Series RLC Circuit with DC Source

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Circuit Description:

We have a series RLC circuit connected to a DC voltage source V0​.

Initial conditions:

  • V0​: DC step voltage

  • R: Resistance (Ω)

  • L: Inductance (H)

  • C: Capacitance (F)

s-Domain Circuit Representation

Apply KVL in the t-domain:

v0=vR+vL+vC=Ri(t)+Ldi(t)dt+1Ci(t)dtv_{0}=v_{R}+v_{L}+v_{C}=Ri(t)+L\frac{d i(t)}{dt}+\frac{1}{C}\int_{}^{}i(t)dt\\

Apply KVL in the s-domain:

V0s=RI(s)+LsI(s)+1sCI(s)\frac{V_{0}}{s}=RI(s)+LsI(s)+\frac{1}{sC}I(s)

Factor out 𝐼(𝑠):

V0s=(R+Ls+1sC)I(s)\frac{V_{0}}{s}=\left( R+Ls+\frac{1}{sC} \right)I(s)

Solve for current 𝐼(𝑠):

I(s)=V0s(R+Ls+1sC)=V0Ls2+Rs+1CI(s)=\frac{V_{0}}{s(R+Ls+\frac{1}{sC})}=\frac{V_{0}}{Ls^{2}+Rs+\frac{1}{C}}

4. Solving for Current in the s-Domain

Let’s define:

  • a=L

  • b=R

  • c=1/C

Then:

I(s)=V0as2+bs+cI(s)=\frac{V_{0}}{as^{2}+bs+c}

Case Example:

Let’s assume:

  • V0=10V

  • R=2 Ω

  • L=1 H

  • C=0.25 F

Then:

I(s)=10s2+2s+4I(s)=\frac{10}{s^{2}+2s+4}

This is a standard second-order system.

5. Inverse Laplace: Time-Domain Current

Use the known inverse Laplace pairs:

L1{ω(s+α)2+ω2}=eα.tsin(ω.t)L^{-1}\left\{ \frac{\omega}{\left( s+\alpha \right)^{2}+\omega^{2}} \right\}=e^{-\alpha.t}sin\left( \omega.t \right)

We complete the square:

s2+2s+4=(s+1)2+3ω=3    and    α=1s^{2}+2s+4=(s+1)^{2}+3\Rightarrow \omega=\sqrt{3}\;\;and\;\;\alpha=1

So:

I(s)=10(s+1)2+3i(t)=10etsin(3t)3I(s)=\frac{10}{(s+1)^{2}+3}\Rightarrow i(t)=10e^{-t}\frac{sin\left( \sqrt{3}t \right)}{\sqrt{3}}

6. Python Simulation Using SymPy

Here’s the full Python code to simulate:

Output

Keywords

RLC circuit, series RLC, Laplace transform, s-domain analysis, DC source, transient response, differential equations, inverse Laplace, time domain, second-order system, underdamped circuit, damping factor, electrical engineering, current response, step input, Python simulation, SymPy, circuit analysis, signal processing, engineering math, nerd cafe

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