If V = I x R then does it still "make sense" to refer to a battery's voltage when not connected to anything? As in I = 0..?

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EUrbanAutoTech

Joined Jan 29, 2023
6
I'm an automotive mechanic that specializes in electrical/drivability diagnosis. I like helping other techs get a better grasp on this aspect of our trade. Now imagine I'm with another tech and I have a battery and a multimeter in front of me. I measure the battery's voltage and get 12V.

Me: The battery is 12 volts.
Tech: Is it still 12 volts?
Me: Well, ya. I just measured it...
Tech: But V = I x R and right now it's not in any kind of circuit so shouldn't I = 0, therefore, V = 0 x R = 0..?
Me: Uhh....

I recently realized that I did not have a satisfactory answer to that. If I only responded with, "They're just different, okay? Don't ask that question and deal with it!" then I feel like I would lose significant credibility with the rest of what I say...

So, at this point, would the battery just have a voltage potential and not really a "true voltage" until it has some non zero current? Would its voltage potential really just be a manifestation of its chemical energy? But it still would have charge separation so it really should have a "true voltage" ... shouldn't it?

I'm just having a hard time getting past my discomfort with something having a non zero voltage but a zero current value. Any clarity would be greatly appreciated.

Many thanks,
Eric
 

ericgibbs

Joined Jan 29, 2010
17,171
hi Eric.

I = 0, therefore, V = 0 x R = 0..? should read V= 0 * ∞, when there is no load.
So, at this point, would the battery just have a voltage potential and not really a "true voltage"

The open circuit voltage is usually referred to as a potential difference.

E
Open-circuit voltage (abbreviated as OCV or VOC) is the difference of electrical potential between two terminals of an electronic device when disconnected from any circuit. There is no external load connected. No external electric current flows between the terminals.
 

Papabravo

Joined Feb 24, 2006
19,854
The circuit consists of the battery and the multimeter. A small amount of current is drawn through a high resistance and that current will imply the voltage of the battery precisely because Ohms law is in play. If you look at a spec sheet for a traditional moving coil meter an impedance of 20,000 Ω/volt would not be uncommon. So connected across a 1.5V "D" cell, the meter would have a resistance of 30,000 Ω and draw a current of 50 μA. Modern multimeters are a significant improvement on that.
 

dl324

Joined Mar 30, 2015
15,511
I'm just having a hard time getting past my discomfort with something having a non zero voltage but a zero current value. Any clarity would be greatly appreciated.
A battery not connected to anything is still a battery. If it's a charged state, it will still have voltage. When you measure the voltage, it's in a circuit. In order for Ohm's Law to work, you have to have a complete circuit; even if that circuit is completed by a meter being used to measure the battery voltage.

Do your techs have a problem when the battery voltage isn't 12V? Does self-discharge/internal resistance confuse them? Discharge paths between the terminals of a dirty battery?
 

Thread Starter

EUrbanAutoTech

Joined Jan 29, 2023
6
hi Eric.

I = 0, therefore, V = 0 x R = 0..? should read V= 0 * ∞, when there is no load.
So, at this point, would the battery just have a voltage potential and not really a "true voltage"

The open circuit voltage is usually referred to as a potential difference.

E
Open-circuit voltage (abbreviated as OCV or VOC) is the difference of electrical potential between two terminals of an electronic device when disconnected from any circuit. There is no external load connected. No external electric current flows between the terminals.

Thank you. I like that. That helps me, "put things in the proper box" so to speak.
 

crutschow

Joined Mar 14, 2008
31,535
V, I, and R are all separate characteristics.
V = I x R is the Ohm's law equation that shows the relation between the three.

If R = infinity (open circuit) then I must be 0, but V (a true voltage or potential) is still there.
It does not necessarily depend on the other two.
 

MrChips

Joined Oct 2, 2009
28,137
Welcome to AAC!

Eric, you asked a valid question. More importantly, instead of giving your fellow tech a frivolous answer, you came to the right place for clarification.

V = I x R

That is correct. But that is not Ohm's Law.

Ohm's Law states that the current flowing through a conductor of resistance R is proportional to the applied voltage V and inversely proportional to the resistance.

Stated mathematically,

I = V / R

Battery voltage V is still 12V.
Open circuit resistance is infinity.

Hence,
I = 12V / infinity = 0

There is no contradiction with the math or with Ohm's Law.
Again, battery voltage is still 12V, as measured by the multimeter which represents a very high resistance R, practically infinity for this application.
 

