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17 December, 02:08

A voltage source of 10.0 V is connected to a series RC circuit where R = 2.80 x 100 2, and C = 3.50 uF. Find the amount of time required for the current in the circuit to decay to 4.00% of its original value. Hint: This is the same amount of time for the capacitor to reach 96.0% of its maximum charge.

a. 31.5 s

b. 309 s

c. 0.143 s

d. 13.6 s

e. 19.1 s

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  1. 17 December, 05:04
    0
    The correct answer is option (a) 31.5 s

    Explanation:

    Given dа ta:

    R = 2.8*10^6

    C = 3.5 uF = 3.5*10^6 F

    Calculating the time constant, we have;

    T = R*C

    = 2.8*10^6 * 3.5*10^-6

    = 9.8 seconds

    Calculating the amount of time required for current in the circuit to decay using the formula;

    Q = Qmax * (1 - e^ (-t/T))

    But Q = 96% of Qmax

    The equation becomes;

    0.96*Qmax = Qmax * (1 - e^ (-t/T))

    0.96 = 1 - e^ (-t/T)

    e^ (-t/T) = 1 - 0.96

    t = - T*ln (0.04)

    = - 9.8*ln (0.04)

    = 31.54 seconds
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