Internal Resistance of a Car Battery

[Reliquo]

Homework Statement

An automobile battery has a terminal voltage of 12.8 V with no load. When the starting motor (which draws 90A) is running, the terminal voltage drops to 11.0 V. What is the internal resistance of the battery?

Homework Equations

V= EMF- Ir

The Attempt at a Solution

I think I'm just completely over thinking the problem and maybe missing something. In order to find EMF:

EMF= I(R+r) and in order to find this I need the R of the load in the circuit. This would be the Motor. The voltage drop is 1.8V so Ohm's Law would give V/I= (1.8)/(90A)=.02 Ohms, which seems way to low.

What am I missing here?

 

[gneill]

Hi Reliquo,

Welcome to Physics Forums!

Your result looks okay. The no-load terminal voltage of the battery will be its EMF. The drop in terminal voltage due to a given current draw will be the drop across its internal resistance, so ΔV/I is the correct way to proceed, as you did.

 

[haruspex]

seems way to low.

It looks a bit lower than typical in the real world, but hey, you have to work with the numbers you are given.

 

[Reliquo]

Hi Reliquo,

Welcome to Physics Forums!

Your result looks okay. The no-load terminal voltage of the battery will be its EMF. The drop in terminal voltage due to a given current draw will be the drop across its internal resistance, so ΔV/I is the correct way to proceed, as you did.

So if the no-load terminal voltage of the battery is the EMF (12.8V) this would give:

V= EMF- Ir

11.0 V = 12.8 V - 90A(r)

-1.8 V= -90Ar
.02 = r (Internal resistance)

So is this correct then?

 

[cnh1995]

So if the no-load terminal voltage of the battery is the EMF (12.8V) this would give:

V= EMF- Ir

11.0 V = 12.8 V - 90A(r)

-1.8 V= -90Ar
.02 = r (Internal resistance)

So is this correct then?

Yes.

 

[OmCheeto]

It looks a bit lower than typical in the real world, but hey, you have to work with the numbers you are given.

It actually looks very close to values I came up with about 10 years ago:

__measured__    calculated    notes
vdc    amps     R-internal
25.5     0       _n/a_       no load (and a guess)
22.6    96       0.030
22.5    92       0.033
22.2    97       0.034

22.4    95       0.032       average

I described the experiment briefly in at least three different threads. Here's one: https://www.physicsforums.com/threads/boiling-water-w-a-car-battery.870832/#post-5468458

The 25.5 vdc no load value is guessed, as I didn't document the initial value.
The two batteries I had in series were also mismatched. One was a marine deep cycle, and the other was a typical car "starting" battery.

If we accept my spreadsheets linear interpolation:

0.02 Ω is just about spot on.

Perhaps you are using a newer fancy battery.

 

[haruspex]

Perhaps you are using a newer fancy battery.

No, just using numbers I found on the net.

 

[mn786]

You are correct. It seems low when a person at the electronics bench fiddling with milliamps all day imagines dumping 12 volts across a .1222 ohm load. The power supplies we use on a daily basis would fault out seeing .12ohms... but when you consider the electric motors efficiency, at perhaps 80%, you have .79kw being used as mechanical energy. Thats the equivalent of roughly 1 horsepower. Could a 1 horsepower pony motor crank your engine any better than that electric motor? Probably not. Its not a low resistance for the application, we are just not used to working with those current levels on a daily basis.

 

[CWatters]

Consider what would happen if it was higher, say 1 Ohm. Could you start the car?