if t = 5
2 e^-(.16*5)=2 e^-.8 = 2/e^.8 = .899 = .90 volts
1 = 2 e^-(.16t)
ln .5 = -.16 t
t = 4.33 seconds
at 0 sec V = 2
at 5 seconds V = .90
V = Vi (r)^t
.9 = 2 r^5
.45 = r^5
log .45 = 5 log r
r = .852
so every second multiply by .85
so decreases by 15% every second
The voltage V across a charged capacitor is given by V(t) = 2e^-0.16t volts, where t is in seconds
(A) what is the voltage V(5) after 5 seconds? Find the time t when the voltage will be 1 volt
Enter your answers to two decimal places
V(5) approximately ___________ volts
t approximately ___________ sec
(B) By what percent B does the voltage decrease each second ?
Enter your answer to two decimal places
B approximately ________ %
1 answer
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