Oswaal CBSE Question Bank Class 12 Physics Chapterwise & Topicwise (For March 2020 Exam) 9389067251, 9789389067255

Table of contents :
Cover
Title page
Copyright page
PREFACE
CONTENTS
Latest Syllabus for Academic Year (2019-20)
Sample Question Paper For 2020 Examination
SOLVED PAPER with Marking Scheme
SOLVED PAPER Topper's Answer 2018
MIND MAPS
Unit – I Electrostatics
Unit – II Current Electricity
Unit – III Magnetic Effects of Current and Magnetism
Unit – Iv Electromagnetic Induction and Alternating Currents
Unit – V Electromagnetic Waves
Unit – VI OPTICS
Unit – VII Dual Nature of Radiation and Matter
Unit – VIII Atoms & Nuclei
Unit – Ix Electronic Devices
Unit – x Communication Systems

Citation preview

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Strictly as per the CBSE Revised Syllabus Dated 29th March, 2019

OSWAAL BOOKS LEARNING MADE SIMPLE

FOR

MARCH 2020 EXAM

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CBSE QUESTION BANK Chapterwise & Topicwise

SOLVED PAPERS An i

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2012-2019 PHYSICS

CLASS 12 with

mind maps

Previous Years’ Solved Papers (All Sets) from 2012 to 2019 (With CBSE Board Marking Scheme Answers 2012 - 2019) Sample Paper for 2020 Examinatton Topper's Handwritten Answer Sheet 2016 - 2018

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13 EDITION, 2019

EDITION

CENTRAL BOARD OF SECONDARY EDUCATION DELHI

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SYLLABUS COVERED

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“978-93-89067-25-5”

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All rights reserved. All rights reserved. No part of this book may be No part of this book may be reproduced, reproduced, in a system, retreival stored stored in a retrieval system, or transmitted, in by any from or transmitted, in any form or any means, or be any means, without written without written permission from the publisher. permission from the publishers.

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COPYRIGHT RESERVED BY THE PUBLISHERS

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PREFACE CBSE always believes in Global Trends of Educational Transformation. The CBSE curriculum gets its lead from National Curriculum Framework – 2005 and Right to Free and Compulsory Education Act – 2009. The aim of CBSE Curriculum is not just to let learners obtain basic knowledge but to make them life-long learners. CBSE always updates and reviews the syllabus to make it more relevant with educational transformation and in last few years the chapters and topics which CBSE has added are very interesting and increase practical knowledge. The latest updation of CBSE curriculum happened on 29th March 2019 applicable for the Academic Year 2019-2020.

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Oswaal Solved Papers are designed as per the latest curriculum and emphasize on nurturing individuality thus enhancing one’s innate potentials which help in increasing the self-study mode for students. Features like Chapterwise and Topicwise presentation strengthen knowledge and attitude related to the subject. Oswaal Solved Papers are designed in such a way that students can set their own goals and can improve their problem solving and thinking skills.

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The journey of this book is never ending as it is reviewed every year. It is thoroughly updated as per the latest guidelines of the CBSE Board thereby strictly following the latest syllabus and pattern of the Board. It contains more than sufficient questions which help students in practicing and completing the syllabus. Quick Review in the beginning of each chapter is an added advantage of using this book. Questions incorporated in this book encompass all the ‘Typologies’ mentioned by CBSE namely Remembering, Understanding, Application, Analysing & Evaluation and Creation. Solutions for these have been checked twice and efforts have been made to align them closely to the Marking Scheme. Practically, this book provides students everything they need to learn and excel.

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Some of the Benefits studying term of Oswaal Solved Papers are :  Strictly based on the latest CBSE Curriculum issued on 29th March 2019 for Academic Year 2019-2020, following the latest NCERT Textbook  Previous Years’ Board Questions for in depth study  Handwritten Toppers’ Answers  Answers strictly as per the CBSE Marking Scheme  All Typology of Questions included based on Bloom’s Taxonomy for cognitive skills development  Multiple Choice Questions included as per the latest design of the Question Paper issued by CBSE  Revision Notes for comprehensive study  Mind Maps in each chapter for remembering facts & figures easily  ‘Most likely Questions’ generated by our Editorial Board with 100+ years of teaching experience.  Suggested videos at the end of each chapter for a Digital Learning Experience. At last we would like to thank our authors, editors, reviewers and specially students who regularly send us suggestions which helps in continuous improvement of this book and makes this book stand in the category as “One of the Best”. Wish you all Happy Learning. –Publisher

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am Hrishikesh Lahudkar Podar Inter. School, Akola

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A good book to read for Boards. Helps to clarify topics and prepare for 12th exams and get good scores

Ravindra Sharma, PGT Teacher, Dhar, M.P.

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Y. L. Nayana Kumari, Teacher, Birla International School, Kishangarh, Ajmer

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I will be appearing for my CBSE 12 boards this year and am indebted to Oswaal Books for the Sample Question Paper series. I have been solving them th since Class 9 and have always found them hugely beneficial!

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We are happy to inform you that your publication / books are of great use and beneficial to our students. Thank you so much.

Oswaal Books are very nice and easy to understand. These books give guidance to students for exams in a very simple way. I suggest students to refer to Oswaal Books for getting maximum marks in exams.

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This book is really nice. By using this book you can have a good idea about the questions asked in the exam. The mindmaps are also helpful in getting the concept cleared.



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Great book. All the questions are good and the solutions are easy and understandable. Tricks for solving the problems is also in this book.



Nishant Sharma Student, St. Paul School, Bhopal, M.P.

Kartik Kashyap, Student, KV DGQA, Chennai, Tamil Nadu

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Great book covering almost all the previous years’ questions with point to point answers and Toppers' answer sheet.



Ankur, Student St. Francis School, Ghaziabad, U.P.

CONTENTS      

Latest Notification and Syllabus issued by CBSE on 29th March 2019 for Academic Year 2019-20 Chapter Navigation and Year Planner Sample Question Paper -2020 Solved Paper - 2019 (Delhi & Outside Delhi) with CBSE Marking Scheme Topper’s Answer 2018 (Issued by CBSE) Mind Maps

9 21 25 32 88 113

- 20 - 22 - 31 - 87 - 112 - 127

In each chapter, for better understanding, questions have been classified according to the typology issued by CBSE as : A - Applying

C - Creating,

AE - Analysing and Evaluating

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1 - 32

33 - 67

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Electric Charges and Fields Topic - 1 : Electric Field and Dipole Topic - 2 : Gauss’s Theorem and its Applications Electrostatic Potential and Capacitance Topic - 1 : Electric Potential Topic - 2 : Capacitance

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UNIT-I : ELECTROSTATICS

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R - Remembering, U - Understanding,

UNIT-II : CURRENT ELECTRICITY

Current Electricity Topic - 1 : Electric Current, Resistance and Cells Topic - 2 : Kirchhoff ’s Laws, Wheatstone bridge and their applications Topic - 3 : Metre Bridge Potentiometer and their Applications

68 - 111

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UNIT-III : MAGNETIC EFFECTS OF CURRENT AND MAGNETISM Moving Charges and Magnetism 112 - 149 Topic - 1 : Magnetic Field & Cyclotron Topic - 2 : Ampere’s Circuital Law and its Applications Topic - 3 : Torque and Galvanometer Magnetism and Matter 150 - 169 Topic - 1 : Magnetic Dipole Topic - 2 : Magnetic Field of a solenoid Earth’s Magnetism & Magnetic Properties of Materials

UNIT-IV : ELECTROMAGNETIC INDUCTION AND ALTERNATING CURRENTS 6. 7.

Electromagnetic Induction Topic - 1 : Magnetic Flux and Faraday’s Laws Topic - 2 : Eddy Currents, Self and Mutual Induction and AC Generator Alternating Currents Topic - 1 : Alternating Current Topic - 2 : LCR Series Circuits Topic - 3 : AC Generator and Transformer (7)

170 - 194

195 - 225



... contd. Contents UNIT-V : ELECTROMAGNETIC WAVES

8. Electromagnetic Waves Topic - 1 : Electromagnetic Waves and Maxwell’s Equations Topic - 2 : Electromagnetic Spectrum

226 - 242

UNIT-VI : OPTICS

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9. Ray Optics and Optical Instruments 243 - 276 Topic - 1 : Reflection by spherical mirrors Topic - 2 : Refraction through Glass Slab, Prism and Lenses and Total Internal Reflection Topic - 3 : Optical Instruments 10. Wave Optics 277 - 308 Topic - 1 : Wave theory and Huygens’ Principle Topic - 2 : Superposition of light waves (Interference and Diffraction) Topic - 3 : Resolving Power of Optical instruments and Polarisation of light

UNIT-VII : DUAL NATURE OF RADIATION AND MATTER 309 - 330

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11. Dual Nature of Radiation and Matter Topic - 1 : Photoelectric Effect Topic - 2 : Dual Nature of Matter

UNIT-VIII : ATOMS & NUCLEI 331 - 347 348 - 366

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12. Atoms 13. Nuclei Topic - 1 : Nucleus and Mass Energy Relation Topic - 2 : Radioactivity and Nuclear Reactor

UNIT-IX : ELECTRONIC DEVICES

14. Semiconductor Electronics : Materials, Devices and Simple Circuits Topic - 1 : Energy Bands Topic - 2 : Semiconductors Diodes and Applications

367 - 390

UNIT-X : COMMUNICATION SYSTEMS 15. Communication Systems* Topic - 1 : Elements of Communication System Topic - 2 : Modulation

391 - 408

Note*:- Other Important Chapter / Topics for Competitive Examinations Preparation.

qq LOOK OUT FOR 'HIGHLY LIKELY QUESTIONS'

These questions are selected by Oswaal Books Proprietary Artificial Intelligence Algorithm. They are highly likely to be asked in the upcoming examinations.

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CENTRAL BOARD OF SECONDARY EDUCATION

F.1001/CBSE-Acad/Curriculum/2019 

March 29th, 2019

 All Heads of Institutions affiliated to CBSE

Cir No Acad-20/2019

Subject: Secondary and Senior Secondary School Curriculum for academic session 2019-20

1.

The curriculum provided by CBSE seeks to provide opportunities for students to achieve excellence in learning is based on National Curriculum Framework-2005 and adopted/ adapted from the NCERT Curriculum.



2.

It is important that schools ensure curriculum transaction as per directions given in the Curriculum document. The subjects to be taught must be as per syllabi given. Before going through the syllabus, the teachers must be well-versed with the strategies given in the initial pages of the curriculum. Therefore, it is desired that the head of the school may circulate the initial pages of the curriculum and the syllabi to all the teachers and ensure that curriculum is transacted as per the directions given therewith for optimal learning and specifically to avoid difficulties for students at the time of the examinations. Teachers may also be advised to refer the Sample Question Papers published by the Board for preparing students for Board’s Examination as and when they are made available on CBSE Website.



