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• Notes

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

$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)$

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