What is the Effect of Voltage Drops on Electrical Potential Energy?

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Voltage drop refers to the decrease in electrical potential energy as current flows through a resistor in a circuit. When current passes through a resistor, the voltage across that resistor decreases, indicating a loss of energy per unit charge. However, the total current remains constant throughout the circuit, similar to water flow in a pipe where pressure differences drive the flow. Even if the voltage appears to drop to zero after a resistor, the current can still return to the battery's positive terminal due to the resistance present in the entire circuit, including wires and connections. Understanding voltage as a relative measurement helps clarify that there is always some potential difference in an active circuit loop.
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I am a fifteen year old teaching myself electricity and electronics I already know Ohms law and all direct current basics as I've read a book on it. Although I understand all of the relations I am still confused as to what a "voltage drop" actually is. When it is said that the voltage "drops" across a resistor, does this mean that the voltage actually decreases after going through a resistor?

This is how I understand voltage thus far. Work is required to cause a difference in the amount of electrons on two bodies. When a conductive path is given, the electrons will flow to the body with a deficiency of electrons and perform work on any resistance within the circuit. A battery is constantly doing work to maintain a potential difference. Here's what I don't get:

If I have a battery with an electrical potential of 10 Joules per Coulomb (volts) and I have a 2 Ohm resistor (R1) and a 3 Ohm resistor (R2) in series, there will be a current of 2 Amps flowing through the circuit. In the 2 Ohm resistor there will be a drop in electrical potential energy of 4 Joules per Coulomb. Therefore, the voltage right after R1 should be 6 Joules per Coulomb. Then, the current will go through R2 and the voltage should drop to 0 Joules per Coulomb. My question is, if there is 0 Joules per Coulomb after R2, how does the current make it to the positive terminal of the battery? It just makes sense that after going through each resistor, the potential energy of each charge (the joules per coulomb; volts) should decrease. Am I understanding this correctly? If not could you please explain electrical potential energy (voltage). Any help would be greatly appreciated. Thank you.

Anthony
 
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aje8127 said:
I am still confused as to what a "voltage drop" actually is. When it is said that the voltage "drops" across a resistor, does this mean that the voltage actually decreases after going through a resistor?
It means there is a voltage difference across the resistor. Think of it like water, current is the amount flowing through a component - you can't gain or lose current in the same way you can't gain or lose water in a pipe. Voltage is like pressure, there must be a pressure difference for the water to flow - all the pressure differences must add up.


My question is, if there is 0 Joules per Coulomb after R2, how does the current make it to the positive terminal of the battery?
In reality the connection has a bit of resistance so the potential is actually zero until it reaches battery.
 
Across any component(s) - including wires and battery - of a (nonsuperconducting) circuit, resistance remains greater than zero. Thus voltage, as a relative measurement, always has nonzero value between any two distinct points in an active circuit loop.
 
It may be shown from the equations of electromagnetism, by James Clerk Maxwell in the 1860’s, that the speed of light in the vacuum of free space is related to electric permittivity (ϵ) and magnetic permeability (μ) by the equation: c=1/√( μ ϵ ) . This value is a constant for the vacuum of free space and is independent of the motion of the observer. It was this fact, in part, that led Albert Einstein to Special Relativity.
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