# Lead acid battery resistance testing.

#### asee2dsee

Joined Nov 24, 2023
9
Hi folks, I’m new to this forum so thought I’d say hi first of all.
I’m learning about electronics and so have been doing some little projects to test what I’ve learned.

I decided to test a car lead acid battery to see if I could work out the ir.

My questions are;

1) Did I use the correct formula?
V1-V2
V2 X R = Ohms

2) The result I got is below, is this a plausible result?

V1 12.99 V2 12.47 R 3.23
Result 0.1346 ohms

I am teaching myself in this as a hobby and also part of my work. Any assistance is much appreciated.

thanks

#### BobTPH

Joined Jun 5, 2013
8,686
V = I R

You have two R’s and no I. So no, it is not right. You cannot multiply a voltage by a resistance and get a resistance as a result.

First, you need to tell us how you did these measurements. What is the circuit? What is the R and what are V1 and V2?

#### Ian0

Joined Aug 7, 2020
9,531
Internal resistance = ΔV/ΔI
ΔI is the load current (it should be the change in current, but if the starting current is zero, then it equals the load current.

#### Irving

Joined Jan 30, 2016
3,813
I think the TS' formula was let down by HTML formatting. He meant:

$$R_{int} = (V_{OC} - V_{load})/V_{load} * R_{load}$$

Which is correct.

$$R_{int} = (12.99 - 12.47)/12.47 * 3.23 = 0.135Ω$$

Which is plausible for an SLA at a low state of charge.

#### Ian0

Joined Aug 7, 2020
9,531
I think the TS' formula was let down by HTML formatting. He meant:

$$R_{int} = (V_{OC} - V_{load})/(V_{load} * R_{load})$$

Which is correct.

$$R_{int} = (12.99 - 12.47)/12.47 * 3.23 = 0.135Ω$$
but the first time I looked at what you put, I misread it as
$$R_{int} = (V_{OC} - V_{load})/(V_{load} * R_{load})$$
so it would be less open to misinterpretation if written as

$$R_{int} = (V_{OC} - V_{load})*R_{load} / V_{load}$$
or you could have gone full Latex and written:
$$R_{int} = \frac{(V_{OC} - V_{load})R_{load}} { V_{load} }$$

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#### BobTPH

Joined Jun 5, 2013
8,686
I would question the 3.23Ω load resistor. That would be 44W and get mighty hot! Was the load a 12V bulb that he measured the cold resistance of?

#### Alec_t

Joined Sep 17, 2013
14,234
A word of warning. A lead-acid battery is capable of supplying an enormous number of Amps, which can cause a lot of damage if they pass through the wrong thing. Sparks can fly, including bits of molten metal.

#### asee2dsee

Joined Nov 24, 2023
9
Hi there, thanks for all the replies.

BobTPH = ok so the circuit is a car battery open circuit for V1. Then I attached a load of a headlamp bulb let the voltage settle and then read V2. I used R = V divided by I to calculate the bulbs resistance.

IanO = I used my amp clamp to measure the headlamp current > 3.9a

Irvine = yeah I did have the V1-V2 above the divide by V2 but when I posted it changed it a bit.

BobTPH = yes the headlamp bulb measured around 44w.

I do have a carbon pile load tester I could use to put more load on the battery.

I’m aiming to learn how measure a car lead acid battery’s resistance as an additional tool to asses the condition of the battery.

Sorry all, I should have been more clear, I’m new to this so don’t understand all the formulas etc. I’m just trying to become a more knowledgeable mechanic.
Thanks

#### asee2dsee

Joined Nov 24, 2023
9
A word of warning. A lead-acid battery is capable of supplying an enormous number of Amps, which can cause a lot of damage if they pass through the wrong thing. Sparks can fly, including bits of molten metal.
haha yes, as an apprentice mechanic years ago I unleashed “the angry pixies” a few times and learnt my lesson. But thanks for the advice. I’m well into my career now but want to be a better mechanic than;

Customer; why

I just want to understand more. I just recently learned about car alternator internals and how to test them so now I’m going more in depth with batteries. Generally in the trade mechanics just load test then condemn or pass. I want to know and go a bit further.
Thanks

#### Tonyr1084

Joined Sep 24, 2015
7,787
First, from all of us here on AAC, welcome to the forum.

Something mentioned by
Was the load a 12V bulb that he measured the cold resistance of?
Yes, cold resistance. As the lamp heats up the resistance goes way up. Off hand I can't tell you what to expect.

From what I think I've heard is that a standard halogen headlamp is 55W @ 12V. So 55W ÷ 12V = 4.58A. I'm not great with the formulas so I tend to break things down in pieces. 12V @ 4.58A = 2.6Ω when fully illuminated. In theory that is. I'm sure there are variations unaccounted for. One variation is the battery voltage. A good and fully charged 12V battery can be 12.7V with no load. What that voltage drops down to during illumination will be less. Maybe 12.6V, maybe 12.2V. Somewhere between those numbers or even lower. So exact numbers will change.

