Quiz Electronics Engineering 12 Oct 2020

Quiz Electronics Engineering
Exam: NIC
Topic: Miscellaneous
Date: 12/10/2020

Each Question carries 1 Mark
Negative Marking: 1/4
Time: 10 Minutes

Q1. In an SSB transmitter one is most likely to find
(a) Class-C audio amplifier
(b) Tuned modulator
(c) Class-B RF amplifier
(d) Class-AB power amplifier

Q2. Secondary breakdown does not occur in
(a) MOSFET but occurs in BJT
(b) Both MOSFET and BJT
(c) BJT but occurs in MOSFET
(d) None of the above

Q3. The resistance between points A and B as shown in figure will be

(a) 100 Ω
(b) 150 Ω
(c) 250 Ω
(d) 300 Ω

Q4. For the circuit shown in Fig. the value of input for Y = 0 is

(a) 00
(b) 10
(c) 01
(d) 11

Q5. To which one of the following difference equations, the impulse response h(n) = 𝛿(n + 2) – 𝛿(n – 2) corresponds?
(a) y(n + 2) = x(n) – x(n – 2)
(b) y(n – 2) = x(n) – x(n – 4)
(c) y(n) = x(n + 2) + x(n – 2)
(d) y(n) = -x(n + 2) + x(n – 2)

Q6. Which flag does not change by the execution of the instruction DCR B in 8085 microprocessor?
(a) Parity
(b) Zero
(c) Carry
(d) Sign

Q7. A charge of 1 C is placed near a grounded conducting plate at a distance of 1 m. What is the force between them?
(a) 1/(16πε_o ) N
(b) 1/(8πε_o ) N
(c) 1/(4πε_o ) N
(d) 4πε_o N

Q8. The DHCP server can provide the _______ of the IP addresses.
(a) dynamic allocation
(b) automatic allocation
(c) static allocation
(d) all of the above

Q9. An intrinsic semiconductor has the following properties:
1. Its electron concentration equals its hole concentration.
2. Its carrier density increases with temperature.
3. Its conductivity decreases with temperature.
Which of these statements are correct ?
(a) 1, 2 and 3
(b) 2 and 3 only
(c) 1 and 2 only
(d) 1 and 3 only

r

Q10. If in the below given figure resistor R_3 become open circuited, the voltmeter will read

(a) Zero
(b) 62.5 V
(c) 125 V
(d) 250 V

SOLUTIONS

S1. Ans.(c)
Sol. In SSB carrier is suppressed. So, Class B- RF amplifier is enough to amplify RF signal.

S2. Ans.(a)
Sol. Secondary breakdown is a failure mode in BJT because it has Negative Temperature Coefficient. In it negative resistance (current concentration) occurs under high-voltage and high-current conditions. MOSFET has Positive Temperatures Coefficient so no secondary breakdown in it.

S3. Ans.(b)
Sol. No current flows through the encircled branch.

⇒ RAB = 50 + 50 + (100 × 100)/(100 + 100)
⇒ RAB = 100 + 10000/200 = 150 Ω
So, RAB = 150 Ω

S4. Ans (d)
Sol. If A =1 and B = 1. Then Y = 1.

S5. Ans.(b)
Sol. Given:
h(n) = 𝛿(n + 2) – 𝛿(n – 2)
We know that,
x(n)*𝛿(n – n_o) □(↔┴( F.T ) ) x(n – n_o)
Here,
y(n) = x(n)*h(n)
⇒ y(n) = x(n)* 𝛿(n + 2) – 𝛿(n – 2)
⇒ y(n) = x(n + 2) – x(n – 2)
and y(n – 2) = x(n) – x(n – 4)

S6. Ans.(c)
Sol. ‘DCR B’ means the contents of the B register is decremented by ‘1’. Except carry flag, all the flag change due to this instruction. Carry Flag is set when carry is generated in 8085 microprocessor.

S7. Ans.(a)
Sol. Using the method of image charge.
A charge of 1 C is placed near a grounded conducting plate at a distance of 1 m so a similar but opposite charge is placed at a distance of 1 m below the conducting plate.
Distance between the two charges = 2 m
From Coulomb’s Law
F = (Q_1 Q_2)/(4πε_0 r^2 )
Now,
F = ( 1 × 1 )/(4πε_0 (2^2))
So, F = 1/(16πε_0 )

S8. Ans.(d)

S9. Ans.(c)
Sol. Conductivity of intrinsic semiconductor increases with temperature.

S10. Ans.(d)
Sol. As Voltmeter has very high internal resistance circuit so when R_3 become open circuited no current flow through voltmeter and thus circuit acts as Open Circuit. No current flow in the circuit so no voltage drop across any resistance hence voltage across the voltmeter is Supply Voltage i.e. 250 Volts.

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