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

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SUMMARY

The discussion clarifies the concept of voltage drops in electrical circuits, specifically addressing the confusion surrounding the behavior of voltage across resistors. When current flows through resistors, a voltage drop occurs, indicating a decrease in electrical potential energy. For example, in a series circuit with a 2 Ohm resistor (R1) and a 3 Ohm resistor (R2) connected to a 10 Volt battery, the voltage after R1 is 6 Volts, and while it may seem that the voltage drops to 0 Volts after R2, the current continues to flow due to the presence of resistance in the circuit. This resistance ensures that voltage remains non-zero across all components, allowing current to return to the battery's positive terminal.

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aje8127
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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.
 

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