Timeline for Two coupled $\text{LC}$ circuits Kirchhoff's law?
Current License: CC BY-SA 4.0
21 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 7 at 8:33 | history | edited | Qmechanic♦ |
edited tags
|
|
| Oct 7 at 8:32 | comment | added | Qmechanic♦ | Hi FriendlyLagrangian. If you haven't already done so, please take a minute to read the definition of when to use the homework-and-exercises tag, and the Phys.SE policy for homework-like problems. | |
| Oct 7 at 8:21 | history | edited | FriendlyLagrangian | CC BY-SA 4.0 |
added 19 characters in body; edited tags
|
| Oct 7 at 8:20 | vote | accept | FriendlyLagrangian | ||
| Oct 7 at 8:13 | answer | added | FriendlyLagrangian | timeline score: 0 | |
| Oct 2 at 16:02 | history | reopened |
FriendlyLagrangian gandalf61 Thomas Fritsch |
||
| Oct 2 at 13:29 | comment | added | Bob D | @FriendlyLagrangian What do you mean I keep closing this question? This is the first time I've seen it and voted to close. Anyway, I see the community is reviewing to reopen, so wait for their decision (I am not part of the review, so don't worry about me) | |
| Oct 2 at 12:50 | review | Reopen votes | |||
| Oct 2 at 16:02 | |||||
| Oct 2 at 12:48 | comment | added | FriendlyLagrangian | @BobD Why do you keep closing this question?! It's not on the site and it is extremely relevant to many fields like superconducting circuits and quantum computing. The moderation of this site is becoming hyper restrictive and will impact it negatively. This is not the $n$-th question about a parabolic trajectory from a lazy high schooler, this is directly related to modern papers. Surely with enough imagination, every question is homework like or check my work type of question no?? | |
| Oct 2 at 12:45 | comment | added | Farcher | With knowledge of the sign convention which you are using it is. not possible to answer your question with certainty. | |
| Oct 2 at 12:27 | history | closed |
Bob D Jon Custer Vincent Thacker |
Not suitable for this site | |
| Oct 2 at 11:42 | review | Close votes | |||
| Oct 2 at 12:27 | |||||
| Oct 2 at 11:26 | history | edited | Bob D |
edited tags
|
|
| Oct 2 at 11:07 | comment | added | FriendlyLagrangian | @hyportnex by off I mean that, if my last equation above was right too, then it would imply that $$L_1( \frac{\ddot{Q_1}}{C_1} - L_2 \dddot{I}_4 ) = 2(\ddot{Q_1} - \ddot{Q_2}),$$ which I don't see how. Also, there has to be an easier way of doing this haha | |
| S Oct 2 at 11:00 | history | suggested | Bml | CC BY-SA 4.0 |
Fixed typos in title and body.
|
| Oct 2 at 10:59 | comment | added | FriendlyLagrangian | @hyportnex I see, so indeed $Q_g \neq Q_1 - Q_2$ (which dispels the worry of their sign). Then, if up to that point my reasoning is correct, my only way of removing $Q_g$ is by finding an equation for it and substituting. That would be, again by Kirchoff: $$L_2 \dot{I}_4 + \frac{Q_g}{C_g} - \frac{Q_1}{C_1} = 0.$$ So $$Q_g = C_g(\frac{Q_1}{C_1} -L_2 \dot{I}_4).$$ We substitute this into $$\frac{Q_{1}}{C_1} + L_1 (\ddot{Q}_g + \ddot{Q}_1) = 0$$ and obtain $$\frac{Q_1}{C_1} + L_1(C_g(\frac{\ddot{Q}_1}{C_1} -L_2 \dddot{I}_4) + \ddot{Q}_1)$$ and similarly for $Q_2$. But this seems off too? | |
| Oct 2 at 10:28 | comment | added | Vincent Thacker | Previous question by OP: physics.stackexchange.com/q/829671/174766 | |
| Oct 2 at 10:21 | review | Suggested edits | |||
| S Oct 2 at 11:00 | |||||
| Oct 2 at 10:18 | comment | added | hyportnex | hint: Charge conservation is not that the sum of charges on the three capacitors is constant but rather that the total charge in the whole circuit is a constant and is expressed by the two node conditions: $I_1=I_2+I_3$ and $I_2=I_4+I_5$. This is because the charges move in and out of the inductors and capacitors. How would you write down charge conservation for a pair a simple parallel LC resonator? | |
| Oct 2 at 10:15 | history | edited | Qmechanic♦ | CC BY-SA 4.0 |
edited tags; edited title
|
| Oct 2 at 10:11 | history | asked | FriendlyLagrangian | CC BY-SA 4.0 |