Question

Find the value of t/τ for which the current in an LR circuit builds up to (a) 90%, (b) 99% and (c) 99.9% of the steady-state value.

Answer 1

Current i in the LR circuit at time t is given by

i = i₀(1 − e^−t/τ)

Here,

i₀ = Steady-state value of the current

(a) When the value of the current reaches 90% of the steady-state value:-

\[i = \frac{90}{100} \times i_0\]

\[\frac{90}{100} i_0\] = i_o(1 − e^−t/τ)

⇒ 0.9 = 1 − e^−t/τ

⇒ e^−t/τ = 0.1

On taking natural logarithm (ln) of both sides, we get

ln (e^−t/τ) = ln 0.1

`-t/tau=-2.3`

`rArr t/tau=2.3`

(b) When the value of the current reaches 99% of the steady-state value:-

\[\frac{99}{100} i_0\] = i₀(1 − e^−t/τ)

e^−t/τ = 0.01

On taking natural logarithm (ln) of both sides, we get

ln e^−t/τ = ln 0.01

`rArr -t/tau=-4.6`

`rArr t/tau=4.6`

(c) When the value of the current reaches 99.9% of the steady-state value:-

\[\frac{99 . 9}{100} i_0\] = i₀(1 − e^−t/τ)

⇒ e^−t/τ = 0.001

On taking natural logarithm (ln) of both sides, we get

ln e^−t/τ = ln 0.001

`rArr -t/tau=-6.9`

`rArr t/tau=6.9`