A total charge of 50 Ah is stored in a 12 V battery of a car. Each
of the bulbs of the four parking lights is labelled 3,6 W / 12 V.
a) Find the current from the battery when only the parking lights are switched on.
b) What is the resistance of one bulb?
c) How long does it take before the battery becomes dead if the parking lights are on?
Assume that the internal resistance of the battery and the resistances of the connecting wires
The Attempt at a Solution
First of all. The given data does not clarify what type of circuit that the bulbs are wired, so I assume that they are wired in parallel.[/B]
(sorry I don't know how to use Tex codes. I just tried some but some work, some don't. So I'm not going to use texcode).
P=VI and I = V/R ======> P= V2/R =====> R = V2/P = 122/ 3,6 =40 (Ω)
Four bulbs are connected in parallel, so: 1/Rtotal = 1/R1+1/R2+1/R3+1/R4=4/R=4/40=10 (Ω)
the current from the battery when only the parking lights are switched on: I = V/Rtotal = 12/10=1.2 (A)
Time it takes before the battery becomes dead: t = 50/1.2 = 41.666666....~41.2 hours.
I wonder if my solution is correct ?
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