A capacitor with capacitance `C = 400 pF` is connected via a resistance `R = 650 Omega` to a source of voltage `V_(0)`. How soon will the voltage developed across the capacitor reach a value `V = 0.90 V_(0)` ?

Explanation

The formula is,
`q = C V_(0) (1 - e^(-1//RC))`
or, `V = (q)/(C) = V_(0) (1 - e^(-1//RC))` or , `(V)/(V_(0)) = 1 - e^(-1//RC)`
or, `e^(-1//RC) = 1 - (V)/(V_(0)) = (V_(0) - V)/(V_(0))`
Hence, `t = RC` In `(V_(0))/(V_(0) - V) = R C` In 10, if `V = 0.9 V_(0)`
Thus `t = 0.6 muS`.