Solving the Car Parking Lights Puzzle: Current, Resistance & Time Calculations

AI Thread Summary
The discussion revolves around calculating the current, resistance, and time for a car's parking lights powered by a 12 V battery with a total charge of 50 Ah. The parking lights, rated at 3.6 W each, are assumed to be connected in parallel, leading to a total power of 14.4 W and a calculated current of 1.2 A from the battery. The resistance of one bulb is determined to be 40 Ω, resulting in a battery life of approximately 41.7 hours when the lights are on. Participants express concerns about the assumptions made regarding the circuit configuration and the applicability of formulas for series versus parallel circuits. The conversation highlights the importance of understanding circuit types in solving electrical problems effectively.
stephenranger
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Homework Statement


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.

Homework Equations


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
are negligible.

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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The approach looks good. The math looks good. Have you any specific doubts?
 
My doubt is whether or not my solution is correct. Otherwise, the given data doesn't say that the 4 bulbs are connected in parallel, so is there any some solution else that does not use formula for R both in series and parallel circuit ? you know what I mean
 
Parallel is a safe assumption for automotive applications. One filament failure doesn't turn off every light on the vehicle.
 
Yes. But what if the problem solver doesn't know that lights in car are wired in parallel. That means that he must come up with a solution that does not apply formulas in both series and parallel circuits. And that is what I'm trying to find. Could you give me a solution that doesn't apply formulas in series and parallel circuits ?
 
stephenranger said:
Could you give me a solution that doesn't apply formulas in series and parallel circuits ?
If you have more than one component in the circuit you're analyzing, you've got to make a choice, or start generating solutions for every possible configuration of series and parallel arrangements of components, and leaving the choice among possible solutions to the end user.

i.e., there is no solution for an undefined problem.
 
There was somebody solved like this:
Ptotal = Px4 = 3.6x4 = 14.4 watt
I = Ptotal/V = 14.4/12 = 1.20 A
R = V2/P = 144/3.6 = 40.0 ohm
t = 50/1.2 = 41,7 hours.

He didn't use the formulas in both series and parallel circuits. Could you tell me what's different between his solution and mine ?
 
stephenranger said:
Ptotal = Px4 = 3.6x4 = 14.4 watt
This is an assumption of parallel circuits.
 
Oh thank you.
If we have series circuit. What is formula for P ?
 
  • #10
stephenranger said:
If we have series circuit. What is formula for P ?
What is R in the series circuit?
 
  • #11
Oh. Ok Thanks. I'm going to do it myself.
 
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