Question

In Fig. 30-63, $\mathbf{R}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{k\Omega }$,$\mathbf{L}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{0}\mathbf{\mu H}$and the ideal battery has$\mathbf{\epsilon }\mathbf{=}\mathbf{20}\mathbf{}\mathbf{v}$. How long after switch S is closed is the current 2.0 mA?

Short answer

For $1.0\times {10}^{-9}s$after switch S is closed, the value for the current is 2.0 mA

Step-by-step solution

Given

$$\begin{gathered} \mathrm{R}=4.0\mathrm{k\Omega }=4.0\times {10}^3\mathrm{\Omega } \\ \mathrm{L}=8.0\mathrm{\mu H}=8.0\times {10}^{-6}\mathrm{H} \\ \mathrm{\epsilon }=20\mathrm{v} \\ \mathrm{I}=2.0\mathrm{mA}=2.0\times {10}^{-3}\mathrm{A} \end{gathered}$$

Understanding the concept

We have the equation for current in the loop. We rearrange the equation to find the required value.

Formula:

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

${\tau }_L=\frac{L}{R}$

Calculate how long after switch S is closed is the current 2.0 mA

We have,

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

But,

$$\begin{gathered} {\tau }_L=\frac{L}{R} \\ {\tau }_L=\frac{8.0\times {10}^{-6}}{4.0\times {10}^3} \\ {\tau }_L=2.0\times {10}^{-9}s \\ 2.0\times {10}^3=\frac{20}{4.0\times {10}^3}\times \left(1-e^{\frac{-t}{2.0\times {10}^{-9}}}\right) \\ \left(1-e^{\frac{-t}{2.0\times {10}^{-9}}}\right)=\frac{2.0\times {10}^{-3}\times 4.0\times {10}^3}{20} \\ \left(1-e^{\frac{-t}{2.0\times {10}^{-9}}}\right)=0.4 \\ {e}^{\frac{-t}{2.0\times {10}^{-9}}}=0.4-1 \\ {e}^{\frac{-t}{2.0\times {10}^{-9}}}=-0.6 \\ {e}^{\frac{-t}{2.0\times {10}^{-9}}}=0.6 \end{gathered}$$

Taking natural log at both the sides, we get

$$\begin{gathered} -\frac{\mathrm{t}}{2.0\times {10}^{-9}}=\mathrm{In}(0.6) \\ \frac{\mathrm{t}}{2.0\times {10}^{-9}}=0.51 \\ \mathrm{t}=2.0\times {10}^{-9}\times 0.51 \\ \mathrm{t}=1.0\times {10}^{-9}\mathrm{s} \end{gathered}$$