Ask Question
27 May, 20:55

A capacitor of capacitance 4 µF is discharging through a 2.0-MΩ resistor. At what time will the energy stored in the capacitor be one-third of its initial value?

+2
Answers (1)
  1. 27 May, 21:08
    0
    Given that,

    The capacitance of the capacitor is

    C = 4 µF

    Discharging through a resistor of resistance

    R = 2 MΩ

    Time the energy stored in the capacitor be one-third of its initial energy

    i. e. u (E) _ final = ⅓u (E) _initial

    U / Uo = ⅓

    Energy stored in a capacitor (discharging) can be determined using

    U = Uo•RC•exp (-t/RC)

    U / Uo = - RC exp (-t/RC)

    U / Uo = ⅓

    RC = 2 * 10^6 * 4 * 10^-6 = 8s

    ⅓ = 8 exp (-t / 8)

    Divide both side by - 8

    1/24 = exp (-t/8)

    Take In of both sides

    In (1/24) = In•exp (-t/8)

    -3.1781 = - t / 8

    t = - 3.1781 * - 8

    t = 25.42 seconds

    It will take 25.42 seconds for he capacitor to be ⅓ of it's initial energy
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “A capacitor of capacitance 4 µF is discharging through a 2.0-MΩ resistor. At what time will the energy stored in the capacitor be one-third ...” in 📗 Physics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers