Hi there!
Recall the equation for the voltage of a discharging capacitor:
[tex]V_C(t) = V_0e^{-\frac{t}{\tau}}[/tex]
Vâ‚€ = Initial voltage of the capacitor (V)
t = Time (s)
Ï„ or RC = Time Constant (s)
With the given information, we can plug in the value for Vâ‚€:
[tex]V_C(t) = 3.2e^{-\frac{t}{\tau}}[/tex]
We are given that at t = 20 ms (0.02 s), the voltage of the capacitor is 0.8V. We can use this to solve for the time constant (Ï„).
[tex]0.8 = 3.2e^{-\frac{0.02}{\tau}}\\\\0.25 = e^{-\frac{0.02}{\tau}}[/tex]
Take the natural log of both sides and solve.
[tex]ln(0.25) = -\frac{0.02}{\tau}\\\\-1.3863 = -\frac{0.02}{\tau}\\\\\tau = \frac{0.02}{1.3863} = 0.0144 s[/tex]
Now, we can use this time constant to solve for the time taken for the voltage to drop from 0.8 V to 0.2 V. Solve for the time taken for the capacitor's voltage to drop to 0.2 V:
[tex]0.2= 3.2e^{-\frac{t}{0.0144}}\\\\0.0625 = e^{-\frac{t}{0.0144}}\\\\ln(0.0625) = -\frac{t}{0.0144}\\\\t = (-2.773)(-0.0144) = 0.04 s[/tex]
Now, subtract the times:
[tex]0.04 - 0.02 = 0.02 = \boxed{20 ms}[/tex]
