Question 3.25: An RLC series circuit is excited by a dc source of 100 V. Fi......
An RLC series circuit is excited by a dc source of 100 V. Find an expression for current and voltage in the elements of the circuit. Take R = 60 Ω, L = 0.2 H and C = 50 µ F. Also draw the initial and final state of the circuits.
The time domain RLC circuit is shown in Fig. 1. Let the switch be closed at t = 0 and i (t) be the current through the circuit.
Here, \mathcal{L}\{i( t )\}= I ( s ) \quad \text { and } \quad \mathcal{L} \{100\}=\frac{100}{ s }
The s-domain RLC circuit is shown in Fig. 2. With reference to Fig. 2, we can write,
I(s)=\frac{\frac{100}{s}}{60+0.2 s+\frac{1}{50 \times 10^{-6} s}} \\ \begin{aligned} =\frac{100}{60 s+0.2 s^2+\frac{1}{50 \times 10^{-6}}}=\frac{100}{0.2\left(s^2+\frac{60}{0.2} s+\frac{1}{0.2 \times 50 \times 10^{-6}}\right)} \\ =\frac{500}{s^2+300 s+10^5} \end{aligned}Let us examine the roots of the denominator polynomial of I(s).
The roots of quadratic s² + 300s + 10^5 = 0 are,
Since the roots are complex conjugate the current i(t) will be damped sinusoid. Let us rearrange the terms of denominator polynomial of I(s) as shown below:
\begin{aligned} I(s)= \frac{500}{\left(s^2+2 \times 150 s+150^2\right)+10^5-150^2}=\frac{500}{(s+150)^2+77500} \\=\frac{500}{(s+150)^2+(\sqrt{77500})^2}=\frac{500}{(s+150)^2+278.4^2} \\ =\frac{500}{278.4} \times \frac{278.4}{(s+150)^2+278.4^2}=1.796 \times \frac{278.4}{(s+150)^2+278.4^2} \end{aligned}Add and subtract 150²
(a+b)^2=a^2+2 a b+b^2Let us take the inverse Laplace transform of I (s).
\mathcal{L} ^{-1}\{ I ( s )\}= \mathcal{L} ^{-1}\left\{1.796 \times \frac{278.4}{( s +150)^2+278.4^2}\right\} \mathcal{L}\left\{e^{-a t} \sin \omega t\right\}=\frac{\omega}{(s+a)^2+\omega^2}
\therefore i( t )=1.796 e ^{-150 t } \sin 278.4 t A …(1)
With reference to Fig. 1, by Ohm’s law, we can write,
\begin{aligned}v_{ R }( t ) & = R \times i( t ) \\ & =60 \times 1.796 e ^{-150 t } \sin 278.4 t \\ & =107.76 e ^{-150 t } \sin 278.4 t V \end{aligned} ….(2)
We know that if i(t) is the current through inductance then voltage across the inductance is given by,
Using equation (1)
d(uv) = u dv − v du
\begin{aligned}v_{ L }( t ) & = L \frac{ d }{ dt } i( t ) \\& =0.2 \frac{ d }{ dt }\left[1.796 e ^{-150 t } \sin 278.4 t \right] \\& =0.2 \times 1.796\left[ e ^{-150 t } \cos 278.4 t \times 278.4+ e ^{-150 t } \times\right (−150) \sin 278.4t ] \\& =100 e ^{-150 t } \cos 278.4 t -53.88 e ^{-150 t } \sin 278.4 t ….. (3) \\ & = e ^{-150 t }[\cos 278.4 t \times 100-\sin 278.4 t \times 53.88]….. (4) \end{aligned}Let us construct a right-angled triangle with 100 and 53.88 as two sides as shown in Fig. 3. With reference to Fig. 3, we can write,
\tan \phi=\frac{53.88}{100}=0.5388 \\ \therefore \phi=\tan ^{-1} 0.5388=28.3^{\circ} \\ Also, \sin \phi=\frac{53.88}{113.592} \Rightarrow 53.88=113.592 \sin \phi \\ \therefore 53.88=113.592 \sin 28.3^{\circ} …..(5) \\ \cos \phi=\frac{100}{113.592} \Rightarrow 100=113.592 \cos \phi \\ \therefore 100=113.592 \cos 28.3^{\circ} …..(6)Using equations (5) and (6), equation (4) can be written as,
