So in physics class, we learned about batteries and voltage. We learned that even though you can connect 8 AA batteries in series to create a 12V battery, you still couldn't put that into say, your car because there's not enough current.
But what happens when you reverse the situation? What if you had...a graphing calculator that was driven by 8 AA batteries. If you hooked up a 12V car battery instead of 8 small AA batteries, would the calculator fry, or would it work normally?
In theory it should work normally. Both the 8 AAs and the car battery provide the same amount of potential. The current that is supplied is dependent on the impedence of the load. In other words, the calculator if it's properly designed won't draw more current than it needs.
If the calculator has all 8 cells in series, which is likely but not certain, then it can be run from a car battery. There is no problem with the current. There are lots of things like the clock and interior lights in cars that take hardly any current and run from the car battery.
There are problems running some 12V stuff from a car as there is a lot of noise in real cars, so odd things like spikes could damage a calculator.
At the risk of sounding pedantic, and for the benefit of the OP.....
Your terminology is incorrect and misleading...
The use of the term impedance doesnt apply to dc circuits, only to AC.
Impedance(Z) in an AC circuits made up from two components, actual ohmic resistance and reactance at a defined frequency...
Ah. So as long as the 8 cells are in series, in theory, it should work. But it might not work in practice because of imperfections like noise. Great! I was expecting it to be a lot more complicated.
At the risk of sounding pedantic, and for the benefit of the OP.....
Your terminology is incorrect and misleading...
The use of the term impedance doesnt apply to dc circuits, only to AC.
Impedance(Z) in an AC circuits made up from two components, actual ohmic resistance and reactance at a defined frequency...