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Homework Help: How long does it take to reach 50% of V1?

  1. Jun 10, 2015 #1
    I'm unsure of the best way to go about finding out the time it takes for the capacitor to reach a voltage of 50% of V1.

    V1: 14V
    Resistor before 2pt switch: 1K ohm
    Resistor after 2pt switch: 220 ohm
    Capacitor: 1.4 micro farads
    (Charging) 1 time constant is 1.7mS
    (Discharging) 1 time constant is 308 micro seconds

    I've attached a picture of the circuit with the values labelled. CR Circuit.png

    Is there an equation for working this out or is there a specific way of working it out? Any help or suggestions are greatly appreciated - I'm not looking for anyone to do the work for me just looking for a jab in the right direction

    I thought I could work it out based on the idea that at 1TC the capacitor has reached 63% of the supply voltage (8.82V) and working out the difference there but kept getting myself in a right muddle.
  2. jcsd
  3. Jun 10, 2015 #2


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    You know that charging a capacitor goes according to [itex] V=V_{s}(1-e^{\frac{-t}{T}})[/itex]? And discharging goes as [itex] V=V_{0}(e^{\frac{-t}{T}})[/itex]?
  4. Jun 10, 2015 #3
    I think I vaguely remember that from class. so for me it would be something like: 7=14(1-e-t/1.7x10-3)? Is that right?
    I'm assuming it's t I'm looking for so how would I go about transposing that to solve for t?
  5. Jun 10, 2015 #4


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    What have you tried?

    There's some algebraic simplifications which can be done the equation before solving for t.
  6. Jun 10, 2015 #5
    So simplified is 7=14-14e-t/1.7x10-3

    The only idea I could find (Until Svein commented) was working it out from the fact that the capacitor is 63% charged when 1 time constant had lapsed. I worked out the voltage from that and thought I could work out the time from the differences in percentage and voltage. I.e a very confusing way of doing it that a friend suggested.
  7. Jun 10, 2015 #6


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    Yep, that is correct.

    Solve for t.

    ln .5 = -t/(1.7E-3)

    t = 1.18 ms.
  8. Jun 10, 2015 #7


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    This sounds like nonsense. Setting the capacitor voltage to 7 volts in the formula is good. What is said above probably isn't going to do you any favors.
    You don't care when it's charged, you just care when it's 7 volts.
  9. Jun 11, 2015 #8


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    The assumption being made is that the switch is moved from its lower position (shown) to its upper position at t=0. Is that what you're told?

    I ask only because a different switching scenario is possible.
  10. Jun 11, 2015 #9
    That is correct, the switch is being moved from the lower position to the upper position at t=0
  11. Jun 11, 2015 #10


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    So .... you're all sorted now?

    The unfamiliar step was that where you need to take natural logarithms of both sides, in the process losing the exponential function.
  12. Jun 11, 2015 #11
    Yes, I am sorted now, thank you very much! I have now completed the work, thanks again!
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