BobTPH

Joined Jun 5, 2013
6,304
Thank you. I like that. That helps me, "put things in the proper box" so to speak.
But I don’t see how it corrects your misunderstanding. Ohm’s law describes the relationship of the voltage across a resistor to the current through same resistor. It does not describe the same for a battery, a capacitor, an inductor, diode, an LED, a transistor, a microprocessor or an electric motor, or myriad other devices.
 

MrChips

Joined Oct 2, 2009
28,137
Let me put it differently.

Ohm's Law is:

I = V / R

There are three unknowns or variables, I, V, and R.
Mathematically, you can calculate one unknown value if you know the other two.

I = V / R
V = I x R
R = V / I

If you know the current I and the resistance R, you can calculate the applied voltage V,

V = I x R

R = 0
I = ∞, the math symbol for infinity
V = 0 x ∞ which does not have a valid answer.

Current I is a derived answer that depends on the values of V and R.
V is the driving force. I is the result of V.
 

Thread Starter

EUrbanAutoTech

Joined Jan 29, 2023
6
Do your techs have a problem when the battery voltage isn't 12V? Does self-discharge/internal resistance confuse them? Discharge paths between the terminals of a dirty battery?
The problem our techs have is from having very vague definitions/understandings. Most techs are only taught VERY crude models of ultra basic circuit theory. Having such vague definitions and crude models makes it hard to rationalize ideas. So they come up with some truly wild ideas.

Case in point. I saw an experienced guy get confused by this car that wouldn't start after only doing an oil change. Nothing had any electrical power. But it drove in fine. It JUST had a battery installed elsewhere. The negative post of the battery still had the protective plastic cap over the battery post. Then the negative terminal cable was placed over top of that. It took the tech a while to wrap his head around it. This should be a half second thought:

"Car gets a battery installed elsewhere. Plastic cap left on negative post by accident. Car wouldn't start. Person thinks "just a dead battery." Boosts car. Car starts and runs fine. (It's running off the alternator). Drive's it to our shop. Tech shuts it off. Now car won't start. It's not starting because the alternator is not running and the battery is not connected to the circuit."

This sorta stuff happens ALL THE TIME in my trade. We call the positive terminal "Power" and the negative "Ground". Then we treat them as independent and different entities. It's not just us either... Household wiring has a "Hot" wire and a "Neutral" wire. This sorta stuff leads to confusion and can mislead you.

So I'm attempting to make a better "automotive electrical course". I'm trying to formalize my definitions to get rid of any ability to attribute "magic properties" to electricity. That's when I came across this thought. To me, the idea of an open circuit battery still having voltage just "made sense". But when I thought of Ohm's law, I realized I didn't have a good enough definition. Let alone one to teach others with...

Eric
 

MrChips

Joined Oct 2, 2009
28,137
The problem our techs have is from having very vague definitions/understandings. Most techs are only taught VERY crude models of ultra basic circuit theory. Having such vague definitions and crude models makes it hard to rationalize ideas. So they come up with some truly wild ideas.

Case in point. I saw an experienced guy get confused by this car that wouldn't start after only doing an oil change. Nothing had any electrical power. But it drove in fine. It JUST had a battery installed elsewhere. The negative post of the battery still had the protective plastic cap over the battery post. Then the negative terminal cable was placed over top of that. It took the tech a while to wrap his head around it. This should be a half second thought:

"Car gets a battery installed elsewhere. Plastic cap left on negative post by accident. Car wouldn't start. Person thinks "just a dead battery." Boosts car. Car starts and runs fine. (It's running off the alternator). Drive's it to our shop. Tech shuts it off. Now car won't start. It's not starting because the alternator is not running and the battery is not connected to the circuit."

This sorta stuff happens ALL THE TIME in my trade. We call the positive terminal "Power" and the negative "Ground". Then we treat them as independent and different entities. It's not just us either... Household wiring has a "Hot" wire and a "Neutral" wire. This sorta stuff leads to confusion and can mislead you.

So I'm attempting to make a better "automotive electrical course". I'm trying to formalize my definitions to get rid of any ability to attribute "magic properties" to electricity. That's when I came across this thought. To me, the idea of an open circuit battery still having voltage just "made sense". But when I thought of Ohm's law, I realized I didn't have a good enough definition. Let alone one to teach others with...

Eric
Hi Eric,
You are not alone. This happens everywhere in every field.

