Question

An L-R-C series circuit has L = 0.400 H, C = 7.00 mF, and R = 320 Ω. At t = 0 the current is zero and the initial charge on the capacitor is $\mathbf{2}\mathbf{.}\mathbf{18}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{C}$. (a) What are the values of the constants A and f in Eq. (30.28)$\mathbf{q}\mathbf{=}{\mathbf{Ae}}^{\mathbf{-}\left(R/2L\right)\mathbf{t}}\mathbf{cos}\left(\sqrt{\frac{1}{\mathrm{LC}}-\frac{R^2}{4L^2}t+\varphi }\right)$? (b) How much time does it take for each complete current oscillation after the switch in thiscircuit is closed? (c) What is the charge on the capacitor after thefirst complete current oscillation?

Short answer

A) $\mathrm{A}=\mathrm{QAnd\varphi }=0$

B) The time taken to complete current oscillation is 0.0142s

C) The charge on the capacitor is$9.75\times {10}^{-7}\mathrm{C}$

Step-by-step solution

Concept of the charge over the capacitor

The charge over the capacitor is given by equation $\mathbf{q}\mathbf{=}{\mathbf{Ae}}^{\mathbf{-}\mathbf{(}\mathbf{R}\mathbf{/}\mathbf{2}\mathbf{L}\mathbf{)}\mathbf{t}}\mathbf{cos}\mathbf{\left(}\sqrt{\frac{\mathbf{1}}{\mathbf{LC}}\mathbf{-}\frac{{\mathbf{R}}^{\mathbf{2}}}{\mathbf{4}{\mathbf{L}}^{\mathbf{2}}}\mathbf{t}\mathbf{+}\mathbf{\varphi }}\mathbf{\right)}$at t =0 the charge is the maximum, that is,q(t-0)=Q which can be written as$\mathbf{Q}\mathbf{=}{\mathbf{Ae}}^{\mathbf{0}}\mathbf{cos}\left(0+\varphi \right)\mathbf{=}\mathbf{Acos}\left(\varphi \right)$

Calculate the time taken to complete current oscillation

In the equation $\mathrm{Q}={\mathrm{Ae}}^0\cos (0+\mathrm{\varphi })=\mathrm{Acos}\left(\mathrm{\varphi }\right)$, RHS is maximum only when$\cos \left(\mathrm{\varphi }\right)=1$so,$\mathrm{A}=\mathrm{Q}$and $\mathrm{\varphi }=0$,

The angular frequency is given as $\mathrm{\varpi }=\sqrt{\frac{1}{\mathrm{LC}}-\frac{{\mathrm{R}}^2}{4{\mathrm{L}}^2}}$.At the need of the oscillation$\mathrm{\varpi t}=2\mathrm{\pi }$ So, $2\mathrm{\pi }=\sqrt{\frac{1}{\mathrm{LC}}-\frac{{\mathrm{R}}^2}{4{\mathrm{L}}^2}\mathrm{t}}$. Substitute the values in the equationrole="math" localid="1664178436894" $\mathrm{t}=\frac{2\mathrm{\pi }}{\sqrt{\frac{1}{\mathrm{LC}}-\frac{{\mathrm{R}}^2}{4{\mathrm{L}}^2}}}$

$$\begin{gathered} \mathrm{t}=\frac{2\mathrm{\pi }}{\sqrt{{\displaystyle \frac{1}{0.400\times 7.00}}-{\displaystyle \frac{\left({320}^2\right)}{4{\left(0.400\right)}^2}}}} \\ =0.0142\mathrm{s} \end{gathered}$$

Therefore, the time taken to complete current oscillation is 0.0142s

Calculate the charge on the capacitor after the first complete current oscillation

At the end of the first oscillation, the cosine term is 1, since$\mathrm{\varphi }=1$and role="math" localid="1664178711109" $\mathrm{\varpi t}=2\mathrm{\pi }$.So,$\mathrm{q}={\mathrm{Qe}}^{-\left(\mathrm{R}/2\mathrm{L}\right)\mathrm{t}}$

Now substitute with t =0.0142 s and the givens to get q

role="math" localid="1664178926937" $$\begin{gathered} \mathrm{q}=\left(2.18\times {10}^{-4}\mathrm{C}\right){\mathrm{e}}^{-\left(320\mathrm{\Omega }\right)\left(0.0142\mathrm{s}\right)\left(0.400\mathrm{H}\right)} \\ =9.75\times {10}^{-7}\mathrm{C} \end{gathered}$$

Therefore, the charge on the capacitor is$9.75\times {10}^{-7}\mathrm{C}$