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Information about this web site => Request for physics Simulations => Topic started by: Pis on October 07, 2009, 05:39:41 pm



Title: RLC Circuits (Second-Order Circuits)
Post by: Pis on October 07, 2009, 05:39:41 pm
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.

(http://www.phy.ntnu.edu.tw/ntnujava/img/4846_1.jpg)
This is the circuit.

(http://www.phy.ntnu.edu.tw/ntnujava/img/4846_2.jpg)
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...



Title: Re: RLC Circuits (Second-Order Circuits)
Post by: Fu-Kwun Hwang on October 08, 2009, 04:48:12 pm
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) (http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=30.0) to find out the simulated solution.


Title: Re: RLC Circuits (Second-Order Circuits)
Post by: Pis on October 09, 2009, 11:46:45 pm
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) (http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=30.0) 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:

(http://img2.pict.com/e7/67/54/1738217/0/1.jpg)
I simplified it. IC=Initial condition.

Then, settings:

(http://img2.pict.com/d9/91/de/1738218/0/2.jpg)
Tick "Skip initial operating..."

Finally, become like this:

(http://img2.pict.com/65/49/43/1738219/0/3.jpg)

Then I just run the simulation for the graph. Must put the equation also for V(t) and Vr(t)...

Already solved, thanks!  ;D




Title: Re: RLC Circuits (Second-Order Circuits)
Post by: Fu-Kwun Hwang on October 10, 2009, 09:31:03 am
It is great that you can solve it by yourself!  :D