At$\mathit{t}\mathbf{=}\mathbf{0}$, a battery is connected to a series arrangement of a resistor and an inductor. If the inductive time constant is 37.0 ms, at what time is the rate at which energy is dissipated in the resistor equal to the rate at which energy is stored in the inductor’s magnetic field?

Inductive time constant is 37 ms

Here we have to use the formula for energy stored by the inductor’s magnetic field. From that, find the rate of energy stored by the inductor’s magnetic field. And by equating that with thermal power, we can solve for time.

Formula:

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

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

$U_b=0.5L{i}^2$

The rate of change in energy stored intheinductor 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

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

The inductive time constant is as follows:

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

Plug these values in the above equation:

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

Rate of change in thermal energy is

$$\begin{gathered} P_{thermal}=i^{2}R \\ {P}_{thermal}={\left(\frac{E}{R}\left(1-e^{-\frac{t}{{\tau }_L}}\right)\right)}^{2}R..............................................................\left(2\right) \end{gathered}$$

Comparing equation 1 and 2,

$$\begin{gathered} \frac{E^2}{R}=\left(1-e^{-\frac{t}{{\tau }_L}}\right)\times \left(e^{-\frac{t}{{\tau }_L}}\right)={\left(\frac{E}{R}\left(e^{-\frac{t}{{\tau }_L}}\right)\right)}^{2}R \\ {e}^{-\frac{t}{{\tau }_L}}=\left(1-e^{-\frac{t}{{\tau }_L}}\right) \\ {e}^{-\frac{t}{{\tau }_L}}=\frac{1}{2} \end{gathered}$$

Now taking natural log on both sides,

$$\begin{gathered} t={\tau }_l\times In\left(2\right) \\ t=37ms\times In\left(2\right) \\ t=25.6ms \end{gathered}$$