An inductor is connected to the terminals of a battery that has an emf of 16.0 V and negligible internal resistance. The current is 4.86 mA at 0.940 ms after the connection iscompleted. After a long time, the current is 6.45 mA. What are(a) the resistance R of the inductor and (b) the inductance L ofthe inductor?

The current flowing in an RL circuit is given by,$\mathrm{i}=\frac{\mathrm{\epsilon }}{\mathrm{R}}\left(1-{\mathrm{e}}^{-\left(\mathrm{R}/\mathrm{L}\right)\mathrm{t}}\right)$and the inductance is given as $\frac{\mathrm{Rt}}{\mathrm{L}}=-\mathrm{In}\left(1-\frac{\mathrm{i}}{{\mathrm{i}}_{\max }}\right)$ where I is the current flowing in the circuit

The inductor is connected to a battery with an emf of E =16.0 V. So, this circuit is equivalent to a simple RL circuit It is given that the current is 4.86 mA at 0.940 ms after the connection is completed and is 6.45 mA after a long time. First, we need to find the resistance of the inductor. The current flowing in an RL circuit is given by,$\mathrm{i}=\frac{\mathrm{\epsilon }}{\mathrm{R}}\left(1-{\mathrm{e}}^{-\left(\mathrm{R}/\mathrm{L}\right)\mathrm{t}}\right)$,at $\mathrm{t}\to \infty$the current is maximum as ${\mathrm{i}}_{\max }=\frac{\mathrm{\epsilon }}{\mathrm{R}}$so, $\mathrm{R}=\frac{\mathrm{\epsilon }}{{\mathrm{i}}_{\max }}$Substitute the values in the above equation we have,

$$\begin{gathered} \mathrm{R}=\frac{16.0\mathrm{V}}{6.45\times {10}^{-3}\mathrm{A}} \\ =2481\mathrm{\Omega } \end{gathered}$$

Therefore,the resistance R of the inductor is$2481\mathrm{\Omega }$

i(t= 0.940 ms) =4.86 mA, solve (1) for L in the equation$\mathrm{L}=-\frac{\mathrm{Rt}}{\mathrm{In}\left(1-{\displaystyle \frac{\mathrm{i}}{{\mathrm{i}}_{\max }}}\right)}$Substitute the values in the above equation we have

$\mathrm{L}=-\frac{2481\times 9.40\times {10}^{-4}}{\mathrm{In}\left(1-{\displaystyle \frac{4.86}{6.45}}\right)}=1.67\mathrm{H}$

Therefore, the inductance L of the inductor is 1.67H