How to solve this driven RLC circuit?

[zenterix]

Homework Statement

I am stuck on the following problem.

Relevant Equations

Determine an expression for $V_C$, the voltage across the capacitor.

Actually, I would like to solve the problem for every variable.

Here is the circuit as it appears in MIT OCW Vibrations and Waves Problem Solving course

Here is my own picture

I wrote equations for the three loops.

Outer Loop Through $C$
$$-V_0\sin{\omega t}+IR+\frac{Q}{C}=0\tag{1}$$

Outer Look Through $L$
$$-V_0\sin{\omega t}+IR=-L\dot{I}_2\tag{2}$$

Small Inner Loop

$$\frac{Q}{C}=L\dot{I}_2\tag{3}$$

Only two of these equations are independent. We can obtain the third equation given any two of the equations.

We also have

$$I=I_1+I_2\tag{4}$$

$$I_1=\dot{Q}\tag{5}$$

I am really not sure what to do at this point. I've tried to manipulate the equations above to get a differential equation just in one of the variables but I cannot. For example

$$-V_0\sin{\omega t}+(\dot{Q}+I_2)R+L(\dot{I}-\ddot{Q})=0\tag{6}$$

$$\ddot{Q}-\frac{R}{L}\dot{Q}-\frac{R}{L}I_2=-\frac{V_0}{L}\sin{\omega t}+\dot{I}\tag{7}$$

I would very much appreciate some help here.

 

[berkeman]

I'm not sure introducing the extra quantity $Q$ will be of any help. Can you re-write your equations using only currents and voltages in the differential equations?

 

[zenterix]

$Q$ only appears in the expression $V_C=\frac{Q}{I}$.

$$-V_0\sin{\omega t}+IR+V_C=0$$

$$V_C=L\dot{I}_2$$

 

[scottdave]

What math tools are you familiar with or able to use? How about Laplace transforms? What can you say about the relationship of voltage across the capacitor and inductor?

 

[zenterix]

What math tools are you familiar with or able to use? How about Laplace transforms? What can you say about the relationship of voltage across the capacitor and inductor?

I know how to apply Laplace transforms to solve differential equations.

I've completed a full course on differential equations.

The voltage across the capacitor is the same as the voltage across the inductor since they are in parallel.

 

[DaveE]

Sorry, I haven't really reviewed your DEs. But here's another approach.

Since the driving function is a sine wave, and this is a linear network, you know the answer will be a sinusoid at the same frequency with some different magnitude and phase. So you could just use the idea of complex valued impedances (phasors) and solve for just the magnitude and phase. Then this problem is a fairly simple voltage divider.

This is similar to the Laplace transform approach, where you set s=jω for the steady state solutions. You really don't have to (or want to) do the inverse transform.

 

[zenterix]

In terms of physics, suppose there is an open switch in the circuit that is closed at time $t=0$.

Then $I(0)=0$.

Suppose $Q=0$.

Suppose we have a DC voltage $\mathcal{E}$ instead of AC.

If there were no inductor but instead just an ideal wire, no charge would accumulate on the capacitor and the circuit would reduce to a very simple circuit containing a DC voltage and a resistor. Current would flow instantaneously at $I=\frac{\mathcal{E}}{R}$.

With the presence of the inductor, however, when the current starts to increase there is an induced back emf generated by the inductor. It is as if there were a difference of potential across the inductor.

The changing current generates a changing magnetic flux through the inductor. This generates a non-electrostatic electric field. This electric field is not path independent and thus the concept of potential difference is not well-defined. This is why I said as if above.

Indeed, the inductor behaves like a battery trying to push current in the opposite direction to the actual current.

At $t=0$ this back emf is at its largest value. For each increase in current, the back emf (and the derivative of current) decreases a little bit. At some point this back emf and the derivative go to zero and the inductor becomes a short circuit.

Why does current flow into the capacitor?

I'm not sure how to explain it physically in details.

 

[zenterix]

So you could just use the idea of complex valued impedances (phasors) and solve for just the magnitude and phase.

I will look into this.

But for right now, I'm not even interested so much in the method of solving the differential equation. I am trying to just find the differential equation to solve.

