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Science Forum Index » Electronics - Basics Forum » Basic ohms law question; batteries in series and in parallel
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| bsd_mike |
 Posted: Sun Jan 28, 2007 11:52 am |
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Guest
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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 |
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| Another Dave |
| Posted: Sun Jan 28, 2007 12:01 pm |
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bsd_mike wrote:
Quote: 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?
Yes
Quote: What about calculating total internal resistance of the batteries?
1/(bat 1 ohms) + 1/(bat 2 ohms) = 1 /r
Yes
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| Jamie |
| Posted: Sun Jan 28, 2007 12:17 pm |
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bsd_mike wrote:
Quote: 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
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.
--
"I'm never wrong, once i thought i was, but was mistaken"
Real Programmers Do things like this.
http://webpages.charter.net/jamie_5 |
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| Chris |
| Posted: Sun Jan 28, 2007 12:19 pm |
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On Jan 28, 9:52 am, "bsd_mike" <bsddo...@gmail.com> wrote:
Quote: 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
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| Thevenin Equivalent
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| ___
| .-------------|___|-----------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%27s_theorem
Chris |
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| chuck |
| Posted: Sun Jan 28, 2007 12:22 pm |
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bsd_mike wrote:
Quote: 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?
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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| ehsjr |
| Posted: Sun Jan 28, 2007 2:39 pm |
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Guest
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bsd_mike wrote:
Quote: 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
You shouldn't put batteries in parallel. If
you must, do it by using diodes to isolate
the batteries.
- + D1
+---[Batt1]--->|---+-----+ <Vcc
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| - + D2 | |
+---[Batt2]--->|---+ [Load]
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+------------------------+
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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