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
