(adsbygoogle = window.adsbygoogle || []).push({}); total charge lifted by the "charge escalator"

1. The problem statement, all variables and given/known data

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).

2. Relevant equations

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

I = dQ / dt

3. 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.

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Total charge lifted by the charge escalator

**Physics Forums | Science Articles, Homework Help, Discussion**