When calculating ohms law over a circuit....batteries in series;
voltages are added together.
What do you do with batteries in parallel? Voltage stays the same,
right?
What about calculating total internal resistance of the batteries?
1/(bat 1 ohms) + 1/(bat 2 ohms) = 1 /r
Thanks a lot.
-Mike
Hi, Mike. Putting batteries in parallel is usually a bad idea,
because the slight differences between voltages will end up being
evened out at the expense of one of the batteries.
As a study question, you've got a fairly simple circuit when you put
two batteries with internal resistances in parallel (view in fixed
font or Courier):
| Original Circuit
| ___
| .-------------|___|-----o-----o
| | Rs(1) | +
| V1 | ___ |
| --- .-----|___|-----' Vo
| - | Rs(2)
| | V2 |
| | ---
| | -
| | |
| | | -
| '-------o---------------------o
|
|
| Thevenin Equivalent
|
| ___
| .-------------|___|-----------o
| Vth | Rth +
| ---
| -
| |
| | -
| '-----------------------------o
(created by AACircuit v1.28.6 beta 04/19/05
www.tech-chat.de)
Theveninzing the circuit, first read the no load voltage at the
output. That's Vth. Then replace the voltage sources with short
circuits and current sources with opens. Use Ohms Law to calculate
the resistance looking into the voltage source. That's Rth. Done.
Good luck with your studies. Here's a link that explains it well in a
little more detail:
http://en.wikipedia.org/wiki/Thevenin's_theorem
Chris