ELECTRONICS-QUIZ 16

ELECTRONICS-QUIZ 16

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 an experiment, the saturation in the plate current in a diode is observed at 240V.

But a student still wants to increase the plate current. It can be done, ifIn an experiment, the saturation

in the plate current in a diode is observed at 240V. But a student still wants to increase the plate current.

It can be done, if

•  The plate voltage is increased further
•  The plate voltage is decreased
•  The filament current is decreased
•  The filament current is increased

Solution
(d) After saturation plate current can be increased by increasing the temperature of filament. It can be done by increasing the filament current

Q2.In the figure, potential difference between A and B is

•  Zero
•  5 V
•  10 V
•  15 V

Solution
(c) Here p-n junction is forward biased. If p-n junction ideal, its resistance is zero. The effective resistance across A and B =(10×10)/(10+10)=5kΩ. Current in the circuit I=V/R=30/(15×10³ )=2/10³ A Current in arm AB=I=2/10³ Potential difference across A and B=2/10³ ×5×10³=10 V.

Q3. Three amplifier stages each with a gain of 10 are cascaded. The overall gain is

•  10
•  30
•  1000
•  100

Solution
(c) In K is a gain of one stage, then total gain of n stages =(K)ⁿ=10³=1000

Q4. On increasing the reverse bias to a large value in a p-n junction, diode current

•  Increases slowly
•  Remains fixed
•  Suddenly increases
•  Decreases slowly

Solution
(c) Under normal reverse voltage, a very little reverse current flows through a p-n junction. However, if the reverse voltage attains a high value, the junction may breakdown with sudden rise in reverse current. If reverse voltage is increased continuously, the kinetic energy of electrons (minority carriers) may become high enough to knock out electrons from the semiconductor atoms. At this stage breakdown of the junction occurs characterised by a sudden rise of reverse current and a sudden fall of the resistance of barrier region. This may destroy the junction permanently.

Q5.The energy band gap is maximum in

•  Metals
•  Superconductors
•  Insulators
•  Semiconductors

Solution
(C)Insulators  

Q6. The coordination number of Cu is

•  1
•  6
• 10
•  12

Solution
(d) Cu has fcc structure, for fcc structure co-ordination number =12

Q7.A semiconductor dopped with a donor impurity is

•  P-type
•  N-type
•  NPN-type
•  PNP-type

Solution
(b)N-type

Q8.The grid in a triode valve is used

•  To increases the thermionic emission
•  To control the plate to cathode current
•  To reduce the inter-electrode capacity
•  To keep cathode at constant potential

Solution
(b) To control the plate to cathode current

Q9.The minimum potential difference between the base and emitter required to switch a silicon transistor ‘ON’ is approximately

•  1 V
•   3 V
•  5 V
•  4.2 V

Solution
(a) Factual

Q10. Assuming the diodes to be of silicon with forward resistance zero, the current I in the following circuit is

•  0
•  9.65 mA
•  10 mA
•  10.36 mA

Solution
(c) Total resistance in the circuit =2kΩ E=20V 

∴ The current=20/2000Ω=10 mA