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Transients Rc circuits.

  1. Dec 29, 2013 #1
    1. The problem statement, all variables and given/known data
    Determine the voltage drop across the resistor when the capacitor has been discharging for 10μs.
    Capacitance=400 pico farad and is charged to 10v through a 50kΩ resistor.

    2. Relevant equations
    Vr= V*e -t/CR


    3. The attempt at a solution Vr = V-Vc at the instant of 10μs
    Vc=Ve-t/CR
    I get 3.935 volts? (without showing all my workings out)
     
  2. jcsd
  3. Dec 29, 2013 #2

    gneill

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    Staff: Mentor

    It's not clear from what you've presented whether the capacitor is charging up from 0V to 10V, or discharging from 10V down to 0V. Does the capacitor start out at 0V or 10V?
     
  4. Dec 29, 2013 #3
    Fair point.The capacitor is fully charged to 10volts. So i believe the capacitor starts out at 10 volts?
     
  5. Dec 29, 2013 #4

    gneill

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    So the circuit looks something like this?

    attachment.php?attachmentid=65200&stc=1&d=1388347772.gif

    The capacitor is initially charged to 10V and at time t = 0 the switch closes allowing current to start to flow, discharging the capacitor?
     

    Attached Files:

  6. Dec 29, 2013 #5
    The capacitor is fully charged to 10volts.Calculate the voltage drop when the capacitor has been discharging for 10 micro seconds? I thought the switch must be open to allow the capacitor to discharge?
     
    Last edited: Dec 29, 2013
  7. Dec 29, 2013 #6

    gneill

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    Okay then according to that statement the diagram fits the scenario. You will want to revisit your Attempt at a Solution now. Can you show more of your working?
     
  8. Dec 29, 2013 #7
    Ok. here is my workings out.

    Vr= V-Vc at the instant of 10μs.
    Vc=V*e-t/CR
    -t/CR = -0.5
    Vc=10*-0.5
    Vc=6.065volts.
    Vr = V-Vc
    Vr= 10-6.065
    Vr = 3.935 volts?
    Hope this helps!
     
  9. Dec 29, 2013 #8

    gneill

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    Can you justify the above equation? Why would Vr = V - Vc?

    That's a good value for Vc.
     
  10. Dec 29, 2013 #9
    This is how i see this: Voltage is 10v from the battery. The resistor is in series with the capacitor. The capacitor drops 6.065 volts so that must mean the resistor drops 3.935 volts which equates to a total of 10 volts dropped from the battery. But i am confused now! . This would be the case in a steady state circuit but not after 10 micro seconds.
     
  11. Dec 29, 2013 #10

    gneill

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    Are you given a circuit diagram for the problem?

    If there is a battery in the circuit and all three components form a series circuit, then it likely that the capacitor is meant to start out uncharged (0 V), and charges up over time.
     
  12. Dec 29, 2013 #11
    No diagram. The capacitor is set to the maximum 400 pico farad and is charged to 10 volt through a 50k resistor. Determine the voltage drop across the resistor when the capacitor has been discharging for 10 micro seconds. This is the info i have recieved. Hope this helps!
     
  13. Dec 29, 2013 #12

    gneill

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    Okay then. So you can justify your equation for Vr by doing KVL around the loop, and your calculations for Vr are correct.
     
  14. Dec 29, 2013 #13
    i thought Vr= V- Vc. At this instant of 10 micro seconds the voltage is 6.065 volts so i see the voltage must be 10 - Vc?? This would be the voltage across the resistor?
     
  15. Dec 29, 2013 #14

    gneill

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    Right. That's KVL for the loop for that instant in time.
     
  16. Dec 29, 2013 #15
    So the voltage drop is 6.065volts because the voltage across the resistor is 3.935 volts??
     
  17. Dec 29, 2013 #16

    gneill

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    Just write KVL for the loop using V for the battery voltage, Vr for the potential drop across the resistor and Vc for the potential drop across the capacitor. KVL applies always, at all times. You'll then have the equation that relates all the potential drops around the circuit.

    attachment.php?attachmentid=65202&stc=1&d=1388357521.gif
     

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