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$$
\begin{array}{l}{\text { A } 35.0-\mathrm{V} \text { battery with negligible internal resistance, a }} \\ {50.0-\Omega \text { resistor, and a } 1.25-\mathrm{mH} \text { inductor with negligible resistance }} \\ {\text { are all connected in series with an open switch. The switch is }} \\ {\text { suddenly closed. (a) How long after closing the switch will the }} \\ {\text { current through the inductor reach one-half of its maximum value? }} \\ {\text { (b) How long after closing the switch will the energy stored in the }} \\ {\text { inductor reach one-half of its maximum value? }}\end{array}
$$

$$
i=\frac{\varepsilon}{R\left(1-e^{\frac{-t}{\tau}}\right)}
,\quad$$$$
\tau=\frac{L}{R}\quad ,
$$$$
u=\frac{1}{2} L i^{2}
$$

(a) $$
i_{\max }=\frac{\varepsilon}{R}
$$$$
\longrightarrow \quad i=\frac{i \max }{2}
$$

$$
i=\frac{\varepsilon}{R\left(1-e^{\frac{-t}{\tau}}\right)}
$$$$
=\frac{i_\text { max} }{2}
$$$$
=\frac{\varepsilon}{R} * \frac{1}{2}
$$

$$
\left(1-e^{\frac{-t}{\tau}}\right)=\frac{1}{2} \longrightarrow
$$$$
e^{\frac{-t}{\tau}}=\frac{1}{2} \rightarrow \frac{-t}{\tau}=\ln \frac{1}{2}
$$

\(∴ t=\frac{L \ln 2}{R}= \frac{1.25 * 10^{-3} * \ln 2}{50}=17.3 Ms \)

(b) $$
U=\frac{1}{2} U_{\max } \rightarrow
$$$$
i=\frac{i _{max}}{\sqrt{2}}
$$$$
1-e^{\frac{-t}{\tau}}=\frac{1}{\sqrt{2}} \rightarrow e^{\frac{-t}{\tau}}=1-\frac{1}{\sqrt{2}}
$$

\(∴ e^{\frac{-t}{\tau}}= =0.2929 \longrightarrow t= \frac{-L \ln (\sqrt{0.29 29}}{R} \)

\(∴ t=30.7Ms \)

$$
\begin{array}{l}{\text { In Fig. } 30.11, \text { switch } S_{1} \text { is closed while switch } S_{2} \text { is kept }} \\ {\text { open. The inductance is } L=0.115 \mathrm{H} \text { , and the resistance is }} \\ {R=120 \Omega \text { . (a) When the current has reached its final value, the }} \\ {\text { energy stored in the inductor is } 0.260 \mathrm{J} \text { . What is the emf } \mathcal{E} \text { of the }}\end{array}
$$

$$
\begin{array}{l}{\text { battery? (b) After the current has reached its final value, } S_{1} \text { is }} \\ {\text { opened and } S_{2} \text { is closed. How much time does it take for the energy }} \\ {\text { stored in the inductor to decrease to } 0.130 \mathrm{J} \text { , half the original value? }}\end{array}
$$

(a) $$
I=\sqrt{\frac{2 U}{L}}
$$$$
U=\frac{1}{2} li^{ 2} \longrightarrow
$$

$$
=\sqrt{\frac{2(0 \cdot 26)}{0.115}}=2.13 \mathrm{A}
$$

$$
\varepsilon=I R=
$$$$
2.13 * 120=256 \mathrm{V}
$$

(b) $$
i=Ie^{-\left(\frac{R}{t}\right) t}
$$

$$
U=\frac{1}{2} L{i}^{2}
$$$$
\rightarrow U=\frac{1}{2}LI^2 e^{-2(\frac{R}{L})t}=\frac{1}{2}U_0
$$

$$
\rightarrow e^{-2\left(\frac{R}{L}\right) t}
$$$$
\frac{1}{2} \longrightarrow t=-\frac{1}{2 R} \ln \left(\frac{1}{2}\right)
$$

\(∴ t=-\frac{0.115}{2(120)} \ln \left(\frac{1}{2}\right)=3.32*10^{-4} \mathrm{s} \)

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