Solved Problem on RLC Circuits
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A series RLC circuit contains a resistor with resistance R = 75 Ω, an inductor with i
nductance L = 10 mH, and a capacitor with capacitance C = 0.20 μF. The initial charge stored
in the capacitor is equal to q0 = 0.4 mC and the current is zero. Determine:
a) The equation of electric charge as a function of time;
b) What is the type of oscillations in this circuit;
c) The graph of charge q versus time t.
a) The equation of electric charge as a function of time;
b) What is the type of oscillations in this circuit;
c) The graph of charge q versus time t.
Problem data:
- Capacitance: C = 0.20 μF;
- Resistance: R = 75 W;
- Inductance: L = 10 mH;
- Initial charge stored in the capacitor: q0 = 0.4 mC;
- Initial current: I0 = 0.
From the initial instant, a current begins to circulate, and the charge on the capacitor decreases while
the current in the circuit increases, in each element of the circuit we have a potential difference
(Figure 1). With this, we write the Initial Conditions of the problem
\[
\begin{gather}
q(0)=q_{0}\\[10pt]
i_{0}=\frac{dq(0)}{dt}=0
\end{gather}
\]
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 B, we have a potential difference in the inductor given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{V_{L}=L\frac{di}{dt}} \tag{II}
\end{gather}
\]
between points C and D, we have a potential difference in the capacitor given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{V_{C}=\frac{q}{C}} \tag{III}
\end{gather}
\]
between points A and C, we have a potential difference in 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 instantaneous current is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{i=\frac{dq}{dt}}
\end{gather}
\]
\[
\begin{gather}
L\frac{d}{dt}\left(\frac{dq}{dt}\right)+R\frac{dq}{dt}+\frac{q}{C}=0\\[5pt]
L\frac{d^{2}q}{dt^{2}}+R\frac{dq}{dt}+\frac{q}{C}=0
\end{gather}
\]
this is a Second-Order Homogeneous Ordinary Differential Equation. Dividing the equation by the
inductance L
\[
\begin{gather}
\frac{d^{2}q}{dt^{2}}+\frac{R}{L}\frac{dq}{dt}+\frac{1}{LC}q=0
\end{gather}
\]
substituting the values given in the problem
\[
\begin{gather}
\frac{d^{2}q}{dt^{2}}+\frac{75}{10\times10^{-3}}\frac{dq}{dt}+\frac{1}{10\times10^{-3}\times0.20\times10^{-6}}q=0\\[5pt]
\frac{d^{2}q}{dt^{2}}+\frac{75}{10^{-2}}\frac{dq}{dt}+\frac{1}{2\times10^{-9}}q=0\\[5pt]
\frac{d^{2}q}{dt^{2}}+7.5\times10^{3}\frac{dq}{dt}+5\times10^{8}q=0 \tag{V}
\end{gather}
\]
Solution of \( \displaystyle \frac{d^{2}q}{dt^{2}}+7.5\times10^{3}\frac{dq}{dt}+5\times10^{8}q=0 \)
The solution to this type of equation is found substituting
Differentiation of the expression (VI) with respect to time
The solution to this type of equation is found substituting
\[
\begin{gather}
q=\operatorname{e}^{\lambda t}\\[5pt]
\frac{dq}{dt}=\lambda \operatorname{e}^{\;\lambda t}\\[5pt]
\frac{d^{2}q}{dt^{2}}=\lambda ^{2}\operatorname{e}^{\lambda t}
\end{gather}
\]
substituting these values into the differential equation
\[
\begin{gather}
\lambda^{2}\operatorname{e}^{\lambda t}+7.5\times10^{3}\lambda \operatorname{e}^{\lambda t}+5\times10^{8}\operatorname{e}^{\lambda t}=0\\[5pt]
\operatorname{e}^{\lambda t}\left(\lambda ^{2}+7.5\times10^{3}\lambda +5\times10^{8}\right)=0\\[5pt]
\lambda^{2}+7.5\times10^{3}\lambda +5\times10^{8}=\frac{0}{\operatorname{e}^{\lambda t}}\\[5pt]
