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    1) A100 µF capacitor is connected in series with an 8000Ω resistor. Determine thetime constant of the circuit.
    If the circuit is suddenlyconnected to a 100 dc supply find

    I. The initial rise of P.D across thecapacitor, (rate of voltage rise per volt to the first time constant)

    T= CR= 100µF * 10¯⁶ *8000Ω = 8 seconds

    63.2% of 100v = 63.2v

    =63.2 / 8 = 7.9v rise per volt to first time constant

    II. The initial charging current, (at zerotime)

    I = 100/ 8000 =0.0125a

    III. The ultimate charge in the capacitor

    Q = C V = 100µF * 10¯⁶ * 100v = 0.01 coulomb

    IV. The ultimate energy stored in the capacitor

    W = 1/2 C V² = 100 * 10^-6 *100^2/2 = 0.5 joules


    Hey kind folks I was wondering if someone can check these questions to see if it is correct. Many thanks.
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    (Original post by sunny1982)
    1) A100 µF capacitor is connected in series with an 8000Ω resistor. Determine thetime constant of the circuit.
    If the circuit is suddenlyconnected to a 100 dc supply find

    I. The initial rise of P.D across thecapacitor, (rate of voltage rise per volt to the first time constant)

    T= CR= 100µF * 10¯⁶ *8000Ω = 8 seconds

    63.2% of 100v = 63.2v

    =63.2 / 8 = 7.9v rise per volt to first time constant

    II. The initial charging current, (at zerotime)

    I = 100/ 8000 =0.0125a

    III. The ultimate charge in the capacitor

    Q = C V = 100µF * 10¯⁶ * 100v = 0.01 coulomb

    IV. The ultimate energy stored in the capacitor

    W = 1/2 C V² = 100 * 10^-6 *100^2/2 = 0.5 joules


    Hey kind folks I was wondering if someone can check these questions to see if it is correct. Many thanks.

    Check your time constant again. The equation is correct but the decimal is in the wrong place.

    What do you mean by '=63.2 / 8 = 7.9v rise per volt to first time constant' ????

    Other than that it's all OK.
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    1) A100 µF capacitor is connected in series with an 8000Ω resistor. Determine thetime constant of the circuit.
    If the circuit is suddenlyconnected to a 100 dc supply find

    I. The initial rise of P.D across thecapacitor, (rate of voltage rise per volt to the first time constant)

    T= CR= 100µF * 10¯⁶ *8000Ω = 0.8s

    63.2% of 100v = 63.2v

    = 63.2/8= 79v rise per volt to first time constant

    II. The initial charging current, (at zerotime)

    I =100/8000=0.0125a

    III. The ultimate charge in the capacitor

    Q = C V = 100µF * 10¯⁶ * 100v = 0.01 coulomb

    IV. The ultimate energy stored in thecapacitor

    W = 1/2 C V² =100*10^-6*100^2/2 = 0.5 joules

    The 63.2/8= the voltage rise to the first time constant, should it be 0.632/0.8 ?
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    T=CR This the distance along the x-axis for 1 time constant.

    The voltage starts rising to 100V at T = 0 seconds. But because the charging rate is an inverse exponential function, it will only get to 63.2% of 100V in that time (0.8 seconds).

    So the rate of change of voltage with time at T = 0 seconds is: rise in y-axis/change in x-axis. i.e. the slope or tangent at T=0 or dv/dt.

    That is @ T=0 the slope is given by V/CR

    i.e. 100/0.8 = 125 volts per second.

    The capacitor will charge to 63.2% of 100V in 0.8 seconds at an initial rate of 125 volts/second.
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    (Original post by uberteknik)
    T=CR This the distance along the x-axis for 1 time constant.

    The voltage starts rising to 100V at T = 0 seconds. But because the charging rate is an inverse exponential function, it will only get to 63.2% of 100V in that time (0.8 seconds).

    So the rate of change of voltage with time at T = 0 seconds is: rise in y-axis/change in x-axis. i.e. the slope or tangent at T=0 or dv/dt.

    That is @ T=0 the slope is given by V/CR

    i.e. 100/0.8 = 125 volts per second.

    The capacitor will charge to 63.2% of 100V in 0.8 seconds at an initial rate of 125 volts/second.
    so the rate of voltage rise per volt to the first time constant is 125v per second?
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    (Original post by sunny1982)
    so the rate of voltage rise per volt to the first time constant is 125v per second?
    Yup.
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    (Original post by sunny1982)
    so the rate of voltage rise per volt to the first time constant is 125v per second?
    I think you have copied the question down incorrectly.
    Is this the original question and are the comments in brackets part of it or something you or someone else has added?

    Surely Q1 is just asking for the initial rate at which the voltage rises on the capacitor. That is, at T=0 as uberteknik has said.

    The bracketed comment after the question "(rate of voltage rise per volt to the first time constant)" makes no sense.

    They are not, I assume, asking for the rate of voltage rise per volt. What does that mean? The initial pd on the capacitor is zero so the initial rate of rise per volt would be infinite!

    They are also not asking for this initial rate of voltage rise "to the first time constant". The rate of voltage rise gets less as the capacitor charges. It isn't uniform. You can find an average, for sure. But that is not what the question asks.
    As uberteknik says, it starts off rising at 125 V per second, but only gets to 63.2V after 0.8s
    (That would be an average of 79 V per s)
 
 
 
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