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?
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