Consider a 12V battery. The "12V" tells us that the potential difference between the positive and negative terminal is 12V after our battery undergoes some chemical process.
Let's consider what this 12V value means. Let's say we have a 1C test charge. If we release our test charge from point B to A, it will have lost 12J of EPE, and gained 12J of KE.
Now, if we apply this logic to a circuit, it seems to me that even if there is 0 resistance, charges starting from the positive terminal and going to the negative will have lost 12V. This is because as the charges get closer and closer to the negative terminal, they are naturally converting potential energy to kinetic energy.
If that doesn't make sense, here is an gravitational analogy. If I lift a ball up h height, it will gain mgh PE. Current status of ball: mgh PE, 0 KE. When I drop it, and it reaches height 0, all PE turns into KE. Now lets say I introduce air resistance. Well now, when I release the ball, when it reaches 0 height it has a different status - 0 PE, less than mgh KE. Air resistance is analogous to a resistor.
Here is where voltage drops come in. From my current understanding, all voltage drops do is convert some of the KE of the charge into other forms of energy (ie light, heat). It has nothing to do with PE and potential difference. Thus, the voltage drops are not necessarily equal to the voltage of the battery. When they are equal, all that means is that all of the KE of the charge got converted to something else; the charge now has 0 velocity.
Can someone explain why I am wrong and what my misunderstanding is?