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  1. For the circuit, let us assume that R = 1 Ω, L = 10-3 H, and V(t) = 10 cos (1000t + 30°)V and the differential equation for I(t) is given by

  2. A.
    7.07 cos (1000t + 15°) A
    B.
    70.7 cos (1000t - 30°) A
    C.
    7.07 cos (1000t - 15°) A
    D.
    7.07 cos (1000t + 15°) A

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    Answer : Option C

    Explanation :

    Replacing current and voltage by their phasors I and 10ejp/b, resp. and by jω = j1000, we obtain the phasor equation

    I + 10-3(j1000I) = 10ejp/6

    or I(1 + j1) = 10ejp/6

    Having determined the value of I, we now find the required solution to be

    I(t) = Rc [I ejωt] = Rc[7.07e -jp/2 ej1000t]

    = 7.07 cos (1000t - 15°)A.


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