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 /rThanks a lot.
-Mike
Yes
Yes
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in theory yes. but in practice it may not be a good idea if one cell is worse than the other. it just gives you more current handling/twice the life in theory.
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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
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:
Chris
If all the batteries have equal no-load voltages, then yes, the voltage stays the same.
If the batteries do not have identical no-load voltages, then the voltage of a parallel combination depends on the internal resistances as well as the no-load voltages.
So put a half-discharged 1.5 volt battery in parallel with a fully charged one and you will see a voltage lower than 1.5 volts.
Chuck
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You shouldn't put batteries in parallel. If you must, do it by using diodes to isolate the batteries.
- + D1 +---[Batt1]--->|---+-----+ |---+ [Load] | | | | +------------------------+
Then the problem is real world, not theory, and Vcc becomes whichever is greater: (Batt1 - voltage drop in D1) versus (Batt2 - voltage drop in D2)
Given some discharge time, the voltages will be identical.
Ed
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