Mismatched voltage wired in parallel. What is resulting voltage?

[Thomas G. Marshall]

Fundamental questions.

Given:

If I wire two 3V 1mA batteries in parallel, the result is 3V 2mA.

If I wire two 3V 1mA batteries in series, the result is 6V 1mA.

BUT:

1. What happens if I wire the following in parallel:

3V 1mA
6V 1mA

2. And similarly, the following in series:

3V 1mA
3V 2mA

Does #1 above average the voltage to 4.5V?

Does #2 above average the current to 1.5mA?

Thanks!

 

[DJ Delorie]

1. What happens if I wire the following in parallel:

3V 1mA
6V 1mA

Does #1 above average the voltage to 4.5V?

This type of circuit depends heavily on the resistance of the wiring
and batteries. What's going to happen is that a LOT of current is
going to want to flow from the 6V battery to the 3V battery, in an
attempt to fast-charge the 3V from the 6V until they're both the same
voltage, or until something fails.

Aside from burning out the batteries and/or wires (unlikely for a 1mA
capacity), the actual voltage at the junction depends on the relative
resistances of the two half-circuits (wiring, battery's internal
resistance, contact resistance, etc). Basically, it's a voltage
dividier, as if you had two ideal voltage sources each with a separate
series resistor.

Guessing (wildly) from the descriptions, the 6V source would have
twice the internal resistance of the 3V source (else it would have a
2mA max), so a 2:1 divider across the loads results in a 4v node
voltage.

Does #2 above average the current to 1.5mA?

No, current ratings are maximums, not absolutes. If those are the
real maximums, the result is a 6V 1mA source. Actual current draw
depends on the load, not the source.

 

[Thomas G. Marshall]

DJ Delorie said something like:

"Thomas G. Marshall"


This type of circuit depends heavily on the resistance of the wiring
and batteries. What's going to happen is that a LOT of current is
going to want to flow from the 6V battery to the 3V battery, in an
attempt to fast-charge the 3V from the 6V until they're both the same
voltage, or until something fails.

Aside from burning out the batteries and/or wires (unlikely for a 1mA
capacity), the actual voltage at the junction depends on the relative
resistances of the two half-circuits (wiring, battery's internal
resistance, contact resistance, etc). Basically, it's a voltage
dividier, as if you had two ideal voltage sources each with a separate
series resistor.

Guessing (wildly) from the descriptions, the 6V source would have
twice the internal resistance of the 3V source (else it would have a
2mA max), so a 2:1 divider across the loads results in a 4v node
voltage.


No, current ratings are maximums, not absolutes. If those are the
real maximums, the result is a 6V 1mA source. Actual current draw
depends on the load, not the source.

Ah...ok. Thanks!

 

[Thomas G. Marshall]

John Popelish said something like:

Batteries are not rated for current, but with current times
time, e.g. mA H for milliampere times hour.

Ok, fair enough.

The higher voltage battery runs down very quickly, trying to
charge the lower voltage battery. The exact process depends
on whether this causes either of the batteries to be
permanently damaged or not.


If those are milliampere hour ratings, then you get a 6 volt
battery for the first milliampere hour output, then you get
a much lower voltage as the larger battery tires to reverse
charge the smaller battery with whatever load the remaining
voltage drives through the load.


Almost certainly, no.


No.

Fair enough.....thanks! Too bad, in addition to two very clear, educating,
and friendly replies, there had to be a rude idiot. After 5000 posts in
usenet, I'm used to seeing all forms of foolishness, but it still amazes me
a little.

 

[Phil Allison]

"Thomas G. Marshall = Asinine Cunthead "



** Go DROP DEAD

- Mr SHIT for BRAINS.





....... Phil

 

[Thomas G. Marshall]

Michael A. Terrell said something like:

Ignore phil. He was breast fed by his dad till he was 16, and he has
never got over it.

Well, he certainly seems set on making a fool of himself.

Service to my country? Been there, Done that, and I've got my DD214 to
prove it.
Member of DAV #85.

I'm one of those sappy types that thanks people for their service. Thank
you.

Do you mind me asking in what way you became disabled?

 

[Phil Allison]

"Thomas G. Marshall = Asinine Cunthead "



** Go DROP DEAD

- Mr SHIT for BRAINS.






