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Charging and discharging a capacitor through a fixed resistor help!! Watch

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    How can current decrease when a capacitor is both charging and discharging? When charging the current decreases to 0 so how can it decrease further? If it decreased into minus values I would understand but the graph in my textbook shows it decreasing from a value above 0. Does that mean it doesn't actually decrease to 0?
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    (Original post by G.Y)
    How can current decrease when a capacitor is both charging and discharging? When charging the current decreases to 0 so how can it decrease further? If it decreased into minus values I would understand but the graph in my textbook shows it decreasing from a value above 0. Does that mean it doesn't actually decrease to 0?
    Discharge case:

    Q = CV

    As the capacitor discharges, the charge stored gets used up and must fall, but the capacitance remains constant. Less charge on the capacitor means the p.d. across the capacitor plates also falls.

    I = V/R

    Since the p.d. is the voltage pressure driving current through the constant resistance load, the rate of discharge (current) must also fall. i.e. no charge left = no p.d. = no current.


    Charge case:

    Q = CV

    As the charge on the capacitor plates increases, so does the p.d. This presents a pressure resistance to the supply providing the new charge, which becomes increasingly more difficult to push further charge onto the plates, until the point where the p.d. across the capacitor equals the supply emf and equilibrium is achieved.

    i.e. supply e.m.f. = capacitor p.d. = no further current.
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    (Original post by G.Y)
    How can current decrease when a capacitor is both charging and discharging? When charging the current decreases to 0 so how can it decrease further? If it decreased into minus values I would understand but the graph in my textbook shows it decreasing from a value above 0. Does that mean it doesn't actually decrease to 0?
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    I hope the graphs can explain what you are looking. As current is the rate of flow of charges, which is I  =\frac{dQ}{dt}, so the magnitude of the current can be seen as decreasing exponentially.

    During the discharge, some authors would define discharging current as
    I  = - \frac{dQ}{dt}
    The minus sign is to show that charge is lost from the capacitor. I think this is what your textbook is using.

    While for the charging current, they would define as

    I  = + \frac{dQ}{dt} .
    The plus sign is to show that charge is added to the capacitor.
 
 
 
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