# Internal Resistance of a Car Battery

• Reliquo
In summary: You might be able to start it with a higher voltage, but you would quickly destroy the battery. Is it really worth it?In summary, the battery has an internal resistance of .02 ohms.
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?

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?

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.

Reliquo said:
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.

gneill said:
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?

CWatters
Reliquo said:
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.

haruspex said:
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:

Code:
__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.

#### Attachments

• 2018.02.04.linear.interpolation.of.R.internal.png
4 KB · Views: 5,784
OmCheeto said:
Perhaps you are using a newer fancy battery.
No, just using numbers I found on the net.

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.

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

## 1. What is internal resistance of a car battery?

The internal resistance of a car battery refers to the resistance that occurs within the battery itself, which can affect its performance and overall lifespan. It is measured in ohms and is a result of the materials and design of the battery.

## 2. Why is internal resistance important in a car battery?

Internal resistance is important in a car battery because it can impact the battery's ability to deliver power to the car's electrical system. A higher internal resistance can result in a weaker and slower startup of the car, as well as a shorter lifespan of the battery.

## 3. How does internal resistance affect the performance of a car battery?

The internal resistance of a car battery can affect its performance by causing a voltage drop and limiting the amount of power that can be delivered to the car's electrical system. This can result in a slower startup, weaker electrical components, and a shorter lifespan of the battery.

## 4. What factors can contribute to increased internal resistance in a car battery?

There are several factors that can contribute to increased internal resistance in a car battery, including age, temperature, and repeated deep discharging. Other factors such as the type of materials used and the design of the battery can also play a role.

## 5. How can I measure the internal resistance of my car battery?

The most accurate way to measure the internal resistance of a car battery is with a specialized battery tester. However, you can also estimate the internal resistance by measuring the voltage drop across the battery while it is under load. A higher voltage drop indicates a higher internal resistance.

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