Woopydalan said:
Homework Statement
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?Homework Equations
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.
'Looks good to me
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,
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.)
but I would think the emf is constant at 12.6 V,
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.
so there can't be anymore potential drawn from it.
.08(35A) + 5(35) is not 12.6 V, so I am stuck
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.
- 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.
- 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).
- 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.]