Is there a way to calculate voltage drop with internal resistance and load values in the equation? Or is it somehow connected to ohms law? If so, how is it done? I have this test circuit below to illustrate what I mean.
The following is what I mean when I say internal resistance:
When designing a circuit with a battery, we often assume that the battery is an ideal voltage source. This means that no matter how much or little load we attach to the battery, the voltage at the source's terminals will always stay the same.
In reality, several factors can limit a battery's ability to act as an ideal voltage source. Battery size, chemical properties, age, and temperature all affect the amount of current a battery is able to source. As a result, we can create a better model of a battery with an ideal voltage source and a resistor in series.
Stated in my previous post: electronics.stackexchange.com I understand that a batteries internal resistance + a resistor will reduce voltage by small degrees. I also know that with no internal resistance, voltage is unchangeable, except from the battery itself. If this phenomenon is somehow linked to ohms law, how is it linked? or is there a separate "law" that covers this?
My second question is very basic: Is the voltage the same thru an entire series circuit?