3.

As far as Secondary and Senior Secondary Curriculum 2019-20 is concerned, the changes notified vide Circular Nos. Acad-3/2019 dated 10.01.2019, Acad- 11/2019 dated 06.03.2019 and Acad-12/2019 dated 08.03.2019 are incorporated in the respective Curriculum documents.



4.

It is retreated that any deviation in transaction of the curriculum may seriously affect holistic assessment of students and their preparation for the Board examinations. Hence, all the teachers and students are cognizant of the Curriculum Document (Initial Pages and Syllabus) in totality, which is available on www.cbseacademic.nic.in under the link - curriculum.

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(Dr. Joseph Emmanuel) Director (Academics)

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CENTRAL BOARD OF SECONDARY EDUCATION

F.1001/CBSE/Dir(Acad)/2019 

March 6, 2019



Circular No. Acad-11/2019

All Heads of CBSE affiliated schools Subject: Strengthening Assessment and Evaluation Practices of the Board.

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As we focus our attention on ‘what our students are learning’ alongwith ‘the skill sets they are acquiring’, it becomes imperative to strengthen current Assessment and Evaluation Practices and align them to the future requirements of the learners. The Board has always stressed that its students must acquire the skills of critical thinking, problem solving, analyzing information, collaboration, effective communication, developing curiosity and imagination as part of the learning process. You are also aware that the Board continues to make small changes in the assessment and evaluation practices almost every year to eventually reach the goal of achieving the aforementioned skills for all its students.

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Report card of the National Assessment Survey (2017-18) has indicated that the performance of CBSE class X students in Mathematics, Science, Social Science, English and Modern Indian Language is 52%, 51%, 53%, 58% and 62% respectively. Although this competency-based survey places the CBSE students above the national average, it indicates that there is ample scope for improvement in their performance.

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Further, the decision by MHRD to participate in PISA (Programme for International Student Assessment) in 2021, has given even more impetus to the requirement of aligning the Board’s assessment system to future requirements. It may also be mentioned here, that since the Board has made it mandatory for all its affiliated schools to adopt the Learning Outcomes vide circular dated 18.01.2019, it is a necessity now that ‘assessment of learning’ must be augmented with ‘assessment as learning’ and ‘assessment for learning’. In view of above, countrywide consultations were held with CBSE stake holders including teachers, students, heads of Institutions and experts in the field to suggest ways to strengthen the Assessment and Evaluation Practices of the Board. It was agreed upon that the School Based /Internal Assessment needs to be strengthened by incorporating more diverse strategies. Further, there is need of exposing the students to different types and formats of questions in the year end/Board examination, so that a large range of learning outcomes can be assessed. Based on this, following few changes are proposed in the Assessment and Examination practices for the year 2019-20 onwards. As the next academic session (2019-20) is going to start from April 2019 onwards, hence a summary view of the proposed changes in Internal Assessment and Year End/ Board Examinations are hereby informed in advance.

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CENTRAL BOARD OF SECONDARY EDUCATION The changes proposed for XI-XII (2019-20) are as under :

Classes XI-XII Existing Maximum marks per subject

Proposed

100

No Change



Internal Assessment

No Change

Assessment



Board Examination



Assessment of Co-scholastic areas (Health and Physical Education including Work Education, General Studies)

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Types of

Political



70:30 (Fine Arts and some other subjects)

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0:100 (Mathematics, Languages, Science, and Legal Studies)

Compulsory portion of Internal Assessment/ Project  20:80 (Humanities / Commerce Based Subjects work/ Practical in all and some other subjects) subjects of at least 20 marks  30:70 (Science based subjects and some other subjects )

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Distribution



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Marks

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Internal Assessment / Practical: Board Exam / Year end Exam

Board Examination/Year-end Examination Board Examination/ Year end Examination

Proposed

Maximum Marks 100/80/70/30 As per nature of subject

80 Marks in Mathematics, languages, Political Science and Legal Studies.  No change in other subjects

Duration

3 hours

3 hours for all subjects of 100/80/70 Marks and 2 hours for papers with less than 70 marks theory portion.

Components of Board Examination Paper

Short Answer/Long Answer (Objective as well as Subjective)

Objective type Minimum 25% including Multiple marks in Year End/ Choice Questions Board Exam

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Marks

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Existing

Subjective –number Maximum 75% of questions will be marks in Year End/ reduced to enable Board Exam student to have enough time to give analytical and creative responses ( 11 )

Internal Assessment/ Project work/ Practical Internal Assessment/ Project/ Practical Marks

Existing

Proposed



0 marks (Mathematics, Languages, Political Science)  20 marks (Humanities / Commerce Based Subjects and some other subjects )  30 marks (Science based subjects and some other subjects)  70 marks (Fine Arts and some other subjects

20 marks (Mathematics, Languages, Political Science and Legal Studies )  No change in remaining subjects 



Proposed

Internally Assessed



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Existing

Internally Assessed

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Assessment of Co-scholastic areas

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School Based Assessment of Co-scholastic Areas (Work Experience, Art Education, Health & Physical Education Discipline)

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It may be noted that the changes introduced in 2019 Board Exams – that is, 33% internal options and section wise format – shall continue henceforth for all subjects. The details of the Year End-Board Examination/ Internal Assessment/School Based Assessment for the academic year 2019-20, shall be reflected in the Curriculum document to be published by the Board. The curriculum document shall be made available to all the schools by April 2019. This circular is being issued in advance to enable schools to prepare for implementing the few proposed changes from next academic session (2019-20) onwards.

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Schools are also hereby informed that the Board is preparing detailed guidelines on how Art can be integrated with the teaching learning process at all levels. This is a part of the Board’s thrust on innovative pedagogy, that will go a long way in achieving the process of ‘assessment as learning’. These guidelines are being issued separately.

(Dr. Joseph Emmanuel) Director (Academics)

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Latest Syllabus for Academic Year (2019-20) PHYSICS (Code No. 042) CLASS–XII

Time : 3 Hours

Total Marks 70 No. of Periods

Unit I

Electrostatics Chapter–1: Electric Charges and Fields

22

Chapter–2: Electrostatic Potential and Capacitance Unit II

Current Electricity Chapter–3: Current Electricity

Unit III

20

Magnetic Effects of Current and Magnetism Chapter–4: Moving Charges and Magnetism

22

Electromagnetic Induction and Alternating Currents

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Chapter–6: Electromagnetic Induction

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Chapter–7: Alternating Current Unit V

Electromagnetic waves Chapter–8: Electromagnetic waves Optics

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Unit VI

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Chapter–5: Magnetism and Matter Unit IV

Unit VII

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Chapter–10: Wave Optics

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Chapter–9: Ray Optics and Optical Instruments Dual Nature of Radiation and Matter

Chapter–11: Dual Nature of Radiation and Matter Unit VIII

20

04

27

08

Atoms and Nuclei

Chapter–12: Atoms

15

Electronic Devices

Chapter–14: Semiconductor Electronics : Materials, Devices and Simple Circuits

12

Total

150

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Unit IX

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Chapter–13: Nuclei

Unit I : Electrostatics

Marks

}

16

} } }

}

17

18

12

7

70

(22 Periods)

Chapter–1 : Electric Charges and Fields Electric Charges; Conservation of charge, Coulomb’s law-force between two point charges, forces between multiple charges; superposition principle and continuous charge distribution. Electric field, electric field due to a point charge, electric field lines, electric dipole, electric field due to a dipole, torque on a dipole in uniform electric fleld. Electric flux, statement of Gauss’s theorem and its applications to find field due to infinitely long straight wire, uniformly charged infinite plane sheet and uniformly charged thin spherical shell (field inside and outside). Chapter–2 : Electrostatic Potential and Capacitance Electric potential, potential difference, electric potential due to a point charge, a dipole and system of charges; equipotential surfaces, electrical potential energy of a system of two point charges and of electric dipole in an electrostatic field. ( 13 )

Conductors and insulators, free charges and bound charges inside a conductor. Dielectrics and electric polarisation, capacitors and capacitance, combination of capacitors in series and in parallel, capacitance of a parallel plate capacitor with and without dielectric medium between the plates, energy stored in a capacitor. Unit II : Current Electricity

20 Periods

Chapter–3 : Current Electricity Electric current, flow of electric charges in a metallic conductor, drift velocity, mobility and their relation with electric current; Ohm’s law, electrical resistance, V-I characteristics (linear and non-linear), electrical energy and power, electrical resistivity and conductivity, Carbon resistors, colour code for carbon resistors; series and parallel combinations of resistors; temperature dependence of resistance. Internal resistance of a cell, potential difference and emf of a cell, combination of cells in series and in parallel, Kirchhoff ’s laws and simple applications, Wheatstone bridge, metre bridge. Potentiometer - principle and its applications to measure potential difference and for comparing EMF of two cells; measurement of internal resistance of a cell.

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Unit III : Magnetic Effects of Current and Magnetism Chapter–4 : Moving Charges and Magnetism

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Concept of magnetic field, Oersted’s experiment.

22 Periods

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Biot - Savart law and its application to current carrying circular loop.

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Ampere’s law and its applications to infinitely long straight wire. Straight and toroidal solenoids (only qualitative treatment), force on a moving charge in uniform magnetic and electric fields, Cyclotron.

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Force on a current-carrying conductor in a uniform magnetic field, force between two parallel currentcarrying conductors-definition of ampere, torque experienced by a current loop in uniform magnetic field; moving coil galvanometer-its current sensitivity and conversion to ammeter and voltmeter. Chapter–5 : Magnetism and Matter

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Current loop as a magnetic dipole and its magnetic dipole moment, magnetic dipole moment of a revolving electron, magnetic field intensity due to a magnetic dipole (bar magnet) along its axis and perpendicular to its axis, torque on a magnetic dipole (bar magnet) in a uniform magnetic field; bar magnet as an equivalent solenoid, magnetic field lines; earth’s magnetic field and magnetic elements.

