Question

In Fig. 30-63, a V = 12.0 V ideal battery, a$\mathit{R}\mathbf{=}\mathbf{20}\mathbf{.}\mathbf{0}\mathit{\Omega }$resistor, and an inductor are connected by a switch at time t = 0 .At what rate is the battery transferring energy to the inductor’s field at$\mathit{t}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{61}{\mathit{\tau }}_{\mathbf{L}}$ ?

Short answer

The rate of battery transferring the energy to the inductor’s field $T=1.61{\tau }_L$ is

$\frac{d{U}_B}{dt}=1.15W$

Step-by-step solution

Given

$$\begin{gathered} V=12.0V \\ R=20.0\Omega \\ t=0 \end{gathered}$$

Understanding the concept

By using the equation of energy stored in the battery which depends on self-inductance and the current, we can take differentiation with respect to time to get the rate of energy transferred by the battery to the inductor.

Formula:

$$\begin{gathered} U_B=\frac{1}{2}L{\mathrm{i}}^2 \\ \mathrm{i}=\frac{\mathrm{\epsilon }}{\mathrm{R}}\left(1-{\mathrm{e}}^{-\frac{\mathrm{Rt}}{\mathrm{L}}}\right) \end{gathered}$$

Calculate the rate of the battery transferring the energy to the inductor’s field  t=1.61τL

By using the equation which is the energy stored in the battery we find the rate of energy to the inductor’s

$U_B=\frac{1}{2}L{i}^2$

Differentiate this with respect to time we can get

$\frac{d{U}_B}{dt}=Li\frac{di}{dt}.........................................................................................................\left(1\right)$

But from equation 30-41 the current through the battery

$i=\frac{\epsilon }{R}\left(1-e^{-\frac{Rt}{L}}\right)$

So,

role="math" localid="1661428000788" $\frac{di}{dt}=\frac{\epsilon }{t}{e}^{-\frac{Rt}{L}}$

So the equation (1) becomes

$$\begin{gathered} \frac{d{U}_B}{dt}=L\frac{\epsilon }{R}\left(1-e^{-\frac{Rt}{L}}\right)L\frac{\epsilon }{t}{e}^{-\frac{Rt}{L}} \\ \frac{d{U}_B}{dt}=\frac{{\epsilon }^2}{R}\left(1-e^{-\frac{Rt}{L}}\right)e^{-\frac{Rt}{L}} \end{gathered}$$

By substituting the value we can find

$\frac{d{U}_B}{dt}=\frac{12.0^2}{20.0}\left(1-e^{-\frac{Rt}{L}}\right)e^{-\frac{Rt}{L}}$

Where${\tau }_L=L/R$so that

$\frac{d{\cup }_B}{dt}=\frac{12.0^2}{20.0}\left(1-e^{-\frac{t}{{\tau }_L}}\right)e^{-\frac{t}{{\tau }_L}}$

By putting the value of $t=1.61{\tau }_L$

role="math" localid="1661428546325" $\frac{d{\cup }_B}{d{d}^t}=\frac{12.0^2}{20.0}\left(1-e^{-\frac{1.61{\tau }_L}{{\tau }_L}}\right)e^{-\frac{1.61{\tau }_L}{{\tau }_L}}$

role="math" localid="1661428556492" $\frac{d{U}_B}{dt}=\frac{12.0^2}{20.0}\left(1-e^{-1.61}\right)e^{-1.61}$

role="math" localid="1661428568772" $\frac{d{U}_B}{dt}=\frac{144}{20.0}\left(1-0.20\right)0.20$

$\frac{d{U}_B}{dt}=7.2\times 0.80\times 0.20$

$\frac{d{U}_B}{dt}=1.15W$