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As batteries discharge, their voltage decreases. The manufacturer of a cell phone battery would like it to maintain 90.0% of its initial voltage. If a battery has an initial voltage of 6.45 V and loses an average of 0.040 V per minute, how long will it remain at 90.0% or more of the initial voltage?

Respuesta :

jufsanabriasa

Answer:

t = 16,1 minutes

Explanation:

For a battery of 6,45V the 90,0% of its voltage is:

6,45V×0,900 = 5,805V

The voltage decreasing of the batteries could be written as:

[tex]V_{t} = V_{i} - 0,040V/min*t[/tex]

Where t is discharge time .

As initial voltage is 6,45V and Vt is 5,805V:

[tex]5,805V = 6,45V - 0,040V/min*t[/tex]

[tex]-0,645V = - 0,040V/min*t[/tex]

t = 16,1 minutes

I hope it helps!

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