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Expert-verified Found in: Page 696 ### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000 # Consider a battery used to supply energy to a cellular phone. Construct a problem in which you determine the energy that must be supplied by the battery, and then calculate the amount of charge it must be able to move in order to supply this energy. Among the things to be considered are the energy needs and battery voltage. You may need to look ahead to interpret manufacturer's battery ratings in ampere-hours as energy in joules.

The supplied charge is 13500 C.

See the step by step solution

## Step 1: Principle and Formulas

The power supplied by a source to a circuit element :

${\mathbf{P}}{\mathbf{=}}{\mathbf{I\Delta V}}{\mathbf{·}}{\mathbf{·}}{\mathbf{·}}{\mathbf{·}}{\mathbf{·}}{\mathbf{·}}{\mathbf{·}}{\mathbf{·}}{\mathbf{·}}{\mathbf{·}}{\mathbf{·}}{\mathbf{·}}{\mathbf{·}}{\mathbf{·}}{\mathbf{·}}{\mathbf{·}}{\mathbf{·}}{\mathbf{·}}{\mathbf{·}}{\mathbf{·}}{\mathbf{·}}{\mathbf{·}}{\mathbf{\left(1\right)}}$

where ${\mathbf{∆}}{\mathbf{V}}$is the potential difference and is I the current across the element.

The power of a process is

${\mathbf{P}}{\mathbf{=}}\frac{\mathbf{\Delta E}}{\mathbf{\Delta t}}{\mathbf{\text{}}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{\left(}}{\mathbf{2}}{\mathbf{\right)}}$

where ${\mathbf{∆}}{\mathbf{E}}$is the change in energy and ${\mathbf{∆}}{\mathbf{t}}$ is change in time.

## Step 2: calculating battery supplied energy.

The Charger’s voltage output is: ${\mathbf{\Delta V}}{\mathbf{=}}{\mathbf{5}}{\mathbf{.}}{\mathbf{0}}{\mathbf{}}{\mathbf{V}}$

The rate of the battery is 3750 mA which means that it can produce a current of 3750 mA for 1 h.

The power supplied by the battery is found from Equation (1):

${\mathbf{P}}{\mathbf{=}}{\mathbf{I\Delta V}}$

Substitute for P from Equation (2):

$\frac{\mathbf{\Delta E}}{\mathbf{\Delta t}}{\mathbf{=}}{\mathbf{I\Delta V}}$

Solve for $∆E$:

${\mathbf{\Delta E}}{\mathbf{=}}{\mathbf{I\Delta V\Delta t}}$

## Step 3: Calculation of the amount of charge supply

The energy supplied by the battery:

Substitute, 3750 mA for / 1 h for $${\rm{\Delta t}}$$, and 5.0 for $∆V$:

$\begin{array}{rcl}{\mathbf{\Delta E}}& {\mathbf{=}}& {\mathbf{\left(}}{\mathbf{3750}}{\mathbf{}}{\mathbf{mA}}{\mathbf{\right)}}\left(\frac{1A}{1000\mathrm{mA}}\right){\mathbf{\left(}}{\mathbf{5}}{\mathbf{.}}{\mathbf{0}}{\mathbf{}}{\mathbf{V}}{\mathbf{\right)}}{\mathbf{\left(}}{\mathbf{1}}{\mathbf{}}{\mathbf{h}}{\mathbf{\right)}}\left(\frac{3600s}{1h}\right)\\ & {\mathbf{=}}& {\mathbf{67500}}{\mathbf{}}{\mathbf{J}}\end{array}$

Converting the ampere-hour rating into coulombs, the amount of charge it must be able to move in order to supply this energy is

$\begin{array}{rcl}{\mathbf{q}}& {\mathbf{=}}& {\mathbf{\left(}}{\mathbf{3750}}{\mathbf{}}{\mathbf{mA}}{\mathbf{\right)}}\left(\frac{1A}{1000A}\right){\mathbf{\left(}}{\mathbf{1}}{\mathbf{}}{\mathbf{h}}{\mathbf{\right)}}\left(\frac{3600s}{1h}\right)\\ & {\mathbf{=}}& {\mathbf{13500}}{\mathbf{\text{}}}{\mathbf{C}}\\ & & \end{array}$

Therefore, the amount of charge to be supplied is 13500 C. ### Want to see more solutions like these? 