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Para-, dia- and ferro - magnetic substances, with examples. Electromagnets and factors affecting their strengths, permanent magnets. Unit IV : Electromagnetic Induction and Alternating Currents

20 Periods

Chapter–6 : Electromagnetic Induction Electromagnetic induction; Faraday’s laws, induced EMF and current; Lenz’s Law, Eddy currents. Self and mutual induction. Chapter–7 : Alternating Current Alternating currents, peak and RMS value of alternating current/voltage; reactance and impedance; LC oscillations (qualitative treatment only), LCR series circuit, resonance; power in AC circuits, power factor, wattless current. AC generator and transformer. Unit V : Electromagnetic Waves

04 Periods

Chapter–8 : Electromagnetic Waves Basic idea of displacement current, Electromagnetic waves, their characteristics, their Transverse nature (qualitative ideas only). Electromagnetic spectrum (radio waves, microwaves, infrared, visible, ultraviolet, X-rays, gamma rays) including elementary facts about their uses. ( 14 )

Unit VI : Optics

27 Periods

Chapter–9 : Ray Optics and Optical Instruments Ray Optics : Reflection of light, spherical mirrors, mirror formula, refraction of light, total internal reflection and its applications, optical fibres, refraction at spherical surfaces, lenses, thin lens formula, lensmaker’s formula, magnification, power of a lens, combination of thin lenses in contact, refraction and dispersion of light through a prism. Scattering of light - blue colour of sky and reddish appearance of the sun at sunrise and sunset. Optical instruments: Microscopes and astronomical telescopes (reflecting and refracting) and their magnifying powers. Chapter–10 : Wave Optics

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Wave optics : Wave front and Huygen’s principle, reflection and refraction of plane wave at a plane surface using wave fronts. Proof of laws of reflection and refraction using Huygen’s principle. Interference, Young’s double slit experiment and expression for fringe width, coherent sources and sustained interference of light, diffraction due to a single slit, width of central maximum, resolving power of microscope and astronomical telescope, polarisation, plane polarised light, Brewster’s law, uses of plane polarised light and Polaroids. 08 Periods

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Chapter–11 : Dual Nature of Radiation and Matter

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Unit VII : Dual Nature of Radiation and Matter 

Dual nature of radiation, Photoelectric effect, Hertz and Lenard’s observations; Einstein’s photoelectric equation-particle nature of light.

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Matter waves-wave nature of particles, de-Broglie relation, Davisson-Germer experiment (experimental details should be omitted; only conclusion should be explained). Chapter–12 : Atoms

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Unit VIII : Atoms and Nuclei

15 Periods

Alpha-particle scattering experiment; Rutherford’s model of atom; Bohr model, energy levels, hydrogen spectrum. Chapter–13 : Nuclei

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Composition and size of nucleus, Radioactivity, alpha, beta and gamma particles/rays and their properties; radioactive decay law. Mass-energy relation, mass defect; binding energy per nucleon and its variation with mass number; nuclear fission, nuclear fusion. Unit IX : Electronic Devices

12 Periods

Chapter–14 : Semiconductor Electronics: Materials, Devices and Simple Circuits Energy bands in conductors, semiconductors and insulators (qualitative ideas only) Semiconductor diode - I-V characteristics in forward and reverse bias, diode as a rectifier; Special purpose p-n junction diodes: LED, photodiode, solar cell and Zener diode and their characteristics, zener diode as a voltage regulator. Practicals

Total Periods 60

The record to be submitted by the students at the time of their annual examination has to include :  Record of at least 15 Experiments [with a minimum of 6 from each section], to be performed by the students.  Record of at least 5 Activities [with a minimum of 2 each from section A and section B], to be demonstrated by the teachers.  The Report of the project to be carried out by the students. ( 15 )

Evaluation Scheme Time Allowed: Three hours

Max. Marks: 30

Two experiments one from each section

8+8 Marks

Practical record [experiments and activities]

6 Marks

Investigatory Project

3 Marks

Viva on experiments, activities and project

5 Marks

Total

30 marks

SECTION–A Experiments : 1. To determine resistance per cm of a given wire by plotting a graph for potential difference versus current.

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2. To find resistance of a given wire using metre bridge and hence determine the resistivity (specific resistance) of its material. 3. To verify the laws of combination (series) of resistances using a metre bridge.

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4. To verify the laws of combination (parallel) of resistances using a metre bridge.

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5. To compare the EMF of two given primary cells using potentiometer.

6. To determine the internal resistance of given primary cell using potentiometer.

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7. To determine resistance of a galvanometer by half-deflection method and to find its figure of merit.

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8. To convert the given galvanometer (of known resistance and figure of merit) into a voltmeter of desired range and to verify the same. 9. To convert the given galvanometer (of known resistance and figure of merit) into an ammeter of desired range and to verify the same. 10. To find the frequency of AC mains with a sonometer. Activities (For the purpose of demonstration only)

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1. To measure the resistance and impedance of an inductor with or without iron core.

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2. To measure resistance, voltage (AC/DC), current (AC) and check continuity of a given circuit using multimeter. 3. To assemble a household circuit comprising three bulbs, three (on/off) switches, a fuse and a power source. 4. To assemble the components of a given electrical circuit. 5. To study the variation in potential drop with length of a wire for a steady current. 6. To draw the diagram of a given open circuit comprising at least a battery, resistor/rheostat, key, ammeter and voltmeter. Mark the components that are not connected in proper order and correct the circuit and also the circuit diagram. SECTION–B Experiments : 1. To find the value of v for different values of u in case of a concave mirror and to find the focal length. 2. To find the focal length of a convex mirror, using a convex lens. 3. To find the focal length of a convex lens by plotting graphs between u and v or between 1/u and 1/v. 4. To find the focal length of a concave lens, using a convex lens. ( 16 )

5. To determine angle of minimum deviation for a given prism by plotting a graph between angle of incidence and angle of deviation. 6. To determine refractive index of a glass slab using a travelling microscope. 7. To find refractive index of a liquid by using convex lens and plane mirror. 8. To draw the I-V characteristic curve for a p-n junction in forward bias and reverse bias. 9. To draw the characteristic curve of a zener diode and to determine its reverse breaks down voltage. 10. To determine the wavelength of a laser beam by diffraction. Activities (For the purpose of demonstration only) 1. To identify a diode, an LED, a resistor and a capacitor from a mixed collection of such items. 2. Use of multimeter to see the unidirectional flow of current in case of a diode and an LED and check whether a given electronic component (e.g., diode) is in working order. 3. To study effect of intensity of light (by varying distance of the source) on an LDR. 4. To observe refraction and lateral deviation of a beam of light incident obliquely on a glass slab.

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5. To observe polarization of light using two Polaroids. 6. To observe diffraction of light due to a thin slit.

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7. To study the nature and size of the image formed by a (i) convex lens, (ii) concave mirror, on a screen by using a candle and a screen (for different distances of the candle from the lens/mirror). 8. To obtain a lens combination with the specified focal length by using two lenses from the given set of lenses.

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Suggested Investigatory Projects

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1. To study various factors on which the internal resistance/EMF of a cell depends. 2. To study the variations in current flowing in a circuit containing an LDR because of a variation in (a) the power of the incandescent lamp, used to ‘illuminate’ the LDR (keeping all the lamps at a fixed distance). (b) the distance of a incandescent lamp (of fixed power) used to ‘illuminate’ the LDR.

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3. To find the refractive indices of (a) water (b) oil (transparent) using a plane mirror, an equi convex lens (made from a glass of known refractive index) and an adjustable object needle.

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4. To design an appropriate logic gate combination for a given truth table. 5. To investigate the relation between the ratio of (i) output and input voltage and (ii) number of turns in the secondary coil and primary coil of a self-designed transformer. 6. To investigate the dependence of the angle of deviation on the angle of incidence using a hollow prism filled one by one, with different transparent fluids. 7. To estimate the charge induced on each one of the two identical styrofoam (or pith) balls suspended in a vertical plane by making use of Coulomb’s law. 8. To study the factor on which the self-inductance of a coil depends by observing the effect of this coil, when put in series with a resistor/(bulb) in a circuit fed up by an A.C. source of adjustable frequency. 9. To study the earth’s magnetic field using a tangent galvanometer.

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Practical Examination for Visually Impaired Students of Class XII Evaluation Scheme Time Allowed: Two hours

Max. Marks: 30

Identification/Familiarity with the apparatus

5 marks

Written test (based on given/prescribed practicals)

10 marks

Practical Record

5 marks

Viva

10 marks Total

30 marks

General Guidelines  The practical examination will be of two hour duration.  A separate list of ten experiments is included here.  The written examination in practicals for these students will be conducted at the time of practical examination of all other students.

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 The written test will be of 30 minutes duration.

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 The question paper given to the students should be legibly typed. It should contain a total of 15 practical skill based very short answer type questions. A student would be required to answer any 10 questions.  A writer may be allowed to such students as per CBSE examination rules.  All questions included in the question papers should be related to the listed practicals.

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Every question should require about two minutes to be answered.

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 These students are also required to maintain a practical file. A student is expected to record at least five of the listed experiments as per the specific instructions for each subject. These practicals should be duly checked and signed by the internal examiner.  The format of writing any experiment in the practical file should include aim, apparatus required, simple theory, procedure, related practical skills, precautions etc.  Questions may be generated jointly by the external/internal examiners and used for assessment.

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 The viva questions may include questions based on basic theory/principle/concept, apparatus/ materials/chemicals required, procedure, precautions, sources of error etc. Class XII

A. Items for Identification/ familiarity with the apparatus for assessment in practicals (All experiments) Meter scale, general shape of the voltmeter/ammeter, battery/power supply, connecting wires, standard resistances, connecting wires, voltmeter/ammeter, meter bridge, screw gauge, jockey Galvanometer, Resistance Box, standard Resistance, connecting wires, Potentiometer, jockey, Galvanometer, Lechlanche cell, Daniell cell [simple distinction between the two vis-à-vis their outer (glass and copper) containers], rheostat connecting wires, Galvanometer, resistance box, Plug-in and tapping keys, connecting wires battery/power supply, Diode, Resistor (Wire-wound or carbon ones with two wires connected to two ends), capacitors (one or two types), Inductors, Simple electric/electronic bell, battery/power supply, Plug-in and tapping keys, Convex lens, concave lens, convex mirror, concave mirror, Core/hollow wooden cylinder, insulated wire, ferromagnetic rod, Transformer core, insulated wire.

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B. List of Practicals 1. To determine the resistance per cm of a given wire by plotting a graph between voltage and current. 2. To verify the laws of combination (series/parallel combination) of resistances by Ohm’s law. 3. To find the resistance of a given wire using a meter bridge and hence determine the specific resistance (resistivity) of its material. 4. To compare the e.m.f of two given primary cells using a potentiometer. 5. To determine the resistance of a galvanometer by half deflection method. 6. To identify a resistor, capacitor, inductor and diode from a mixed collection of such items. 7. To understand the principle of (i) a NOT gate (ii) an OR gate (iii)an AND gate and to make their equivalent circuits using a bell and cells/battery and keys /switches 8. To observe the difference between (i) a convex lens and a concave lens (ii) a convex mirror and a concave mirror and to estimate the likely difference between the power of two given convex /concave lenses.

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9. To design an inductor coil and to know the effect of

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(i) change in the number of turns

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(ii) Introduction of ferromagnetic material as its core material on the inductance of the coil. 10. To design a (i) step up (ii) step down transformer on a given core and know the relation between its input and output voltages.

