I seem to recall a question like this in an introductory physics textbook. The basic idea was that your car has a 12-volt battery that is used to start the engine, right? Well, 8 D-Cell batteries in series will also make a 12-volt battery. However, not all 12-volt batteries are the same. The D-cells just won't do the job of starting your car.
But why? Also, just how many D-cell batteries would you need to get your car started?
Internal Resistance
The problem is that as you increase the current out of a battery, the voltage across this battery does not stay constant but instead decreases. For example, take a D-cell and connect the negative and positive terminals with a plain copper wire. This wire has a ridiculously low resistance. Suppose the wire had a resistance of just 1 Ω, With a 1.5 volt potential difference, you would have 1.5 amps of current going through this wire. That's pretty high for a single D-cell battery.
Then how can we model the actual voltage across a battery? The key is to model the battery as a constant change in potential along with a built in resistor in series with this. Here is my diagram of a D-cell battery.
A couple of notes about this model.
- This is just a model. Battery makers don't intentionally put a resistor inside of their batteries to prevent you from using them to start your car. Instead, the resistor is a way to model the limitations of the chemical reaction inside the battery.
- I list two potentials in the diagram, Vi and Vb. Both of these are actually changes in potential. However, it is customary to drop the Δ symbol in front of the V. I don't know why, but all the cool physicists seem to just say "potential".
- Maybe it is already clear, but Vi is the "internal voltage" of the battery and Ri is the internal resistance. Vb is the actual voltage across the terminals of the battery that you would measure (if you measured it).
Here is how the model works. As the current from the battery increases, the voltage across the battery would be:
This also shows why your D-cell won't start a car. Suppose you need 200 amps going to the starter motor for it to function correctly. Even if the internal resistance is just 1 Ω, that would be a 200-volt drop inside the battery. Of course, that is larger voltage drop than the internal battery itself. Bummer.
There is a way to get 200 amps out of a D-cell - have a whole bunch of D-cells in parallel. Lets say I have 200 sets of 8 D-cells in parallel. In that case, each D-cell would only need to produce 1 amp of current. With them all together, this would be 200 amps.
Measuring Internal Resistance
There are a couple of ways to determine the internal resistance of a battery. It is easy to determine the "internal" voltage. All you need to do is measure the voltage across the battery while it has no load (not connected to anything). Next, hook the battery up to some low (but known) resistance. Like this.
Using the voltage loop rule, I can write:
In these two equations, there are two things I don't know. I don't know the current and I don't know the internal resistance. If I solve one equation for I and substitute that into the other equation, I can solve for Ri. I get:
Just a check. This does have the correct units. Also, if the battery has a really low internal resistance, the voltage across the battery will be about the same as the "internal" voltage. This is essentially what this equation says.
The only thing is that to use this method, you need to know the resistance value of the load you put on the battery. There is another way.
What if I measure the current and the voltage across the battery for a variety of resistance loads? In this, case, I would expect the following relationship.
From this, I can plot the voltage across the battery as a function of the current coming out of the battery. This way I can put whatever resistors I like for a load and it just doesn't even matter what these values are. The nice thing about this other method is that it uses more data points and ignores the resistance. If I take a 10Ω resistor and hook it up to a battery, it might not actually be 10Ω. Resistors don't always have a constant resistance.
Now for some data. I looked at a D-cell, a AA-cell, and my 5-cell penny battery. Here is the data for just the AA and D-cell.
I am not sure why the D-cell data looks so much nicer than the AA data. For the linear fits for both sets of data, I get an internal resistance of 0.204Ω for the D-cell and 2.258Ω for the AA. I didn't plot the data for the penny since it was way off scale. Here it is by itself.
From this data, I get an internal resistance of 66.89Ω. In general, the smaller the battery, the higher the internal resistance.
Starting a Car
What kind of current can I get from one of these batteries? Well, what if I had a perfect short circuit across one of these batteries. Suppose I have a wire with zero resistance to short the battery. In this case, the loop around the circuit would just include the internal battery and the internal resistance. The loop equation would be:
This is the highest theoretical current from this battery. Of course, this assumes two important things. First, that the "internal resistance" model is valid at this extreme range. Second, that you could actually find a zero-resistance wire. It would have to be a superconductor, I guess. Oh, notice that in this case the voltage across the real battery would be zero.
If I use my values for the D-cell battery, I have a maximum current of 0.7 Amps 1.5v/.204 Ohms (Hat tip to @weirdnoise for catching my error) = 7.35 Amps. Of course, I can't get that high of a value with my starter motor since it isn't a superconductor and has a non-zero resistance. Let me just say that I will get half this value of current - so 0.35 3.7 Amps. In this case the voltage across my actual battery would drop from 1.6 volts to 0.8 volts.
How many of these D-cells would I need in series to get up to my estimated current of 200 Amps? If each battery by itself would produce 0.35 3.7Amps, I would need 571 54 D-cell batteries in parallel. But don't forget, I need 12 volts to start the car. So this would be 571 54 sets of 8 batteries for a total of over 4,500 432 D-cell batteries. Hint: That's a lot.
How much space would this take up? Well, according to Wikipedia a D-cell has diameter of 33.2 mm and a height of 61.5 mm. If I stack these batteries 8 high and in rows 23 by 24 7 by 8 they would take up a volume of about 0.03 m3. Not as big as I thought. Still it wouldn't be so cheap to buy 4,500 432 batteries.
I will leave the penny batteries and the AA batteries calculation for homework.