 

[berkeman]

Why does current flow into the capacitor?

The impedance of the inductor creates a voltage across it (that decreases over time after t=0), so that voltage generates a current into the capacitor.

 

[berkeman]

I am trying to just find the differential equation to solve.

I prefer KCL node equations over KVL loop equations.

I would ground the lower left node between the AC signal generator and the resistor to provide a reference node, and then I would number the node between the R and the LC as "1" and the right node as "2".

Sum currents leaving Node 1:
$$ \frac{V_1(wt)}{R} = C \frac{d(V_1(wt)-V_2(wt))}{dt} + \frac{1}{L} \int(V_1(wt)-V_2(wt))dt $$

Where $V_2(wt) = V_0 sin(wt)$
So you need to solve for $V_1(wt)$

Does that help at all?

EDIT -- fixed typos in LaTeX.

 

[DaveE]

Your DEs are correct, so one take on this is just a Math problem. You have a bunch of choices for the variables to solve for (charge, currents, voltages, etc.) but you do need a complete independent set. There's no point choosing capacitor voltage and capacitor charge for example, they are trivially related and don't span the solution space. Normally, a simple rule to create a complete description of the state space is to choose each capacitor voltage and each inductor current. These are the quantities that "remember" the history of the system (state) and all other quantities can be calculated from them. But, you could choose others, like capacitor current in place of voltage, or even more complex combinations.

So here is the work you've already done reformatted in a canonical state space formulation:

Note first that I'll use $v_c(t)$ and $i_l(t)$, the cap voltage and inductor current as my state variables. I'll also name the driving voltage $v_s$, just to save extra typing.

Then some equivalence relations to map to your equations: $I_1 = C \dot{v_c}$ and $I = i_l + C \dot{v_c}$

Then the outer KVL loop equation is $R(C \dot{v_c} + i_l) + v_c - v_s =0 ~~ \Rightarrow ~~ \dot {v_c} = \frac{-1}{RC} v_c - \frac{1}{C} i_l + v_s$

The small loop is simply $\dot {i_l} = \frac{1}{L} v_c$

The canonical form is then

$\frac{d}{dt} \begin{bmatrix} v_c(t) \\ i_l(t) \end{bmatrix} = \begin{bmatrix} \frac{-1}{RC} & \frac{-1}{C} \\ \frac{1}{L} & 0 \end{bmatrix} \begin{bmatrix} v_c(t) \\ i_l(t) \end{bmatrix} + \begin{bmatrix} 1 \\ 0 \end{bmatrix} v_s(t)$

PS: you should check my work, since I didn't.

 

[DaveE]

Also, once you've entered into the "lumped element" circuit domain of EE, you would be best off to ignore many EM concepts, like induced voltages not being conservative and such. Lumped element schematics are a gross simplification of the real world, which is why they are so valuable. Just use the component definitions, like $v_l(t) = L \dot{i_l(t)}$, as the model constraint. Then ask yourself is this a good model or do I need Maxwell's Equations.

 

[zenterix]

I'm going to read a chapter from my circuits and electronics book and I'll post back here in a while.

 

[vela]

I wrote equations for the three loops.

Outer Loop Through $C$
$$-V_0\sin{\omega t}+IR+\frac{Q}{C}=0\tag{1}$$

Outer Look Through $L$
$$-V_0\sin{\omega t}+IR=-L\dot{I}_2\tag{2}$$

Small Inner Loop

$$\frac{Q}{C}=L\dot{I}_2\tag{3}$$

Only two of these equations are independent. We can obtain the third equation given any two of the equations.

We also have

$$I=I_1+I_2\tag{4}$$

$$I_1=\dot{Q}\tag{5}$$

In terms of$I_1$ and $I_2$, you write (2) as
$$(I_1+I_2)R + L\dot{I}_2 = V_0 \sin\omega t.$$ Differentiating (3) with respect to time gives you
$$\frac{I_1}{C} = L\ddot{I}_2.$$ Combining those two equations, you get
$$(LC \ddot{I}_2+I_2)R + L\dot{I}_2 = V_0 \sin\omega t,$$ which is a second-order differential equation in just $I_2$.