Quiz Electronics Engineering 23 July 2020

Quiz Electronics Engineering
Exam: NIC
Topic: Miscellaneous
Date: 23/07/2020

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

Q1. Heat generated across a resistance is
(a) directly proportional to current flowing through it
(b) inversely proportional to the value of resistance
(c) directly proportional to the square of current flowing through it
(d) directly proportional to square of resistance

Q2. In the saturated region, the transistor acts like a_________
(a) closed switch
(b) amplifier
(c) open switch
(d) poor transistor

Q3. The 2’s compliment of the binary number (01010101)_2 is
(a) (10000000)_2
(b) (11111111)_2
(c) (01010110)_2
(d) (10101011)_2

Q4. A Butterworth filter is known as…….
(a) An ideal high pass filter
(b) An ideal bond pass filter
(c) an ideal low pass filter
(d) None of these

Q5. The pulse communication system that is highly immune to noise is
(a) Pulse Amplitude Modulation (PAM) system
(b) Pulse Code Modulation (PCM) system
(c) Pulse Width Modulation (PWM) system
(d) Pulse Position Modulation (PPM) system

Q6. A quarter wave transformer matching a 75Ω source with a 300Ω load should have a characteristic impedance of
(a) 150 Ω
(b) 100 Ω
(c) 50 Ω
(d) 200 Ω

Q7. How many entries will be in the truth table of a 3 input NAND gate?
(a) 3
(b) 8
(c) 6
(d) 9

Q8. The impulse response of a linear time invariant system is given as g(t) = e^(-t), t > 0.
The transfer function of the system is equal to
(a) 1/s
(b) 1/(s(s+1))
(c) 1/((s+1))
(d) s/((s+1))

Q9. Transformer transforms
(a) power
(b) voltage
(c) current
(d) frequency

Q10. A 1 kW carrier is Amplitude Modulated to a depth of 60%. The total power in the modulated carrier is
(a) 1 kW
(b) 1.06 kW
(c) 1.6 kW
(d) 1.18 kW

SOLUTIONS

S1. Ans. (c)
Sol. Heat generated across a resistance is directly proportional to the square of current flowing through it. Heat generated is
H = i^2Rt

S2. Ans.(a)
Sol. In saturation mode, A transistor acts like a short circuit between collector and emitter. Both emitter and collector are forward biased. The transistor now acts like a closed switch.

S3. Ans. (d)
Sol. For 2’s compliment we use 1st compliment and then add 1.
Step 1: Write the compliment of the given binary number
01010101 ⟶ 10101010
Step 2: Add 1 to the complemented number
10101010 + 1 = 10101011

Therefore, The 2’s compliment of the binary number (01010101)_2 is (10101011)_2.

S4. Ans.(c)
Sol. The Butterworth filter has a flat frequency response. It has Flat Magnitude filter from 0 Hz to cut-off frequency without any ripples. So, it is also called as an ideal Low Pass filter.

S5. Ans(b)
Sol. In PCM modulation, the analog signal is sampled at regular intervals and after that it is quantized and transmitted serially as data words in pulse code i.e., digital signal. It is highly immune to noise hence highly efficient. PAM, PWM and PPM are more affected by noise.

S6. Ans.(a)
Sol. Given:
ZS = 75Ω
ZL = 300Ω
We know that, in case of quarter wave transformer,
Characteristic Impedance(ZO) = √(Z_S Z_L )
Now, ZO = √(75 x 300) = √22500
So, ZO = 150 Ω

S7. Ans(b)
Sol. Total number of Entries in the truth table of a NAND gate with n number of inputs is
⇒ Y = 2n

Here, n = 3,
So, Y = 2^3 = 8
Therefore, there will be 8 entries in the truth table of a 3 input NAND gate.

S8. Ans.(c)
Sol. Given:
g(t) = e^(-t), t > 0.
Its Laplace transform is –
G(s) = 1/(s+1)

S9. Ans.(a)
Sol. By varying the winding turns of the transformer, both magnitude of voltage and current can be varied. Since P = VI so transformer transforms the power.

S10. Ans(d)
Sol. Given:
Carrier Power(P_c) = 1 kW
Modulation depth(µ) = 60% = 0.6
We know that, Total power is
P_t=P_c (1+μ^2/2)

Now, P_t = 1(1+〖0.6〗^2/2) kW
=1 (1+0.18) kW = 1.18 kW.

Therefore, The total power in the modulated carrier is 1.18 kW.

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