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How long can the battery power the lightbulb?

  1. Sep 30, 2015 #1
    1. The problem statement, all variables and given/known data
    A 2.9 V battery rated at 695.5 A(h) has been powering the same light bulb for 44.2 hours. The total resistance in the circuit has averaged out to be 5.3 Ohms.

    The light bulb has unexpectedly burned out, and was replaced with a new one. The total resistance in the circuit with the new light bulb averages out to be 2.2 Ohms.

    How many more hours of light can we expect (assuming the new light bulb does not burn out?)

    2. Relevant equations
    ΔV=IR



    3. The attempt at a solution
    If current (I)=ΔV/R, then the initial current for the first bulb should be 0.547 Amps (2.9V/5.3Ω). At that current the battery should have used 24.18 Amps in 44.2 hours (0.547A x 44.2 hr) . At that battery rating (695.5 A*h) there should be 671.3 Amps left. With the new bulb resistance the new current should be 1.32 Amps (2.9V/2.2Ω). With 671.3 Amps left it should power the bulb for 508.6 more hours (671.3 Amp*hr /1.32 Amp).

    Logically speaking, I don't see how a 2.9 V battery could power a lightbulb for almost 23 days, BUT it is a homework question and I've seen crazier answers. Thanks for any help.
     
  2. jcsd
  3. Sep 30, 2015 #2

    gneill

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    Staff: Mentor

    Hi TravvyM, Welcome to Physics Forums!

    A comment on your attempt: A load doesn't "use" Amps in the way you've implied; current is not a stored item. The load uses Amp-hours, which units reduce to charge (Coulombs). Think of the battery as a reservoir of charge that happens to be made available at a given potential (2.9 V in this case).

    So for the first run of the circuit 0.547 A (or 547 mA) were drawn for 44.2 hours, for a total usage of 24.2 A-hr. Note that at no time over the 44.2 hour period was the current anything other than 0.547 Amps.

    Otherwise your calculations look good. You might want to keep some additional significant figures in intermediate values so that rounding and truncation errors don't creep into subsequent calculations. Never round intermediate values except for presentation purposes.

    I agree with you that a 2.9 V, 695.5 A-hr battery would be a pretty big item. I wouldn't think it would be "button cell" size :smile:
     
  4. Sep 30, 2015 #3
    gneill,

    Thanks for the reply. Yeah I had a feeling my explanation of current being a "stored" and depleted was wrong, but I couldn't think of how else to describe my train of thought. I'm not sure what happened, but when I logged back in the site that I submit the answers to, it accepted the answer of 508.6 more hours.
     
  5. Sep 30, 2015 #4

    gneill

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    Staff: Mentor

    No doubt the program is designed to accept results within some margin of error, and if they're not being picky about significant figures at this point, then your value was deemed to be fine.
     
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