Solved Problem on RLC Circuits
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For a series RLC circuit, determine:
a) The equation for the oscillations given by the charge as a function of time q(t);
b) The solution for the equation of the circuit, in the case of subcritical damping, and the angular frequency of the oscillations.
a) The equation for the oscillations given by the charge as a function of time q(t);
b) The solution for the equation of the circuit, in the case of subcritical damping, and the angular frequency of the oscillations.
Solution
a) Applying Kirchhoff's Second Law (Figure 1)
\[
\begin{gather}
\bbox[#99CCFF,10px]
{\sum_{i=1}^{n}V_{i}=0} \tag{I}
\end{gather}
\]
Between points A and C, we have an emf in the inductor given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{V_{L}=L\frac{di}{dt}} \tag{II}
\end{gather}
\]
between points B and D, we have the emf on the capacitor given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{V_{C}=\frac{q}{C}} \tag{III}
\end{gather}
\]
between points A and B, we have the emf on the resistor given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{V_{R}=Ri} \tag{IV}
\end{gather}
\]
substituting expressions (II), (III), and (IV) into expression (I)
\[
\begin{gather}
V_{L}+V_{R}+V_{C}=0\\[5pt]
L\frac{di}{dt}+Ri+\frac{q}{C}=0
\end{gather}
\]
the current is the rate of change in charge relative to the time
\[
\begin{gather}
\bbox[#99CCFF,10px]
{i=\frac{dq}{dt}} \tag{V}
\end{gather}
\]
\[
\begin{gather}
L\frac{d}{dt}\left(\frac{dq}{dt}\right)+R\frac{dq}{dt}+\frac{q}{C}=0\\[5pt]
L\frac{d^{2}q}{dt^{2}}+R\frac{dq}{dt}+\frac{q}{C}=0
\end{gather}
\]
this is a Second-Order Homogeneous Ordinary Differential Equation. Dividing the equation by the
inductance L
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{\frac{d^{2}q}{dt^{2}}+\frac{R}{L}\frac{dq}{dt}+\frac{1}{LC}q=0}
\end{gather}
\]
b) In the equation of the previous item, we will make the following definitions
\[
\begin{gather}
2\gamma \equiv \frac{R}{L} \tag{VI}\\[10pt]
\omega ^{2}\equiv\frac{1}{LC} \tag{VII}
\end{gather}
\]
\[
\begin{gather}
\frac{d^{2}q}{dt^{2}}+2\gamma \frac{dq}{dt}+\omega ^{2}q=0 \tag{VIII}
\end{gather}
\]
The solution to this type of equation is found substituting
\[
\begin{array}{l}
q=\operatorname{e}^{\lambda t}\\[5pt]
\dfrac{dq}{dt}=\lambda\operatorname{e}^{\;\lambda t}\\[5pt]
\dfrac{d^{2}q}{dt^{2}}=\lambda^{2}\operatorname{e}^{\lambda t}
\end{array}
\]
substituting these values in the equation of item (a)
\[
\begin{gather}
\lambda ^{2}\operatorname{e}^{\;\lambda t}+2\gamma\lambda \operatorname{e}^{\lambda t}+\omega^{2}\operatorname{e}^{\lambda t}=0\\[5pt]
\operatorname{e}^{\lambda t}\left(\lambda ^{2}+2\gamma \lambda +\omega^{2}\right)=0\\[5pt]
\lambda^{2}+2\gamma \lambda +\omega^{2}=\frac{0}{\operatorname{e}^{\lambda t}}\\[5pt]
