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    Can someone please tell me on this attatchment what the 'o' stands for in the 'Vo' 'Io' & 'Qo' on the discharging column please.

    Is it the voltage/current/charge at 'o' seconds?
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    (Original post by picklesmartyn)
    Can someone please tell me on this attatchment what the 'o' stands for in the 'Vo' 'Io' & 'Qo' on the discharging column please.

    Is it the voltage/current/charge at 'o' seconds?
    Yes.

    The '0' subscript for the given variable means 'at time t=0'.

    i.e. these are the initial conditions when t=0 seconds.

    Hence

    Q_0 is the initial charge at t=0 etc.
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    Thankyou for clarifying
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    so if I had a question, calculate (i) the time constant for the circuit above. (ii) the length of time it will take for the capacitor to reach a voltage of 25 volts (iii) if the capacitor were fully charged and then the battery disconnected and switch 1 closed, the time it would take for the voltage to drop to 25 volts.

    does 'Qo/C' stand for the voltage at the capacitor at t=0 because Q/C=V. And because we are now calculating from the capacitor as a source, this voltage will change as it discharges hence its Q/C voltage and not the source voltage

    so the full charge on this capacitor will be 50 x (15x10^-6)= 7.5x10^-4 coulombs







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    (Original post by picklesmartyn)


    so if I had a question, calculate (i) the time constant for the circuit above. (ii) the length of time it will take for the capacitor to reach a voltage of 25 volts (iii) if the capacitor were fully charged and then the battery disconnected and switch 1 closed, the time it would take for the voltage to drop to 25 volts.

    does 'Qo/C' stand for the voltage at the capacitor at t=0 because Q/C=V. And because we are now calculating from the capacitor as a source, this voltage will change as it discharges hence its Q/C voltage and not the source voltage

    so the full charge on this capacitor will be 50 x (15x10^-6)= 7.5x10^-4 coulombs

    This question is about understanding what is meant by initial conditions and dependent on what each section of the question is asking.

    Yes. There are two instances where t = 0.

    The first is when the capacitor starts from a fully discharged state Q_0 = 0 and the switch is closed at t=0 to begin charging.

    The p.d. developed across the capacitor will follow the relationship:

    V_C_{(t)} = V_{0}(1 - e^{[\frac{-t}{CR}]})

    where

    V_0 = V_{supply} = 50V

    Part ii)

    As  t \Rightarrow \infty the capacitor will be fully charged. Infinite (t) is a very long time! For practical purposes and as a general rule-of-thumb, the capacitor will achieve >99% full charge in 5 time constants. i.e. 5CR time periods. You need to make a note of this.

    In order to answer, you will need to rearrange the above expression to make (t) the subject and then substitute the initial and final conditions.

    HINT: V_C_{(t)} = 25V \mathrm {and \  }   V_0 = 50V You will also need to use natural logarithms to isolate (t) from the exponential.

    part iii)

    The fully charged capacitor will indeed hold a theoretical maximum charge of

    Q = CV

    ......and when fully charged, the p.d. across the capacitor will equal the supply voltage.

    NB

    The second t=0 condition occurs when the fully charged capacitor starts discharge as the switch is changed over at the new t=0 such that:

    V_C_{(t)} = V_0(e^{[\frac{-t}{CR}]})

    (Note the difference between the charge and discharge equations)

    So the initial conditions at the new t=0 for part iii). are V_0 = 50V \mathrm{ and \  } Q_0 = CV_s = 0.75mC

    V_C_{(t)} = 25V

    Once again (t) needs to be isolated using the same technique.


    Have a go and please do post your answers here so that I can check you have understood fully and also for the benefit of others who will no doubt be watching with interest.
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    Here are my answers sorry I dont know how to write equations on computer.
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    (Original post by picklesmartyn)
    Here are my answers sorry I dont know how to write equations on computer.
    No worries.

    Your answers are correct in both cases.

    This is an important result and it says that the time taken to rise or fall by the same voltage will be identical for both charge and discharge cases.

    i.e. the time to rise to say 12V would be identical to the time taken to fall by 12V (fall to 38V) etc.

    Are you OK with the concepts now?
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    Yes I cant thank you enough. The lecturer I have at the minute is very poor and im growing increasingly frustrated as all I want to do is learn and im having to use the internet to teach myself.

    So can I just ask a couple more questions. sorry

    1) so am I right in saying that Qo/C = the voltage of the fully charged capacitor at t=0, which is the same as (99% the nearest it can be to) the voltage supply?

    2)when you put V(t), this means at any instance of time?

    3) in the equation for Ic (discharging), why is there now an 'R' for Qo/RC. could you elaborate a bit more as to why the 'R' is there now

    I promise I wont ask any more questions on this matter I just want to get it right
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    (Original post by picklesmartyn)
    Yes I cant thank you enough. The lecturer I have at the minute is very poor and im growing increasingly frustrated as all I want to do is learn and im having to use the internet to teach myself.
    Wow, that's a bummer. But hey, if you can teach yourself, then that is a priceless skill which will last you throughout your life and is truly a gift.

    (Original post by picklesmartyn)
    So can I just ask a couple more questions. sorry

    1) so am I right in saying that Qo/C = the voltage of the fully charged capacitor at t=0, which is the same as (99% the nearest it can be to) the voltage supply?
    As long as the switch is closed during, the capacitor will keep charging towards the supply voltage. Beyond 5CR the voltage will get closer and closer to that supply voltage (>>99%). The 5CR time period is used in engineering as a rule-of-thumb and enables quick estimates to be made without resorting to the lengthy equations. It's a similar trick to rounding exponents up or down to the nearest multiple of 3. The final exponent is then easy to calculate in ones head and convert into milli, micro. mega, nano, pico etc.


    (Original post by picklesmartyn)
    2)when you put V(t), this means at any instance of time?
    Yes. Exactly that and you will need this when the equations become differentials w.r.t time t for a.c. analysis.

    (Original post by picklesmartyn)
    3) in the equation for Ic (discharging), why is there now an 'R' for Qo/RC. could you elaborate a bit more as to why the 'R' is there now.
    Simply from ohms law. I = V/R

    Q = CV

    so

    V = Q/C which is then substituted into I = V/R to give I = Q/RC

    Which stacks up since current is defined as the rate of change of charge. The units of charge are Coulombs and the units of R x C are seconds, so Q/RC is in fact Coulombs per second.


    (Original post by picklesmartyn)
    I promise I wont ask any more questions on this matter I just want to get it right
    Keep asking. It's what TSR is here for!
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    Thanks again
 
 
 
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