LC oscillations - two capacitors and an inductor

[palaphys]

Homework Statement

picture below

Relevant Equations

1/2Li^2, Q=CV

I tried:
Initial energy= $E_{C1} + E_{C2}= 1/2C(4V_0) +1/2 (3C)(9V_0)= 31CV_0^2/2$
this energy will be completely converted to magnetic energy in the inductor when the current through it is maximum.
Final energy= $1/2 Li_{max}^2$
Equating the two and solving, I get
$i_{max}= \sqrt{31C/L }V_0$
which is not the correct answer
As for the second part, I felt that when the current in the circuit is inductor, the charge stored in the capacitor will be zero. Hence the potential difference (Q=CV) will be zero as well across both the capacitors. This seems to be the incorrect answer as well.

Not sure where I am wrong.

 

[TensorCalculus]

What is the correct answer? I just tried part a) and got the same as you. Although I am also a student so I might have got it wrong as well...

 

[palaphys]

here is the solution given:

 

[vela]

Are you sure both capacitors are discharged completely when the current through the inductor is at a maximum? That’s true in the case where you have a single capacitor in the LC circuit, but the situation is more complicated here because you have two capacitors that start off with unequal charges.

 

[palaphys]

Are you sure both capacitors are discharged completely when the current through the inductor is at a maximum? That’s true in the case where you have a single capacitor in the LC circuit, but the situation is more complicated here because you have two capacitors that start off with unequal charges.

not really. Not sure how the situation would look like in the capacitor when the current is at a maximum in the inductor. I am used to solving such problems with single capacitors, so this one stumped me a bit.

 

[TSny]

Consider the section colored blue. This is a single piece of conductor.

What is the initial total charge on this piece?

Does the total charge on this piece change with time?

Is there ever a time when there is zero total energy in the capacitors?

Edit: I see that @vela already addressed this point while I was constructing my post.

 

[palaphys]

View attachment 362903

Consider the section colored blue. This is a single piece of conductor.

What is the initial total charge on this piece?

Does the total charge on this piece change with time?

Is there ever a time when there is zero total energy in the capacitors?

initial charge is $-C_1V_1+ C_2V_2$ (on this piece)
this piece seems like an isolated part of a conductor. I assume the charge on it is constant.
So if my previous statement is true, then there will never be a time where the energy is 0 in the capacitors.

 

[TSny]

initial charge is $-C_1V_1+ C_2V_2$ (on this piece)
this piece seems like an isolated part of a conductor. I assume the charge on it is constant.
So if my previous statement is true, then there will never be a time where the energy is 0 in the capacitors.

Yes. Good.

 

[palaphys]

Yes. Good.

I tried something else

I considered the two capacitors as though they are in series, and combined them into a single capacitor of capacitance 3C/4, and the potential difference between its terminals is $5V_0$
now, if I apply the standard method,$1/2 (3C/4)(5V_0)^2 = 1/2 Li_{max}^2$
we get $i_{max}= 5V_0/2\sqrt{3C/L}$ which is the correct ans.

How can this result be explained? is this a coincidence or is it a correct way?

 

[palaphys]

After some thinking, I think I have solved this.

Consider the function for energy of the capacitor system at any time t, where the the charge on the 3C capacitor is $7CV_0-q$ and the charge $q$ on the C capacitor, so the energy $E(q)= (7CV_0 -q)^2/6C +q^2/2C$.

When we minimize this function, we get
$q=7CV_0/4$
Hence, $E_{min}= 49CV_0^2/8$
When the energy in the capacitor system is minimum, the energy in the inductor reaches a maximum.

Subtracting this from the total energy of the system we get
$1/2 Li_{max}^2= 31CV_0^2/2 - 49CV_0^2/ 8$
hence we get $i_{max}= 5V_0/2\sqrt{3C/L}$
which is the correct ans.
Looking forward for more approaches (if possible)

 

[DaveE]

Another quick approach is to replace the two capacitors with one equivalent. The voltages just add to 5V_o, the capacitance adds like parallel resistors to (3/4)C. Then it's a simple LC transient response. Then the energy approach works well. But if you work with this basic circuit enough you will have memorized the answer that the maximum current is the maximum capacitor voltage divided by the characteristic impedance, in this case $I_{max}=5V_o\sqrt{\frac{3C}{4L}}$

PS: I'll just add that this is the typical way analog EEs approach these problems. Simplify the circuit before you resort to equations. More simple equations is usually easier and safer than fewer complex equations.

 

[TSny]

After some thinking, I think I have solved this.

Consider the function for energy of the capacitor system at any time t, where the the charge on the 3C capacitor is $7CV_0-q$ and the charge $q$ on the C capacitor, so the energy $E(q)= (7CV_0 -q)^2/6C +q^2/2C$.

When we minimize this function, we get
$q=7CV_0/4$
Hence, $E_{min}= 49CV_0^2/8$
When the energy in the capacitor system is minimum, the energy in the inductor reaches a maximum.

Subtracting this from the total energy of the system we get
$1/2 Li_{max}^2= 31CV_0^2/2 - 49CV_0^2/ 8$
hence we get $i_{max}= 5V_0/2\sqrt{3C/L}$
which is the correct ans.
Looking forward for more approaches (if possible)

I think this is OK.

I’m not sure of your sign conventions for the charges on the capacitors. For example, does your charge $q$ represent the charge on the left plate of the $C$ capacitor, or does it represent the charge on the right plate of the $C$ capacitor?

Likewise, which plate of the $3C$ capacitor has the charge $7CV_0 – q$?

My approach to the problem was to set up and solve the differential equation for the current as a function of time. The maximum current agrees with the given answer.

 

[palaphys]

I’m not sure of your sign conventions for the charges on the capacitors. For example, does your charge $q$ represent the charge on the left plate of the $C$ capacitor, or does it represent the charge on the right plate of the $C$ capacitor?

honestly I don't think this matters, as the charge terms are squared anyway. I agree I have not defined it consistently.

 

[DaveE]

honestly I don't think this matters, as the charge terms are squared anyway. I agree I have not defined it consistently.

Polarity matters when you have to consider if energy (charge) is being increased or decreased with current flow.

 

[TSny]

honestly I don't think this matters, as the charge terms are squared anyway. I agree I have not defined it consistently.

But I think it's interesting to determine the polarity of the 4 plates at the instant when the current first reaches a maximum (assuming the initial polarities are as shown in the diagram).

 

[vela]

To fill in a step that wasn't immediately clear to me in the solution in post #3: when the current $i$ is at a maximum, the voltage across the inductor vanishes. Applying the Kirchhoff's voltage law tells you that the voltages across the capacitors must add to 0 so that $V_{C1} = -V_{C2}$. So the potential differences aren't the same, which sounded wrong to me: they differ in sign, which makes more sense.