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    Physics Circuits: charging time, voltage, current flow, discharge

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    Please see the attached problem.

    Consider the following transient RC-circuit and answer these questions in the given order.

    1) Find charging the constant.

    2) Find full-charged time.

    3) Find charging equations.

    4) Find voltage across and current flow for capacitor at t = 4 second/charging.

    5) Find discharging time constant.

    6) Find full-discharged time.

    7) Find discharging equations.

    8) Find voltage across and current flow for capacitor at t = 16 seconds.

    9) Plot voltage across and current flow for capacitor during time period o =< t =< 16

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    https://brainmass.com/physics/ohms-law/physics-circuits-charging-time-voltage-current-flow-discharge-286486

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    1) Charging time constant of the R1C circuit = τC = R1C = 50x103x40x10-6 = 2 sec

    2) Full charge time: Charging of the capacitor takes place as per the equation: Q = Q0(1- e-t/τ) where Q0 is the final charge on the capacitor. Strictly, the capacitor takes infinite time to charge fully. However, in practice after a period of 5 times the time constant, the capacitor is over 99% charged and 5τ is taken as the time to charge the capacitor fully. Hence, time to charge the capacitor fully = 5 x 2 = 10 sec.

    3) Voltage across the capacitor VC = Q/C

    By Kirchhoff's loop law: R1I + Q/C = 20

    Current I = dQ/dt. Hence, R1(dQ/dt) + Q/C = 20

    dQ/dt + Q/R1C = 20/R1

    dQ/dt = (20/R1 - Q/R1C)

    dQ/(20/R1 - Q/R1C) = dt

    Integrating: ∫dQ/(20/R1 - Q/R1C) = ∫dt

    Substituting (20/R1 - Q/R1C) = x, - (1/R1C) dQ = dx or dQ = - R1C dx

    - R1C∫dx/x = t

    - R1C logex = t + D

    - R1C loge(20/R1 - Q/R1C) = t + D where D is the constant of integration

    Assuming the capacitor is fully uncharged initially, substituting t = 0, Q = 0 we get:
    D = - R1C loge(20/R1)

    Substituting for D: - R1C loge(20/R1 - Q/R1C) = t - R1C loge(20/R1)

    R1C [loge(20/R1 - Q/R1C) - loge(20/R1)] = - t

    loge(20/R1 - Q/R1C) - loge(20/R1) = - t/R1C

    loge[(20/R1 - Q/R1C)/(20/R1)] = - t/R1C

    (20/R1 - Q/R1C)/(20/R1) = e-t/R1C

    (20/R1 - Q/R1C) = (20/R1)e-t/R1C

    Q/R1C = 20/R1[1 - e-t/R1C]

    Q = 20C[1 - e-t/R1C] ............(1)

    In the above ...

    Solution Summary

    Using diagrams and step by step equations, this solution provides answers for charging the constant, full-charged time, charging equations, voltage, discharging time constant, full-discharged time, discharging equations, and voltage.

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