....... Phil

 

[Thomas G. Marshall]

John Fields said:

(something like a 12 year old)

<PLONK>

{yawn}

 

[Thomas G. Marshall]

DJ Delorie said something like:

....[snip]...

No, current ratings are maximums, not absolutes. If those are the
real maximums, the result is a 6V 1mA source. Actual current draw
depends on the load, not the source.

Hmmmmm.....

If I connect an ammeter (from my handy multimeter) to a battery, what is it
measuring? The amount of amps allowed to pass through the wires, etc. of
the multimeter? If it depends entirely on load, then why wouldn't the
ammeter always measure the same regardless of what I connect it to?

I'm sorry, but my electrical background is limited to *some* digital
electrical engineering: A year of TTL circuits, karnaugh maps, up/down
counters, and the like from my computer science degree in college. There
are *huge* gaps in my understanding of the analog world.

 

[DJ Delorie]

If I connect an ammeter (from my handy multimeter) to a battery,
what is it measuring? The amount of amps allowed to pass through
the wires, etc. of the multimeter?

If you connect an ideal battery to an ideal ammeter, it will read
infinite amps.

However, the battery has internal resistance, wires have some
resistance, the ammeter has internal resistance.

So if you connect a real battery to a real ammeter, you get the
battery's voltage divided by the actual total resistance
(Rbatt+Rwire+Rmeter).

A battery which can only provide 1mA would have a high internal
resistance, perhaps 3000 ohms. A car battery can provide hundreds of
amps at 12v, closer to 0.1 ohms.

If it depends entirely on load, then why wouldn't the ammeter always
measure the same regardless of what I connect it to?

In normal circuits, the resistance of the battery is negligible.
You're making up abnormal circuits, so it isn't. Plus, amps depends
on the volts as well; for a *given* battery, the amps depends on the
load. For *all* batteries, it also depends on the voltage of the
load. V=IR, as long as you account for all those tiny internal
resistances.

 

[Peter Bennett]

DJ Delorie said something like:

...[snip]...

No, current ratings are maximums, not absolutes. If those are the
real maximums, the result is a 6V 1mA source. Actual current draw
depends on the load, not the source.

Hmmmmm.....

If I connect an ammeter (from my handy multimeter) to a battery, what is it
measuring? The amount of amps allowed to pass through the wires, etc. of
the multimeter? If it depends entirely on load, then why wouldn't the
ammeter always measure the same regardless of what I connect it to?

The resistance of an ammeter is very low, so a very large current
would flow. The current would be limited by the resistance of the
wires, ammeter, and the internal resistance of the battery.


--
Peter Bennett, VE7CEI
peterbb4 (at) interchange.ubc.ca
new newsgroup users info : http://vancouver-webpages.com/nnq
GPS and NMEA info: http://vancouver-webpages.com/peter
Vancouver Power Squadron: http://vancouver.powersquadron.ca

 

[Phil Allison]

"Thomas G. Marshall = ASD Fucked Cunthead "



** Go DROP DEAD

- Mr SHIT for BRAINS.




....... Phil

 

[Don Kelly]

---
No, it isn't.
---


---
No, it isn't.
---


---
The 6V battery will discharge into the 3V battery until there is no
voltage difference between the batteries.
---



---
No. There will be no current until a load is attached.
---



---
**** you.

Since when are you the moderator here who can tell _everyone_ what
to do? Just ask your stupid questions, wait for your answers, and
then just go away instead of trying to run the show.

What was your reason for your last comment? Is there something I missed? You
had given good answers 'til then. Possibly his comments were aimed at Phil
who was playing jackass - if so you missed the target.

 

[Thomas G. Marshall]

Don Kelly said something like:

What was your reason for your last comment? Is there something I
missed? You had given good answers 'til then. Possibly his comments
were aimed at Phil who was playing jackass - if so you missed the
target.

One of my rotating signatures is the comment above about relaxing about
crossposting. For some reason he equated this with an attempt at
self-establishing myself as a moderator or something. Agree or disagree
with my signature, and that's fine, but sputtering that kind of childish
crap got him killfiled.

 

[Don Kelly]

----------------------------

Don Kelly said something like:

One of my rotating signatures is the comment above about relaxing about
crossposting. For some reason he equated this with an attempt at
self-establishing myself as a moderator or something. Agree or disagree
with my signature, and that's fine, but sputtering that kind of childish
crap got him killfiled.