Question

The manufacturer of a 1.5 V D flashlight battery says that the battery will deliver 9 {\rm mA} for 37 continuous hours. During that time the voltage will drop from 1.5 {\rm V} to 1.0 {\rm V} . Assume the drop in voltage is linear with time.Part A How much energy does the battery deliver in this 37h interval?Express your answer to two significant figures and include the appropriate units.

Answer 1

1.42 KJ

Step-by-step explanation:

solution:

power in beginning
`p_(0)`=(1.5 V).(9×
`10^(-3)` A)

= 13.5 mW

after continuous 37 hours it drops to

`p_(37)`=(1 V).(9×
`10^(-3)` A)

=9 mW

When the voltage will drop energy will not remain the same but the voltage drop will always remain same if the voltage was drop to for example from 5 V to 4.5 V the drop will remain the same.

37 hours= 37.60.60

=133200‬ s

w=(9×
`10^(-3)` A×133200‬ )+
`(1)/(2)(13.5.10^(-3)-9.10^(-3))(133200)`

=1.42 KJ

NOTE:

There maybe a calculation error but the method is correct.