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Electronics and Communication Engineering - Exam Questions Papers

Exercise : Exam Questions Papers - Exam Paper 5
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 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.

22.
Find equivalent capacitance if each capacitance is 1 F.
56 F
6/5 F
6/10 F
5/12 F
Answer: Option
Explanation:

Use similar method. But for capacitor take inverse of value of capacitance and then apply procedure same as in resistance case. After getting the final answer take inverse of result to get Ceq

∴

In Cuboid case,

For resistance

For inductance

For capacitance.

23.
The Z inverse of the given Z transform is :
unit step
unit impulse
unit ramp
unit parabola
Answer: Option
Explanation:

24.
The RMS value of the signal given below is:
Answer: Option
Explanation:

RMS value .

25.
Use block diagram reduction methods to obtain the equivalent T.F from R to C.
Answer: Option
Explanation:

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