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LC circuit

  • Thread starter arpon
  • Start date
  • #1
236
16

Homework Statement


upload_2015-1-2_19-52-17.png
[/B]
[itex]C[/itex] and [itex]2C[/itex] represent the capacitance of the respective capacitors. [itex]L[/itex] and [itex]2L[/itex] represent the inductance of the respective inductors. Let, the charge on the capacitors GM and PM be [itex]q_1[/itex] and [itex]q_2[/itex] which are variable. [itex]t[/itex] represents time. Initially (when the switch k was closed and [itex]t = 0[/itex]) , [itex]q_1 = \frac{2q_0}{3} ; q_2 = \frac{q_0}{3}[/itex]. Would it be,
[itex]i_1 = - \frac{dq_1}{dt} ; i_2 = - \frac{dq_2}{dt} [/itex] ?
Explain your answer.


Homework Equations


That is conceptual question. No equation is needed.

The Attempt at a Solution


The actual homework was different. But I need to clear my concept before solving the actual problem. My confusion is, the MGHJ segment (which has the current [itex]i_1[/itex]) is connected with both the capacitors. So, won't both the capacitor affect the current [itex]i_1[/itex] ?
 

Answers and Replies

  • #2
gneill
Mentor
20,792
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The short circuit between M and J when the switch is closed effectively creates two isolated loops as no potential can develop across the short. In other words, M and J become a single node. No current from either loop can stray into the other loop past that short circuit.
 
  • #3
236
16
The short circuit between M and J when the switch is closed effectively creates two isolated loops as no potential can develop across the short. In other words, M and J become a single node. No current from either loop can stray into the other loop past that short circuit.
So, if I change the circuit into the following , I think [itex]i_1[/itex] and [itex]i_2[/itex] won't change. Am I right?
upload_2015-1-2_21-11-15.png
 
  • #4
gneill
Mentor
20,792
2,770
Right.
 

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