Electronics and Communication Engineering - Exam Questions Papers

36. 

Find the value of ix if each resistance is of 3 Ω and each battery is of 6 V.

A. 1.1428 V
B. 1.432 A
C. 0.2892 A
D. O A

Answer: Option C

Explanation:

Using Mesh analysis

Loop 1 : 6 = 9i1 - 3i4 ...(i)

Loop 2: 6 + 6 = - 6i2

i2 = - 2A ...(ii)

Loop 3: 6 = 9i3 - 3i4 ...(iii)

Loop 4: 6 = 9i4 - 3i3 - 3i1 ...(iv)

Solving the above equations simultaneously

i1 = i3 = 1.1428 A

i4 = 1.432 A

ix = i4 - i3 = 0.2892 A.


37. 

An image uses 512 x 512 picture elements. Each of the picture elements can take any of the 8 distinguishable intensity levels. The maximum entropy in the above image will be

A. 2097152 bits
B. 786432 bits
C. 648 bits
D. 144 bits

Answer: Option B

Explanation:

Picture elements (pixel) = 512 x 512 = 262144

Information per picture element N = log2 M

M = 2N

8 = 2N

N = 3

Information for 262144 pixels = 262144 x 3 = 786432 bits

Entropy is same as average information per message.


38. 

Find RTH

A. 200 Ω
B. 0 Ω
C. 400 Ω
D. None of these

Answer: Option C

Explanation:

Since there is no independent source in the network

VTH = O V

To find RTh apply : A current source of 1 A across A and B.

Apply KCL at node n1

Apply KCL at node Vn2

Vn1 = 2Vx put in (i)

Vn2 = 4vx

put in (ii)

100 + Vx = 2Vx

100 = Vx

Vn2 = 4Vx = 400 V

.


39. 

If then for this to be true x(t) is __________ .

A. exp (j2pf/t)
B. exp (-j2pf/t)
C. exp (j2pft)
D. exp (-j2pft)

Answer: Option C

Explanation:

Consider x(t) = ej2pft

y(t) = h(t) * x (t)

y(t) = h(t). ej2pf(t - t) . dt

= ei2pfth(t) . e- j2pft . dt

= ej2pft x H(f) = x(t) x H(f)

.


40. 

The dual transform of the given network as

A.
B.
C.
D. none of the above

Answer: Option C

Explanation:

Dual of R → G