The manufacturer of a 9V dry-cell flashlight battery says that the battery will deliver 20 mA for 80 continuous hours. During that time the voltage will drop from 9 V to 6 V. Assume the drop in voltage is linear with time. How much energy does the battery deliver in this 80 h interval

17280 J or 17.28 kJ

Explanation:

Given that the voltage drop,

U = U2 - U1

U = 9 - 6

U = 3V

Also, we're told that the current, I is equal to 20 mA with the discharge time, t being 80 hrs.

Converting the time from h oi urs to seconds, we have

t = 80 * 3600

t = 288000

Now, to find the energy needed, we're going to use the formula

w = pt, where p = U * I

p = 3 * 20*10^-3

p = 60*10^-3

w = 60*10^-3 * 288000

w = 17280 J or 17.28 kJ

Therefore, the total energy the battery delivers in the 80 hrs is 17.28 kJ

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