\lambda ^{2}+7.5\times10^{3}\lambda +5\times10^{8}=0
\end{gather}
\]
this is the Characteristic Equation that has a solution
\[
\begin{gather}
\Delta=b^{2}-4ac=\left(7.5\times10^{3}\right)^{2}-4.5\times10^{8}=5.63\times10^{7}-2.00\times10^{9}=-1.94\times10^{9}
\end{gather}
\]
for Δ<0 the roots are complex of the form a+bi, where
\( i=\sqrt{-1\;} \)
\[
\begin{gather}
\lambda =\frac{-b+\sqrt{\Delta\;}}{2a}=\frac{-7.5\times10^{3}+\sqrt{-1.94\times10^{9}\;}}{2\times1}=-{\frac{-7.5\times10^{3}\pm4.4\times10^{4}i}{2}}\\[5pt]
\lambda_{1}=-3.75\times10^{3}+2.20\times10^{4}i\qquad \text{e}\qquad \lambda_{2}=-3.75\times10^{3}-2.20\times10^{4}i
\end{gather}
\]
The solution to the differential equation will be
\[
\begin{gather}
q=C_{1}\operatorname{e}^{\lambda_{1}t}+C_{2}\operatorname{e}^{\lambda_{2}t}\\[5pt]
q=C_{1}\operatorname{e}^{\left(-3.75\times10^{3}+2.20\times10^{4}i\right)t}+C_{2}\operatorname{e}^{\left(-3.75\times10^{3}-2.20\times10^{4}i\right)t}\\[5pt]
q=C_{1}\operatorname{e}^{\left(-3.75\times10^{3}t+2.20\times10^{4}it\right)}+C_{2}\operatorname{e}^{\left(-3.75\times10^{3}t+2.20\times10^{4}it\right)}\\[5pt]
q=C_{1}\operatorname{e}^{-3.75\times10^{3}t}\operatorname{e}^{2.15\times10^{4}i\;t}+C_{2}\operatorname{e}^{-3.75\times10^{3}t}\operatorname{e}^{-2.15\times10^{4}it}\\[5pt]
q=\operatorname{e}^{-3.75\times10^{3}t}\left(C_{1}\operatorname{e}^{2.20\times10^{4}it}+C_{2}\operatorname{e}^{-2.20\times10^{4}it}\right)
\end{gather}
\]
where C1 and C2 are constants of integration, using
Euler's Formula
\( \operatorname{e}^{i\theta }=\cos \theta +i\sin \theta \)
\[
\begin{gather}
q=\operatorname{e}^{-3.75\times10^{3}t}\left[C_{1}\left(\cos2.20\times10^{4}t+i\sin 2.20\times10^{4}t\right)+C_{2}\left(\cos2.20\times10^{4}t-i\sin 2.20\times10^{4}t\right)\right]\\[5pt]
q=\operatorname{e}^{-3.75\times10^{3}t}\left(C_{1}\cos2.20\times10^{4}t+iC_{1}\sin 2.20\times10^{4}t+C_{2}\cos2.20\times10^{4}t-iC_{2}\sin 2.20\times10^{4}t\right)\\[5pt]
q=\operatorname{e}^{-3.75\times10^{3}t}\left[\left(C_{1}+C_{2}\right)\cos2.20\times10^{4}t+i\left(C_{1}-C_{2}\right)\sin 2.20\times10^{4}t\right]
\end{gather}
\]
defining two new constants α and β in terms of C1 and
C2
\[
\begin{gather}
\alpha \equiv C_{1}+C_{2}\\[5pt]
\text{e}\\[5pt]
\beta \equiv i(C_{1}-C_{2})
\end{gather}
\]
\[
\begin{gather}
q=\operatorname{e}^{-3.75\times10^{3}t}\left(\alpha \cos 2.20\times10^{4}t+\beta\sin 2.20\times10^{4}t\right)
\end{gather}
\]
multiplying and dividing this expression by
\( \sqrt{\alpha^{2}+\beta^{2}\;} \)
\[
\begin{gather}
q=\operatorname{e}^{-3.75\times10^{3}t}\left(\alpha \cos2.20\times10^{4}t+\beta\sin 2.20\times10^{4}t\right).\frac{\sqrt{\alpha^{2}+\beta^{2}\;}}{\sqrt{\alpha^{2}+\beta^{2}\;}}\\[5pt]
q=\sqrt{\alpha^{2}+\beta^{2}\;}\operatorname{e}^{-3.75\times10^{3}t}\left(\frac{\alpha}{\sqrt{\alpha^{2}+\beta^{2}\;}}\cos 2.20\times10^{4}t+\frac{\beta}{\sqrt{\alpha^{2}+\beta^{2}\;}}\sin 2.20\times10^{4}t\right)
\end{gather}
\]
settings
\[
\begin{gather}
A\equiv \sqrt{\alpha^{2}+\beta^{2}\;}\\[5pt]
\cos\varphi \equiv \frac{\alpha}{\sqrt{\alpha^{2}+\beta{2}\;}}\\[5pt]
\sin \varphi \equiv \frac{\beta}{\sqrt{\alpha^{2}+\beta^{2}\;}}
\end{gather}
\]
\[
\begin{gather}
q=A\operatorname{e}^{-3.75\times10^{3}t}\left(\cos \varphi \cos2.20\times10^{4}t+\sin \varphi\sin 2.20\times10^{4}t\right)
\end{gather}
\]
From the trigonometric identity
\( \cos(a+b)=\cos a\cos b+\sin a\sin b \).
