Vc=Vd\$!!! So it would be equivalent to say that \$ V_{1}-Vr-Vd=0\$. And now we have two independent equations. If you solve the system of equations from the current and the voltage analysis you will find a solution. Then you work backwards and find any value (current, voltage, power) you want.