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22 December, 07:47

The quantity t1/2=τ ln 2 is called the half-life of an exponential decay, where τ=RC is the time constant in an RC circuit. The current in a discharging RC circuit drops by half whenever t increases by t1/2. For a circuit with "R=1.0 kΩ and C=4.0 μF, if the current is 6.0 mA at t=4.0 ms", at what time (in ms) will the current be 3.0 mA?

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  1. 22 December, 09:29
    0
    6.77 ms

    Explanation:

    τ=RC

    = 1 x 10³ x 4 x 10⁻⁶

    4 X 10⁻³

    t1/2=τ ln 2

    = 4 X 10⁻³ X ln 2

    = 2.77 x 10⁻³

    2.77 ms

    Half life = 2.77 ms

    Current is 6 mA AT t = 4 ms

    and half life is 2.77 ms ie current becomes half after every 2.77 ms

    so current will reduce from 6 mA to 3 mA in further 2.77 ms

    or at t = 4 + 2.77 = 6.77 ms current will become 3 mA.
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