Question

The current in an RL circuit builds up to one-third of its steady-state value in 5.00 s . Find the inductive time constant.

Short answer

${\tau }_L=12.3s$

Step-by-step solution

Given

Understanding the concept

$i_m-\mathrm{current}\mathrm{at}\mathrm{steady}\mathrm{value}$If a constant emf $\mathbf{\epsilon }$is introduced into a single-loop circuit containing a resistance Rand an inductance L, the current rises to an equilibrium value of$\mathbf{\epsilon }\mathbf{/}\mathbf{R}$as

$\mathbf{i}\mathbf{=}\frac{\mathbf{\epsilon }}{\mathbf{R}}\left(1-e^{-\frac{t}{{\tau }_L}}\right)$

Formulae:

$\mathrm{i}=\frac{\mathrm{\epsilon }}{\mathrm{R}}(1-{\mathrm{e}}^{-\frac{\mathrm{t}}{{\mathrm{\tau }}_{\mathrm{L}}}})$

Here,$\mathrm{\epsilon }-\mathrm{emf};\mathrm{R}-\mathrm{resistance};\mathrm{t}-\mathrm{time};{\mathrm{\tau }}_{\mathrm{L}}-\mathrm{inductive}\mathrm{time}\mathrm{constant}\mathrm{of}\mathrm{circuit}$

Given:

At $t=5.0s,i=i_m/3$

Calculate the inductive time constant

We have current in the inductive circuit as\

$\mathrm{i}=\frac{\mathrm{\epsilon }}{\mathrm{R}}(1-{\mathrm{e}}^{-\frac{\mathrm{t}}{{\mathrm{\tau }}_{\mathrm{L}}}})$

We know that steady state current of RL circuit is

${\mathrm{i}}_{\mathrm{m}}=\mathrm{\epsilon }/\mathrm{R}$

Now plug in this value in equation:

$$\begin{gathered} \frac{{\mathrm{i}}_{\mathrm{m}}}{3}=\frac{\mathrm{\epsilon }}{\mathrm{R}}\left(1-e^{-\frac{t}{{\tau }_L}}\right) \\ \frac{\mathrm{\epsilon }}{3\mathrm{R}}=\frac{\mathrm{\epsilon }}{\mathrm{R}}\left(1-e^{-\frac{t}{{\tau }_L}}\right) \\ {e}^{-\frac{t}{{\tau }_L}}=1-\frac{1}{3}=2/3 \end{gathered}$$

Hence,

$$\begin{gathered} -\frac{t}{{\tau }_L}=In\left(\frac{2}{3}\right) \\ -\frac{-0.5}{{\tau }_L}=-0.4054 \end{gathered}$$

Therefore,

${\tau }_L=12.3s$

The inductive time constant is 12.3s