v_L(t) = e^{−150t} [\cos 278.4 t × 113.592 \cos 28.3° − \sin 278.4 t × 113.592 \sin 28.3° ]= e ^{−150t} 113.592 [\cos 278.4 t \cos 28.3° − \sin 278.4 t \sin 28.3° ]
= 113.592 e −150t \cos (278.4 t + 28.3° )
= 113.592 e ^{−150t} \sin (278.4 t + 28.3° + 90° )
= 113.592 e^ {−150t} \sin (278.4 t + 118.3° ) V
\cos (A+B)=\cos A \cos B-\sin A \sin B
\cos \theta=\sin \left(\theta+90^{\circ}\right)
With reference to Fig. 1, by KVL, we get,
v_{ R }( t )+v_{ L }( t )+v_{ C }( t )=100\therefore v_C(t) =100-v_R(t)-v_L(t) \\ =100-107.76 e^{-150 t} \sin 278.4 t-100 e^{-150 t} \cos 278.4 t+ 53.88 e ^{-150 t } \sin 278.4 t \\ =100-53.88 e^{-150 t} \sin 278.4 t-100 e^{-150 t} \cos 278.4 t \\ =100-e^{-150 t}[\sin 278.4 t \times 53.88+\cos 278.4 t \times 100] …..(7)
Let us construct a right-angled triangle with 53.88 and 100 as two sides as shown in Fig. 4. With reference to Fig. 4, we get,
\begin{aligned} & \tan \phi=\frac{100}{53.88}=1.856 \\ & \therefore \phi=\tan ^{-1} 1.856=61.7^{\circ} \end{aligned} \\ \text { Also, } \quad \cos \phi=\frac{53.88}{113.592} \Rightarrow 53.88=113.592 \times \cos \phi\\ \therefore \quad 53.88=113.592 \cos 61.7^{\circ} …(8) \\ \sin \phi=\frac{100}{113.592} \Rightarrow 100=113.592 \sin \phi \\ \therefore 100=113.592 \sin 61.7^{\circ} …(9)
Using equations (8) and (9), equation (7) can be written as,
\begin{aligned}v_{ C }( t ) & =100- e ^{-150 t }\left[\sin 278.4 t \times 113.592 \cos 61.7^{\circ}+\cos 278.4 t \times 113.592 \sin 61.7^{\circ}\right] \\ & =100-113.592 e ^{-150 t }\left[\sin 278.4 t \cos 61.7^{\circ}+\cos 278.4 t \sin 61.7^{\circ}\right] \\ & =100-113.592 e ^{-150 t } \sin \left(278.4 t +617^{\circ}\right) \end{aligned} \sin (A+B)=\sin A \cos B+\cos A \sin BIn summary
\begin{aligned} i( t ) & =1.796 e ^{-150 t } \sin 278.4 t A \\ v_{ R }( t ) & =107.76 e ^{-150 t } \sin 278.4 t \ V \\ v_{ L }( t ) & =113.592 e ^{-150 t } \sin \left(278.4 t +118.3^{\circ}\right) \ V \\ v_{ C }( t ) & =100-113.592 e ^{-150 t } \sin \left(278.4 t +61.7^{\circ}\right) \ V \end{aligned}Initial state circuit
At t =0^{+}, i( t )=i\left(0^{+}\right)=1.796 \times e ^0 \times \sin 0=0\\ At t =0^{+}, v_{ R }( t )=v_{ R }\left(0^{+}\right)=107.76 \times e ^0 \times \sin 0=0\\ At \ t =0^{+}, v_{ L }( t )=v_{ L }\left(0^{+}\right)=113.592 \times e ^0 \times \sin 118.3^{\circ}=100.0152 V \approx 100 V \\ At \ t =0^{+}, v_{ C }( t )=v_{ C }\left(0^{+}\right)=100-113.592 \times e ^0 \times \sin 61.7^{\circ}\\ =100-100.0152 \approx 0
From the above analysis we can say that at t = 0^+ , the inductance behaves as an open circuit and the capacitance behaves as a short circuit. Final state circuit.
Final state circuit
At t =\infty, i( t )=i(\infty)=1.796 \times e ^{-\infty} \times \sin (\infty)=0\\ At t =\infty, v_{ R }( t )=v_{ R }(\infty)=107.76 \times e ^{-\infty} \times \sin (\infty)=0\\ At t =\infty, v_{ L }( t )=v_{ L }(\infty)=113.592 \times e ^{-\infty} \times \sin (\infty)=0\\ At t =\infty, v_{ C }( t )=v_{ C }(\infty)=100-113.592 \times e ^{-\infty} \times \sin (\infty)=100-0=100 \ V
From the above analysis we can say that at t = ∞, i.e., at steady state, the inductance behaves as a short circuit and the capacitance behaves as an open circuit.
