For the circuit of Figure, assume that $\mathbf{\epsilon }\mathbf{=}\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{V}\mathbf{,}\mathbf{R}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{70}\mathbf{\Omega }\mathbf{,}\mathbf{}\mathbf{and}\mathbf{}\mathbf{L}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{50}\mathbf{H}$. The ideal battery is connected at time$\mathit{t}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}$. (a) How much energy is delivered by the battery during the first 2.00 s? (b) How much of this energy is stored in the magnetic field of the inductor? (c) How much of this energy is dissipated in the resistor?

i) Emf of battery V = 10.0 V

ii) Resistance R = 6.70$\Omega$

iii) Inductance L = 5.50 H

We need to use the formula of the energy delivered by the battery to find energy delivered by the battery during the first. Then, using the formula of the energy stored in the inductors magnetic field, we can find theenergy stored in the magnetic field of the inductor. Using these values, we can find the energy dissipated in the resistor considering theconservation of total energy.

Formula:

$$\begin{gathered} E={\int }_0^{t}Pdt \\ P=\frac{V^2}{R}\left(1-e^{\frac{Rt}{L}}\right) \\ {U}_B=\frac{1}{2}L{i}^2 \\ i=\frac{V}{R}\left(1-e^{\frac{Rt}{L}}\right) \end{gathered}$$

The energy delivered by the battery is given by

$$\begin{gathered} E={\int }_0^{t}Pdt \\ \mathrm{But}, \\ \mathrm{P}=\frac{{\mathrm{V}}^2}{\mathrm{R}}\left(1-{\mathrm{e}}^{\frac{\mathrm{Rt}}{\mathrm{L}}}\right) \\ \mathrm{Therefore}, \\ \mathrm{E}={\int }_0^{\mathrm{t}}\frac{{\mathrm{V}}^2}{\mathrm{R}}\left(1-{\mathrm{e}}^{\frac{\mathrm{Rt}}{\mathrm{L}}}\right)\mathrm{dt} \\ \mathrm{E}=\frac{{\mathrm{V}}^2}{\mathrm{R}}\left({\int }_0^{\mathrm{t}}\mathrm{dt}-{\int }_0^{\mathrm{t}}{\mathrm{e}}^{\frac{\mathrm{Rt}}{\mathrm{L}}}\mathrm{dt}\right) \\ \mathrm{E}=\frac{{\mathrm{V}}^2}{\mathrm{R}}\left[\mathrm{t}+\frac{\mathrm{L}}{\mathrm{R}}\left({\mathrm{e}}^{\frac{\mathrm{Rt}}{\mathrm{L}}-1}\right)\right] \\ \mathrm{E}=\frac{{\left(10.0\mathrm{V}\right)}^2}{6.70\mathrm{\Omega }}\left[\left(2.00\mathrm{s}\right)+\frac{\left(5.50\mathrm{H}\right)}{\left(6.70\mathrm{\Omega }\right)}\right]\left({\mathrm{e}}^{\frac{\left(6.70\mathrm{\Omega }\right)\left(2.00\mathrm{s}\right)}{\left(5.50\mathrm{H}\right)}}-1\right) \\ \mathrm{E}=18.7\mathrm{J} \end{gathered}$$

Therefore, the energy delivered by the battery during the first 2.00 s is 18.7 J.

The energy stored in the magnetic field of the inductor is given by

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

But,

$$\begin{gathered} i=\frac{V}{R}\left(1-e^{\frac{Rt}{L}}\right) \\ \mathrm{Therefore}, \\ {\mathrm{U}}_{\mathrm{B}}=\frac{1}{2}\mathrm{L}{\left[\frac{\mathrm{V}}{\mathrm{R}}\left(1-{\mathrm{e}}^{\frac{\mathrm{Rt}}{\mathrm{L}}}\right)\right]}^2 \\ {\mathrm{U}}_{\mathrm{B}}=\frac{1}{2}\mathrm{L}{\left(\frac{\mathrm{V}}{\mathrm{R}}\right)}^2\left[1-2\left(1\right){\left({\mathrm{e}}^{\frac{\mathrm{Rt}}{\mathrm{L}}}\right)}^2\right] \\ {\mathrm{U}}_{\mathrm{B}}=\frac{1}{2}\left(5.50\mathrm{H}\right){\left(\frac{10.0\mathrm{V}}{6.70\mathrm{\Omega }}\right)}^2\left[1-2\left(1\right)\left({\mathrm{e}}^{\frac{\left(6.70\mathrm{\Omega }\right)\left(2.00\mathrm{s}\right)}{\left(5.50\mathrm{H}\right)}}\right)+{\left({\mathrm{e}}^{\frac{\left(6.70\mathrm{\Omega }\right)\left(2.00\mathrm{s}\right)}{\left(5.50\mathrm{H}\right)}}\right)}^2\right] \\ {\mathrm{U}}_{\mathrm{B}}=5.10\mathrm{J} \end{gathered}$$

Therefore, the energy stored in the magnetic field of the inductor is5.10 J.

Energy delivered by the battery during the first 2.00 s is 18.7 J and energy stored in the magnetic field of the inductor is 5.10 J

Therefore, the amount of energy dissipated in the resistor is

E = 18.7 J-5.10 J = 13.6 J

Therefore, the energy dissipated in the resistor is 13.6 J.