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New answer posted

10 months ago

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P
Pallavi Pathak

Contributor-Level 10

Chapter 1 Electric Charges and Fields NCERT Exemplar goes beyond the basic NCERT textbook and contains application-based and advanced-level questions. It is designed to enhance students' problem-solving skills, improve their understanding of electrostatics, and prepare them for their CBSE board exams and competitive exams like NEET and JEE.

New answer posted

10 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (a, d)

Explanation- In reverse biasing, the minority charge carriers will be accelerated due to reverse biasing, which on striking with atoms cause ionization resulting secondary electrons and thus more number of charge carriers.

When doping concentration is large, there will be large number of ions in the depletion

region, which will give rise to a strong electric field.

New answer posted

10 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- a, c, d

Explanation-ripple factor is inversely proportional to RL, C  and frequency. so to reduce ripple factor all these should be increase.

New answer posted

10 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer-b, d

Explanation- During regulation action of a Zener diode, the current through the Rs changes and resistance offered by the Zener changes. The current through the Zener changes but the voltage across the Zener remains constant.

New answer posted

10 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- a, b, d

Explanation- The space-charge regions on both the sides of p-n junction which has immobile ions and entirely lacking of any charge carriers will form a region called depletion region of a diode.

The number of ionized acceptors on the p -side equals the number of ionized donors on the n-side.

New answer posted

10 months ago

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A
alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- b, c

Explanation- IC= 10mA

Ic= 95/100  Ie

Ie= 10 * 100 95 = 10.53mA

Ib=Ie-IC= 10.53-10= .53mA

New answer posted

10 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar 

Answer- a, c

Explanation- Here emitter-base junction is forward biased i.e., the positive pole of emitter base battery is connected to base and its negative pole to emitter. Also, the collector base junction is reverse biased, i.e., the positive pole of the collector base battery is connected to collector and negative pole to base.

Thus, electron move from emmiter to base and cross over from emitter to collector.

New answer posted

10 months ago

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A
alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (a, c)

Explanation- when we apply temperature across semiconductor then electron will starts from lower energy to high energy level that is from valence band to conduction band. When electron goes from lower to higher then holes that is left behind they goes to lower energy levels.

New answer posted

10 months ago

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A
alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- c

Explanation- C=A.B and D=A'B

E= C+D = (A.B)+ (A'.B)

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B

A'

C=A.B

D=A'B

E=C+D

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New answer posted

10 months ago

0 Follower 9 Views

A
alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer-b

Explanation- r1=5kohm and r2= 5kohm and both are in series

Then V-0.3 = [ (r1+r2)103] * [0.2 * 10-3]

 V-0.3= 2

V= 2.3V

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