Quiz Electronics Engineering
Each Question carries 1 Mark
Negative Marking: 1/4
Time: 10 Minutes
Q1. Fastest memory cell is –
(a) Bubble memory
(b) Semiconductor RAM
(c) Superconductor memory
(d) Semiconductor memory
Q2. Which one of the following must be satisfied if a signal is to be periodic for -∞ < t < ∞?
(a) x(t + T_o) = x(t)
(b) x(t + T_o) = dx(t)/dt
(c) x(t + T_o) = ∫_t^(T_o)▒〖x(t)〗dt
(d) x(t + T_o) = x(t) + kT_o
Q3. In the OSI model, as a data packet moves from the lower to the upper layers, headers are _______ .
(d) Random placement
Q4. A pn junction diode’s dynamic conductance is directly proportional to
(a) the applied voltage
(b) its current
(c) the thermal voltage
(d) the temperature
Q5. What is the value of Thevenin voltage V_Th in the given Thevenin’s equivalent circuit?
Q6. Which of the following processor structure is pipelined?
Q7. Assertion(A): Net charge within a conductor is always zero.
Reason(R): The conductor has a very large number of free electrons.
(a) Both A and R are True and R is the correct explanation of A.
(b) Both A and R are True but R is not the correct explanation of A.
(c) A is True but R is False.
(d) A is False but R is True.
Q8. After counting 0, 1, 10, 11 the next binary number is
Q9. If you have to send multimedia data over SMTP it has to be encoded into _______
Q10. v(t) = 5[cos(106πt) – sin(103πt) x sin(106πt)] represents
(a) NBFM signal
(b) AM signal
(c) SSB upper sideband signal
(d) DSB-SC signal
Sol. Semiconductor memory is made up of flip-flop so both its access time and cycle time is very less. Hence, It is the fastest memory cell.
Sol. Header contains the control information and parameters agreed between the applications. It is only sent at the beginning of an operation. In OSI model, when data packet moves from lower layers to higher layer, headers get removed and vice versa.
Sol. In a pn junction diode,
r = (ηV_T)/I
Transconductance = 1/resistance
So, g_m = I/(ηV_T )
Therefore, A pn junction diode’s dynamic conductance is directly proportional to its current.
Sol. For calculating Thevenin voltage, load resistance is removed and calculated across AB.
V_Th is Open circuit voltage at terminal AB
No current flows through DA branch due to open circuit. Net resistance is 10 Ω.
Current in open circuit is –
I = 50/10=5 A
So, Thevenin voltage is calculated across 7 Ω resistance.
⇒ V_Th = 5 × 7
So, V_Th = 35 volts
Sol. 8086 microprocessor is a microprocessor which works on the pipeline technique. It works in two stages – Fetch Stage and Execution Stage.
Sol. Both the statements are correct.
Net charge within a conductor is always zero because due to law of electrical charge neutrality.
S8. Ans (c)
Sol. The sequence is as following:
⇒ 02 = 010
⇒ 12 = 110
⇒ 102 = 210
⇒ 112 = 310
⇒ 1002 = 410
Sol. Only 7-bit ASCII codes are transmitted through SMTP. So, it is mandatory to convert binary multimedia data to 7-bit ASCII before it is sent using SMTP.
v(t) = 5[cos(106πt) – sin(103πt) x sin(106πt)]
⇒ v(t) = 5cos(106πt) – 5/2cos(106 – 103) πt + 5/2cos(106 + 103) πt
From above expression of signal, we get that carrier and upper side band frequency components are in phase but the lower side band frequency component is out of phase.
Therefore, The given expression represents a NBFM signal.