Calling it a built-in voltage is something of a misnomer. People usually think of "voltage" as "what you measure with a voltmeter". So "voltage" is normally synonymous with "electrochemical potential of electrons" (in stat mech terminology) and with "difference in fermi level" (in semiconductor terminology). Under this definition, the built-in "voltage" is not actually a voltage.
Then what is it? It's what chemists call "galvani potential", and some physicists call "electrostatic potential". It's the line-integral of electric field. (Maybe you should call it "built-in potential", not "built-in voltage".)
Voltage / fermi level measures the total "happiness" of electrons, the sum of all influences on the electron. The electric field (galvani potential) is just one of those many influences. Other influences include diffusion (entropy), the kinetic energy of the electron's wave function, etc. etc. But it's the sum of all influences that determines how the electron moves. That's why it is the voltage, not the galvani potential, that determines the most important things like current flow and energy dissipation.
So to summarize: The "voltage" across a p-n junction is zero, when the word "voltage" is defined in the most common and sensible and intuitive way. After all, the junction is in equilibrium; an electron is equally happy to be on either side.
For more details see my other answer: Fermi level alignment and electrochemical potential between two metals
Going around a loop, both the voltage differences and the galvani potential differences sum to zero. But only the former is really important. For the galvani potential differences, most of them are unobservable, like the volta potential at the junction when you solder an aluminum wire to a copper wire. It is possible to figure out the galvani potential differences everywhere in a p-n junction circuit, including at the wire contacts, at the voltmeter, and so on. If you do figure them out, and add them all up, you'll get zero! But since none of those parameters matter for the circuit behavior, people rarely think about them or try to figure them out.
Voltage is similar to height. It plays the same role for electric charge as height*gravity does for a ball on a hill. So high voltage means high potential energy the same way a ball being high up on a hill means high potential energy.
Voltage is not potential energy, the same way height is not energy. However, if you have a certain amount of charge $q$, you can multiply it to the voltage to get the potential energy, which his $Vq$. This is similar to the way you can multiply height to mass*gravity to get $mgh$ for the potential energy of a ball on the hill. So voltage is potential energy per unit charge the same way height*gravity is potential energy per unit mass.
Voltage must be measured between two points for the same reason height must be. When someone says "the height here is 1000 feet", they are actually comparing it to a point at sea level. In electronics, "sea level" often gets replaced with "ground". So if someone says, "this fence is electrified at 10,000 Volts", they mean there is a 10,000 Volt difference between the fence and the ground, the same way they mean that there is a 1,000 foot drop between the current elevation and the ocean. However, you can use any two points to measure height differences. If you drop a ball, it makes more sense to talk about height above the floor of the room you're in than to talk about sea level. Similarly, if you want to look at a single resistor, it makes the most sense just to talk about the voltage change across that resistor.
The work done on a charge as it moves from point to point is the quantity of charge times the voltage difference. This is just like the work done on a ball as it slides down a hill is the mass of the ball times the height of the hill times gravity.
A single battery cell can only produce a couple of volts. That's how much the potential changes for a single electron in the chemical reaction in the cell. This is a bit like the way a pump that works via suction can only lift water about 30 feet into the air, since that's the potential energy from buoyancy from the entire atmosphere. You can stack multiply batteries on top each other to get a higher total voltage drop (as is done in 9V or 12V batteries) the same way that you could use multiple pumps to suck water higher than 30 feet.
If you increase the voltage across a circuit element, in general the behavior might be quite complicated. This is like saying that if you tilt a ramp to a steeper angle, you will change the way that objects slide down the ramp. In many materials, we find that the behavior simple: current = voltage/resistance. So if you double the voltage, you double the current. This is called Ohm's Law. An accurate description of why it is true is probably a bit too advanced for right now. You will do okay for intuition if you start thinking of electrical current as being like water flowing through a tube. Then Ohm's Law says that if you're powering the flow by having the water flow downhill, if you make the downhill flow twice as steep, the water flows twice as fast. Yes, you can think of it as saying that the electrons are going faster.
Adding resistors in series is like adding several pipes to go through. If you try to push the water through more pipes, it will become more difficult. If you were letting water flow down a hill through a series of pipes, the more pipes you have, the less each pipe can be pointed downhill. That means that adding more pipes makes the water flow more slowly everywhere. Similarly, adding more resistors in series reduces the current everywhere.
The quantity you actually measure when it comes to current is the total flow - number of electrons per second passing through. If you have a 1-ohm, 5-ohm, 1-ohm resistor series, they will all have the same current going through them. This is because if they did not the current would start building up somewhere, and that would change the flow. (This actually happens, just very quickly because the wires have very low capacitance.) The way they all get the same current is they have different voltages. Most of the voltage drop for the entire circuit will be across the 5-Ohm resistor. This is like setting up pipes so that a skinny pipe goes down a steep portion of a hill while two fat pipes go down shallow portions of the hill. The total water going through each pipe per second would be the same. In this case, the water would move faster through the skinny pipe (the high-resistance portion). This is just because the total flow is the same, so if the cross-sectional area is less, the velocity is higher to compensate. This sort of picture roughly works with electrons as well. It is called the Drude model. It is the easiest to visualize, but it is not true to the quantum picture of modern physics.
Batteries do die slowly, yes. That is why flashlights, for example, grow dimmer and dimmer before turning off entirely.
To say a circuit component has a voltage is just saying that there is a certain voltage drop across that element. It is like saying that each pipe in a series of pipes running down a hill has a certain height difference, and that the height difference for the entire system of pipes is the sum of all the height differences of the individual pipes.
If two resistors are in parallel, they have the same voltage drop. This is like saying that two pipes side by side have the same height difference. The one with 1-Ohm resistance will have five times as much current going through as the one with 5-Ohm resistance.
Best Answer
In short: No, not all voltages sum to zero, unless you have a closed path
Take an easy counterexample and put a number $n$ resistors in parallel to a voltage source. The voltage drop over all resistors is just the voltage of the source. If you sum them up, you get something like $(n+1)V_0$. Only voltages in a closed circle sum to 0, which are your equations
BandC. As check you might use e.g. $ 40 I_3 - 40 I_2 = 10$, which is just the outer circle.When thinking about voltages i like to remind myself that theres an electrical potential behind. Every point in the circuit has some potential, like every point on a landscape has a height. Voltages are nothing than differences in potential (or differences in height). If you walk around the circuit you walk sometimes up, sometimes down, but wherever you walk, once you get to the point where you started you are at the same height again. And adding all voltages you passed (watch the sign!) sums to 0. If you end up at a different point however, it can sum to anything.
So as tests qualify all paths that return to the starting point. E.g. also walking from A via 10V soucre, lower resistor, 20V source, middle resistor, upper resistor, lower resistor, 10V source, middle resistor, 20V source, lower resistor and 10V source back to A. Totally overcomplicated, but it adds to 0 if you got the signs right. The easier one is the outer circle which i used in the first paragraph.
The second possible test is like your equation
A, that on each point the sum of all currents has to be 0 (again, watch the sign). You have a second point at the right side, even though thats the same equation just with opposite signs.