Electronics Quiz-2

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. The transistors provide good power amplification when they are used in
  •  Common collector configuration
  •  Common emitter configuration
  •  Common base configuration
  •   None of these

Q2.The voltage gain of the following amplifier is

  •  10
  •  100
  •  1000
  •  9.9
(b) Voltage gain A+Vo/Vi=Rf/Ri=(100 kΩ)/(1 kΩ)=100 

Q3.  The truth table given below is for (A and B are the inputs, Y is the output)
  •  NOR gate
  •  AND gate
  •  XOR gate
  •  NAND gate
(d) The output Y is a combination of AND + NOT gate. Hence, the truth table is for NAND gate.

Q4. The circuit shown in the figure contains two diodes each with a forward resistance of 30 Ω and with infinite backward resistance. If the battery is 3V, the current through the 50 Ω resistance (in ampere) is
  •  0
  •  0.01
  •  0.02
  •  0.03
(c) In the circuit the upper diode D1 is reverse biased and the lower diode D2 is forward biased. Thus there will be no current across upper diode junction. The effective circuit will be as shown in figure.

 Total resistance of circuit R\=50+70+30=150 Ω Current in circuit, I=V/R=3/150=0.02 A.

Q5.The potential in depletion layer is due to
  •  Electrons
  •  Holes
  •  Ions
  •  Forbidden Band
 (c) The potential in depletion layer is due to ions. It appears as if some fictitious battery is connected across the junction with its negative pole connected to p-region and positive pole connected to n-region. The potential difference developed across the junction due to migration of majority charge carries in potential barrier.

Q6. Pure Si at 500 K has equal number of electron (ne) and hole (nh) concentrations of 1.5×〖10〗16 m-3. Doping by indium increases neh to 4.5×1022 m-3. The doped semiconductor is of
  •  n-type with electron concentration ne=2.5×1023m-3
  •  p-type having electron concentration ne=5×109m-3
  • p-type having electron concentration ne=5×1022m-3
  •  n-type with electron concentration ne=2.5×1010m-3

Q7.A p-n photodiode is made of a material with a band gap of 2.0 eV. The minimum frequency of the radiation that can be absorbed by the material is nearly
  •  10×1014 Hz
  •  5×1014 Hz
  •  1×1014 Hz
  •  20×1014 Hz
(b) p-nphotodiode is a semiconductor diode that produces a significant current when illuminated. It is reversed biased but is operated below the breakdown voltage. Energy of radiation = band gap energy

Q8.In an intrinsic semiconductor, the Fermi level is
  •  Nearer to valency band than conduction band
  •  Equidistance from conduction band and valency band
  •  Nearer to conduction band than valency band
  •  Bisecting the conduction band
(b) In intrinsic semiconductor of Fermi level is near the middle of the forbidden gap.

Q9. A common emitter amplifier is designed with NPN transistor (α=0.99). The input impedance is 1 KΩ and load is 10 KΩ. The voltage gain will be
  •  9.9
  •  99
  •  990
  •  9900
(c) Voltage gain =β× Resistance gain β=α/(1-α)=0.99/((1-0.99))=99 Resistance gain =(10×103)/103=99 ⇒ Voltage gain =99×10=990

Q10. Find VAB
  •  10 V
  •  20V
  •  30V
  • None of these
(a) Diode is in forwards biasing hence the circuit can be redrawn as follows 

 VAB=30/((10+5))×5=10 V

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