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20 September, 18:56

A capacitor of 20.0 μ F and a resistor of 90.0 Ω are quickly connected in series to a battery of 8.00 V. What is the charge Q on the capacitor 0.00200 s after the connection is made? Q =

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  1. 20 September, 19:09
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    Q=107μC

    Explanation:

    Given an RC series circuit

    Input voltage V=8V

    Resistor = 90Ω

    And the capacitor=20 μF

    Charge Q? After t=0.002

    In an RC circuit,

    The capacitor voltage is given as

    Vc=V (1-e^-t/τ)

    Where τ is time constant

    τ = RC

    τ = 90*20*10^-6

    τ=0.0018s

    Therefore,

    Vc=V (1-e^-t/τ)

    Given that V=8V and t=0.002

    Then,

    Vc=8 (1-e^-0.002/0.0018)

    Vc=8 (1-e^-1.111)

    Vc=8 (1-0.329)

    Vc=8*0.671

    Vc=5.37Volts

    Then the voltage across the capacitor is 5.37V

    From the relation

    Q=CV, we can calculate the charge

    Q=20*10^-6*5.37

    Q=1.07*10^-4 C

    Q=107μC
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