Help finding the differential equations from an L + R in parallel with C

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
4 replies · 2K views
AMSA
Messages
3
Reaction score
0
Hi there guys.

Homework Statement



I am trying to solve a circuit that has a constant voltage source (V) in series with an inductor (L), then that inductor is connected to a resistor (R) in parallel with a capacitor (C).

Homework Equations



-------



The Attempt at a Solution



I am trying to solve that for the current iL.

I came up with this equations:

iL = iC + iR;

V = vL + vC;

L diL/dt + 1/C ∫ iC dt = V

1/C ∫ iC dt = R iR

I want to solve those equations in order to get the current in the branch where the inductor is, so iL.

I've tried many combinations and I don't get any coherent result.

I tried to solve that, through iL = iC + iR and what I got was the differential equation in terms of vC, and I want it in terms of iL!

Regards and thanks in advance.
 
Physics news on Phys.org
AMSA said:
Hi there guys.

Homework Statement



I am trying to solve a circuit that has a constant voltage source (V) in series with an inductor (L), then that inductor is connected to a resistor (R) in parallel with a capacitor (C).

Homework Equations



-------



The Attempt at a Solution



I am trying to solve that for the current iL.

I came up with this equations:

iL = iC + iR;

V = vL + vC;

L diL/dt + 1/C ∫ iC dt = V

1/C ∫ iC dt = R iR

I want to solve those equations in order to get the current in the branch where the inductor is, so iL.

I've tried many combinations and I don't get any coherent result.

I tried to solve that, through iL = iC + iR and what I got was the differential equation in terms of vC, and I want it in terms of iL!

Regards and thanks in advance.

A couple of points:

You seem to be talking about using a constant DC voltage source, in which case the problem simplifies considerably. An inductor is a DC short, and a capacitor is a DC open, so you are just left with V=IR as a solution.

If instead you mean a constant amplitude AC voltage source with amplitude V, then you can write the differential equations and solve them.

But what you have written for equations doesn't make sense to me. You should write a KCL equation for the node between the inductor and the parallel RC combination.

Can you re-try using the KCL approach?
 
Hi,

Thanks for your reply.

I'll try to draw the circuit here:

|--- \------L------------A
|........|
|........|
|...... ---------
VDC.....|...|
|......C...R
|.......|...|
|.......---------
|........|
|........|
|-------------------------

The KCL to the node A is: (http://www.physics.uoguelph.ca/tutorials/ohm/Q.ohm.KCL.html)

iL = iC + iR

Before the switch is off. Then at t=0 we close the switch.

Now I want to write down the equations that describes the behavior of the circuit. I came up wit those in the preview post.

vL + vC = V

L diL/dt + 1/C ∫ iC dt = V

vC = vR

1/C ∫ iC dt = R iR

I want to solve that in order to get the diferential equation for d^2 iL /dt + diL/ dt ...
 
AMSA said:
Hi,

Thanks for your reply.

I'll try to draw the circuit here:
If you wish to try ASCII graphics, enclose your whole composition between [code] and [/code] instructions and that will force the use of a monospaced font for that block.

But I think you have succeeded in getting the idea across on this occasion. :smile:
 
berkeman said:
A couple of points:

You seem to be talking about using a constant DC voltage source, in which case the problem simplifies considerably. An inductor is a DC short, and a capacitor is a DC open, so you are just left with V=IR as a solution.
In the steady state, yes. But poster might be seeking the transient response to switch closure.