Question

A certain capacitor, in series with a resistor, is being charged. At the end of 25 ms its charge is half the final value. The time constant for the process is about:

Answer 1

τ = 36 10⁻³ s

Step-by-step explanation:

The charge of a circuit Rc is given by the expression

q = q₀ (1 -
`e^(-t/ \tau )`)

q / q₀ = 1 - e^{-t/ \tau }

in this exercise they indicate that

q / q₀ = ½

for a time t = 25 10⁻³ s

we substitute

½ = 1 -
`e^{- 25 \ 10^(-3) / \tau}`

e^{- 25 \ 10^{-3} / \tau} = 1 -½

- 25 10⁻³ /τ = ln 0.5

-25 10⁻³ / ln 0.5 = τ

τ = 36 10⁻³ s