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 Pages:    Go Down Author Topic: RLC Circuits (Second-Order Circuits)  (Read 21776 times) 0 Members and 1 Guest are viewing this topic. Click to toggle author information(expand message area).
Pis
Newbie  Offline

Posts: 2 « Embed this message on: October 07, 2009, 05:39:41 pm » posted from:Kuala Lumpur,Kuala Lumpur,Malaysia I using LTSPice and PSpice to do a simulation eith RLC circuits. This is my first time using LTSpice for rlc simulation. The problem is, I don't know to handle RLC circuits using LTSpice. PSpice, i haven't try yet. This is the circuit. This is my drawing using LTSPice

I already run the simulation, but, I think, my setting is wrong, for RLC circuits, is different, i cannot directly just input the dc value right?....i confused, please help me...anyone...

1) I want to find V(t) and Vr(t) for t>0 for R=10, R=20, and R=30 Ohm...so i want to verify my answer with this simulation using LTspice...

2) Then i want determine v(t) for t=0.8s...

Thank you, this is my first time, using ltspice for rlc circuits... Logged
Fu-Kwun Hwang
Hero Member      Offline

Posts: 3085   « Embed this message Reply #1 on: October 08, 2009, 04:48:12 pm » posted from:Taipei,T'ai-pei,Taiwan Your problem can be solved analytically.
The switch between capacitor and 2Ω opened at t=0, it means that the capacitor is charged to 12V*2/3=8V at t=0.
The problem reduced to RLC circuit with initial conditions:
t=0;
I=0, Vc=8V, VR=0V, VL=2V (VL=10-Vc-VR )
The differential equation need to be solved is $10=L \frac{dI}{dt}+IR+\frac{1}{C}\int I dt$
or $L\frac{d^Q}{dt^2}+I\frac{dQ}{dt}+\frac{Q}{c}=0$

The solution is similar to a spring with constant k, attached with mass m and damping constant b. $m\frac{d^2x}{dt^2}+b\frac{dx}{dt}+kx=0$

The analytical solution can be found in standard textbook.

You are also welcomed to check out RLC circuit simulation (DC Voltage source) to find out the simulated solution. Logged
Pis
Newbie  Offline

Posts: 2 « Embed this message Reply #2 on: October 09, 2009, 11:46:45 pm » posted from:Serdang,Negeri Sembilan,Malaysia Your problem can be solved analytically.
The switch between capacitor and 2Ω opened at t=0, it means that the capacitor is charged to 12V*2/3=8V at t=0.
The problem reduced to RLC circuit with initial conditions:
t=0;
I=0, Vc=8V, VR=0V, VL=2V (VL=10-Vc-VR )
The differential equation need to be solved is $10=L \frac{dI}{dt}+IR+\frac{1}{C}\int I dt$
or $L\frac{d^Q}{dt^2}+I\frac{dQ}{dt}+\frac{Q}{c}=0$

The solution is similar to a spring with constant k, attached with mass m and damping constant b. $m\frac{d^2x}{dt^2}+b\frac{dx}{dt}+kx=0$

The analytical solution can be found in standard textbook.

You are also welcomed to check out RLC circuit simulation (DC Voltage source) to find out the simulated solution.

Oh, ok, thanks, this problem already solved.

I must put in the initial condition for capacitor, 8V.

So my circuit will become like this: I simplified it. IC=Initial condition.

Then, settings: Tick "Skip initial operating..."

Finally, become like this: Then I just run the simulation for the graph. Must put the equation also for V(t) and Vr(t)...

Already solved, thanks!  Logged
Fu-Kwun Hwang
Hero Member      Offline

Posts: 3085   « Embed this message Reply #3 on: October 10, 2009, 09:31:03 am » posted from:Taipei,T\'ai-pei,Taiwan It is great that you can solve it by yourself!  Logged
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"Progress, therfore, is not and accident,¡K" ..."Herbert Spencer(1820-1903, British philosopher)"
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