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

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SUMMARY

The discussion centers on calculating the charge on a capacitor in a two-loop RC circuit after a switch is opened. The circuit consists of resistors R1 = R2 = 22 Ω, R3 = 91 Ω, R4 = 129 Ω, a capacitor with capacitance C = 40 μF, and a battery voltage of V = 12 V. The charge on the capacitor at time topen = 790 μs after the switch opens is determined using the equation Q(t) = Qmax(1 - e^(-t/(R*C))). The correct approach involves identifying the total resistance seen by the capacitor and recognizing that the charge decreases exponentially over time.

PREREQUISITES
  • Understanding of RC circuits and time constants
  • Familiarity with exponential decay functions
  • Knowledge of Kirchhoff's laws for circuit analysis
  • Ability to apply the formula Q(t) = Qmax(1 - e^(-t/(R*C)))
NEXT STEPS
  • Calculate the total resistance in a two-loop RC circuit after the switch opens
  • Learn about the behavior of capacitors in discharging circuits
  • Explore the implications of time constants in RC circuits
  • Investigate the effects of varying capacitance and resistance on charge decay
USEFUL FOR

Students studying electrical engineering, circuit designers, and anyone interested in understanding the dynamics of RC circuits and capacitor discharge behavior.

  • #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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