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Simple Capacitor Concept

  1. Mar 20, 2009 #1
    Hi Guys,

    I have the following circuit I want to analyze:

    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!
  2. jcsd
  3. Mar 20, 2009 #2

    Are your measurements at t=0s or some other time? I would suggest adding a switch to your schematic.
  4. Mar 20, 2009 #3


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    Science Advisor
    Gold Member

    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.
  5. Mar 20, 2009 #4
    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.
  6. Mar 20, 2009 #5


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    Staff: Mentor

    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.
  7. Mar 20, 2009 #6
    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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