Discussion :: Exam Questions Paper
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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
A.
7.07 cos (1000t + 15°) A
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B.
70.7 cos (1000t - 30°) A
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C.
7.07 cos (1000t - 15°) A
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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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