Electronics and Communication Engineering - Exam Questions Papers

Exercise :: Exam Questions Papers - Exam Paper 10

16.

A conductor in x-y plane and having length 1 m is moving with a velocity V = (2i + 3j + k) m/sec. A magnetic induction field B = (i + 2j) Wb/m² is applied to the conductor. The potential difference between the ends of the conductor is

A.	0
B.	4.88 V
C.	6 V
D.	None of these

Answer: Option C

Explanation:

= - (V x B) = - (2i + 3j + k) x (i + 2j)

∴ || = (E²_x + E²_y + E²_z) = 6 v/m

∴ Potential Difference across 1 m length = 6.

17.

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 passing the wave through this circuit?

A.	0.4
B.	0.2
C.	0.27
D.	0.554

Answer: Option C

Explanation:

Resulting depth of modulation is given by :

when δ =

f_c = 1000 x 10³ Hz

f_m = 4 x 10³ Hz

δ =

m₀ =

= = 0.27 .

18.

The V_TH at terminals A and B is equal to,

A.	0 V
B.	5 V
C.	- 5 V
D.	None of these

Answer: Option B

Explanation:

Apply KVL to loop 1

5V - 3I₁ - 2(I₁ - I₂) - 5V = 0

∴ I₂ = 0

5 - 3I₁ - 2I₁ - 5V = 0

∴ I₁ = 0

∴ V_AB = 5V.

19.

If , then y has a

A.	maximum at x = e
B.	minimum at x = e
C.	maximum at x = e^-1
D.	minimum at x = e^-l

Answer: Option A

Explanation:

.

20.

If . dt, then F[y(t)] is

A.	- jsgn(f). x(f)
B.	j2pf . e^-jptf . x(f)
C.
D.

Answer: Option A

Explanation:

⇒ F{Y(t)} = Y(f) = X(f) x

sin (t) ⇌

By duality :⇌ sgn + (- f)

⇒ ⇌ jsgn (- f)

∴⇌ - j sgn (f)

∵ sgn(- f) = - sgn(f)

⇒ Y(f) = X(f). [- j sgn(f)].

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