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**total charge lifted by the "charge escalator"**

## Homework Statement

A 1.5 V flashlight battery is connected to a wire with a resistance of 5.00 ohms. The figure shows the battery's potential difference as a function of time. The slope is linear going from (0, 1.5 V) to (2 hrs, 0).

## Homework Equations

I = [tex]\Delta[/tex]V / R

I = dQ / dt

## The Attempt at a Solution

I = 1.5 V / 5 ohms = 0.3 A

0.3 A = dQ / (2 hrs x 60 x 60) --> dQ = 2160 C

This is not the correct answer, and I don't understand why. Thanks for your help.