Question

A coil is connected in series with a $\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{k}\mathbf{\Omega }$resistor. An ideal 50.0 Vbattery is applied across the two devices, and the current reaches a value of 2.00 mAafter 5.00 ms.(a) Find the inductance of the coil. (b) How much energy is stored in the coil at this same moment?

Short answer

a)$L=97.9H$

b)$U_b=0.196\mathrm{mJ}$

Step-by-step solution

Given

$$\begin{gathered} R=10k\Omega =10\times {10}^3\Omega \\ E=50.0V \\ I=2.00mA=2.00\times {10}^{-3}A \\ t=5.00ms=5\times {10}^{-3}s \end{gathered}$$

Understanding the concept

Here we have to use the formula for current through RL circuit to find inductance. Then by using formula for energy stored by inductor’s magnetic field, we can calculate the energy stored in the coil.

Formula:

$$\begin{gathered} i=\frac{E}{R}\left(1-e^{\frac{t}{{}^{t}L}}\right) \\ {U}_b=0.5L{i}^2 \end{gathered}$$

(a) Find the inductance of the coil

Here, the rise of current for RL circuit is as follows:

$$\begin{gathered} i=\frac{E}{R}\left(1-e^{-\frac{t}{{}^{t}L}}\right) \\ i=\frac{E}{R}\left(1-e^{-\frac{t\times R}{L}}\right) \\ 2.00\times {10}^{-3}=\frac{50}{10\times {10}^3}\left(1-e^{-\frac{5\times {10}^{-3}\times 10\times {10}^3}{L}}\right) \\ 0.4=\left(1-e^{\frac{50}{-L}}\right) \\ 0.6=e^{-50/L} \\ L=-\frac{50}{In\left(0.6\right)} \\ L=97.9H \end{gathered}$$

(b) Calculate how much energy is stored in the coil at this same moment

The inductor’s magnetic field stores energy and is given by

$$\begin{gathered} U_b=0.5L{i}^2 \\ {U}_b=0.5\times 97.9\times (2.00\times {10}^{-3}{)}^2 \end{gathered}$$

$U_b=0.196\mathrm{mJ}$