# series(R) -parallel(LC)- RLC

Discussion in 'Homework Help' started by hiking bengal, Jun 29, 2012.

1. ### hiking bengal Thread Starter New Member

Jun 29, 2012
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I have a similar RLC circuit I need the DE for except the resistor is in series with the inductor and capacitor in parallel. Vout is along the capacitor and not the inductor.

here is how i solved it
I(R)=I(L)+I(C)
I(R)=(V-Vc)/R
I(L)=(1/L)*∫(Vc dt)
I(C)=C*d(Vc)/dt

Substituting all the equations, and differentiating with respect to time I got
0=(d^2(Vc)/dt)+ 1/(RC)*(d(Vc)/dt)+(1/LC)*Vc

This is exactly the same as if all the components were in parallel, am I right or where did I go wrong. Please help.

Last edited: Jun 29, 2012
2. ### WBahn Moderator

Mar 31, 2012
18,087
4,917

How is this possible, if the capacitor and inductor are in parallel and, by definition, therefore have the same voltage?

THANK YOU! THANK YOU!! THANK YOU!!!

You have just made a wonderful first impression on lots of people here by simply posting what you have done in an effort to solve the problem.

THANK YOU!!!!

This is correct except for one thing. What happened to Vin? If Vin = 0, then all three (surviving) components ARE in parallel! But there is no driving signal, so you just have a transient due to initial conditions that decays as a second order system.

If all of the components are in parallel, then wouldn't Vout = Vin?

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3. ### hiking bengal Thread Starter New Member

Jun 29, 2012
5
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I said that the output voltage was along the capacitor as want to denote it as Vout=Vc. Since the inductor and capcitor ar in parallel then yes VL=Vc=Vout, sorry about the poor wording. Vin is a DC source simply noted as V.

When I substituted in the equation I got this form:
I(R)=I(L)+I(C)
=
(1/R)*(V-Vc)=(1/L)*∫(Vc dt)+(C*(d(Vc)/dt)

When differentiating with respect to time, wouldn't the source voltage term (V/R) be considered a constant, so its derivative zero.

This is most likely where I going wrong, but I do know that the resistor is in series with the source voltage, and the inductor & capacitor in parallel.

Last edited: Jun 29, 2012
4. ### WBahn Moderator

Mar 31, 2012
18,087
4,917
Thank you. It's a minor thing, but a big part of engineering is about precision in communication, and not only to convey thoughts to others. If we are sloppy in our descriptions, there is a real good change that, from time to time at least, our thinking about the problem will get sloppy and follow the way we have described things, not the way they actually are. So I tend to be a stickler for clear and correct descriptions. I try to hold myself to the same standard, but like everyone I'm human and don't always succeed. All you, me, or anyone can do is to consistently try to do it right and do it better in the future.

I see your point. I scratched my head for awhile because I came at the problem from a KVL standpoint and the Vin remains with no problems):

(1) v_in = v_L + v_C

(2) v_L = L(di/dt) (i=i_L is the total current)

(3) v_C = ∫(i - i_R)dt / C

(4) i_R = v_C/R

Substituting (2) into (1)

(5) v_in = L(di/dt) + v_C

Substituting (4) into (3)

(3) v_C = ∫(i - v_C/R)dt / C

Differentiating (3), we get

dv_C/dt = (i - v_C/R) / C

Differentiaing again, we get

d²v_C/dt² = (di/dt)/C - (dv_C/dt)/RC

Rearranging to get di/dt

(6) di/dt = C(d²v_C/dt²) - (dv_C/dt)/R

Substituting (6) into (5)

(5) v_in = L(C(d²v_C/dt²) - (dv_C/dt)/R) + v_C

Rearranging into normal DE format

LC(d²v_C/dt²) - (L/R)(dv_C/dt) + v_C = v_in

Yes.

The explanation for the apparent contradiction is that we have to take into account initial conditions and what happens at t=0. By assuming the Vin was a constant and, therefore, has a time derivative of zero, you are assuming that Vin has been applied at t=-∞, remains unchanged through t=0, and continues on unchanged through t=+∞. If that's the case, then your initial conditions for your DE are the final state values and there will be no transient response (it died out at t=-∞). The same should happen to mine if I make the same assumption when it comes time to determine my initial conditions.

