Question 5.15: Series RLC Circuit The DC voltage source of the circuit show...
Series RLC Circuit
The DC voltage source of the circuit shown in Figure 5.38 is connected to the series RLC circuit by closing the switch at t = 0. The initial conditions are: i_L (0) = 0 and v_C (0) = 0. Find the voltage and the current across the capacitor if:
a. R = 60 Ω
b. R = 40 Ω
c. R = 30 Ω
To find the circuit’s conditions, calculate parameters α, \omega_0, and ζ :
a. In this case:
\begin{aligned} \alpha&=\frac{R}{2 L}=\frac{60}{2 \times 4 \times 10^{-3}}=7500 \\\omega_0&=\frac{1}{\sqrt{L C}}=\frac{1}{\sqrt{4 \times 10^{-3} \times 10 \times 10^{-6}}}=5000 \\\zeta&=\frac{\alpha}{\omega_0}=\frac{7500}{5000}=1.5\end{aligned}Because ζ > 1, the circuit is overdamped and the roots of the characteristic function can be calculated using Equations (5.54) and (5.55) as follows:
\begin{aligned}&s_1=-\alpha+\sqrt{\alpha^2-\omega_0^2}=-\alpha\left(1-\sqrt{1-\frac{\omega_0^2}{\alpha^2}}\right)=-\alpha\left(1-\sqrt{1-\frac{1}{\zeta^2}}\right) (5.54) \\&s_2=-\alpha-\sqrt{\alpha^2-\omega_0^2}=-\alpha\left(1+\sqrt{1-\frac{\omega_0^2}{\alpha^2}}\right)=-\alpha\left(1+\sqrt{1-\frac{1}{\zeta^2}}\right) (5.55)\end{aligned} \begin{aligned}&s_1=-\alpha+\sqrt{\alpha^2-\omega_0^2}=-1.91 \times 10^3 \\&s_2=-\alpha-\sqrt{\alpha^2-\omega_0^2}=-1.31 \times 10^4\end{aligned}Substituting these values in Equation (5.53):
V_{C}(s)=\frac{V_{s} \times \omega_0^2}{s\left(s^2+2 \alpha s+\omega_0^2\right)} (5.53)
\begin{aligned}&V_{C}(s)=\frac{20 \times 25 \times 10^6}{s\left(s^2+15 \times 10^3 s+25 \times 10^6\right)}=\frac{5 \times 10^8}{s\left(s+1.91 \times 10^3\right)(s+1.31 \times 10^4)} \\&V_{C}(s)=\frac{20}{s}-\frac{23.42}{s+1.91 \times 10^3}+\frac{3.42}{s+1.31 \times 10^4} \end{aligned}The corresponding inverse Laplace transform is:
v_{C}(t)=20-23.42 e^{-1.91 \times 10^3 t}+3.42 e^{-1.31 \times 10^4 t}The current flowing through the circuit is given by:
\begin{aligned}i_{C}(t)=C \frac{dv_{C}(t)}{dt}=& 0-10 \times 10^{-6} \times 23.42 \times-1.91 \times 10^3 e^{-1.91 \times 10^3 t}+10 \times 10^{-6} \\& \times 3.42 \times-1.31 \times 10^4 e^{-1.31 \times 10^4 t} \\i_{C}(t)=&0.447 e^{-1.91 \times 10^3 t}-0.447 e^{-1.31 \times 10^4 t}\end{aligned}Plots of capacitor voltage and current are shown in Figure 5.39.
b. When the resistor changes to 40 Ω:
\begin{gathered}\alpha=\frac{R}{2 L}=\frac{40}{2 \times 4 \times 10^{-3}}=5000 \\\omega_0=\frac{1}{\sqrt{L C}}=\frac{1}{\sqrt{4 \times 10^{-3} \times 10 \times 10^{-6}}}=5000\\ \zeta=\frac{\alpha}{\omega_0}=\frac{5000}{5000}=1 \end{gathered}Because ζ = 1, the circuit is critically damped and the roots of the characteristic function are equal to -α. Therefore:
s_1=s_2=-5 \times 10^3Substituting these values in Equation (5.58):
V_{C}(s)=\frac{V_{s} \times \omega_0^2}{s(s+\alpha)^2}=\frac{V_s}{s}+\frac{k_1}{(s+\alpha)}+\frac{k_2}{(s+\alpha)^2} (5.58)
V_{C}(s)=\frac{20 \times 25 \times 10^6}{s\left(s+5 \times 10^3\right)^2}=\frac{V_{s}}{s}+\frac{k_1}{\left(s+5 \times 10^3\right)}+\frac{k_2}{\left(s+5 \times 10^3\right)^2}The residues k_1 and k_2 are computed as follows (see Appendix B):
\begin{aligned}&k_2=\left.\left(s+5 \times 10^3\right)^2 Y(s)\right|_{s=-5 \times 10^3}=\frac{20 \times 25 \times 10^6}{-5 \times 10^3}=\frac{5 \times 10^8}{-5 \times 10^3}=-10^5 \\&k_1=\left.\frac{d}{ds}\left[\left(s+5 \times 10^3\right)^2 Y(s)\right]\right|_{s=-5 \times 10^3}=-\frac{20 \times 25 \times 10^6}{\left(-5 \times 10^3\right)^2}=-20\end{aligned}As a result:
V_{C}(s)=\frac{20}{s}-\frac{20}{\left(s+5 \times 10^3\right)}-\frac{10^5}{\left(s+5 \times 10^3\right)^2}The inverse Laplace transform is:
\begin{aligned} &v_C(t)=20-20 e^{-5 \times 10^3 t}-10^5te^{-5 \times 10^3 t} \\i(t)=C \frac{dv_{C}(t)}{dt} &=0-10 \times 10^{-6} \times 20 \times-5 \times 10^3 e^{-5 \times 10^3 t}-10 \times 10^{-6} \times 10^5 e^{-5 \times 10^3 t} \\&-10 \times 10^{-6} \times 10^5 \times-5 \times 10^3 t e^{-5 \times 10^3 t} \\i(t)=&e^{-5 \times 10^3 t}-e^{-5 \times 10^3 t}+5 \times 10^3 t e^{-5 \times 10^3 t}=5 \times 10^3 t e^{-5 \times 10^3 t}\end{aligned}Plots of capacitor voltage and current are shown in Figure 5.40.
