Basic theory question

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Kyle

Active member
Joined
Nov 18, 2005
Messages
27
Location
Las Vegas
I'm trying to work my way through a "Teach Yourself Electricity and Electronics" book, but they didn't address something I've seen as a given in even the most basic primers.

Why does short circuiting a battery do what it does? Say you have a 9 volt battery and you connect the terminals. The battery heats up and drains quickly. Is that just because the power needs to dissipate somewhere and doesn't have anywhere to go? If you put a low-value resistor in series with the terminals, why doesn't more-or-less the same thing happen? Or does it in some way I'm just not noticing.

If someone could shed some light on this issue, I'd appreciate it.

-Kyle
 
A battery is not a theoretical perfect voltage source. A simple equivalent circuit of a real battery is a voltage source with a small resistor in series. In the case of a 9V battery, it's actually six little batteries (and their "resistors") in series, but these can be reduced to a single voltage source and single series resistance in the equivalent circuit.

Hash out the rest with Ohm's law and the answer becomes apparent.
 
Approximately what sort of resistor values are in this series inside the battery? I *think* I get what you're saying though- If there was, say, 100 ohms over 9 volts you've got like 11 amps and 99 watts coming out of a little battery.
 
[quote author="Kyle"]If you put a low-value resistor in series with the terminals, why doesn't more-or-less the same thing happen? Or does it in some way I'm just not noticing.
[/quote]

With a low value resistor it will still happen. It's just not as easy to notice.

The resistance inside the battery is sometimes called "internal resistance". It varies from battery to battery, especially from design to design (eg. dry cells, lead-acid).

A 9V battery may have an internal resistrance of ~ a few ohms. You can find the actual value from a datasheet for the battery (if one exists).

As NYD says, use your Ohm's law.

I=V/R . Say the internal resistance is 9 Ohms. Then the current (I) would equal 9/9 = 1 Amp! That would be a lot for a 9V battery.

Power (P)=I*I*R
 
> If you put a low-value resistor in series with the terminals

What is "low value"?

A cheap 9V battery may have 90Ω internal resistance. With a dead-short, 9V/90Ω= 0.1 Amps flows. The power dissipated is 9V*0.1A= 0.9 Watts. Since we assumed a dead-short, all of this power must be dissipated in the battery. 0.9W in the size of a 9V battery is pretty warm.

I stupidly put two 9V batts in my back pocket, and they jiggled around until their snaps meshed. This is actually the same situation, but doubled. It didn't burn my butt, but it was awful hot back there.

The total capacity of a 9V battery is about 0.25 Amp-Hours. If we suck 0.1 Amps, it will go flat in about 0.25AH/0.1A= 2.5 hours. (Actually, at this rate and the resulting heat, it will not last this long.)

Now let's add the "low value" resistor. Say 1Ω. Now we have 9V/91Ω, which is nearly 0.1 Amps, almost 0.9 Watts, most of which appears in the battery. Put that in your abacus, you will see that the result is nearly the same.

OTOH, a car battery has about 0.01Ω internal resistance. Now a 1Ω resistor is not a "low value". It is actually about the power of the heater-fan, and the big low-R car battery hardly heats at all with a 1Ω load. The 0.1Ω of the starter motor pulls >200A, and the battery heats a bit. A dead-short on a 12V 0.01Ω car battery can pass 12V/0.01Ω= 1,200 Amps, 14,400 Watts, and it will heat up pretty good. (Kids, do NOT try this at home. You don't have a piece of metal big enough to qualify as a "dead-short" at this high current, and if you do, acid will explode in your face.)

Around 9V battery circuits, 1KΩ is a "low value". A fuzz-box might use a 10K load resistor. Put 1KΩ on 9V, the actual resistance is 90Ω+1KΩ= 1,090Ω. The current is 9V/1,090Ω= 0.00825 Amps. That means the 1K resistor will drop 8.25V, the battery 0.75V. Now the heat in the battery is 0.75V*0.008257A= 0.0062 Watts, the resisistor 8.25V*0.008257A= 0.068 Watts. Heat in the battery is negligible, over 90% of total battery power is going into the load where it is presumably doing good work for us.
 
[quote author="PRR"] A dead-short on a 12V 0.01Ω car battery can pass 12V/0.01Ω= 1,200 Amps, 14,400 Watts, and it will heat up pretty good. (Kids, do NOT try this at home. You don't have a piece of metal big enough to qualify as a "dead-short" at this high current, and if you do, acid will explode in your face.)
[/quote]

Done that.

At school, my physics teacher wanted to show me what would happen and proceeded to connect the terminals with a cable with crocodile clips. Bad idea. He attached the second clip (which he was planning on removing immediately) but had trouble pulling it off and ended up with it attached for probably around 5 seconds. It was pretty smoking.

I would definitely not recommend trying it. Even for a laugh.

An interesting point he made was the effects of cold weather on starting the engine which I had never considered.

Basically, the increased internal resistance has the obvious effect of limiting the amount of current you can draw when the battery is very cold.
 
> connect the terminals with a cable with crocodile clips.

That's not a DEAD-short on a car battery. In fact the cable, even if battery-cable, may have more resistance than the battery, and the clips give "high"-resistance contact to the battery posts.

