Electromotive force of automobile battery

1. Mar 29, 2013

Woopydalan

1. The problem statement, all variables and given/known data
An automobile battery has an emf of 12.6 V and an internal
resistance of 0.080 0 Ω. The headlights together
present equivalent resistance 5.00 Ω (assumed constant).
What is the potential difference across the headlight bulbs
(a) when they are the only load on the battery and

(b) when the starter motor is operated, taking an additional
35.0 A from the battery?

2. Relevant equations

3. The attempt at a solution

I did 12.6 V = 5.08I and found I=2.48 A, then multiplied 2.48A(5) = 12.4 V for part a.

I am stuck on part b, I seem to not be understanding the physics of the question. Now there is a 35 A that goes through the battery, but I would think the emf is constant at 12.6 V, so there can't be anymore potential drawn from it.

.08(35A) + 5(35) is not 12.6 V, so I am stuck

2. Mar 29, 2013

collinsmark

'Looks good to me

Wait, hold on. I don't interpret the problem that way.

The way I interpret the problem, the starter motor is taking exactly 35 A from the battery. And that's in addition to whatever current the headlights happen to be drawing at this point in time.

(I agree, the problem is worded slightly ambiguously. I don't think the problem is trying to say the battery's current is 35 A in addition to what the headlights drew before the starter motor was turned on. Rather I think it's simply proposing is that the starter motor is drawing 35 A. And that 35 Amps is is being sourced by the battery in addition to the current also going to the headlights, whatever that is.)

Yes, the battery does have a constant emf of 12.6 V. But don't forget there is also a voltage drop across the internal resistance.

That's not quite right. The current through the battery, and thus the internal resistance, is more than 35 A. Remember, a full 35 A is going to the starter motor. There is some more current in addition to that going to the headlights.

Here are a couple of hints.
1. Draw out the circuit. Model the starter motor as a resistor. This resistor is in parallel with the resistor used to model the headlights. You already know the current going through this starter motor resistor: 35 A.
2. Find a KVL that includes the battery and the headlight resistor; find the current through the headlights (this will take a little algebra, but you already know the current through the battery is 35 A more than the current through the headlights).
3. From there it shouldn't be much trouble to find the voltage across the headlight resistor.

[Edit: edited the sentence in Dark Red for clarity.]

Last edited: Mar 29, 2013
3. Mar 29, 2013

rude man

I see no ambiguity.

Current thru the starter is 35A so there are 2 equations and 2 unknowns: R_s and V where R_s is the starter resistance and V is the voltage across the starter and headlamps.

4. Mar 29, 2013

Woopydalan

Maybe this is a problem of me not knowing what a starter motor is. Is that something different from the battery? Or how you are saying there is more than 35 A going through the internal resistance. I can't see how it's not just 35 A

I really don't understand the physicality of the problem. It seems like 35 A are going through the battery and the headlights

5. Mar 30, 2013

collinsmark

That's understandable. If you're not familiar with automotive, internal combustion engines, you might not be familiar with a starter motor. (And as a point of interest, perhaps in several decades if cars are primarily electric or fuel cell, most people won't be familiar with this subject either).

If you've ever been in a gasoline powered automobile, you might have noticed that before the car will move, the driver must first "start the engine up." Usually this involves turning the key past its normal "on" position, but only for a short period of time. During this time, a voltage is applied to the starter motor (and solenoid, which is related, but not so important for this problem). Once the engine is able to "idle," there's no more need for the starter motor.

That's a simple synopsis, I admit. But you can do some more research on that if you wish.

But for this problem, just model the starter motor as a resistor in parallel with the car's headlights. And through this starter motor resistor, 35 Amps are flowing. (and ignore the solenoid. Or consider it part of the starter motor. Whatever. Just forget about that for now.)

35 A is not the current going through the battery. It's more than that. 35 Amps is the current going through the "starter motor" resistor. The current going through the battery is 35 Amps + the current going through the headlights.

6. Mar 30, 2013

CWatters

Somthing to remember while solving this one... Don't assume the headlamp current in "b" is the same as in part "a". The extra load from the starter will drag down the battery voltage.

7. Mar 30, 2013

Staff: Mentor

Sometimes it helps to sketch the circuit:

The potential at the node labelled Vh would be the voltage across the headlamps when the starter motor is drawing 35 A.

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8. Mar 31, 2013

Woopydalan

Thanks everyone for the help. I was curious, why is information about the resistance of the starter motor not needed to solve this problem?

9. Apr 1, 2013

rude man

Because when you solve the two equations I mentioned earlier you don't need to solve for the starter resistance if you don't want to.

10. Apr 1, 2013

PeterO

When calculating part (a) you used the resistance of the globes (plus internal resistance of the battery) to find the current through the battery.

You could then have used that current to find the Drop in PD inside the battery [2.48 x 0.08 = 0.1984 ≈ 0.2] thus of the original 12.6 Emf, there is only 12.4 left to drop across the globes.

For part (b) you were told the current the starter motor would draw, so you could already calculate that 35 x 0.08 = 2.8V would be dropped inside the battery due to the influence of the starter. SO already the PD across the battery is down to 9.8 V. the globes will mean a further ≈ 0.2 V so the final PD is a little over 9.6 V.

Had you been calculating what was "lost" in the battery, rather than what was dropping over the globes you would see that you didn't need the resistance of the starter (since the current drawn was already given).

11. Apr 2, 2013

Woopydalan

Last question, is the voltage at "V_h" in post #7 the same as the voltage across the resistor with 5 ohms? If so, doesn't the current need to pass through the resistor to drop, yet at that point labeled V_h, the current hasn't reached the resistor yet.

Thank you

12. Apr 2, 2013

collinsmark

Yes.

Yes, it has. Assume the entire circuit is in steady-state.

Last edited: Apr 2, 2013
13. Apr 2, 2013

PeterO

If you are thinking of the current as leaving the battery and getting to the V_h label, it has already passed through the 0.08 Ω resistor. Thats why is has dropped from the 12.6 at the exit of the cell to the remaining 9.6 or so.

The remaining 9.6 will drop as the charges pass through either the globe or the starter.