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31 May, 12:06

A 1.40 mH inductor and a 1.00 µF capacitor are connected in series. The current in the circuit is described by I = 14.0 t, where t is in seconds and I is in amperes. The capacitor initially has no charge.

(a) Determine the voltage across the inductor as a function of time. mV

(b) Determine the voltage across the capacitor as a function of time. (V/s2) t2

(c) Determine the time when the energy stored in the capacitor first exceeds that in the inductor.

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  1. 31 May, 15:44
    0
    Inductance L = 1.4 x 10⁻³ H

    Capacitance C = 1 x 10⁻⁶ F

    a)

    current I = 14.0 t

    dI / dt = 14

    voltage across inductor

    = L dI / dt

    = 1.4 x 10⁻³ x 14

    = 19.6 x 10⁻³ V

    = 19.6 mV

    It does not depend upon time because it is constant at 19.6 mV.

    b)

    Voltage across capacitor

    V = ∫ dq / C

    = 1 / C ∫ I dt

    = 1 / C ∫ 14 t dt

    1 / C x 14 t² / 2

    = 7 t² / C

    = 7 t² / 1 x 10⁻⁶

    c) Let after time t energy stored in capacitor becomes equal the energy stored in capacitance

    energy stored in inductor

    = 1/2 L I²

    energy stored in capacitor

    = 1/2 CV²

    After time t

    1/2 L I² = 1/2 CV²

    L I² = CV²

    L x (14 t) ² = C x (7 t² / C) ²

    L x 196 t² = 49 t⁴ / C

    t² = CL x 196 / 49

    t = 74.8 μ s

    After 74.8 μ s energy stored in capacitor exceeds that of inductor.
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