Capacitor problem

Discussion in 'Homework Help' started by metelskiy, Nov 19, 2010.

  1. metelskiy

    Thread Starter Member

    Oct 22, 2010
    66
    3
    a) The capacitor is uncharged when the switch is thrown into position 1. The switch remains in position 1 for 10msec and then is thrown into position 2, where it remains indefinitely. Draw the complete waveform for the capacitor voltage.
    b) If the switch is thrown back to position 1 after 5ms in position 2, and then is left in position 1, how will the waveform appear?
    P1070141 [1280x768].JPG
    I'm not sure how to proceed with this problem.
    I was thinking this way: while switch is in position 1 for 10msec i calculated voltage of capacitor at 10msec. First i found time constant at position 1, RC=(R1+R2)*1μF=55msec
    Then i calculated Vc at 10msec:
    Vc=(Vfinal-Vinitial)(1-e^-t/RC)
    Vc=20V(1-e^-10/55)=3.32V
    So Capacitor voltage is 3.32V, then switch thrown into position 2.
    I think that when switch thrown in pos 2, capasitor starts dischargin so I'd use discharging formula. First I found pos 2 time constant
    RC=(R2||R3)*1uF=7.67msec
    Did i proceed in right way till this point? If so, how do i find in what time capacitor discharged to 0V? And how can i draw waveform for capacitor voltage?
    I didn't start on b) part yet.
     
  2. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
    6,357
    718
    If you have graphed the charge cycle, which is an exponential graph, the discharge should pick up where the charge graph stopped, and discharge began, it will almost be a mirror image of the charge curve, flipped on the Y axis, though the decline will be a bit quicker.

    Have you plotted the formulas?
     
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  3. mik3

    Senior Member

    Feb 4, 2008
    4,846
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    The time constant at position 2 is (R2+R3)C sec. Theoretically, the voltage will reach zero at an infinite time. However, assume that it reaches zero after 5(R2+R3)C sec.
     
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  4. metelskiy

    Thread Starter Member

    Oct 22, 2010
    66
    3
    Thanks for help guys, I figured it out after mik3 pointed out that RC at position 2 will be (R2+R3)C but i don't understand why R2 is in series with R3 if by looking at the graph R3 is in parallel with R2 and capacitor, please can someone explain that to me.

    On b) part i figured out that capacitor at pos 1 charges for 10msec to 3.32V, then discharges for 5msec to 2.96V and then when put back to pos 1 and remains there, C charges all the way to 105msec or would I be right if i say that C charges to 20v at an infinite time?
     
  5. mik3

    Senior Member

    Feb 4, 2008
    4,846
    63
    To find the resistance seen by the capacitor, you remove the capacitor and replace all the voltage sources and all the current sources by the internal resistance. Then you calculate the total resistance between the points where the capacitor was connected on. If the voltage and current sources are ideal then you replace them by short and open circuits respectively.

    This is like you apply Thevenins theorem.
     
  6. mik3

    Senior Member

    Feb 4, 2008
    4,846
    63
    If it is a theoretical question then you can say it charges to 20V at an infinite time. In practise you would say it charges to 20V at a time of 5RC. If it is a question for a school, then the answer depends on what the teacher wants too.
     
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