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Note : The above practicals may be carried out in an experiential manner rather than recording observations.

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QUESTION PAPER DESIGN Physics (Code No. 042) Class XII (2019-20)

Board Examination–Theory Maximum Marks : 70

1

–

9

12%

Understanding : Demonstrate understanding of facts and ideas by organizing, comparing, translating, interpreting, giving descriptions, and stating main ideas.

6

2

21

30%

Applying : Solve problems to new situations by applying acquired knowledge, facts, techniques and rules in a different way.

6

Analysing and Evaluating : Examine and break information into parts by identifying motives or causes. Make inferences and find evidence to support generalizations Present and defend opinions by making judgments about information, validity of ideas, or quality of work based on a set of criteria.

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2

1

2

1

2

23

33%

6

1

2

–

14

20%

Creating : Compile information together in a different way by combining elements in a new pattern or proposing alternative solutions.

–

–

1

–

3

5%

TOTAL

20×1=20

7×2= 14

7×3= 21

3×5= 15

70

100%

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2

5.

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2

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3.

Perce ntage

Remembering : Exhibit memory of previously learned material by recalling facts, terms, basic concepts, and answers.

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2.

Typology of Questions

SA LA-I LA-II Total (2 marks) (3 marks) (5 marks) Marks

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1.

VSA Objective Type (1 mark)

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S. No.

Duration : 3 hours

Practical : 30 Marks Note : 1. Internal Choice : There is no overall choice in the paper. However, there will be at least 33% internal choice. 2. The above template is only a sample. Suitable internal variations may be made for generating similar templates keeping the overall weightage to different form of questions and typology of questions same. qq

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Detailed Answer

Marking Scheme

Source Based Question

Syllabus

Topper's Answer

Developed by Oswaal Editorial Board

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Segregation

Topic wise

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1

Focused Study

1

1

1

• Discrete random variable : It is a random variable which can take only finite or countably infinite number of values. • Continuous random variable : A variable which can take any value between two given limits is called a continuous random variable.

Know the Terms

A random variable is a real valued function defined over the sample space of an experiment. In other words, a random variable is a real-valued function whose domain is the sample space of a random experiment.

1. RANDOM VARIABLE :

Facilitate Concept Clarity

Easy to Learn

1

2016 Delhi OD Foreign

Revision Notes

Definition Simplify

Know the Term

Map Question (Social Science)

V Short, Short-I, Short-II, Long & Practical Based (Science)

Typologies

Various Question

1

2015 Delhi OD Foreign

Gist of the Chapter

VSA (1 Mark) SA (2 Mark) LA-I (4 Mark) LA-II (6 Mark)

Revision Notes

Chapter Analysis

What you need to study ?

Prescribed by CBSE

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Commonly Made Error

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Suggested Videos

Scan the Code

Answering Tips

Tips to improve your Score

Exam Oriented Preparation Tool

Scan the given QR Code

Types of relations : Reflexive, symmetric, transitive and equivalence relations. One-to-one and onto functions, composite functions, inverse of a functions. Binary operations.

Syllabus

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CBSE



PHYSICS, Class-XII

Sample Question Paper For 2020 Examination

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eneral Instructions : G 1. All questions are compulsory. There are 37 questions in all. 2. This question paper has four sections : Section A, Section B, Section C, Section D. 3. Section A contains twenty questions of one mark each, Section B contains seven questions of two marks each, Section C contains seven questions of three marks each, Section D contains three question of five marks each. 4. There is no overall choice. However, internal choices have been provided in six question of one mark, two question of two marks, two questions of three marks and two questions of five marks weightage. You have to attempt only one of the choices in such questions. 5. You may use the following values of physical constants wherever necessary : h = 6.63 × 10–34 Js c = 3 × 108 m/s, e = 1.6 × 10–19 C, μ0 = 4p × 10–7 T mA–1

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1 = 9 × 109 Nm2C–2 4 0

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e0 = 8.854 × 10–12C2N–1m–2,

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Section ‘A’

mass of neutron = 1.675 × 10–27 kg Avogadro’s number = 6.023 × 1023 per gram mole

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(mass of electron) me = 9.1 × 10–31 kg, mass of proton = 1.673 × 10–27 kg, Boltzmann constant = 1.38 × 10–23 JK–1

1. A point charge +q, is placed at a distance d from an isolated conducting plane. The field at a point P on the other side of the plane is (a) directed perpendicular to the plane and away from the plane. (b) directed perpendicular to the plane but towards the plane. (c) directed radially away from the point charge. (d) directed radially towards the point charge. 2. A positively charged particle is released from rest in an uniform electric field. The electric potential energy of the charge (a) remains constant because the electric field is uniform. (b) increases because the charge moves along the electric field. (c) decreases because the charge moves along the electric field. (d) decreases because the charge moves opposite to the electric field. 3. Which of the following characteristics of electrons determine the current in a conductor? (a) Drift velocity alone (b) Thermal velocity alone (c) Both drift velocity and thermal velocity (d) Neither drift nor thermal velocity.

26 ]

Oswaal CBSE Chapterwises & Topicwise Question Bank, PHYSICS, Class–XII

4. A circular current loop of magnetic moment M is in an arbitrary orientation in an external magnetic field B. The work done to rotate the loop by 30° about an axis perpendicular to its plane is 3 MB . 2

(b)

(a) MB. MB . (c) 2

(d) zero.

OR A paramagnetic sample shows a net magnetization of 8 Am–1 when placed in an external magnetic field of 0.6 T at a temperature of 4 K. When the same sample is placed in an external magnetic field of 0.2 T at a temperature of 16 K, the magnetization will be

(a) 32 Am–1

(b)



(c) 6 Am–1

(d) 2.4 Am–1

2 Am–1 3

B = B0 ( 2ˆi + 3ˆj + 4kˆ) Tesla, where B0 is constant. The magnitude of flux passing through the square is



(a) 2B0L2Wb.

(b) 3B0L2Wb.



(c) 4B0L2Wb.

(d)

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6. Carbon, silicon and germanium have four valence electrons each. These are characterised by valence and conduction bands separated by energy band gap respectively equal to (Eg )C ,(Eg )Si and (Eg )Ge Which of the following statements is true?

(b) (Eg )C (Eg )Si

(c) (Eg )C >(Eg )Si >(Eg )Ge



(d) (Eg )C =(Eg )Si =(Eg )Ge

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(a) (Eg )Si I 3 > I 2 > I 1

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5. Which part of the electromagnetic spectrum is used in RADAR ? Give its frequency range.

1

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OR

How are electromagnetic waves produced by accelerating charges ?

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SECTION – B

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6. A capacitor made of two parallel plates, each of area 'A' and separation 'd' is charged by an external dc source. Show that during charging, the displacement current inside the capacitor is the same as the current charging the capacitor. 2 7. A photon and a proton have the same de-Broglie wavelength l. Prove that the energy of the photon is (2m lc/h) times the kinetic energy of the proton. 2 8. A photon emitted during the de-excitation of electron from a state n to the first excited state in a hydrogen atom, irradiates a metallic cathode of work function 2 eV, in a photo cell, with a stopping potential of 0.55 V. Obtain the value of the quantum number of the state n. 2 OR A hydrogen atom in the ground state is excited by an electron beam of 12.5 eV energy. Find out the maximum number of lines emitted by the atom from its excited state. 2 9. Draw the ray diagram of an astronomical telescope showing image formation in the normal adjustment position. Write the expression for its magnifying power.2 OR Draw a labelled ray diagram to show image formation by a compound microscope and write the expression for its resolving power. 2 10. Write the relation between the height of a TV antenna and the maximum range up to which signals transmitted by the antenna can be received. How is this expression modified in the case of line of sight communication by space waves ? In which range of frequencies, is this mode of communication used?* [Out of syllabus]2 11. Under which conditions can a rainbow be observed ? Distinguish between a primary and a secondary rainbow.2 12. Explain the following : (a) Sky appears blue. (b) The Sun appears reddish at (i) sunset, (ii) sunrise.2

38 |

Oswaal CBSE Chapterwise & Topicwise Question Bank, PHYSICS, Class – XII

SECTION – C 13. A capacitor (C) and resistor (R) are connected in series with an ac source of voltage of frequency 50 Hz. The potential difference across C and R are respectively 120 V, 90 V, and the current in the circuit is 3 A. Calculate (i) the impedance of the circuit (ii) the value of the inductance, which when connected in series with C and R will make the power factor of the circuit unity. 3 OR The figure shows a series LCR circuit connected to a variable frequency 230 V source. 40  80 F

230 V

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5.0 H

(a) Determine the source frequency which drives the circuit in resonance.

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(b) Calculate the impedance of the circuit and amplitude of current at resonance.

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(c) Show that potential drop across LC combination is zero at resonating frequency. 14. Give reason to explain why n and p regions of a Zener diode are heavily doped. Find the current through the Zener diode in the circuit given below : (Zener breakdown voltage is 15 V).

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> nh ¾¾ Impurity atom in n-type semiconductor is called donor which generates new energy level below the conduction band, known as ED

Energy bands for n-type semiconductor at T > 0 K

p-type Semiconductors : ¾¾ If Boron atom with 3 valence band electrons is added, it will accepts e– and give extra holes (h+) to move freely which leaves behind negatively charged nucleus.

¾¾ The crystal is electrically neutral known as “p-type” silicon in which concentration of acceptor atoms ~1028 cm–3 where hole movement needs breaking of bond thereby giving low mobility, where, μp ≈ 500 cm2/V

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Oswaal CBSE Chapterwise & Topicwise Question Bank, PHYSICS, Class – XII

Energy Band Diagram of p-Type Semiconductor ¾¾ On doping a semiconductor with trivalent impurity like Indium (In) or Gallium (Ga), extrinsic semiconductor so obtained is known as p-type. ¾¾ p-type semiconductor has large number of holes known as majority (charge) carriers where number of free electrons is small known as minority (charge) carriers. nh >> he ¾¾ Impurity atom in p-type semiconductor is known as acceptor atom. ¾¾ In p-type, extra holes in band gap allow excitation of valence band electrons which leaves mobile holes in valence band. ¾¾ Large number of holes in covalent bond is created in crystal will trivalent impurity.

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¾¾ In extrinsic semiconductors ne ¹ nh but ne.nh = ni2

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Energy bands for n-type semiconductor or T > 0 K

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Objective Type Questions

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¾¾ Energy band : Range of energies that an electron may possess in an atom. ¾¾ Valence Band : Range of Energy possessed by valence electrons. ¾¾ Conduction Band : Range of energy possessed by electrons. ¾¾ Forbidden Band : Energy band in between the conduction band and valence band.