As for ESR (Equivalent Series Resistance, what you refer to as "ir" or Internal Resistance) that's something I'm learning myself. So I can't really offer any real substantive information other than the cursory stuff I mentioned.

Happy learning.

#### asee2dsee

Joined Nov 24, 2023
9
First, from all of us here on AAC, welcome to the forum.

Something mentioned by

Yes, cold resistance. As the lamp heats up the resistance goes way up. Off hand I can't tell you what to expect.

From what I think I've heard is that a standard halogen headlamp is 55W @ 12V. So 55W ÷ 12V = 4.58A. I'm not great with the formulas so I tend to break things down in pieces. 12V @ 4.58A = 2.6Ω when fully illuminated. In theory that is. I'm sure there are variations unaccounted for. One variation is the battery voltage. A good and fully charged 12V battery can be 12.7V with no load. What that voltage drops down to during illumination will be less. Maybe 12.6V, maybe 12.2V. Somewhere between those numbers or even lower. So exact numbers will change.

As for ESR (Equivalent Series Resistance, what you refer to as "ir" or Internal Resistance) that's something I'm learning myself. So I can't really offer any real substantive information other than the cursory stuff I mentioned.

Happy learning.
Thanks for the welcome.

Ah ok that makes sense. A bulb probably wasn’t the best load to use. A battery website did suggest a minimum load of 50a for an “engine starting battery”, but there’s a lot of conflicting info between some websites.

Would I be better using my carbon pile load tester to put a load on the battery? Or does the carbon pile resistance vary to much as it heats.

I also tried a second formula that is more simple and I got a very similar result to the first formula. The second formula I used is
V1-V2 /I = R

V1 = 12.99 battery open circuit
V2 = 12.47 battery with load

12.99v - 12.47v = 0.52v / 3.9a = 0.13333333 ohms

thanks

#### MrChips

Joined Oct 2, 2009
30,518
If you measure the current through the load you can calculate the internal resistance of the battery without needing to determine the resistance of the load.

#### BobTPH

Joined Jun 5, 2013
8,686
Okay, if you measured the voltage and current of the operating bulb, then you got the correct resistance.

#### Ian0

Joined Aug 7, 2020
9,531
Thanks for the welcome.

Ah ok that makes sense. A bulb probably wasn’t the best load to use. A battery website did suggest a minimum load of 50a for an “engine starting battery”, but there’s a lot of conflicting info between some websites.

Would I be better using my carbon pile load tester to put a load on the battery? Or does the carbon pile resistance vary to much as it heats.

I also tried a second formula that is more simple and I got a very similar result to the first formula. The second formula I used is
V1-V2 /I = R

V1 = 12.99 battery open circuit
V2 = 12.47 battery with load

12.99v - 12.47v = 0.52v / 3.9a = 0.13333333 ohms

thanks
Nothing at all wrong with using a lamp as a load, provided that you measure the current through it (which you did), rather than inferring it from its resistance.
To get 50A, just mount ten lampholders on a board. You can still get 12V 60W GLS lamps with B22 or E27 bases because they have escaped the government ban, by being "specialist"

#### asee2dsee

Joined Nov 24, 2023
9
If you measure the current through the load you can calculate the internal resistance of the battery without needing to determine the resistance of the load.
Okay, if you measured the voltage and current of the operating bulb, then you got the correct resistance.
Nothing at all wrong with using a lamp as a load, provided that you measure the current through it (which you did), rather than inferring it from its resistance.
To get 50A, just mount ten lampholders on a board. You can still get 12V 60W GLS lamps with B22 or E27 bases because they have escaped the government ban, by being "specialist"
Thanks for all the replies. I understand it much better now. I think the second formula is the way to go as it simplifies it, which makes it easier for me to teach to apprentices. I did read somewhere that more information can be gained by using a small load for the test and then a larger load to compare the difference in resistance between the two conditions. I’ll read about that again next I think and may have some questions later. I also have a question about emp’s but I’ll start another thread on that.

thanks again folks

#### MrChips

Joined Oct 2, 2009
30,518
The battery’s internal resistance is not constant.
It will vary depending on the load current. The internal resistance will be higher at higher currents.

#### Ya’akov

Joined Jan 27, 2019
8,975
or you could have gone full Latex and written:
$$R_{int} = \frac{(V_{OC} - V_{load})R_{load}} { V_{load} }$$
Or, at least for my eye, even better:

$$\large{ \mathsf{R_{\it{int}} = \frac{(V_{\it{OC}} - V_{load})R_{\it{load}}} { V_{\it{load}}}}}$$

 $$\large{ \mathsf{R_{\it{int}} = \frac{(V_{\it{OC}} - V_{load})R_{\it{load}}} { V_{\it{load}}}}}$$

by using a \mathsf{} around it, some \it{} action for strategic effect. I find the default $$italic\space font$$ hard to read and while the $$\mathsf{ maths\space font }$$ isn't necessarily so great itself, in contrast it wins. The \large{} can also help make things easier to read.

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