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-   -   Why is there a voltage drop across an impedance? (http://www.physicsforums.com/showthread.php?t=675017)

 CraigH Feb27-13 08:29 PM

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!

 CraigH Feb27-13 08:33 PM

Re: Why is there a voltage drop across an impedance?

And if this is true, is there a way to calculate a scalar value for the divergence of charge at a particular point in a circuit made of wires and resistors?

 willem2 Feb28-13 01:12 AM

Re: Why is there a voltage drop across an impedance?

When you connect a battery to a resistor, small charges will form on each side of the resistor, that will produce the electric field that will drive the electrons through the resistor. To get a net field, there has to be more charge on one side of the resistor than on the other side.
This charge is very small. about 10^(-11) coulomb would be enough to produce the field of 600 V/m that you would get if you put 6V across a resistor with a length of 1 cm.

This charge only influences the current by producing an electric field. It's only a very small part of the free charges in a conductor, or in the resistor.

 Menaus Feb28-13 02:24 AM

Re: Why is there a voltage drop across an impedance?

Quote:
 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 depends on whether the resistors are parallel or are in series. If they are in parallel, the voltage is the same across both resistors (12v each), if they are in series then the voltage is shared across the two resistors (6 for each).

In a series connection, the voltage is shared across all loads, this is how series circuits work. I believe there is a voltage gradient across the conductor which makes this happen.

Steinmetz threw out the notion of charge on a conductor because it made misconceptions and confusion arise, instead he said the electric circuit arose from the conjugate energies of the dielectric and magnetic fields.

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