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## Homework Statement

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?)

## Homework Equations

ΔV=IR[/B]

## 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.