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Q. 1. In an n-type silicon, which of the following statement is true : (a) Electrons are majority carriers and trivalent atoms are the dopants. (b) Electrons are minority carriers and pentavalent atoms are the dopants. (c) Holes are minority carriers and pentavalent atoms are the dopants. (d) Holes are majority carriers and trivalent atoms are the dopants. [NCERT Exemplar] Ans. Correct option : (c) Explanation : In an n-type silicon the holes are the minority carriers. An n-type semiconductor is obtained when pentavalent atoms, such as phosphorus, are doped in silicon atoms. Q. 2. Which of the statements given in is true for p-type semiconductors? [NCERT Exemplar] Ans. Correct option : (d) Explanation : In a p-type semiconductor, the holes are the majority carriers, while the electrons are the minority carriers. A p-type semiconductor is obtained when trivalent atoms, such as aluminium, are doped in silicon atoms. Q. 3. Carbon, silicon and germanium have four valence electrons each. These are characterised by valence and conduction bands separated by energy band gap respectively equal to ( E g )C ,( E g )Si and ( E g )Ge Which of the following statements is true? ( E g )Si < ( E g )Ge < ( E g )C . (a) (b) ( E g )C < ( E g )Ge > ( E g )Si .

(1 mark each)

(c) ( E g )C > ( E g )Si > ( E g )Ge . (d) ( E g )C = ( E g )Si = ( E g )Ge .

[NCERT Exemplar]

ns. Correct option : (c) A Explanation  : Above mentioned three given elements, the energy band gap of carbon is the maximum and that of germanium is the least. The energy band gaps of these elements are related as : ( Eg )C > ( Eg )Si > ( Eg )Ge

Q. 4. In an unbiased p-n junction, holes diffuse from the p-region to n-region because (a) free electrons in the n-region attract them. (b) they move across the junction by the potential difference. (c) hole concentration in p-region is more as compared to n-region. (d) All the above. [NCERT Exemplar] Ans Correct option : (c) Explanation : The diffusion of charge carriers across a junction takes place from the regions of higher concentration to lower concentration. In this case, the p-region has greater concentration of holes than the n-region. Hence, in an unbiased p-n junction, holes diffuse from the p-region to the n-region. Q. 5. When a forward bias is applied to a p-n junction, it (a) raises the potential barrier. (b) reduces the majority carrier current to zero. (c) lower the potential barrier. (d) None of the above [NCERT Exemplar]

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ns. Correct option : (b) A Explanation : Atoms of semiconductor are binding by covalent bonds between the atoms of same or different type. Due to thermal agitation when an electron leaves its position and become free, then it leaves a vacancy of electron, this vacancy in the bond (covalent) is called hole. Q. 9. The output of the given circuit in Figure A Vm sin t

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(a) would be zero at all times. (b) would be like a half wave rectifier with positive cycles in output. (c) would be like a half wave rectifier with negative cycles in output. (d) would be like that of a full wave rectifier.  [NCERT Exemplar] Ans. Correct option : (c) Explanation : When positive cycle is at A, diode will be in forward bias and resistance due to diode is approximately zero so potential across diode will be about zero. Similarly, when there is negative half cycle at A, diode will be in reverse bias and resistance will be maximum so potential difference across diode is Vmsinwt with negative at A. So we get only negative output at A so it behaves like a half wave rectifier with negative cycle at A in output. Q. 10. In the circuit shown in Figure, if the diode forward voltage drop is 0.3  V, the voltage difference between A and B is

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ns. Correct option : (c) A Explanation : When a forward bias is applied to a p-n junction, it lowers the value of potential barrier. In the case of a forward bias, the potential barrier opposes the applied voltage. Hence, the potential barrier across the junction gets reduced. Q. 6. The conductivity of a semiconductor increases with increase in temperature because (a) number density of free current carriers increases. (b) relaxation time increases. (c) both number density of carriers and relaxation time increase. (d) number density of current carriers increases; relaxation time decreases but effect of decrease in relaxation time is much less than increase in number density. [NCERT Exemplar] Ans. Correct option : (d) Explanation : In semiconductor, the density of charge carriers (electron, holes) are very small, so its resistance is high. When temperature increases the charge carriers (density) increases which increases the conductivity. As temperature of semiconductor increases, the speed of free electrons increases which decreases the relaxation time. As the density of charge carrier is small so there is small effect on decrease of relaxation time. Q. 7. In Figure, Vo is the potential barrier across a p-n junction, when no battery is connected across the junction 1 2

Vo0



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(a) 1 and 3 both correspond to forward bias of junction. (b) 3 corresponds to forward bias of junction and 1 corresponds to reverse bias of junction. (c) 1 corresponds to forward bias and 3 corresponds to reverse bias of junction. (d) 3 and 1 both correspond to reverse bias of junction. [NCERT Exemplar] Ans. Correct option : (b) Explanation : when p-n junction is in forward bias, it compresses or decreases the depletion layer so potential barrier in forward bias decreases and in reverse layer so potential barrier in forward bias decrease and in reverse bias potential barrier increases. Q. 8. Hole is (a) an anti-particle of electron. (b) a vacancy created when an electron leaves a covalent bond. (c) absence of free electrons. (d) an artificially created particle.  [NCERT Exemplar]

(a) 1.3 V (b) 2.3 V (c) 0 V (d) 0.5 V  [NCERT Exemplar] Ans. Correct option : (b) Explanation : In the middle right of the circuit the capacitor behaves like an open circuit for dc 0.2 mA current so current will flow from A to B only. Let potential across A and B is V, so by Kirchhoff’s loop law, VAB = (5000 × 0.2 × 10−3) + 0.3 + 5000 × 0.2 × 10−3 VAB = 1 V + 0.3 V + 1 V

VAB = 2.3 V

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Oswaal CBSE Chapterwise & Topicwise Question Bank, PHYSICS, Class – XII

Short Answer Type Questions-I Q. 1. Write two points of difference between intrinsic and extrinsic semiconductors. [Foreign 2017] OR Distinguish between ‘intrinsic’ and ‘extrinsic’ semiconductors. [Delhi I, II, III 2015] Ans. Any two differences

Q. 2. Draw energy band diagrams of n-type and p-type semiconductors at temperature T > 0 K. Mark the donor and acceptor energy levels with their energies. [Foreign I 2014] Ans. Energy bands of

1+1

n-type at T > 0 K

p-type at T > 0 K

Extrinsic

(i) Pure semiconductor.

(i) Doped or impure.

(ii) ne = nh.

(ii) ne ¹ nh.

(iii) Low conductivity at (iii) Higher conducroom temperature. tivity at room temperature.



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Commonly Made Error

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 Many students couldn't draw these diagrams correctly.

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Short Answer Type Questions-II Q. 1. (i) Distinguish between n-type and p-type semiconductors on the basis of energy band diagrams. (ii) Compare their conductivities at absolute zero temperature and at room temperature. R [Delhi I, II, III 2015] Ans. (i) p-type semiconductor

In this, extra energy band known as acceptor energy level is produced above the top of valence band.

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n-type semiconductor In this, an extra energy level known as donor energy level is produced below the bottom of the conduction band.

In this, most of the In this, density of electrons come from holes in valence band is predominantly due the donor impurity. to impurity in extrinsic semiconductors.

nh >ne 1

[CBSE Marking Scheme 2014] 1 + 1



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(iv) Conductivity de- (iv) Conductivpends on temperature. ity does not depend significantly on temperature. [CBSE Marking Scheme 2017]

ne >nh

Energy bands of

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Intrinsic

(2 marks each)



(3 marks each)

[Any one of the above, or any one, other correct distinguishing feature] (ii) At absolute zero temperature conductivities of both type of semiconductors will be zero. 1 For equal doping, an n-type semi conductor will have more conductivity than a p-type semiconductor, at room temperature. 1 Q. 2. (i) Draw the energy band diagrams of (a) n-type (b) p-type semiconductor at temperature, T > 0 K. (ii) In the case n-type semiconductor, the donor energy level is slightly below the bottom of conduction band whereas in p-type semiconductor, the acceptor energy level is slightly above the top of the valence band. Explain what role these energy levels play in conduction and valence bands. U [OD I, II, III 2015] Ans. (i) Try yourself, Similar to Q. 2 Short Answer Type-I. 1 (ii) In case of n-type semiconductor, electrons from donor impurity atoms will move into conduction band with very small supply of energy. Hence, the conduction band will therefore have electrons as majority charge carriers. 1 In case of p-type semiconductor, very small supply of energy can cause an electron from its valence band to jump to the acceptor energy level. Hence, the valence band will have dominant density of holes which shows that holes are the majority charge carriers in p-type semiconductor. 1

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Ev Ec

Valence band (a)



Electron energies



(Eg  0)

Eg > 3 eV Ev Valence band



(b) Ec

Eg < 3 eV Ev

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Ans.

Empty conduction band

Ec

Overlapping conduction band

Electron energies

Electron energies

Q. 3. Write any two distinguishing features between conductors, semiconductors and insulators on the basis of energy band diagrams. U [Foreign 2014; OD I, II, III 2014]

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(c)



(a) metals, (b) insulators and (c) semiconductors



Two distinguishing features :

(½ each)



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(i) In conductors, the valence band and conduction band tend to overlap (or nearly overlap), while in insulators they are separated by a large energy gap and in semiconductors they are separated by a small energy gap. 1



(ii) The conduction band of a conductor, has a large number of electrons available for electrical conduction. However, the conduction band of insulators is almost empty while that of the semi-conductor has only a (very) small number of such electrons available for electrical conduction. [CBSE Marking Scheme 2014] ½

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OR

Insulators

Conductors

Semiconductors

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Materials in which there is large Materials in which there is no Materials in which there is a very energy difference between valence and difference of energy among valence small energy difference between conduction band. and conduction band, as these bands valence and conduction bands. get overlapped.

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Forbidden gap is very large between There is no forbidden gap between Forbidden gap between valence band and conduction band is very valence and conduction band, due to valence and conduction band. which it will not conduct electricity. small.

Electron energies

Eg > 3 eV

Overlapping conduction band (Eg  0) Ev Ec

Valence band

Ev Valence band (b)

(a)

Electron energies

Empty conduction band

Ec

Electron energies

Conduction band is empty and electrons Electrons jump from valence band to Electrons in this can easily jump by in valence band acquires large amount of conduction band. getting small amount of energy. energy to jump in conduction band and become free.

