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• Notes

$\begin{array}{l}{\text { An } L-R-C \text { series circuit has } L=0.450 \mathrm{H}, C=2.50 \times} \\ {10^{-5} \mathrm{F}, \text { and resistance } R . \text { (a) What is the angular frequency of the }} \\ {\text { circuit when } R=0 ? \text { (b) What value must } R \text { have to give a } 5.0 \%} \\ {\text { decrease in angular frequency compared to the value calculated in }} \\ {\text { part (a)? }}\end{array}$

(a) $w_{0}=\frac{I}{\sqrt{L c}}$                        when $R=0$

$= \frac{I}{0.45 * 2.5 * 10^{-5}}$$=298 \mathrm{rad/s}$

(b) $\frac{w^{\prime}}{w_{0}}$$=0.95 \longrightarrow$$\frac{\left(\frac{I}{LC}-\frac{R^{2}}{4 L^{2}}\right)}{\frac{I}{L C}}$$=I-\frac{R^{2} c}{4 L}=(0.95)$

$R \rightarrow ??$            $R=\sqrt{\frac{4 L}{c}\left(1-0.95^{2}\right)}=$

$\sqrt{\frac{4* 0 \cdot 45}{2.5 * 10^{-5}}(0.0975)}$$=83.8 \Omega$

$\begin{array}{l}{\text { For the circuit of Fig. } 30.17 \text { , let } C=15.0 \text { nF, } L=} \\ {22 \mathrm{mH}, \text { and } R=75.0 \Omega \text { . (a) Calculate the oscillation frequency of }} \\ {\text { the circuit once the capacitor has been charged and the switch has }} \\ {\text { been connected to point } a \text { . (b) How long will it take for the ampli- }} \\ {\text { tude of the oscillation to decay to } 10.0 \% \text { of its original value? }} \\ {\text { (c) What value of } R \text { would result in a critically damped circuit? }}\end{array}$

(a) $w^{\prime}=\sqrt{\frac{1}{L c}-\frac{R^{2}}{4L^{2}}}$$=\sqrt{\frac{1}{22 * 10^{-3} * 15 * 10^{-9}}-\frac{(75)^{2}}{4\left(22 * 10^{-3}\right)^{2}}}$$=5.5*10^{4} \mathrm{rad/}$

$f=\frac{\omega}{2 \pi}=$$\frac{5.5 * 10^{4}}{2 \pi}$$=8.76 * 10^{3} \mathrm{Hz}$

$=8.76 \mathrm{kHz}$

(b) $A(x)=A_{0} e^{-\left(\frac{R}{2 L}\right) t}$

$\rightarrow t=$$\frac{-2 l \ln \left(\frac{A}{A_{0}}\right)}{R}=$$\frac{-2\left(22*10^{-3}\right) \ln (0 \cdot 1)}{75}$$=1.35*10^{-3} s$

$=1.35 \mathrm{ms}$

(c) $R=\sqrt{\frac{4 L}{C}}$$=\sqrt{\frac{4\left(22*10^{-3}\right)}{15 * 10^{-9}}}$$=2420 \Omega$