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    Variation of Voltage Across a Capacitor

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    See attached file for full problem description. Please answer only question 1.2

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    https://brainmass.com/engineering/power-engineering/variation-voltage-across-capacitor-113555

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    SOLUTION

    Volts
    I(t)
    6
    +
    150nF V(t) 4 5V
    -
    2
    μsec
    -4 -2 2 4 6
    -4V

    a) Charge Q on a capacitor, its capacitance C and the voltage V across it are related as :

    Q = CV

    Current flowing through the capacitor is equal to the rate of change of charge Q.

    I(t) = dQ/dt = C dV/dt .............(1)

    Thus, current flowing during different time intervals is calculated using (1) as follows :

    i) Upto t = -2 μsec : The voltage is constant at -4 V. Hence, dV/dt = 0. From (1), I(t) = 0

    ii) t = -2 μsec to t = 4 μsec : Change in voltage = 4 V and change in time = 6μsec = 6x10-6 sec

    dV/dt = 4/(6x10-6)

    From (1) I(t) = 150 x 10-9 x 4/(6x10-6) = 0.1 A

    iii) t = 4 μsec to t = 5 μsec : Change in voltage = 5 ...

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