But if we assume that v_in is zero up until t=0, then your assumption is incorrect and it DOES have a time derivative (an impulse multiplied by v_in) at t=0 which will yield the form that I have.

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5. ### hiking bengal Thread Starter New Member

Jun 29, 2012
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I see, but I'm confused about equations 2 and 3.
I thought since the resistor is in series with the input voltage wouldn't the total current i=i_R=(v_in)/R ?

KVL around the first inner loop would be:
v_in=v_R+v_L

and the outer-most loop

v_in=v_R+v_C

Maybe I did not explain the circuit correctly, so I maybe diagram of it.

6. ### WBahn Moderator

Mar 31, 2012
18,087
4,917
You are correct. I had the diagram that you referred to up in another browser window and forget as I was writing that post that you were working with a different circuit.

However, note that the homogeneous DE is the same. This is because the homogeneous equation for all of them is the same since, with no driving signal, all three components are in parallel.

The correct DE is

RC(d²v_C/dt²) - (R/L)(dv_C/dt) + v_C = d(v_in)/dt = Vinδ(t)

NOTE: The above is wrong as a result of a transciption error. See how the units don't work out! The correct one (hopefully) is

RC(d²v_C/dt²) + (dv_C/dt) + (R/L)v_C = d(v_in)/dt = Vinδ(t)

Notice how this is quite different, but if the RHS is set to zero (the homogenous DE), then they are the same. What this says is that the steady state solutions are different, as you would expect. Also, while the homogeneous equation governing the transient response is the same, the initial conditions for the two circuits would be different (specifically, the capacitor voltage would be Vin in the original and 0V in the one you are working) and, hence, the transient solutions would be different.

Last edited: Jun 30, 2012
7. ### hiking bengal Thread Starter New Member

Jun 29, 2012
5
0
I cannot come to this same DE, here is my process.

V_in=R*i_R+V_c (1)
where
i_R=i_c+i_L (2)
then i_R becomes
i_R=(C*d(V_c)/dt)+((1/L)*∫(V_c dt)) (3)

plugging in (3) into (1) I get
V_in=R* ((C*d(V_c)/dt)+((1/L)*∫(V_c dt))) + V_c (4)

Equation for then simplifies to
V_in=(R*C)*d(V_c)/dt)+(R/L)*∫(V_c dt) + V_c (5)

differentiating (5) with respect to time I got

RC(d²V_c/dt²) + (dV_c/dt) + (R/L)V_c = d(V_in)/dt (6)

The coefficients of (6) are different.

8. ### WBahn Moderator

Mar 31, 2012
18,087
4,917
You are correct. On my paper I had

d(v_in)/dt = RC(d²v_C/dt²) + (R/L)v_C + (dv_C/dt) = Vinδ(t)

The same as you.

It was at this point that I verified that my units worked out (namely each term having units of V/s - the impluse function, δ(t), is basically 1/0s).

When I transcribed them into the post, I wanted to order them according to the order of the derivative and simply messed up (I have no idea how the minus sign ended up there). Had I checked my units again, I would have seen that they no longer worked out and, therefore, would have known that what I was posting was absolutely wrong.

When I saw your post, the first thing I did was check your units, figuring I would find something. When I didn't, I checked my units and immediately felt my jaw drop. That is very uncharacteristic of me. I panicked a bit because I couldn't believe I had not checked the units in my work and made such a glaring error. So I dug the paper out of the trash and was relieved to see that I hadn't done so, just got sloppy in transcribing and then (unforgiveably) failed to check the units on the finished product. My bad. Learn from it.

In the future, feel free to say, "I didn't get quite what you got. Mine may not be right, but I KNOW your's is wrong because the units don't work out!"

Last edited: Jun 30, 2012
9. ### hiking bengal Thread Starter New Member

Jun 29, 2012
5
0
I feel like such slob because the notion of making sure my units were correct never came to me as a check. I tried solving it different way using mesh current, KVL, and KCL as a form of check, but not checking my units first is knock against me as an engineering student.