Question

A supercapacitor is an electrical energy storage device. A supercapacitor, initially charged to 2.1 thousand millivolts, supplies power to an emergency beacon. The power draw from the load causes the capacitor to lose 9.9% of its charge every minute. When the charge drops below 8 hundred millivolts the beacon will cease functioning. How long will t

Answer 1

Answer:

the time taken t is 9.25 minutes

Explanation:

Given the data in the question;

The initial charge on the supercapacitor = 2.1 × 10³ mV = 2.1 V

now, every minute, the charge lost is 9.9 %  

so we need to find the time for which the charge drops below 800 mV or 0.8 V

to get the time, we can use the formula for compound interest in basic mathematics;

A = P × ( (1 - r/100 )ⁿ

where A IS 0.8, P is 2.1, r is 9.9

so we substitute

0.8 = 2.1 × ( 1 - 0.099 )ⁿ

0.8/2.1 = 0.901ⁿ

0.901ⁿ = 0.381

n = 9.25 minutes

Therefore, the time taken t is 9.25 minutes