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29 August, 01:49

An uncharged 5.0-µF capacitor and a resistor are connected in series to a 12-V battery and an open switch to form a simple RC circuit. The switch is closed at t = 0 s. The time constant of the circuit is 4.0 s. What is the charge on either plate after one time constant has elapsed?

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  1. 29 August, 02:57
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    For a direct current resistor-capacitor circuit where the capacitor is initially uncharged, the charge stored on one of the capacitor's plates is given by:

    Q (t) = Cℰ (1-e^{-t / (RC) })

    Q (t) is the charge, t is time, ℰ is the battery's terminal voltage, R is the resistor's resistance, and C is the capacitor's capacitance.

    The time constant of the circuit τ is the product of the resistance and capacitance:

    τ = RC

    Q (t) can be rewritten as:

    Q (t) = Cℰ (1-e^{-t/τ})

    We want to know how much charge is stored when one time constant has elapsed, i. e. what Q (t) is when t = τ. Let us plug in this time value:

    Q (τ) = Cℰ (1-e^{-τ/τ})

    Q (τ) = Cℰ (1-1/e)

    Q (τ) = Cℰ (0.63)

    Given values:

    C = 5.0*10⁻⁶F

    ℰ = 12V

    Plug in these values and solve for Q (τ):

    Q (τ) = (5.0*10⁻⁶) (12) (0.63)

    Q (τ) = 3.8*10⁻⁵C
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