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

Answer:

1.42 KJ

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 )+$\frac{1}{2}(13.5.10^{-3}-9.10^{-3})(133200)$

                                 =1.42 KJ

NOTE:

There maybe a calculation error but the method is correct.