\[
\begin{gather}
q(t)=A\operatorname{e}^{-3.75\times10^{3}t}\cos \left(2.20\times10^{4}t-\varphi\right)\tag{VI}
\end{gather}
\]
where A and φ are constants determined by the Initial Conditions.
Differentiation of the expression (VI) with respect to time
\[
\begin{gather}
q=\underbrace{A\operatorname{e}^{-3.75\times10^{3}t}}_{u}\underbrace{\cos(2.20\times10^{4}t-\varphi)}_{v}
\end{gather}
\]
using the Product Rule for the differentiation of functions
\[
\begin{gather}
(uv)'=u'v+uv'
\end{gather}
\]
where
\( u=A\operatorname{e}^{-3.75\times10^{3}t} \)
and
\( v=\cos (2.20\times10^{4}t-\varphi) \),
the function v is a composite function, using the Chain Rule
\[
\begin{gather}
\frac{dv[w(t)]}{dt}=\frac{dv}{dw}\frac{dw}{dt}
\end{gather}
\]
with
\( v=\cos w \)
and
\( w=2.20\times10^{4}t-\varphi \)
\[
\begin{gather}
\frac{dx}{dt}=\frac{du}{dt}v+u\frac{dv}{dt}\\[5pt]
\frac{dx}{dt}=\frac{du}{dt}v+u\frac{dv}{dw}\frac{dw}{dt}\\[5pt]
\frac{dx}{dt}=\frac{d\left(A\operatorname{e}^{-3.75\times10^{3}t}\right)}{dt}\left[\cos(2.20\times10^{4}t-\varphi)\right]+\left(A\operatorname{e}^{-3.75\times10^{3}t}\right)\frac{d\left(\cos w\right)}{dw}\frac{d\left(2.20\times10^{4}t-\varphi\right)}{dt}\\[5pt]
\frac{dx}{dt}=-3.75\times10^{3}\operatorname{e}^{-3.75\times10^{3}t}\cos(2.20\times10^{4}t-\varphi)+\left(A\operatorname{e}^{-3.75\times10^{3}t}\right)\left(-\sin w\right)\left(2.20\times10^{4}\right)\\[5pt]
\frac{dx}{dt}=-3.75\times10^{3}A\operatorname{e}^{-3.75\times10^{3}t}\cos(2.20\times10^{4}t-\varphi)-2.20\times10^{4}A\operatorname{e}^{-3.75\times10^{3}t}\sin (2.20\times10^{4}t-\varphi)\\[5pt]
\frac{dx}{dt}=-A\operatorname{e}^{-3.75\times10^{3}t}\left[3.75\times10^{3}\cos(2.20\times10^{4}t-\varphi)+2.20\times10^{4}\sin (2.20\times10^{4}t-\varphi)\right] \tag{VII}
\end{gather}
\]
Substituting the Initial Conditions into expressions (VI) and (VII)
\[
\begin{gather}
q(0)=4\times10^{-4}=A\operatorname{e}^{-\gamma\times 0}\cos(\omega\times 0-\varphi)\\
4\times10^{-4}=A\cos (-\varphi)
\end{gather}
\]
since cosine is an even function we have
\( \cos (-\varphi)=\cos \varphi \)
\[
\begin{gather}
4\times10^{-4}=A\cos \varphi \\[5pt]
A=\frac{4\times10^{-4}}{\cos\varphi} \tag{VIII}
\end{gather}
\]
\[
\begin{gather}
\frac{dq(0)}{dt}=0=-A\operatorname{e}^{-3.75\times10^{3}.0}\left[3.75\times10^{3}\cos(2.20\times10^{4}\times0-\varphi)+2.20\times10^{4}\sin (2.20\times10^{4}\times0-\varphi)\right]\\[5pt]
0=-A.1\left[3.75\times10^{3}\cos (0-\varphi)+2.20\times10^{4}\sin (0-\varphi)\right]\\[5pt]
0=-A\left[3.75\times10^{3}\cos (-\varphi)+2.20\times10^{4}\sin (-\varphi)\right]
\end{gather}
\]
as cosine is an even function and sine is an odd function
\( \sin (-\varphi)=-\sin \varphi \)
