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
$$\begin{gathered} -\frac{t}{\tau }=-2.3 \\ \Rightarrow \frac{t}{\tau }=2.3 \end{gathered}$$
(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
⇒ - $\frac{t}{\tau }$ = − 4.6
⇒ $\frac{t}{\tau }$ = 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 ge
ln e^−t/τ = ln 0.001
⇒ ^-$\frac{t}{\tau }$ = − 6.9
⇒ $\frac{t}{\tau }$ = 6.9