please answer this :

Question

A charged capacitor is being discharged through a resistor. At the end of one time constant

the charge has been reduced by (1 – 1/e) = 63% of its initial value. At the end of two time

constants the charge has been reduced by what percent of its initial value?

A) 82%

B) 86%

C) 100%

D) between 90% and 100%

E) need to know more data to answer the question

Answer 1

If Q₀ is the initial charge held by the capacitor and t is the time since it started discharging, the total charge, Q, on the capacitor as a function of time will be

Q(t) = Q₀e^-t/τ where τ is the time constant (by the way for an RC circuit, τ will equal RC where R is the resistor's resistance and C will be the capacitor's capacitance.

Now the LOSS of charge will be

Q₀ - Q₀e^-t/τ = Q₀(1 - e^-t/τ)

The percentage of loss in charge will be (Q(t)/Q₀) = (1 - e^-t/τ)

That loss percentage for t equal to one time constant, τ, will be (1 - e^-1) or (1 - 1/e)

At t equals 2, the loss will be (1 - e^-2) = (about) 0.865 (corresponding to about 87% or answer b or close to it)