\lambda^{2}+2\gamma \lambda +\omega^{2}=0
\end{gather}
\]
this is the Characteristic Equation that has as a solution
\[
\begin{array}{l}
\Delta=b^{2}-4ac=\left(2\gamma \right)^{2}-4.1.\omega^{2}=4\gamma ^{2}-4\omega ^{2}=4\left(\gamma^{2}-\omega^{2}\right)\\[5pt]
\lambda_{1}=\dfrac{-b+\sqrt{\Delta\;}}{2a}=\dfrac{-2\gamma +\sqrt{4\left(\gamma^{2}-\omega^{2}\right)\;}}{2.1}=-{\dfrac{2\gamma }{2}}+\dfrac{2\sqrt{\gamma^{2}-\omega^{2}\;}}{2}=-\gamma +\sqrt{\gamma^{2}-\omega^{2}\;}\\[5pt]
\lambda_{2}=\dfrac{-b-\sqrt{\Delta\;}}{2a}=\dfrac{-2\gamma-\sqrt{4\left(\gamma^{2}-\omega^{2}\right)\;}}{2.1}=-{\dfrac{2\gamma}{2}}-\dfrac{2\sqrt{\gamma^{2}-\omega^{2}\;}}{2}=-\gamma -\sqrt{\gamma^{2}-\omega ^{2}\;}
\end{array}
\]
For the circuit to oscillate critically damped, we must have ω2>γ2,
the term in the square root will be
\[
\begin{gather}
\sqrt{-1.\left(\omega^{2}-\gamma^{2}\right)}=\sqrt{-1}.\sqrt{\left(\omega ^{2}-\gamma^{2}\right)}=i\sqrt{\left(\omega ^{2}-\gamma ^{2}\right)}
\end{gather}
\]
where \( i=\sqrt{-1} \)The solution of expression (VIII) is written as
\[
\begin{gather}
q=C_{1}\operatorname{e}^{\lambda_{1}t}+C_{2}\operatorname{e}^{\lambda_{2}t}\\[5pt]
q=C_{1}\operatorname{e}^{\left(-\gamma +i\sqrt{\omega^{2}-\gamma^{2}\;}\right)t}+C_{2}\operatorname{e}^{\left(-\gamma-i\sqrt{\omega^{2}-\gamma^{2}\;}\right)t}\\[5pt]
q=C_{1}\operatorname{e}^{\left(-\gamma t+i\sqrt{\omega^{2}-\gamma^{2}\;}\;t\right)}+C_{2}\operatorname{e}^{\left(-\gamma t-i\sqrt{\omega^{2}-\gamma ^{2}\;}\;t\right)}\\[5pt]
q=C_{1}\operatorname{e}^{-\gamma t}\operatorname{e}^{\left(i\sqrt{\omega^{2}-\gamma^{2}\;}\;t\right)}+C_{2}\operatorname{e}^{-\gamma t}\operatorname{e}^{\left(-i\sqrt{\omega ^{2}-\gamma^{2}\;}\;t\right)}\\[5pt]
q=\operatorname{e}^{-\gamma t}\left[C_{1}\operatorname{e}^{\left(i\sqrt{\omega ^{2}-\gamma^{2}\;}\;t\right)}+C_{2}\operatorname{e}^{\left(-i\sqrt{\omega^{2}-\gamma^{2}\;}\;t\right)}\right]
\end{gather}
\]
where C1 and C2 are constants of integration, using
Euler's Formula
\( \operatorname{e}^{i\theta}=\cos \theta +i \operatorname{sen}\theta \)
\[
\begin{split}
q=&\operatorname{e}^{-\gamma t}\left[C_{1}\operatorname{e}^{\left(i\sqrt{\omega^{2}-\gamma^{2}\;}\;t\right)}+C_{2}\operatorname{e}^{\left(-i\sqrt{\omega^{2}-\gamma^{2}\;}\;t\right)}\right]\\[8pt]
q=&\operatorname{e}^{-\gamma t}\left\{C_{1}\left[\cos \left(\sqrt{\omega^{2}-\gamma^{2}\;}\;t\right)+i\operatorname{sen}\left(\sqrt{\omega ^{2}-\gamma^{2}\;}\;t\right)\right]\right.+\\
&\left.+C_{2}\left[\cos\left(\sqrt{\omega ^{2}-\gamma^{2}\;}\;t\right)-i\operatorname{sen}\left(\sqrt{\omega ^{2}-\gamma^{2}\;}\;t\right)\right]\right\}\\[8pt]
q=&\operatorname{e}^{-\gamma t}\left\{C_{1}\cos \left(\sqrt{\omega ^{2}-\gamma^{2}\;}\;t\right)+iC_{1}\operatorname{sen}\left(\sqrt{\omega^{2}-\gamma ^{2}\;}\;t\right)\right.+\\
