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Homework Help: CR Transient Problem

  1. May 25, 2012 #1
    1. The problem statement, all variables and given/known data

    The circuit shown in Figure 5 contains two emfs of opposite polarity E1 = 40 V,
    and E2 = -20 V. The circuit also contains a switch, resistors R1 and R2, and a
    capacitor C.The capacitor is initially discharged.

    The switch is connected to position ‘a’ for a period of 8 seconds, after which it is
    returned to the unconnected position ‘b‘ for a period of 12 seconds before being
    connected to position ‘c’ for a further 8 seconds.

    What are the time constants for each of the three time intervals?

    Calculate the potential difference across the capacitor, VC, at the end of
    each time interval.


    2. Relevant equations

    3. The attempt at a solution

    From 0-8 seconds the time constant is T=CR = 5.44 s
    From 8-20 seconds T is 12 seconds
    From 20- 28 seconds T is 5.44 s

    Capacitor voltage at point A : Vc=40(1-e^(-8/5.44)) = 30.808 V ( Charging )

    Voltage at B : ( Discharging) Vc' = 30.808e^(-12/12) = 11.33 V

    Voltage at C : I dont know what to do . At first the capacitor has 11.33 V but because the polarity of this source is reversed then the capacitor will have to discharge ( go to 0 Volts ) . Is that right ? I am really confused here . Because these 8 seconds from 20-28 it will take some time to discharge from 11.33 to 0 and then recharge again with the other polarity .. WHAT DO I DO ?!

    Thank you :)
  2. jcsd
  3. May 25, 2012 #2


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    Staff: Mentor

    0V isn't a special value; It's all in how you write the equations. If you have a starting voltage and a "final" target voltage, either of which may be positive or negative, what's important is the size of the ΔV that the exponential is traversing. Find the magnitude of that ΔV and use it to write the appropriate decay portion:

    ##\Delta V e^{-t/\tau}##

    then 'adjust' the function by adding an appropriate offset so that it begins at the correct starting voltage.

    Regarding the voltage source E2, I see by the diagram that it's given a value E2 = -20 V, but that the symbol for the cell is also oriented with its negative terminal upwards. By strict interpretation then, what should be the potential at point c?
  4. May 25, 2012 #3
    Hmm , 20 volts ?

    But why is it a decay ? It is supposed to be charging , no ?
  5. May 25, 2012 #4


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    Staff: Mentor

    That would be my take on it, strictly interpreting the evidence.
    I was going by your first interpretation of the net polarity of point c and describing how one might go about accounting for a capacitor voltage change that begins at +11.33V and heads to -20V. If it's really +20V at c, then it will be charging from +11.33V towards +20V.
  6. May 25, 2012 #5
    Okay so :

    Vc = E - e^(-t/T)*(E-Vc') so Vc = -20 -e^(-8/5.44)*(-20-11.33) = -12.8 V , Is that correct ?

    A friend of mine said that Vc = 11.33*e^(-8/12) -20*(1-e^(-8/5.44)) . He said that it's discharging through two resistors ... i am even more confused now !
  7. May 25, 2012 #6


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    Staff: Mentor

    Your result looks okay if the potential at c is in fact -20V.
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