Ec Eg < 3 eV Ev

(c)

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Oswaal CBSE Chapterwise & Topicwise Question Bank, PHYSICS, Class – XII

Long Answer Type Question

(5 marks)

(ii) When the semiconductor is doped with an acceptor impurity there is an additional energy level which is little above the top of the valence band. Also, electrons from the valence band, easily jump over to the acceptor level, leaving ‘holes‘ behind, so ‘holes‘ become the majority charge carriers. 1½ (iii) The donor impurity results in an additional energy level which is little below the bottom of the conduction band. Electrons from donor level, easily ‘jump over‘ to the conduction band. Hence, electrons become the majority charge carriers. 1½ [CBSE Marking Scheme 2016]

Q. 1. (i) Draw “Energy bands’, diagrams for (a) pure semiconductor, (b) insulator. (ii) How does the energy band, for pure semiconductor get affected when this semiconductor is doped with (a) acceptor impurity (b) donor impurity ? (iii) Hence discuss why the ‘holes, and the ‘electrons‘ respectively, become the ‘majority charge carriers‘ U [Foreign, 2016] in these two cases ? Ans. (i) Try yourself, Similar to Q. 3, Short Answer Type Questions-II. 2

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Semiconductor Diodes and Applications

Revision Notes

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Diode ¾¾ Diode is an electronic device consisting of a junction of semiconductors p-type and n-type. It is represented as : n

Semiconductor diode ¾¾ Semiconductor diodes were first semiconductor electronic devices which is common type of diode that is made of crystalline piece of semiconductor material with p–n junction across its terminals. P N Cathode

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Anode

Depletion region

¾¾ When a p-type semiconductor material is suitably joined to n-type semiconductor, the contact surface is called a p-n junction. ¾¾ It is an electrical device that allows current only in one direction The direction. of arrow is the direction of current when it is forward biased. ¾¾ At junction, electrons and holes diffuse and form the diffusion current. ¾¾ p-n junction layer is also called the depletion layer. Potential barrier is created at junction due to diffusion current. It acts as barrier for majority carriers. ¾¾ The potential barrier helps the minority carrier to flow. Thus a drift current forms, which is opposite in direction of the diffusion current. ¾¾ Under equilibrium condition, diffusion current is equal to the drift current and hence net current is zero as both are in opposite direction. There are many types of semiconductor diode such as : Avalanche diodes, Gunn diodes, light emitting diodes (LED), Photodiodes etc. ¾¾ Semiconductor diode can be made either from Silicon or Germanium and each differs in size and properties. Forward Bias ¾¾ When an external voltage is applied, where negative of battery is connected to n side while positive of battery is connected to p side, then barrier potential will get reduced and more current can flow across the junction that decreases the p-n junction width.

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¾¾ The positive terminal of battery repels majority carriers, holes in p-region while negative terminal repels electrons in n-region which pushes them towards the junction. ¾¾ Here, an increase in concentration of charge carriers near the junction is observed, where recombination takes place thereby reducing width of depletion region.

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¾¾ Due to rise in forward bias voltage, depletion region will continue to reduce its width which results in more and more carriers recombination.

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Reverse Bias

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¾¾ If an external voltage is applied in reverse direction where positive of battery is connected to n side while negative of battery is connected to p side, then barrier potential will increase and minority charge carriers will flow across junction. Anode

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Cathode

P

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–––––––––– –––––––––– –––––––––– –––––––––– ––––––––––

++++++++++ ++++++++++ ++++++++++ ++++++++++ ++++++++++

+ –

R

¾¾ In this, the current will be quiet small and is independent of external voltage.

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¾¾ Beyond certain voltage, diode will break down with Avalanche breakdown mechanism or Zener breakdown mechanism.

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¾¾ Here, negative terminal of battery will attract majority carriers, holes in p-region and positive terminal attracts electrons in n-region which pulls them away from the junction. ¾¾ As a result of this, there will be decrease in concentration of charge carriers near junction which increases the width of depletion region. Reverse Voltage (V) –10 –8 –6 –4 –2

0

breakdown voltage

2 4 6 8

Reverse (mA) Current

0 C

D

¾¾ A small amount of current flows due to minority carriers known as reverse bias current or leakage current and with rise in reverse bias voltage, depletion region continues to increase in width without any increase in flow of current. I-V Characteristics of Diode ¾¾ In I-V characteristics of diode, on voltage axis, “Reverse Bias” is an external voltage potential that increases the potential barrier while external voltage that decreases the potential barrier is in “Forward Bias” direction.

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Oswaal CBSE Chapterwise & Topicwise Question Bank, PHYSICS, Class – XII

VIN

VOUT

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Rectifier

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¾¾ Biasing of Diode can be Forward Biasing or Reverse Biasing. Diode as rectifier ¾¾ Rectifier is a circuit which converts AC supply into unidirectional DC supply.

¾¾ With rectification, alternating current (AC) gets converted to direct current (DC) through a rectifier.

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¾¾ The bridge rectifier circuits uses semiconductor diode for converting AC as it allows the current to flow in one direction only.

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Half wave rectification

Primary

Vac

Secondary

Vdc

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Transformer ¾¾ The half-wave rectifier with single diode, allows current to flow in one direction. ¾¾ Here, AC power source Vac is connected to primary side of transformer while secondary terminals of transformer are connected to diode and resistor in series. ¾¾ If Vac is in positive cycle, a positive voltage is produced on secondary side of transformer. ¾¾ The positive voltage will forward bias the diode and diode will start passing the current as a result, the voltage will drop across the load. ¾¾ If Vac is in negative cycle, then secondary side will have negative voltage where diode is reverse biased and will not pass any current. ¾¾ Voltage waveform across load resistor is shown below, where positive side of sinusoidal cycle is present while negative side sinusoidal cycle has been clamped off.

¾¾ The output voltage Vdc is similar to the output of battery which is always positive. ¾¾ The positive waveform is bumpy as single diode is applied to produce half wave rectification where one half of AC wave is removed that will not pass through the diode. Full wave rectification ¾¾ For rectifying AC power for using both half cycles of sine wave, full wave rectification is used.

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D1 AC voltage source

Centretap Load Output

D2

¾¾ A simple kind of full wave rectifier uses centre tap transformers with two diodes. ¾¾ In full wave rectification, in first half-cycle, when source voltage polarity is positive (+) on top and negative (-) on bottom, then only top diode will conduct while bottom diode blocks the current When source voltage polarity is negative (–) on top and positive (+) on bottom then only bottom diode will conduct while the top diode blocks the current. Special purpose p-n junction diodes Apart from simple p-n junction diodes, there are many more types of diodes which are used in various specific applications that take advantage of the behaviour and features. Anode

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Light–emitting diode Anode

Anode

Cathode

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Cathode

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Anode

Cathode

Photo diode

Anode

Cathode

Zener diode

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Schottky diode

Cathode

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LED ¾¾ Light Emitting Diode or LED is most widely used semiconductor diodes among all the different types of semiconductor diodes available today. ¾¾ It emits visible light or invisible infrared light when forward biased. ¾¾ The LEDs which emit invisible infrared light are used for remote controls. ¾¾ In this, diode in forward biased will make electrons and holes to move fast across the junction and helps in combining constantly by removing one another. ¾¾ Electrons which move from n-type to p-type silicon will combine with holes and give energy in the form of light.

¾¾ Recombination of electrons and holes in depletion region decreases the width of the region which allows more charge carriers to cross the p-n junction. ¾¾ Here, some of the charge carriers from p-side and n-side will cross the p-n junction before they recombine in depletion region.

I-V characteristics

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Oswaal CBSE Chapterwise & Topicwise Question Bank, PHYSICS, Class – XII

Photodiode ¾¾ Photodiode is a transducer which takes light energy and converts it to electrical energy.

¾¾ It is a p-n junction which consumes light energy to generate electric current. Anode

Cathode

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¾¾ It is referred to as photo-detector, photo-sensor or light detector. ¾¾ It is specially designed to operate in reverse bias condition where p-side is connected to negative terminal of battery and n-side connected to positive terminal of battery. ¾¾ It is sensitive to light as when light or photons fall on it, it easily converts light into electric current. ¾¾ In photodiode circuit, current flows from the cathode to anode when exposed to light. ¾¾ Photodiode is capable of converting light energy to electrical energy and can be expressed as a percentage known as Quantum Efficiency (Q.E.).

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mA

volts

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I1 I2 I3 I4

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Reverse current

Reverse bias

µA

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I4>I3>I2>I1

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I-V characteristics Solar cell ¾¾ Solar cell is an electronic device which absorbs sunlight and generates emf. ¾¾ In this, there are n-type silicon and p-type silicon layers that generates electricity using sunlight to make electrons in order to jump across the junction between different types of silicon material.

n-type 2 p-type 3

¾¾ When sunlight shines on solar cell, photons bombard the upper surface and generates electron-hole pairs. ¾¾ They get separated due to voltage barrier at junction, electrons are swept to n side & holes are swept to p-side. ¾¾ Metal contacts hold these electron-hole pairs. Thus p side becomes positive and n side becomes negative and hence it act as photo voltage cell. I Open circuit voltage (Voc) V

O

ISC

Short circuit current

I-V characteristics Zener diode and its characteristics ¾¾ Zener Diode is an electronic component which can be used to make very simple voltage regulator circuit.

Symbol of Zener diode

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¾¾ They are special type of semiconductor diodes which allow the current to flow in one direction when exposed to high voltage. ¾¾ It is a p-n junction semiconductor device which is designed to operate in reverse breakdown region. ¾¾ The breakdown voltage of zener diode is set by controlling the doping level. ¾¾ Zener Diode allows electric current in forward direction similar to normal diode and also allows electric current in reverse direction when applied reverse voltage is more than zener voltage. ¾¾ It is always connected in reverse direction since it is specifically designed to work in reverse direction. ¾¾ Zener Diode circuit enables a fixed stable voltage to be taken from an unstable voltage source like battery that fluctuates as per state of charge of battery.

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¾¾ The circuit of Zener Diode has a resistor in series with diode which limits the output current. ¾¾ In Zener Diode, there are two breakdown mechanisms : Zener Breakdown or Avalanche Breakdown mechanism. ¾¾ I-V Characteristics Curve of zener diode shows current-voltage relationship.

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I-V characteristics ¾¾ In I-V curve, in right half side, zener diode receives forward voltage which is positive voltage across its anode to cathode terminals. ¾¾ In right half side of zener diode characteristics curve, diode is forward biased and current is more. ¾¾ In left half side of I-V curve, zener diode will receive positive voltage across its cathode to anode terminals where diode is reverse biased. ¾¾ At reverse voltage, current will be very small which is known as leakage current that flows through the diode. ¾¾ After hitting breakdown voltage, avalanche current will sharply increase. ¾¾ At breakdown voltage point, when voltage of zener diode reaches, it remains constant in spite of increase in current making zener diode suitable for voltage regulation. Zener diode as voltage regulator ¾¾ Voltage regulation is a measure of ability of circuit to maintain constant voltage output under variation either in input voltage or load current. ¾¾ In zener diode voltage regulator circuit : • resistor Rs is used to limit reverse current through diode to safer value. • Vs and Rs are selected such that diode operates in breakdown region. • series resistor Rs absorbs output voltage fluctuations to maintain voltage across load to constant value. Rss IIss IL R IL Supply Voltage (Vs )

IZ VZ

IL RL

Vout

¾¾ Zener diode maintains constant voltage across load as long as supply voltage is more than zener voltage. ¾¾ When input voltage increases, current through Zener diode also increases keeping the voltage drop constant. ¾¾ Current in the circuit increases the voltage drop across the resistor which increases by an amount equal to difference between the input voltage and zener voltage of the diode.

