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28 July, 22:15

A 30 μFμF capacitor initially charged to 30 μCμC is discharged through a 1.9 kΩkΩ resistor. How long does it take to reduce the capacitor's charge to 5.0 μCμC?

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  1. 29 July, 02:00
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    Given Information:

    Resistance = R = 1.9 k Ω

    Capacitance = C = 30 uF

    Initial charge = 30 uC

    Final charge = 5 uC

    Required Information:

    Time taken to reduce the capacitor's charge to 5.0 μC = ?

    Answer:

    t = 0.101 seconds

    Explanation:

    The voltage across the capacitor is given by

    V = V₀e^ (-t/τ)

    Where τ is time constant, V₀ is the initial voltage, V is the voltage after some time t

    τ = RC

    τ = 1900*30x10⁻⁶

    τ = 0.057 sec

    The initial voltage across the capacitor was

    V₀ = Q/C

    V₀ = 30/30

    V₀ = 1 V

    Voltage to reduce the charge to 5 uF

    V = 5/30

    V = 0.167 V

    V = V₀e^ (-t/τ)

    0.167 = 1*e^ (-t/0.057)

    take ln on both sides

    ln (0.167) = ln (e^ (-t/0.057))

    -1.789 = - t/0.057

    t = 1.789*0.057

    t = 0.101 seconds
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