Quiz Electronics Engineering

Exam: NIC

Topic: Miscellaneous

Date: 20/07/2020

Each Question carries 1 Mark

Negative Marking: 1/4

Time: 10 Minutes

Q1. Which is the dominant mode in rectangular waveguides?

(a) TE_10

(b) TE_11

(c) TM_01

(d) TM_11

Q2. The signal to quantization noise ratio in an n-bit PCM system

(a) depends upon the sampling frequency employed

(b) is independent of the value of â€˜nâ€™

(c) increasing with increasing value of â€˜nâ€™

(d) decreases with the increasing value of â€˜nâ€™

Q3. The light emitted diodes consist of:

(a) Si

(b) GaAs

(c) Ge

(d) Diamond

Q4. Which of the following flip-flop is used as latch?

(a) JK flip-flop

(b) D flip-flop

(c) RS flip-flop

(d) T flip-flop

Q5. What is the power of periodic non-sinusoidal voltage and currents?

(a) The sum of the root mean square power of the sinusoidal components including the fundamental

(b) The sum of the average powers of the harmonics excluding the fundamental

(c) The sum of the average powers of the sinusoidal components including the fundamental

(d) The average power of the fundamental components alone.

Q6. Four capacitors are connected as shown in figure. What is the total capacitance?

(a) 9 Î¼F

(b) 20 Î¼F

(c) 10 Î¼F

(d) 36 Î¼F

Q7. Phase velocity â€˜v_pâ€™ and the group velocity â€˜v_gâ€™ in a waveguide (â€˜câ€™ is velocity of light) are related as

(a) v_p Ã— v_g = c^2

(b) v_p + v_g = c

(c) v_p/v_g = a constant

(d) v_p + v_g = a constant

Q8. Four messages signal band limited to W, W, 2W and 3W respectively are to be multiplexed using Time Division Multiplexing(TDM). The minimum bandwidth required for transmission of this TDM signal is

(a) W

(b) 3W

(c) 6W

(d) 7W

Q9. In which of these is reverse recovery time nearly zero?

(a) Zener diode

(b) Tunnel diode

(c) Schottky diode

(d) PIN diode

Q10. If a parallel resonant circuit is shunted by a resistance then

(a) circuit impedance is decreased

(b) the gain of the circuit is increased

(c) â€˜Qâ€™ of the circuit is increased

(d) None of these

SOLUTIONS

S1. Ans.(a)

Sol. Dominant mode is that mode for which the cut-off wavelength (Î»_C) achieves a maximum value. it has lowest cut off frequency which is equal to c/2a. So, it has maximum cut-off wavelength.

S2. Ans.(c)

Sol. Signal to quantization noise in an n bit PCM is

S/N_Q = (SQNR) = 3/2.22n â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(1)

Where, n is number of bits

It can also be written in dBs

(SQNR)dB = (S/N_Q )_dB= 1.76 + 6nâ€¦â€¦â€¦â€¦â€¦â€¦â€¦(2)

From above equations, we get that (S/N) increases with increasing value of â€˜nâ€™.

S3. Ans.(b)

Sol. A Light Emitting Diode (LED) is a semiconductor electronic components. It emits light when current flows through it. Electrons in the semiconductor recombine with electron holes, releasing energy in the form of photons.

S4. Ans.(c)

Sol. RS flip-flop is used as a latch

S5. Ans.(a)

S6. Ans.(a)

Sol. All the three 6 Î¼F capacitors are connected in parallel.

Net capacitance is

â‡’ 6 + 6 + 6 = 18 Î¼F

Now, Net capacitance 18 Î¼F is in series with 18 Î¼F.

The equivalent capacitance is

C_eq=(18 Ã— 18 )/(18 +18 )=(18 Ã— 18)/36=9Î¼F

S7. Ans.(a)

S8. Ans.(d)

Sol. Given:

fm1 = W

fm2 = W

fm3 = 2W

fm4 = 3W

In TDM, Minimum bandwidth is

BWmin = R_b/2

For minimum bandwidth, n = 1, that means Rb = nfs

Sampling frequency of each message signal is

â‡’ fs1 = 2fm1 = 2 x W = 2W

â‡’ fs2 = 2fm2 = 2 x W = 2W

â‡’ fs3 = 2fm3 = 2 x 2W = 4W

â‡’ fs4 = 2fm4 = 2 x 3W = 6W

âˆ´ fs = fs1 + fs2 + fs3 + fs4 = 14W

Now,

âˆ´ Rb =1 x 14W = 14W

So, BWmin = 14W/2 = 7W

Therefore, The minimum bandwidth required for transmission of this TDM signal is 7W.

S9. Ans. (c)

Sol. When switching from the conducting to the blocking state, a diode or rectifier has stored charge which must be discharged at first before the diode blocks reverse current. This discharging or reverse flow of current, takes a finite amount of time known as the Reverse Recovery Time.

S10. Ans.(a)

Sol. Net impedance,

Z = âˆš(R Â² +XÂ² )

where,

R = resistance of circuit

X = reactance of circuit

When parallel resonant circuit is shunted by a resistance then reactance of the circuit decreases hence circuit impedance is decreased.