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Oswaal CBSE Chapterwise & Topicwise Question Bank, PHYSICS, Class – XII

Objective Type Questions Q. 1. In Figure, assuming the diodes to be ideal,

(1 mark each) Q. 2. A 220 V A.C. supply is connected between points A and B in Figure. What will be the potential difference V across the capacitor?



(a) 220 V. (c) 0 V.

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(b) 110 V. (d) 220 2 V.

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 [NCERT Exemplar] ns. Correct option : (d) A Explanation : Potential difference across capacitors will be peak voltage when diode is in forward bias. Diode will be in forward bias when end A is at positive potential of cycle. So potential at C=peak value of V = Vrms 2 = 220 2 V.

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(a) D1 is forward biased and D2 is reverse biased and hence current flows from A to B (b) D2 is forward biased and D1 is reverse biased and hence no current flows from B to A and vice versa. (c) D1 and D2 are both forward biased and hence current flows from A to B. (d) D1 and D2 are both reverse biased and hence no current flows from A to B and vice versa.  [NCERT Exemplar] Ans. Correct option : (b) Explanation : In circuit, A is at −10 V and B is at 0 V. So B is positive than A. So D2 is in forward bias and D1 is in reverse bias so no current flows from A to B or B to A.

Very Short Answer Type Questions Q. 1. Name the junction diode whose I-V characteristics are drawn below :

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I

V

Ans. Solar cell

1 [CBSE Marking Scheme 2017]

Detailed Answer : I

Open circuit voltage VOC O

V

energy more than band gap energy gets incident on the junction creating the electron-hole pairs that moves in opposite directions thus producing photovoltage. ½ Q. 2. Write the two processes involved in the formation of p-n junction. R [Foreign 2016] Ans. (i) Diffusion (ii) Drift

R [Delhi I,II,III 2017; NCERT Exemplar]

½

I SC Short circuit current The V-I Characteristics curve shown is of Solar cell which is a junction diode , that converts light energy into electrical energy. Here photons with

(1 mark each)

½ ½ [CBSE Marking Scheme 2016]

Q. 3. Why is photodiode used in reverse bias ? R [OD Comptt. I, II, III 2013] Ans. In forward bias, current is mainly due to major carriers while in reverse bias, current is due to minor carriers. With such fractional change in reverse current resulting from photo effects, it can be easily measured as in case forward bias current. Hence, the photodiodes will operate in reverse bias. 1 [CBSE Marking Scheme 2013] Q. 4. The graph shown in the figure represents a plot of current versus voltage for a given semiconductor. Identify the region, if any, over which the semiconductor has a negative resistance.   U [OD I, II, III 2013]

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Ans. The resistance of a material can be found out by the slope of the curve V vs. I. Part BC of the curve shows the negative resistance as with the increase in voltage current decreases. 1 [CBSE Marking Scheme 2013] Detailed Answer : It is known that negative resistance is a property of an electrical circuits/devices where an increase in voltage across the device terminals results in decrease in electric current through it. From the graph, it is observed that region BC experiences negative resistance. 1

Current (mA)

B

A C Voltage (V)

Short Answer Type Questions-I Q. 1. Explain with the help of a circuit diagram, the working of a p-n junction diode as a half-wave rectifier. U [OD I, II, III 2014]

Secondary

am

gr



RL B

Y

Working : During one half of the input a.c., the

diode is forward biased and a current flows through RL.

[CBSE Marking Scheme 2012]



Q. 4. Assuming that the two diodes D1 and D2 used in the electric circuit shown in the figure are ideal,

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Primary

A

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Transformer

Ans. Even though the current in forward bias has a larger magnitude, the change due to the changes in light intensity, in the minority carrier dominates reverse bias current, which is more and is, therefore, more easily detectable. 2

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Ans.

(2 marks each)

find out the value of the current flowing through 2.5 Ω resistor.

A [Delhi Comptt. I, II, III 2013]

1

During the other half of the input a.c., the diode is

reverse biased and no current flows through the load RL.

½

Hence, the given ac input is rectified.





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½

An i

[CBSE Marking Scheme 2014]

Q. 2. Mention the important considerations required while fabricating a p-n junction diode to be used as a Light Emitting Diode (LED). What should be the order of band gap of an LED if it is required to emit light in the visible range ?  R [Delhi I, II, III 2013] Ans. (i) It is a heavily doped p-n junction diode. (ii) The reverse breakdown voltages of LEDs are very low. (iii) The semiconductor used for fabrication of visible LEDs must at least have a band gap of 1.8 eV. (Any two) ½ + ½

The order of band gap is about 3 eV to 1.8 eV.





Ans. In the circuit, if D1 is open is D2 is short then equivalent circuit will result as : 3Ω

3Ω

½

10 V D2 is reverse biased \ D1 conducts

2.5 Ω



3

1

[CBSE Marking Scheme 2013]

Q. 3. The current in the forward bias is known to be more (~ mA) than the current in the reverse bias (~ mA). What is the reason, then why photodiodes is to operate in reverse bias ? A [Delhi I, II, III 2012]

½

10 V

\,



2.5 

10 10 = = 1.8 A I = 3 + 2.5 5.5

1

382 ]

Oswaal CBSE Chapterwise & Topicwise Question Bank, PHYSICS, Class – XII

Q. 5. For the circuit shown here, find the current flowing through the 1 W resistor. Assume that the two diodes D1 and D2 are ideal diodes.

Hence the resistances of (ideal) diodes D1 and can be taken as zero and infinity respectively.   The given circuit can therefore be redrawn shown in the figure.



2Ω

D1

D2

+



2

2Ω 1

6V

1Ω

–

6V

A [SQP 2013] Ans. Diode D1 is forward biased, while diode D2 is reverse biased. ½

\ Using Ohm’s law, 6 A = 2A I = + 1) ( 2

½

\ Current flowing in the 1 W resistor, is 2 A.



Short Answer Type Questions-II

+5V



1

tS ha _o ffi

(ii) Diagram of full wave rectifier





1

(ii) Diagram of full wave rectifier :



An i

ke





Working : The diode D1 is forward biased during one half cycle and current flows through the resistor, but diode D2 is reverse biased and hence no current flows through it. During the other half of the signal, D1 gets reverse biased and no current passes through it, D2 gets forward biased and current flows through it. In both half cycles current, through the resistor, flows in the same direction. 1 (Note : If the student just draws the following graphs (but does not draw the circuit diagram), award ½ mark only.)

am gr

Output waveform (across RL)

1

ci

Ans. (i) The nature of biasing

al

(ii) Draw the circuit diagram of a full wave rectifier

Waveform at B

te le

U [OD I, II, III 2017]

and state how it works.

(3 marks each) Waveform at A

Q. 1. (i) In the following diagram, is the junction diode forward biased or reverse biased ?

Working (i) Reverse Biased

D2 ½ as ½

t

(i)

t

(ii)



(a)

Due to Due to Due to Due to D1 D2 D1 D2

(b)

1 [CBSE Marking Scheme 2017]



Detailed Answer :

(i) The diode shown in diagram is reverse biased as in case of reverse bias p junction is connected to -ve of battery and n junction with positive which causes effective barrier voltage enhancement VB + V with high resistance. 1 Q. 2. (i) In the following diagram, which bulb out of B1 and B2 will glow and why ? D1

D2

+9V B1

B2

(ii) Draw a diagram of an illuminated p-n junction solar cell. (iii) Explain briefly the three processes due to which generation of emf takes place in a solar cell. U [O.D. II 2017]

[ 383

SEMICONDUCTOR ELECTRONIC : MATERIALs, DEVICEs and simple circuits

Ans. (i) Identification of the bulb and reason ½ + ½ (ii) Diagram of solar cell ½ (iii) Names of the processes ½+½+½ (i) Bulb B1 glows ½ Diode D1 is forward biased ½ (ii) Diagram :

(ii) Photodiode diagram hν > Eg

RL

p-side

n-side

µA

IL –

½

Separation : Electric field of the depletion layer separates the electrons and holes. ½

Commonly Made Error

tS ha _o ffi

 Some common errors made by students are : (i) A.C. input voltages at A & B are not shown. (ii) Complete circuit was not drawn. (iii) Resistor in o/p circuit (RL) was not shown. (iv) The circuit was not labelled.

ci

Collection : Electrons and holes are collected at the n and p side contacts. ½ [CBSE Marking Scheme, 2017]

am



(iii) Generation : Incident light generates electronhole pairs. ½

gr

Depletion layer



te le

n

1 When the photodiode is illuminated with light (photons) (with energy (hn) greater than the energy gap (Eg) of the semiconductor), then electronhole pairs are generated due to the absorption of photons. Due to junction field, electrons and holes are separated before they recombine. Electrons are collected on n-side and holes are collected on p-side giving rise to an emf. ½ When an external load is connected, current flows. I-V Characteristics of the diode :

al

p

ke

Q. 3. (i) In the following diagram ‘S’ is a semiconductor. Would you increase or decrease the value of R to keep the reading of the ammeter A constant when S is heated ? Give reason for your answer.

An i

V

–

+

S

A R

(ii) Draw the circuit diagram of a photodiode and explain its working. Draw its I/V characteristics. U [OD III 2017] Ans. (i) Correct Choice of R ½ Reason ½ (ii) Circuit Diagram 1 Working ½ I-V characteristics ½ (i) R would be increased. ½ Resistance of S (a semiconductor) decreases on heating. ½

RL

+

½

[CBSE Marking Scheme 2017]

Q. 4. A zener diode is fabricated by heavily doping both p and n sides of the junction. Explain, why? Briefly explain the use of zener diode as a dc voltage regulator with the help of a circuit diagram. U [Delhi II 2017] Ans. Explanation of heavily doping of both p and n sides of Zener diode. 1 Circuit diagram of Zener diode as a dc voltage regulator. 1 Explanation of the use of Zener diode as a dc voltage regulator. 1 By heavily doping both p and n sides of the junction, depletion region formed is very thin, i.e., < 10–6 m. Hence, electric field, across the junction is very high (5 × 106 V/M) even for a small reverse bias voltage. This can lead to a 'breakdown' during reverse biasing. 1

384 ]

Oswaal CBSE Chapterwise & Topicwise Question Bank, PHYSICS, Class – XII



An i

ke

am

tS ha _o ffi

ci

Ans. Circuit diagrams of p-n Junction under forward bias reverse bias ½+½ Explanation of p-n junction working for forward and reverse bias ½+½ Characteristic curves for the two cases ½+½

gr

Explain briefly with the help of necessary Q. 5. diagrams, the forward and the reverse biasing of a p-n junction diode. Also draw their characteristic U [Delhi III 2017] curves in the two cases.

