I've read a few books on electronics including the one on this site and they all do a good job of providing information on voltage, but it seems like the information presented on current is a side thought. As a result I just can't seem to grasp the concept.

If I have a 9V battery and say I put it in series with a .01 ohm resistor (if I had such a thing) ohms law would state I should get 900 amps.

Also what is driving the current? If I wire the 9V battery in a short circuit what will my amperage be and what limits it?

It also seems like if something tries to draw more amperage then the source can supply at a given voltage - the voltage drops.

First off why is the load drawing an amperage?
Why does the voltage drop?

I apologize for this questions, but I'm hoping someone can help and clarify this in a way I can understand.

 

It helps some people to think of voltage being similar to a pressure difference in a hydraulic system.

If there is a pressure difference across such a system a current (your "amperage")will flow. The magnitude of the current being limited by the diameter of the pipe/tubing. The smaller the diameter the higher its resistance to current flow.

If the pressure is increased the current increases proportionally.

A perfect voltage, or pressure, source has no internal resistance but practical ones have, especially dry cells and batteries.

In the case of a short circuit across a 9 volt battery the current would be limited by the battery's internal resistance. The relationship between the voltage, resistance and current would still obey Ohm's Law.

 

Originally posted by chomper@Mar 29 2005, 04:46 PM
I've read a few books on electronics including the one on this site and they all do a good job of providing information on voltage, but it seems like the information presented on current is a side thought. As a result I just can't seem to grasp the concept.

If I have a 9V battery and say I put it in series with a .01 ohm resistor (if I had such a thing) ohms law would state I should get 900 amps.

Also what is driving the current? If I wire the 9V battery in a short circuit what will my amperage be and what limits it?

It also seems like if something tries to draw more amperage then the source can supply at a given voltage - the voltage drops.

First off why is the load drawing an amperage?
Why does the voltage drop?

I apologize for this questions, but I'm hoping someone can help and clarify this in a way I can understand.

[post=6539]Quoted post[/post]​

The reason for the voltage drop has to do with a concept known as voltage division. If you put 9v across 2 equal resistors you get 4.5v in the middle of them. For the most part, batteries and voltage sources have an internal resistance that is quite low. Usually less than 1 ohm, even down to a tenth or a hundreth of an ohm depending on the quality of the source. When you connect a load to this source, your loads resistance gets put inline with the sources internal resitance and you get that concept known as the voltage divider here too. Normally, the loads resistance is so high compared to the internal resistance of the source that the voltage division is miniscule. But when your load approaches your internal resistance, you start to get this division, or your loading down your source.

Ohms law goes in a few ways as you can control all parts of the equasion. Its not like a gravity equasion or something from physical mechanics as many of those relationships you can't control all parts of it. I.e., when have you been able to change gravity? But for ohms law, you have control over all 3 if you want it as long as you maintain their relationship. You usually end up setting 1 of them constant, usually the resistance, and you control 1, usually the voltage and your result is the current. This is the typical use since most components work off of voltage sources.