View Single Post

## Why is there a voltage drop across an impedance?

Firstly, there are lots of ways to word the definition of voltage, but the one that intuitively makes the most sense to me is as follows:

"One volt is when:
The difference in (electrical potential energy per unit charge (q)) between two places equals one."

*per unit charge meaning the electrical potential energy a unit charge would have when put into this electric field, which was probably created by a big group of charge

In class we have been taught taught that there is a voltage drop across a component with an impedance because it either uses (transfers) energy (e.g a resistor dissipates heat energy) or it stores energy in another form (e.g an inductor stores energy in a magnetic field)

But this explanation doesn't really explain why there is a potential difference (a voltage) using the definition I gave at the start. If the definition I gave was correct that means that there is a difference in electrical potential energy per unit charge at each side of the component, and I cannot see how this can be true.

For example, say there is a 12 V battery and a 50 Ohm resistors in a simple circuit. the voltage across the resistor will be 12 V, I can see how this is true, as the way you measure the voltage across the resistor will be the same as the way you measure the voltage across the battery, and a battery has a different electrical potential energy per unit charge at each side because a chemical reaction causes there to be more electrons at the anode side of the battery meaning that there is more electrical potential per unit charge at this side.

However if there are two 50 Ohm resistors there will be 6 volts across each one. That means that at each side of one the resistors there is a different electrical potential energy per unit charge. This then means one of either 3 things; There is more charge at one side of the resistor than there is on the other, the charge is more spread out on one side of the resistor, or the charge is a different shape at one side of the resistor. Since charge in equals charge out of a resistor, and the charge manifests itself on the surface of the wire on each side of the resistor that means the answer is that the charge is more spread out on one side of the resistor.

This last paragraph is what confuses me. When I picture charge going through a circuit I do not picture it going faster or slower when coming out of components, but this must be the case in order for the charge to be more spread out on one side of the component. Is this correct?

Thanks!
 PhysOrg.com physics news on PhysOrg.com >> Scientists moving 15-ton magnet from NY to Chicago>> California scientist still reinventing the wheel at 94>> Romania, EU launch works on world's most powerful laser