\[
\begin{gather}
0=-3.75\times10^{3}A\cos \varphi +2.20\times10^{4}A\sin \varphi \tag{IX}
\end{gather}
\]
and substituting expression (VIII) into expression (IX)
\[
\begin{gather}
0=-{\frac{4\times10^{-4}}{\cos \varphi}}3.75\times10^{3}\cos\varphi +\frac{4\times10^{-4}}{\cos \varphi}2.20\times10^{4}\sin \varphi\\[5pt]
-3.75\times10^{3}.4\times10^{-4}+2.20\times10^{4}.4\times10^{-4}\tan \varphi=0\\[5pt]
2.20\times10^{4}.4\times10^{-4}\tan \varphi=3.75\times10^{3}4\times10^{-4}\\[5pt]
2.20\times10^{4}\tan \varphi=3.75\times10^{3}\\[5pt]
\tan \varphi=\frac{3.75\times10^{3}}{2.20\times10^{4}}\\[5pt]
\varphi=\arctan\left(0.17\right)\\[5pt]
\varphi \simeq 0.17
\end{gather}
\]
Substituting the value of φ into the expression (VIII)
\[
\begin{gather}
A=\frac{4\times10^{-4}}{\cos0.17}\\[5pt]
A=\frac{4\times10^{-4}}{0,99}\\[5pt]
A\simeq 4\times10^{-4}
\end{gather}
\]
substituting the constants A and φ into expression (VI)
\[
\begin{gather}
q=4\times10^{-4}\operatorname{e}^{-3.75\times10^{3}t}\cos\left(2.20\times10^{4}t-0.17\right)
\end{gather}
\]
Equation of charge
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{q(t)=4\times10^{-4}\operatorname{e}^{-3.75\times10^{3}t}\cos\left(2.20\times10^{4}t-0.17\right)}
\end{gather}
\]
b) As Δ<0 this is an underdamped RLC circuit oscillator.
c) Plotting the graph of
\[
\begin{gather}
q(t)=4\times10^{-4}\operatorname{e}^{-3.75\times10^{3}t}\cos\left(2.20\times10^{4}t-0.17\right)
\end{gather}
\]
The function q(t) is the product of two functions,
\( f(t)=4\times10^{-4}\operatorname{e}^{-3.75\times10^{3}t} \)
and
\( g(t)=\cos (2.20\times10^{4}t-0.17) \).
To find the roots we set q(t) = 0, as
q(t) = f(t)g(t) we have f(t) = 0 or
g(t) = 0.
- For g(t) = 0
\[
\begin{gather}
g(t)=\cos (2.20\times10^{4}t-0.17)=0
\end{gather}
\]
the function cosine is equal to zero when its argument
\( 2.20\times10^{4}t-0.17 \)
is equal to
\( \frac{\pi}{2} \),
\( \frac{3\pi}{2} \),
\( \frac{5\pi}{2} \),...,
\( \frac{(2n+1)\pi}{2} \),
with n = 0, 1, 2, 3,...,
\[
\begin{gather}
2.20\times10^{4}t-0.17=\frac{(2n+1)\pi}{2}\\[5pt]
\frac{220\times10^{4}}{100}t-\frac{17}{100}=\frac{(2n+1)\pi}{2}\\[5pt]
\frac{220\times10^{4}}{100}t-\frac{17}{100}=\frac{(2n+1)\pi}{2}\times\frac{50}{50}\\[5pt]
\frac{220\times10^{4}}{100}t-\frac{17}{100}=50\frac{(2n+1)\pi}{100}\\[5pt]
220\times10^{4}t-17=50(2n+1)\pi \\[5pt]
t=\frac{50}{220\times10^{4}}(2n+1)\pi+\frac{17}{220\times10^{4}}\\[5pt]
t=\frac{1}{220\times10^{4}}\left[50(2n+1)\pi+17\right]
\end{gather}
\]
for these values of t, we have the roots of the cosine function, the first four values for
n = 0, 1, 2, and 3 will be t = 0.79×10−4;
2.22×10−4; 3.65×10−4 and 5.08×10−4
(Graph 1).
