How Is Charge Calculated in a Two Loop RC Circuit After the Switch Opens?

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Homework Help Overview

The discussion revolves around a two-loop RC circuit involving resistors, a capacitor, a battery, and a switch. Participants are exploring the charge on the capacitor after the switch is opened, specifically at a time of 790 μs. The circuit parameters include resistances and capacitance values, with a focus on how these affect the charge dynamics.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the charge on the capacitor at various times, questioning the maximum charge and the time constant. There is exploration of the behavior of the capacitor as it discharges after the switch is opened, with some participants attempting to clarify the relevant equations and the role of resistance in the circuit.

Discussion Status

There is an ongoing exploration of the equations governing the charge on the capacitor, with some participants providing insights into the nature of exponential decay. Multiple interpretations of the circuit's behavior are being examined, particularly regarding the initial conditions and the effective resistance seen by the capacitor.

Contextual Notes

Participants note the absence of a textbook for reference and the reliance on class notes, which may limit their understanding of the concepts involved. There is also a discussion about which resistors are relevant when the switch is open, indicating potential confusion about the circuit's configuration.

  • #31
gneill said:
Is there a complete path (a circuit) though R1 and R2 to the capacitor when the switch is open?

Okay, so for this part (figuring out R for the time constant) do i just include the resistors that are in the new circuit with the capacitor. Let me try to explain, when the switch is open i can basically ignore that wire and everything that feeds off only that wire. So, in this case the battery, R4, and R1. Is that correct logic?
 
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  • #32
Gee Wiz said:
Okay, so for this part (figuring out R for the time constant) do i just include the resistors that are in the new circuit with the capacitor. Let me try to explain, when the switch is open i can basically ignore that wire and everything that feeds off only that wire. So, in this case the battery, R4, and R1. Is that correct logic?

Yes. Current can only flow when there is a closed loop.
 
  • #33
Awesome. Thank you so much. The whole talk through was very helpful, and it was better than just getting the right answer.
 
  • #34
Gee Wiz said:
Awesome. Thank you so much. The whole talk through was very helpful, and it was better than just getting the right answer.

I'm very happy to have helped :smile:
 
  • #35
Gee Wiz said:
Well immediately after it would just be C=Q(max)/V...?

Did we ever come up with the right Qmax across C? Last I saw was Qmax = CV but V is the battery so that equation is incorrect.
 
Last edited:
  • #36
rude man said:
Did we ever come up with the right Qmax across C? Last I saw was Qmax = CV but V is the battery so that equation is incorrect.

When I saw that equation I assumed that here V was meant to be the voltage across the capacitor; The OP did find the initial charge on the capacitor (180.496uC), which leads one to believe that he had a handle on the voltage, too.
 
  • #37
gneill said:
when i saw that equation i assumed that here v was meant to be the voltage across the capacitor; the op did find the initial charge on the capacitor (180.496uc), which leads one to believe that he had a handle on the voltage, too.

10-4.
 

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