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?(adsbygoogle = window.adsbygoogle || []).push({});

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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# Voltage Drop

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