&\left.+C_{2}\cos\left(\sqrt{\omega ^{2}-\gamma^{2}\;}\;t\right)-iC_{2}\operatorname{sen}\left(\sqrt{\omega^{2}-\gamma ^{2}\;}\;t\right)\right\}\\[8pt]
q=&\operatorname{e}^{-\gamma t}\left\{\left(C_{1}+C_{2}\right)\cos \left(\sqrt{\omega^{2}-\gamma^{2}\;}\;t\right)+i\left(C_{1}-C_{2}\right)\operatorname{sen}\left(\sqrt{\omega^{2}-\gamma^{2}\;}\;t\right)\right\}
\end{split}
\]
defining two new constants α and β in terms of C1 and C2
\[
\begin{gather}
\alpha \equiv C_{1}+C_{2} \quad \text{and} \quad \beta \equiv \text{i}(C_{1}-C_{2})
\end{gather}
\]
\[
\begin{gather}
q=\operatorname{e}^{-\gamma t}\left\{\alpha \;\cos \left(\sqrt{\omega^{2}-\gamma ^{2}\;}\;t\right)+\beta\;\operatorname{sen}\left(\sqrt{\omega^{2}-\gamma^{2}\;}\;t\right)\right\}
\end{gather}
\]
multiplying and dividing this expression by
\( \sqrt{\alpha ^{2}+\beta ^{2}\;} \)
\[
\begin{gather}
q=\operatorname{e}^{-\gamma t}\left\{\alpha \;\cos\left(\sqrt{\omega ^{2}-\gamma ^{2}\;}\;t\right)+\beta\;\operatorname{sen}\left(\sqrt{\omega ^{2}-\gamma^{2}\;}\;t\right)\right\}\frac{\sqrt{\alpha ^{2}+\beta^{2}\;}}{\sqrt{\alpha ^{2}+\beta ^{2}\;}}\\[5pt]
q=\sqrt{\alpha^{2}+\beta ^{2}\;}\;\operatorname{e}^{-\gamma t}\left\{\frac{\alpha}{\sqrt{\alpha ^{2}+\beta ^{2}\;}}\;\cos \left(\sqrt{\omega ^{2}-\gamma^{2}\;}\;t\right)+\frac{\beta }{\sqrt{\alpha^{2}+\beta^{2}\;}}\;\operatorname{sen}\left(\sqrt{\omega^{2}-\gamma^{2}\;}\;t\right)\right\}
\end{gather}
\]
setting
\[
\begin{array}{l}
A\equiv \sqrt{\alpha ^{2}+\beta ^{2}\;}\\[5pt]
\cos \varphi \equiv \dfrac{\alpha }{\sqrt{\alpha ^{2}+\beta ^{2}\;}}\\[5pt]
\operatorname{sen}\varphi \equiv \dfrac{\beta }{\sqrt{\alpha ^{2}+\beta^{2}\;}}
\end{array}
\]
\[
\begin{gather}
q=A\;\operatorname{e}^{-\gamma t}\left\{\cos \varphi \;\cos\left(\sqrt{\omega^{2}-\gamma^{2}\;}\;t\right)+\operatorname{sen}\varphi\;\operatorname{sen}\left(\sqrt{\omega ^{2}-\gamma^{2}\;}\;t\right)\right\}
\end{gather}
\]
From the trigonometric identity
\( \cos (a-b)=\cos a\cos b+\operatorname{sen}a\operatorname{sen}b \)
\[
\begin{gather}
q=A\;\operatorname{e}^{-\gamma t}\;\cos \left(\sqrt{\omega ^{2}-\gamma^{2}\;}\;t-\varphi \right)
\end{gather}
\]
the angular frequency ω0 is given by
\[
\begin{gather}
\omega_{0}=\sqrt{\omega ^{2}-\gamma^{2}\;}
\end{gather}
\]
using the definitions made in (VI) and (VII) for ω2 and γ
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{\omega_{0}=\sqrt{\frac{1}{LC}-\left(\frac{R}{2L}\right)^{2}\;}}
\end{gather}
\]
The solution to the equation of charge is written as
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{q(t)=A\;\operatorname{e}^{-\gamma t}\;\cos \left(\omega_{0}t-\varphi\right)}
\end{gather}
\]
where A and φ are constants determined by the initial conditions of the problem.
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Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .