IndiaBIX

Electronics and Communication Engineering - Exam Questions Papers

Exercise : Exam Questions Papers - Exam Paper 5
16.
In the following circuit, the switch S is closed at t = 0 The rate of change of current (0 +) is given by
0
∞
Answer: Option
Explanation:

At t = 0, inductor acts like a open circuit.

We know at t = 0+, inductor path is open circuit

V = Is.Rs (at t = 0+)

Thus

17.
convert into spherical co-ordinates.
A = 2 sinθ[sinθ tanΦa r + sinΦa θ + a Φ]
A = 2 sinΦ[sinΦ tanΦa r + sinΦa θ + a Φ]
A = 2 sinΦ[sinΦ tanΦa r - sinΦa θ - a Φ]
none of these
Answer: Option
Explanation:

Spherical :

Ar = A. a r =.sinθ.sinφ =

Aθ = A. a θ =.cosφsinφ

= 2 sin2φ

Aφ = Aa φ =cosφ = 2 sinφ

∴ A = a r + 2sin2φa θ + 2sinφaφ

A = 2sinφ[sinθ tanφ a r + sinφa θ + aφ].

18.
Consider the following program segment written for 8085 based system
LXI SP, FFFFH
MVI A, FFH
MOV BB, A
K: DCR B
PUSH HL
JNZ K
INR A
LXI H, FFCEH
CPI 20H
LHLD
The contents of accumulator is :
CE
20
indeterminate
none of these
Answer: Option
Explanation:

Since contents of HL register pair are unknown and are continuously ''PUSH''ed on the stack; retrieval of the same will result in ''Indeterminate'' data.

19.
The reduced form of the function
Y(A, B, C, D, E, F) = ∑m(0, 1, 2, 3, 4, 6, 7, 8, 9, 12, 14, 15, 16, 17, 18, 19, 20, 22, 23, 24, 25, 28, 30, 31) is given by:
A(C D + E F + D E) + D E
(A + C + D + E)(D + E + F)
(C + D + E + A)(A + D + E + F)
A C D + A E F + A(D ⊕ E)
Answer: Option
Explanation:

Y = (A + D + E + F)(A + C + D + E).

20.
A 1000 kHz carrier wave modulated 40% at 4000 Hz is applied to a resonant circuit tuned to a carrier frequency and having Q = 140. What is the degree of modulation after transmission through this circuit?
0.40
0.20
0.27
0.54
Answer: Option
Explanation:

Resulting depth of modulation

Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers