How Do Basic Capacitor Concepts Affect Voltage in Simple Circuits?

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Discussion Overview

The discussion revolves around the behavior of capacitors in a circuit, specifically focusing on the voltage across capacitors after a switch is closed. Participants explore concepts related to charge conservation, energy loss, and the impact of resistance in the circuit over time.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant expresses confusion about why the voltage across both capacitors is 3.75V after the switch is closed, given that the initial charge on one capacitor was 5V.
  • Another participant questions whether the voltage measurements are taken at the initial time or at a later time, suggesting the addition of a switch to the schematic for clarity.
  • A different participant notes that the presence of a resistor implies that the voltage will eventually decay to zero, regardless of the initial conditions.
  • One participant clarifies that they are measuring the voltage at 60 seconds after the switch is closed and reiterates the observed voltage of 3.75V on both capacitors.
  • Another participant explains that the charge on the first capacitor is conserved when the switch is closed, leading to a voltage drop due to the combined capacitance of the capacitors in parallel.
  • One participant calculates the initial and final stored energy in the circuit, noting a loss of energy due to resistance when the switch is closed.

Areas of Agreement / Disagreement

Participants express differing views on the interpretation of voltage measurements and the implications of resistance in the circuit. There is no consensus on the understanding of the voltage behavior after the switch is closed.

Contextual Notes

Participants reference specific times for measurements and the effects of resistance, but there are unresolved assumptions regarding the circuit's configuration and the timing of the measurements.

skybox
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Hi Guys,

I have the following circuit I want to analyze:
29v0war.jpg


I do not understand by the DC voltage on both capacitors is 3.75V each. The initial charge on capacitor C1 is 5V. I am just trying to understand a basic switched capacitor concept and am stuck here. Any insight would be greatly appreciated!
 
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Hello,

Are your measurements at t=0s or some other time? I would suggest adding a switch to your schematic.
 
Yes, you must be measuring at some intermediate time.
The fact that there is dissipation (a resistor) but no source in the circuit means that the voltage will eventually go to zero regardless of the initial state.
 
I am measuring the voltage at t = 60s. I am just wondering how it is 3.75V. I am modeling this circuit after a switch has closed.
 
skybox said:
I am measuring the voltage at t = 60s. I am just wondering how it is 3.75V. I am modeling this circuit after a switch has closed.

You initially had 5V on the left 3uF cap, and after the switch is closed (presumably the 1 Ohm resistor), you have 3.75V on both the 3uF and 1uF caps. Seems pretty straightforward. Do a time plot from 0s to 60s of the voltages on the caps, and you'll see the chage distribution change from initial to final.
 
Before the switch is closed, there is 15 microcoulombs on the capacitor C1 (3 uF), which is charged to 5 V. After the switch is closed, you have 15 microcoulombs on 4 uF (C1 + C2 in parallel), so the voltage should be reduced to 3.75 volts. So charge is conserved. Initially the total stored energy in the circuit is (1/2)CV^2 = 37.5 microjoules, and after the switch is closed, the stored energy is only 28.1 microjoules, so some energy was lost in resistance R1 when the switch was closed.
 

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