Question

5. A capacitor is discharging through a resistor. The initial voltage across the capacitor is 3.2 V at t = 0. The voltage across the capacitor at time t = 20 ms is 0.8 V. The time it takes for the voltage across the capacitor to drop from 0.8 V to 0.2 V is ​

Answer 1

Hi there!

Recall the equation for the voltage of a discharging capacitor:

`V_C(t) = V_0e^{-(t)/(\tau)}`

V₀ = Initial voltage of the capacitor (V)
t = Time (s)
τ or RC = Time Constant (s)

With the given information, we can plug in the value for V₀:

`V_C(t) = 3.2e^{-(t)/(\tau)}`

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 (τ).

`0.8 = 3.2e^{-(0.02)/(\tau)}\\\\0.25 = e^{-(0.02)/(\tau)}`

Take the natural log of both sides and solve.

`ln(0.25) = -(0.02)/(\tau)\\\\-1.3863 = -(0.02)/(\tau)\\\\\tau = (0.02)/(1.3863) = 0.0144 s`

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:

`0.2= 3.2e^{-(t)/(0.0144)}\\\\0.0625 = e^{-(t)/(0.0144)}\\\\ln(0.0625) = -(t)/(0.0144)\\\\t = (-2.773)(-0.0144) = 0.04 s`

Now, subtract the times:

`0.04 - 0.02 = 0.02 = \boxed{20 ms}`