The voltage V across a charged capacitor is given by Vt6e07t

The voltage V across a charged capacitor is given by V(t)=6e0.7t where t is in seconds.

(a) What is the voltage after 4 seconds? (round to the nearest 0.001 volts)

(b) When will the voltage be 1? (round to the nearest 0.01 sec.)

(c) By what percent does the voltage decrease each second? (round to the nearest 0.001%)

Solution

a) V(t) = 6e^-0.7t

at t = 4sec; V(4) = 6e^(-0.7*4) = 6*0.755 = 4.53 V

b) V(t) = 1

1 = 6e^-0.28t

1/6 = e^-0.28t

Take natural log on both sides:

ln(1/6) = -0.7t

t =2.56 sec

c) For each second : V(1) = 6e^-0.7

= 2.97 V

So, % change = (3.03/6)*100 = 50.500%