- For f(t) = 0
\[
\begin{gather}
f(t)=4\times10^{-4}\operatorname{e}^{-3.75\times10^{3}t}=0\\[5pt]
\operatorname{e}^{-3.75\times10^{3}t}=\frac{0}{4\times10^{-4}}\\[5pt]
\operatorname{e}^{-3.75\times10^{3}t}=0
\end{gather}
\]
as there is no t that satisfies this equality, the function f(t) does not intersect the t-axis.
For any real value of t the function will always be positive, f(t) > 0.Differentiation of the expression f(t)
\[
\begin{gather}
\frac{df}{dt}=4\times10^{-4}(-3.75\times10^{3})\operatorname{e}^{-3.75\times10^{3}t}\\[5pt]
\frac{df}{dt}=-0,15\operatorname{e}^{-3.75\times10^{3}t}
\end{gather}
\]
for any real value of t, the derivative will always be negative
\( \left(\frac{df(t)}{dt}<0\right) \)
and the function always decreases. Setting
\( \frac{df(t)}{dt}=0 \)
we find the maximum and minimum points of the function.
\[
\begin{gather}
\frac{df}{dt}=-0,15\operatorname{e}^{-3.75\times10^{3}t}=0\\[5pt]
\operatorname{e}^{-3.75\times10^{3}t}=\frac{0}{-0,15}\\[5pt]
\operatorname{e}^{-3.75\times10^{3}t}=0
\end{gather}
\]
as there is no t that satisfies this equality, there are no maximum or minimum points of the function.The second derivative of the expression f(t)
\[
\begin{gather}
\frac{d^{2}f}{dt^{2}}=-0,15(-3.75\times10^{3})\operatorname{e}^{-3.75\times10^{3}t}\\[5pt]
\frac{d^{2}f}{dt^{2}}=5.63\times10^{2}\operatorname{e}^{-3.75\times10^{3}t}
\end{gather}
\]
For any value of real t, the second derivative will always be positive
\( \left(\frac{d^{2}f(t)}{dt^{2}}>0\right) \)
and the function is concave upward. Setting
\( \frac{d^{2}f(t)}{dt^{2}}=0 \)
we find inflection points in the function.
\[
\begin{gather}
\frac{d^{2}f}{dt^{2}}=5.63\times10^{2}\operatorname{e}^{-3.75\times10^{3}t}=0\\[5pt]
\operatorname{e}^{-3.75\times10^{3}t}=\frac{0}{5.63\times10^{2}}\\[5pt]
\operatorname{e}^{-3.75\times10^{3}t}=0
\end{gather}
\]
as there is no t that satisfies this equality, there are no inflection points in the function.For t = 0 the value f(0) will be
\[
\begin{gather}
f(0)=4\times10^{-4}\operatorname{e}^{-3.75\times10^{3}.0}\\[5pt]
f(0)=4\times10^{-4}\operatorname{e}^{-0}\\[5pt]
f(0)=4\times10^{-4}\times1\\f(0)=4\times10^{-4}
\end{gather}
\]
As the variable t represents time, we do not calculate negative values, t < 0, for t tending
to infinity
\[
\begin{gather}
\lim_{t\rightarrow \infty }f(t)=\lim_{t\rightarrow \infty}4\times10^{-4}\operatorname{e}^{-3.75\times10^{3}t}=\lim_{t\rightarrow \infty}{\frac{4\times10^{-4}}{\operatorname{e}^{3.75\times10^{3}t}}}=\lim_{t\rightarrow \infty }{\frac{4\times10^{-4}}{\operatorname{e}^{\infty }}}=0
\end{gather}
\]
From the above analysis, we plotted the graph of f versus t (Graph 2).
As q(t) = f(t)g(t), the combination of graphs produces a curve that oscillates like the cosine function damped by the exponential (Graph 3).
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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 .