Hi Sean: I guess I am getting a little knotted up by this too. If we have a 100V system, I take it that this would be considered the "nominal" voltage of the system. However, if we have a 2A, 3Ohm wire in the system, wouldn't the voltage of the wire be calculated in determining the nominal voltage of the system?
Intuitively I understand why V drops (I think): As resistance increases, volts must drop. But, I am not sure how V in Ohms Law relates to V-system.
Is there a good book on electricity basics like "Electricity for Dummies with Electric Personalities?
We need to understand that there is a difference between voltage drop and voltage.
Voltage drop is the voltage lost over a long thin wire.
Voltage is the voltage of the system.
Let's use your numbers for an example:
System Voltage = 100V
Current = 2A
Resistance = 3 ohms
First I am going to calculate the voltage drop:
Vdrop = I x R
Vdrop = 2A x 3 ohms = 6V
Now if we wanted to find out our voltage drop percentage:
6V dropped / 100V system voltage = 0.06
Turn decimal into a percentage:
0.06 x 100% = 6% voltage drop
That means that at the source side of the wire, you can measure 100V and at the load side of the wire you would measure 6% less or 94V, since you lost 6V.
Volts x Amps = Watts
and your Amps are constant, then voltage drop percentage is the same as power loss percentage, which means in the example, you would lose 6% of your power.
This power loss translates to heat generated in the wire. If the wire is too thin, it will heat up.
Make sure to not mix up voltage and voltage drop. I was confused by this at first myself.
As for a book on electricity for dummies with electric personalities, that is funny. If you don't mind I will steal that idea from you.
Some other ideas are Community College classes on dc circuits or electricity.
Here is a cool website I just found that fits the bill:
Here is another example of why you can not make voltage our of thin air with resistance.
V = I x R
Think of the equation above and if you downsized your wire (more resistance) is that going to give you more voltage? If it did, we would just use small wires to make power. Doesn't make any sense to think of it that way.
Vdrop = I x R
With this equation above here it does make sense, because if we use a smaller wire, then we lose more voltage.
I did a little search on Amazon for something simple:
The price is the same as the resistance in the example.
Hope this helps.