c. Finally, for R = 30 Ω:
\begin{aligned}\alpha &=\frac{R}{2 L}=\frac{30}{2 \times 4 \times 10^{-3}}=3750 \\\omega_0 &=\frac{1}{\sqrt{L C}}=\frac{1}{\sqrt{4 \times 10^{-3} \times 10 \times 10^{-6}}}=5000 \\\zeta &=\frac{\alpha}{\omega_0}=\frac{3750}{5000}=0.75\end{aligned}Because ζ < 1, the circuit is underdamped. The natural frequency is calculated using Equation (5.62), which is:
\omega_{n}=\sqrt{\omega_0^2-\alpha^2}=3307The roots of the characteristic function can be calculated using Equations (5.60) and (5.61).
\begin{aligned}&s_1=-\alpha+j \sqrt{\omega_0^2-\alpha^2}=-\alpha+j \omega_{n} (5.60) \\&s_2=-\alpha-j \sqrt{\omega_0^2-\alpha^2}=-\alpha-j \omega_{n} (5.61)\end{aligned} \begin{aligned}&s_1=-\alpha+j \omega_{n}=-3750+j 3307 \\&s_2=-\alpha-j \omega_{n}=-3750-j 3307\end{aligned}This is an underdamped case, using Equation (5.63) results in:
\begin{aligned}V_{C}(s) &=\frac{\omega_0^2 V_{s}}{s\left(s+\alpha-j \omega_{n}\right)\left(s+\alpha+j \omega_{n}\right)} \\&=\frac{V_{s}}{s}+\frac{\omega_0^2 V_{s} /\left(j 2 \omega_{n}\left(-\alpha+j \omega_{n}\right)\right)}{s+\alpha-j \omega_{n}}-\frac{\omega_0^2 V_{s} /\left(j 2 \omega_{n}\left(-\alpha-j \omega_{n}\right)\right)}{s+\alpha+j \omega_{n}} \\&=\frac{V_{s}}{s}-\frac{\omega_0^2 V_{s}(s+2 \alpha) /\left(\alpha^2+\omega_{n}^2\right)}{s^2+2 \alpha s+\alpha^2+\omega_{n}^2}, \omega_0^2=\alpha^2+\omega_n^2 \\&=\frac{V_{s}}{s}-\frac{V_{s}(s+2 \alpha)}{(s+\alpha)^2+\omega_{n}^2} \\&=V_{s}\left(\frac{1}{s}-\frac{s+\alpha}{(s+\alpha)^2+\omega_{n}^2}-\frac{\alpha}{(s+\alpha)^2+\omega_{n}^2}\right) \\&=V_{s}\left(\frac{1}{s}-\frac{s+\alpha}{(s+\alpha)^2+\omega_{n}^2}-\frac{\alpha}{\omega_{n}} \frac{\omega_{n}}{(s+\alpha)^2+\omega_{n}^2}\right) (5.63)\end{aligned} V_{C}(s)=\frac{5 \times 10^8}{s(s+3750+j 3307)(s+3750-j 3307)}and
V_{C}(s)=\frac{20}{s}-\frac{20(s+3750)}{(s+3750)^2+3307^2}-\frac{20 \times 3750}{3307} \frac{3307}{(s+3750)^2+3307^2}The corresponding inverse Laplace transform is:
v_{C}(t)=20-20 e^{-3750 t} \cos (3307 t)-22.68 e^{-3750 t} \sin (3307 t)The corresponding current that flows through the circuit is given by:
\begin{aligned}i_{C}(t)=& 0.75 e^{-3750 t} \cos (3307 t)+0.66 e^{-3750 t} \sin (3307 t)+0.85 e^{-3750 t} \sin (3307 t) \\&-0.75 e^{-3750 t} \cos (3307 t) \\i_{C}(t)=& 1.51 e^{-3750 t} \sin (3307 t)\end{aligned}Plots of capacitor voltage and current are shown in Figure 5.41.
