Electronics-Quiz-4

Electronics Quiz-4

Dear Readers,

As per analysis for previous years, it has been observed that students preparing for NEET find Physics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Biology due to their medical background. 

Furthermore, sections such as Physics are dominantly based on theories, laws, numerical in comparison to a section of Biology which is more of fact-based, life sciences, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced. 

Q1. In a common emitter transistor amplifier β=60,R₀=5000Ω and internal resistance of a transistor is 500Ω. The voltage amplification of amplifier will be

•  500
•  460
•  600
•  560

Solution
(c) A_V=β R₀/R_i=60× 5000/500=600.

Q2.The truth table shown in figure is for

•  XOR
•  AND
•  XNOR
•  OR

Solution

Q3.  The valency of the impurity atom that is to be added to germanium crystal so as to make it a N-type semiconductor, is

•  6
•  5
•  4
•  3

Solution
(B)

Q4. Carbon, silicon and germanium have four valence electrons each. At room temperature which one of the following statements is more appropriate?

•  The number of free conduction electrons is significant in C but small in Si and Ge
•  The number of free conduction electrons is negligibly small in all the three
•  The number of free electrons for conduction is significant in all the three
•  The number of free electrons for conduction is significant only in Si and Ge but small in C

Solution
(d) The number of free electrons for conduction is significant only in Si and Ge but small in C, as C is an impurity.

Q5.In a common emitter amplifier, using output resistance of 5000 Ω and input resistance of 2000 Ω, if the peak value of input signal voltage is 10 mV and β=50, then peak value of output voltage is

•  5×10^-6 V
•  12.50×10^-6 V
•  1.25 V
•  125 V

Solution
 (c) A_v=(ΔV₀)/(ΔV_i )=βR₀/R_i ; So,∆V₀=∆V_i×β R₀/R_i ∴ ∆V₀=10×50×500/2000=1250 mV=1.25 V.

Q6. At room temperature, a P-type semiconductor has

•  Large number of holes and few electrons
•  Large number of free electrons and few holes
• Equal number of free electrons and holes
•  No electrons and holes

Solution
(a) In P-type semi conductor, holes are majority charge carriers

Q7.The output of a NAND gate is 0

•  If both the inputs are zero
•  If one input is 0 and the other input is 1
•  If both the inputs are 1
•  Either if both inputs are 1 or if one of the inputs is 1 and the other 0

Solution
(c) If inputs are A and B then output for NAND gate is Y=(AB) ̅ ⇒ If A=B=1,Y=(1.1) ̅=1 ̅=0

Q8.For a junction diode the ratio of forward current (I_F) and reverse current (I_r) is [I_e=electronic charge, V=voltage applied across junction, k= Boltzmann constant, T=temperature in kelvin]

•  e^(-V/kT)
•  e^(V/kT)
•  e^(-eV/kT)+1
•  e^(V/kT)-1

Solution
(d) Current in junction, V is positive; In reverse bias V is negative. Then I_r=I₀ I_F/I_r =(I₀(e^(eV/kT)-1))/I₀=(e^eV/kT-1)

Q9.What is the name of the level formed due to impurity atom in p-type in the forbidden gap?

•  Donor level
•  Acceptor level
•  Conduction Level
•  Forbidden level

Solution
(b) In p-type doping, the impurities have left holes in the atomic structure which tend to attract and hold free electrons. This pulls the fermi level down until it gets close to the valence bond. Hence, name of level formed due to impurity atom is p-type in the forbidden gap is acceptor level. 

Q10. Which of the following gates will have an output of 1

•  
  
  
  
•  
  
  
  
•  
  
  
  
• 
  
  

Solution
(c) For ‘NAND’ gate (option c), output =(0.1) ̅=0 ̅=1