There are "teachers" who try to teach a subject that they know little about and they unfortunately pass on misinformation.

Then there are "teachers" who know their subject fully but are not able to convey it in a meaningful manner to people not conversant in the field.

It is rare breed and a delightful bit of fresh air to find a good teacher who knows the topic fully and can explain it to the uninitiated.

Search around and you will be able to find online courses and tutorials specifically targeted to automotive electricity and electronics. This is absolutely essential. As you are well aware, everything in automobiles are controlled by electronics today.
 

Thread Starter

EUrbanAutoTech

Joined Jan 29, 2023
6
But I don’t see how it corrects your misunderstanding. Ohm’s law describes the relationship of the voltage across a resistor to the current through same resistor. It does not describe the same for a battery, a capacitor, an inductor, diode, an LED, a transistor, a microprocessor or an electric motor, or myriad other devices.
I guess I was just looking for a more discrete reason for why it doesn't apply to batteries et al. My intuition told me it very clearly shouldn't apply to batteries. But if anyone asked me why it didn't apply, I wouldn't be able to give a better answer than, "because!" I feel like that could to lost confidence. It's very hard to teach if you don't have a student's confidence and trust.

My take away is this:

When current passes through a resistor I and R both have finite, non zero values, therefore, it makes sense for V to have a finite, non zero value.

But for an open circuit battery I = 0 and R = ∞. To me: "V = 0 / ∞" shows clearly enough the limit of Ohm's law and that we would need a different tool/formula to understand open circuit voltage of a battery.

In this case I'm not concerned with finding a definition for describing open circuit voltage of a battery. I just wanted to find an adequate reason for why Ohm's law didn't apply to batteries other than just, "because."

This may seem overly pedantic but it's been my experience, in my trade, that vague ideas lead to techs being confused and mislead which leads to things being misdiagnosed which leads to revenue and customers lost.
 

ericgibbs

Joined Jan 29, 2010
17,171
hi Eric,
It is important to remember all batteries have some value of internal resistance due to their chemistry.
It is a VERY low resistance for a lead acid battery.

E
 

MrChips

Joined Oct 2, 2009
28,137
I guess I was just looking for a more discrete reason for why it doesn't apply to batteries et al. My intuition told me it very clearly shouldn't apply to batteries. But if anyone asked me why it didn't apply, I wouldn't be able to give a better answer than, "because!" I feel like that could to lost confidence. It's very hard to teach if you don't have a student's confidence and trust.

My take away is this:

When current passes through a resistor I and R both have finite, non zero values, therefore, it makes sense for V to have a finite, non zero value.

But for an open circuit battery I = 0 and R = ∞. To me: "V = 0 / ∞" shows clearly enough the limit of Ohm's law and that we would need a different tool/formula to understand open circuit voltage of a battery.

In this case I'm not concerned with finding a definition for describing open circuit voltage of a battery. I just wanted to find an adequate reason for why Ohm's law didn't apply to batteries other than just, "because."

This may seem overly pedantic but it's been my experience, in my trade, that vague ideas lead to techs being confused and mislead which leads to things being misdiagnosed which leads to revenue and customers lost.
Here again we see an example of misinformation.

My intuition told me it very clearly shouldn't apply to batteries.
I just wanted to find an adequate reason for why Ohm's law didn't apply to batteries other than just, "because."
This would be a misconception.
There is no contradiction with Ohm's Law.
Ohm's Law still applies to batteries.

The error is in the analysis of the mathematical formula:

V = I x R

If you apply this formula,
V = 0 x ∞

The formula is still correct and any value of V satisfies the formula.
But V is the driving force of the battery and is 12V. You cannot change that. V = 12V is a constant value in this application.
I is the unknown quantity that depends on the value of R.

Thus the error here is not in the understanding of how batteries work.
The error lies in the proper application of math when working with values of zero and infinity.
 

MrChips

Joined Oct 2, 2009
28,137
Beg to differ.
It is mathematically Ohm's law with the variables re-arranged.
It is still commonly called Ohm's law (the relation between voltage, resistance, current).
That is a gross error.
You can differ as much as you please but that is blatant misinformation which we are desperately trying to correct in this valid query.

Ohm's Law is

I = V / R

The corollary of Ohm's Law (or application of Ohm's Law) would be

V = I x R
R = V / I

Mathematically speaking, I is the dependent variable, V and R are the independent variables.
(We are trying not to confuse the TS more than he already is.)
 
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