Question

A coil with an inductance of2.0 H and a resistance of $\mathbf{10}\mathit{\Omega }$is suddenly connected to an ideal battery with $\mathit{\epsilon }\mathbf{=}\mathbf{100}\mathbf{V}$. At 0.10 safter the connection is made, (a) what is the rate at which energy is being stored in the magnetic field? (b) what is the rate at which thermal energy is appearing in the resistance? (c) what is the rate at which energy is being delivered by the battery?

Short answer

$$\begin{gathered} a)\frac{d{U}_b}{dt}=2.4\times {10}^{2}W \\ b)P_{thermal}=1.5\times {10}^{2}W \\ c)P=3.9\times {10}^{2}W \end{gathered}$$

Step-by-step solution

Given

$$\begin{gathered} L=2.0H \\ R=10\Omega \\ E=100V \end{gathered}$$

Understanding the concept

Here we have to use the formula for energy stored by inductor’s magnetic field to calculate energy. Then calculate thermal power from the current and resistance. Then add both to find the energy delivered by the battery.

Formula:

$$\begin{gathered} U_b=0.5L{i}^2 \\ i=\frac{𝛏}{R}\left(1-e^{-\frac{t}{{\tau }_L}}\right) \\ {P}_{thermal}=i^{2}R \end{gathered}$$

(a) Calculate the rate at which energy is being stored in the magnetic field

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

$U_b=0.5L{i}^2$

The rate of change in energy stored in the inductor is

$$\begin{gathered} \frac{d{U}_b}{dt}=\frac{d\left(0.5L{i}^2\right)}{dt} \\ \frac{d{U}_b}{dt}=0.5L\frac{d\left(i^2\right)}{dt} \\ \frac{d{U}_b}{dt}=Li\times \frac{di}{dt} \end{gathered}$$

Here, current is given as

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

Plug these values in the above equation:

role="math" localid="1661856003053" $$\begin{gathered} \frac{d{U}_b}{dt}=L\frac{E}{R}\left(1-e^{-\frac{t}{{\tau }_L}}\right)\times \frac{d\left({\displaystyle \frac{𝛏}{R}}\left(1-e^{-\frac{t}{{\tau }_L}}\right)\right)}{dt} \\ \frac{d{U}_b}{dt}=L\frac{E}{R}\left(1-e^{-\frac{t}{{\tau }_L}}\right)\times \frac{E}{R}\left(\frac{e^{-\frac{t}{{\tau }_L}}}{{\tau }_l}\right) \\ \frac{d{U}_b}{dt}=\frac{E^2}{R}\left(1-e^{-\frac{t}{{\tau }_L}}\right)\times \frac{E}{R}\left(\frac{e^{-\frac{t}{{\tau }_L}}}{{\tau }_l}\right) \end{gathered}$$

Here, ${\tau }_L=\frac{L}{R}=\frac{2.0}{10}=2.0s$

$$\begin{gathered} \frac{d{U}_b}{dt}=\frac{{100}^2}{10}\times \left(1-e^{-\frac{-0.10}{0.20}}\right)\times \left(e^{-\frac{0.10}{0.20}}\right) \\ \frac{d{U}_b}{dt}=2.4\times {10}^2\mathrm{W} \end{gathered}$$

(b) Calculate the rate at which thermal energy is appearing in the resistance

$$\begin{gathered} P_{thermal}=i^{2}R \\ {P}_{thermal}={\left(\frac{E}{R}\left(1-e^{-\frac{t}{{\tau }_L}}\right)\right)}^{2}R \\ {P}_{thermal}=\frac{E}{R}{\left(\left(1-e^{-\frac{t}{{\tau }_L}}\right)\right)}^2 \\ {P}_{thermal}=\frac{{100}^2}{10}{\left(\left(1-e^{-\frac{0.10}{0.20}}\right)\right)}^2 \\ {P}_{thermal}=1.5\times {10}^2\mathrm{W} \end{gathered}$$

(c) Calculate the rate at which energy is being delivered by the battery

Energy being delivered by battery is as follows:

$$\begin{gathered} P=P_{thermal}+\frac{d{U}_b}{dt} \\ P=2.4\times {10}^2+1.5\times {10}^2 \\ P=3.9\times {10}^2\mathrm{W} \end{gathered}$$