As you might guess from the amount of power dissipated in the clip: enough to melt metal right away.

If you could get a really good contact, a foot of battery cable will start smoking, and could possibly melt violently. They will handle (barely) the 200 Amps of the starter, not the 1,200 Amps that the battery can dump if you give it free flow.

A really good dead-short is probably the diameter of the battery-post, an inch or so. A really good clamp is bigger than the old-type cast clamps (and a lot heftier than the sheet-metal battery clamp on my Honda).

Get a 1"x2"x12" bar of copper. Drill two tapered holes exactly as far apart as the battery's terminals. Taper the battery posts exactly the same as the holes. Drop the bar very firmly on the battry posts and run like hell.

> the increased internal resistance has the obvious effect of limiting the amount of current you can draw when the battery is very cold.

The copper resistance is actually lower when cold.

The battery resistance is a lot about electrochemical reactions that run slow when cold, so the effective resistance is higher when cold.

But the big variable is the oil. If the engine turns freely, the starter comes up to speed, the back EMF rises, the current gets small. If the engine is hard to turn, speed is low, back EMF is low, current is high. Since even 10-40 oils get thick when cold (though less than a straight 30W), turning a cold engine is hard work and heavy current. Using a warm battery on a cold engine helps a bit. Using a warm starter actually hurts a bit. But it's that oil in engine bearings and pistons that really makes the battery work hard.
 
That means the 1K resistor will drop 8.25V, the battery 0.75V. Now the heat in the battery is 0.75V*0.008257A= 0.0062 Watts, the resisistor 8.25V*0.008257A= 0.068 Watts. Heat in the battery is negligible, over 90% of total battery power is going into the load where it is presumably doing good work for us.

Is this why most circuits I've seen seem to start with a resistor?

-Kyle
 
[quote author="Kyle"]
That means the 1K resistor will drop 8.25V, the battery 0.75V. Now the heat in the battery is 0.75V*0.008257A= 0.0062 Watts, the resisistor 8.25V*0.008257A= 0.068 Watts. Heat in the battery is negligible, over 90% of total battery power is going into the load where it is presumably doing good work for us.

Is this why most circuits I've seen seem to start with a resistor?

[/quote]

I think that's too broad a question. The "start"? Depends what you define as the start.

Also, things vary from one circuit type to another.

You are probably on the right track though.

Assuming your electronics book is quite good, you should learn about opamps soon (which at first glance are simple) and also voltage dividers. These two subjects will help you understand simple circuits a lot better.
 
[quote author="PRR"]Get a 1"x2"x12" bar of copper. Drill two tapered holes exactly as far apart as the battery's terminals. Taper the battery posts exactly the same as the holes. Drop the bar very firmly on the battry posts and run like hell. [/quote]
Yep
or simply drop a large spanner on the terminals
bulls eye
each end seemed to weld themselves tight and bang
yes the battery blew up
after the smoke cleared the terminals where still welded to the spanner and bits of the battery where all over the OB Garage ... we were still finding them months later.

incident two
after twenty years of putting the OB truck batteries on charge ... he says he always did it the same way ... one day the whole lot blew up in the fellows face. (he was OK)
I understand it was a gas thing and the battery charger was ON when the clips where attached.

So my work place has two Occupation Health and Safety directives sprung from the above two incidents.

Batteries do have an internal resistance and so don't give infinite current BUT as batteries are constantly improving and their ability to deliver BIG current is amazing.

Has anyone ever seen on of those RC electric cars lose a 7.2 battery pack in a catastrophic explosion.
I lost one model car that way and this was ten years ago. the capacity of these small batteries has near doubled.
BANG
 
Car batteries...

An acquintance of mine once touched the -ve pole with his gold wrist watch whilst unscrewing the +ve terminal with an uninsulated screw driver. Very foolish of course, but accidents like these often happen when people underestimate the danger of things...

I haven't witnessed it personally, but as I've heard it, the result was a large blue spark followed by a big puff of smoke, and his watch melted into his wrist. Awfully painful... He still bears the scars of it...
 
[quote author="Frank Eleveld"].. the result was a large blue spark followed by a big puff of smoke, and his watch melted into his wrist. Awfully painful... He still bears the scars of it...[/quote]
aarrgh !!
:shock:
YUK

I don't wear any jewellery ... I'm may be boring but I can still count to 10
A fiend lost his finger while loading concrete re-enforcing mesh on to a truck.
It was a ring and just a freaky set of circumstances.

AND
I have a continual war (Union Rep stuff) with management about ID tags being worn around our necks. I wear the tag on a short workshop make zip tie clip on my hip and so do some of my maint tech team.
The neck tags can fall into gear and cause problems to you and to the gear.
We damaged a Dig Beta Head ... many thousands to replace.

sorry to hiJack the thread
shut up Kev
 
I have to remember to remove the two rings I wear customarily, when I work on electronics, or disaster awaits.

When Gates (battery co., not Bill) started to make some really low internal R lead acid cells they had a lot of cautionary material about rings and bracelets etc. and the possibility of contacting the cell terminals. I think one of the demos to reinforce these admonitions was a piece of wire between the terminals of something not much larger than a big D cell, incandescing.
 
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