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My reasoning:

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?

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  • $\begingroup$ Possible duplicate of I don't understand what we really mean by voltage drop $\endgroup$ – sammy gerbil Mar 16 '18 at 2:43
  • $\begingroup$ If you have no resistance, then you'll have shorted your battery, which isn't well-defined in circuit notation. In order to talk about things in the circuit model, you have to have at least some load between two terminals of a voltage source. $\endgroup$ – probably_someone Mar 16 '18 at 2:47
  • $\begingroup$ Even if you short a battery, there is an internal resistance which limits the current. $\endgroup$ – Bill N Mar 29 '18 at 14:21
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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

By Ohm's law (I suppose you accept Ohm's law since you mention resistance), the voltage across a zero ohm resistance is zero for any finite current through.

If you accept this, it follows, (by applying logic), that it isn't the case that charge flowing through the external circuit "will have lost 12V", since there is no voltage across a zero ohm resistance to 'lose'.

Now, if the 12V battery is a physical battery, i.e., can only supply finite current into a zero ohm load, there is no contradiction; the internal resistance of the battery will 'drop' the 12V (the battery will get hot) and the terminal voltage will be zero (the short circuit will not get hot).

If, by 12V battery, you mean an ideal 12V voltage source, then you have a genuine contradiction, i.e., 12V = 0V. This is why, as is well known in ideal circuit theory, ideal voltage sources cannot be connected in parallel with ideal wires.

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