Why do the headlights on the car become dim when the car is starting?

[zenterix]

Homework Statement

From the perspective of electric circuits and electromagnetism, why do the headlights on the car become dim when the car is starting?

Relevant Equations

I'm not sure how a car engine works electrically.

So I would like to think about this problem in a simpler context.

Suppose we have 12V battery.

Though I can solve electric circuits, there are some very simple practical concepts that are gaping holes in my knowledge.

What does it mean for a circuit device to "pull a lot of current"?

What is a simple circuit element that when I connect it to the battery it will "pull" a lot of current?

If I connect a resistor to the terminals of the battery, then current flows according to Ohm's law. Thus, current depends on the voltage of the battery and resistance of the resistor.

The smaller the resistance, the larger the current.

If the resistance is extremely small, a huge amount of current can flow.

Suppose the resistor is the filament of a light bulb. It lights up.

Suppose the light bulb is connected in parallel to nine other light bulbs, but there is an open switch in the wire that leads to the nine other light bulbs.

The moment we close the switch, the equivalent resistance becomes $R/10$ and ten times more current flows, but the current to the first light bulb is still the same.

I imagine that the charge on the positive terminal of the battery is depleted rapidly. I imagine it takes time for the chemical reactions to re-establish the difference in potential.

Thus, the potential difference created by the battery decreases.

If $i_1$ is the current through the first light bulb, then initially $i_1=\frac{\epsilon_0}{R}$. After the switch is closed, if the emf were to stay the same then the total current in the circuit would be $i=\frac{\epsilon_0}{R_{eq}}=\frac{\epsilon_0}{R/10}=10i_1$ and $i_1=\frac{\epsilon_0}{R}$ is the same as before.

But if $\epsilon_0$ decreases, then the total current is somewhat smaller than $10i_1$ and the current to the first light bulb is smaller than $i_1$. Thus, it's brightness decreases.

Is this roughly why the lights in a car dim when we turn the car on? In other words, it is essentially because the battery electromotive force decreases temporarily because of the presence of extra small resistances in parallel with the headlights?

 

[phinds]

Is this roughly why the lights in a car dim when we turn the car on? In other words, it is essentially because the battery electromotive force decreases temporarily because of the presence of extra small resistances in parallel with the headlights?

Yes.

Starter motors take a large amount of current, but only very briefly, so the lights dim very briefly.

 

[hutchphd]

So I would like to think about this problem in a simpler context.

Suppose we have 12V battery.

Though I can solve electric circuits, there are some very simple practical concepts that are gaping holes in my knowledge.

What does it mean for a circuit device to "pull a lot of current"?

An electric motor (the starter motor ), under a heavy mechanical load (turning the gas engine) , presents a very low resistance to the battery, thus "demanding" a lot of current to satisfy ohm's law.
Real batteries can be modeled as a Thevenin equivalent source: an "ideal" voltage source in series with a resistor. For a 12V car battery this "internal resistance" will be R_int≈0.1 Ω So ordinary circuit theory tells us that when the starter motor needs 30 A, the terminal voltage on the battery will be $$V_{terminal}=12V-30A(0.1 \Omega) =9V$$ which is what my car tells me every morning when I crank it.
An undercharged or damaged battery will exhibit a higher internal resistancen and fail to meet the voltage requirement under heavy load. Did I mention temperature? You get the picture I think.

 

[berkeman]

Just to add to the above good replies...

You can learn more about the internal resistance of lead-acid car batteries here:

https://www.google.com/url?sa=t&rct...usg=AOvVaw1G_V2UaR1GsCtSrrNJxr5P&opi=89978449

But this is also a bit of a trick question as stated in your schoolwork problem. Some cars and most motorcycles actually turn off the headlights while the starter motor is cranking. This is done to provide more energy to the starter motor during the short time that you are starting the car or motorcycle.

 

[hutchphd]

Sometimes I think a battery should be modeled by a Norton Equivalent (for the physics I mean......obviously equivalent means what it says for circuit purposes.)