Electronics and Communication Engineering - Exam Questions Papers - Discussion

Discussion Forum : Exam Questions Papers - Exam Paper 5 (Q.No. 21)

Exam Questions Papers

21.

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

7.07 cos (1000t + 15°) A

70.7 cos (1000t - 30°) A

7.07 cos (1000t - 15°) A

7.07 cos (1000t + 15°) A

Answer: Option

Explanation:

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

I + 10^-3(j1000I) = 10e^jp/6

or I(1 + j1) = 10e^jp/6

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

I(t) = R_c [I e^jωt] = R_c[7.07e^-jp/2 e^j1000t]

= 7.07 cos (1000t - 15°)A.

Discussion:

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