Question

The manufacturer of a 6V dry-cell flashlight battery says that the battery will deliver 15 mA for 60 continuous hours. During that time the voltage will drop from 6V to 4V. Assume the drop in voltage is linear with time, how much energy does the battery deliver in this 60h interval?

Answer 1

Answer:

16200 J

Explanation:

t = Time the battery is on = 60 hours

I = Current = $15\times 10^{-3}\ A$

Average voltage

$V=\dfrac{6+4}{2}=5\ V$

Energy is given by

$E=V\times I\times t$

$\\\Rightarrow E=5\times 15\times 10^{-3}\times 60\times 3600$

$\\\Rightarrow E=16200\ J$

The energy delivered in the given time is 16200 J

Answer 2

The required energy is 16200 J.

Given, that the battery will deliver 15 mA for 60 continuous hours.

Since the voltage drops linearly from 6V to 4V, the average is 5V.

The energy delivered is,

$V\times A\times t = 5\times .015\times(60\times 3600)\\ = 16,200 J.$

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