Ans. Explanation of two processes 1+1 Definition of barrier potential 1 Diffusion : It is the process of movement of majority charge carriers from their majority zone (i.e., electrons from n ® p and holes from p ® n) due to the electric field developed at the junction. 1 Drift : Process of movement of minority charge carriers (i.e., holes from n ® p and electrons from p ® n) due to the electric field developed at the junction. 1 Barrier potential : The loss of electrons from the n-region and gain of electrons by p-region causes a difference of potential across the junction, whose polarity is such as to oppose and then stop the further flow of charge carriers. This (stopping) potential is called Barrier potential. 1 [CBSE Marking Scheme 2017]

te le

[CBSE Marking Scheme 2017]

1 [CBSE Marking Scheme 2017] Explain the two processes involved in the Q. 6. formation of a p-n junction diode. Hence define the term ‘barrier potential’. U [Delhi I, II, III Comptt. 2017]

al

If the input voltage increases/decreases, current through resister Rs, and Zener diode, also increases/decreases. This increases/decreases the voltage drop across Rs without any change in voltage across the Zener diode. 1 This is because, in the breakbown region, Zener voltage remains constant even though the current through the Zener diode changes. 1

½+½

In forward bias, applied voltage does not support potential barrier. As a result, the depletion layer width decreases and barrier height is reduced. Due to the applied voltage, electrons from n side cross the depletion region and reach p side. Similarly holes from p side cross the junction and reach the n side. The motion of charge carriers, on either side, give rise to current. In reverse bias, applied voltage support potential barrier. As a result, barrier height is increased, depletion layer widens. This suppresses the flow of electrons from n ® p and holes from p ® n, thereby decrease the diffusion current. The electric field direction of the junction is such that if electrons on p side or holes on n side in their random motion come close to the junction. They will be swept to its majority zone. This drift of carriers give rise to the current called reverse current. 1

Q. 7. State the reason, why the photodiode is always operated under reverse bias. Write the working principle of operation of a photodiode. The semiconducting material used to fabricate a photodiode, has an energy gap of 1.2 eV. Using calculations, show whether it can detect light of wavelength of 400 nm incident on it. U [OD II Comptt., 2017] Ans. Reason for use in reverse bias Working Principle Whether it can detect

1 1 1



The fractional change, due to photo effects, on the minority charge carrier dominates reverse bias current, which is much more than the fractional change in the forward bias current and can be easily detected Hence, photodiode is used in reverse bias. 1



Working principle of photodiode :



i. Generation of e-h pairs due to light close to junction. ½



ii. Separation of electrons and holes due to electric field of the depletion region. ½

[ 385

SEMICONDUCTOR ELECTRONIC : MATERIALs, DEVICEs and simple circuits



Detection is possible if Ep > Eg







Ep =





½

hc J λ

=

hc eV eλ

=

6.63 × 10 −34 × 3 × 10 8 1.6 × 10 −19 × 400 × 10 −9











= 3.1 eV(>Eg)



Detailed Answer : (i) Characteristics of LED

½

\ It can detect this light.

½

[CBSE Marking Scheme, 2017]



10 1

2

60 40

3

4

V

20







Advantages (any two)

0.2 0.4 0.6 0.8 1.0

Vm



1

(i) Low operational voltage. (ii) Less power consumption.

½

(iii) Long life

½

(iv) Ruggedness

[or any other]

(a) Energy band gap controls the wavelength of light emitted. ½ (b) Forward current controls the intensity of emitted light. ½ [CBSE Marking Scheme 2017]

am

850-940 nm

Infra-red

1.2 V

gr

Wavelength

Vf @ 20 mA

GaAsP

630-660 nm

Red

1.8 V

GaAsP

605-620 nm

Amber

2.0 V

GaAsP

585-595 nm

Yellow

2.2 V

AlGaP

550-570 nm

Green

3.5 V

SiC

430-505 nm

Blue

3.6 V

GalnN

450 nm

White

4.0 V

al

5

ke

OR

80

GaAs

ci

20

0

100

Yellow Green Blue White

Red Amber

Infared

30

An i



40

Colour

Semiconductor Material

1 1 1

50



Typical LED Characteristics

tS ha _o ffi

Forward current I (mA)

Ans. I-V characteristics Two advantages Factors



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Q. 8. Draw the I-V characteristic of an LED. State two advantages of LED lamps over conventional incandescent lamps. Write the factor which controls (a) wavelength of light emitted, (b) intensity of light emitted by an LED. U [OD III Comptt. 2017]

1 (ii) Advantages : (a) LEDs provide instantaneous turn ON and have no issues with frequent switching (b) LEDs consume less power and can operate effectively on low-voltage electrical systems. (c) LEDs are able to operate at virtually any percentage of their rated power (0 to 100%) (d) LEDs have good Colour Rendering Index (CRI) 1 (iii) Factors : (a) Wavelength of light emitted by LED depends on forbidden energy gap of semiconductor material which is used for making LED. The wavelength of light emitted by LED is inversely proportional to forbidden energy gap. 1

(b) Forward current increases as intensity of light increases that reaches to maximum value which on further increase in forward current will lead to decrease in light intensity. Q. 9. The graph of potential barrier versus width of depletion region for an unbiased diode is shown in A. In comparison to A, graphs B and C are obtained after biasing the diode in different ways. Identify the type of biasing in B & C and justify U [CBSE SQP 2016] your answer.

386 ]

Oswaal CBSE Chapterwise & Topicwise Question Bank, PHYSICS, Class – XII B

C W

V (x)

V (x)

V (x)

A

p

x B

C

V (x)

V (x)

E x

x

V (x)

A

n

Va





½



(ii) Transformer x C

x

ci

x

1½

tS ha _o ffi

Ans. Diode B is reverse biased.

When diode is reverse based, the barrier height



increases as the direction of applied voltage (V)

and the direction of barrier potential (V0) is same. The effective barrier height under reverse biased is (V0 + V).

X

Secondary

RL B

Y

½ Working : If an alternating voltage is applied as input, a diode in series with a load, a pulsating voltage will appear across the load only during that half cycle of the ac input during which the diode is forward biased. Therefore, in the positive half cycle of ac input there is current through the load resistor RL and we get an output voltage whereas there is no output during the negative half cycle. Thus, the output voltage is restricted to only one direction and is said to be rectified. 1½ [Note : If the student draws only the input and output wave form, then award ½ mark only] [CBSE Marking Scheme 2016]

al



te le

gr

Primary

V (x)

V (x)

B

A

x

am

x

Diode C is forward biased.

1½



As the direction of applied voltage is opposite

ke



An i

to the barrier potential, therefore, the effective barrier height is reduced to (V0 – V). [CBSE Marking Scheme 2016]

Q. 10. (i) Explain with the help of a diagram the formation of depletion region and barrier potential in a p-n junction. (ii) Draw the circuit diagram of a half wave rectifier and explain its working. U [Outside Delhi I, II, III 2016] Ans. (i) Due to diffusion and drift, the electrons and



Q . 11. Give reason for the following : (i) High reverse voltage does not appear across a LED. (ii) Sunlight is not always required for the working of a solar cell. (iii) The electric field, of the junction of a Zener diode, is very high even for a small reverse bias voltage of U [Delhi Comptt. 2016] about 5 V. Ans. (i) It is because reverse break down voltage of LED is very low i.e., nearly 5 V.

1

holes move across the junctions, creating a final

(ii) Because solar cell can work with any light whose

stage in which a region is created across the

photon energy is more than the band gap energy.

junction wall, which gets devoid of the mobile

1

charge carriers. This region is called depletion

(iii) The heavy doping of p and n sides of p-n junction,

region; the potential difference across the region

makes the depletion region very thin, hence for

½

a small reverse bias voltage, electric field is very

is called Barrier potential.

high.

[CBSE Marking Scheme 2016] 1

[ 387

SEMICONDUCTOR ELECTRONIC : MATERIALs, DEVICEs and simple circuits

Q. 12. With what considerations in view, a photodiode is fabricated ? State its working with the help of a suitable diagram. Even though the current in the forward bias is known to be more than in the reverse bias, yet the photodiode U [Delhi I, II, III 2015] works in reverse bias. What is reason ?  Ans. It is fabricated with a transparent window to allow light to fall on diode.

1½

When the photodiode is illuminated with photons of energy (hn > Eg) greater than the energy gap of the semiconductor, electron hole pairs are generated. These get separated due to the junction electric field (before 1

they recombine) which produce an emf. hn

mA p-side +

am

–

1

n-side

al

te le

gr

Reason : It is easier to observe the change in the current, with change in light intensity, if a reverse bias is applied. Alternatively, The fractional change in the minority carrier current, obtained under reverse bias, is much more than the corresponding fraction change in majority carrier current obtained under forward bias. [CBSE Marking Scheme 2015] 1 OR

An i

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tS ha _o ffi

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Ans.





388 ]

Oswaal CBSE Chapterwise & Topicwise Question Bank, PHYSICS, Class – XII





[Topper's Answer 2015]

Q. 13. (i) How is a photodiode fabricated ? (ii) Briefly explain its working. Draw its I–V characteristics for two different intensities of illumination. U [Foreign 2014; Delhi I, II, III 2013]

gr

am

(ii) Circuit diagram

Ans. (i) Photodiode is fabricated with a transparent

Unregulated voltage(VL )

te le

window to allow light to fall on the diode.

1

(ii) (a) Working : When a reverse biased photodiode

IL

Load RL

Regulated voltage (Vz )

1 Any increase / decrease in the input voltage results in increase / decrease of the voltage drop across RL, without any change in the voltage across the Zener diode. [CBSE Marking Scheme 2014] 1

ci

than the forbidden energy gap (Eg), the electron-

al

is illuminated with the light of energy greater

Iz

Rs

tS ha _o ffi

hole pairs are generated in, or near, the depletion

region. Due to junction field, electrons are collected on the n-side and holes on p-side, giving 1

An i

ke

rise to a potential difference. (b)

[CBSE Marking Scheme 2013] 1 Q . 14. (i) Why is zener diode fabricated by heavily doping both p-and n-sides of the junction ? (ii) Draw the circuit diagram of zener diode as a voltage regulator and briefly explain its working. U [Foreign 2014] Ans. (i) Due to the heavy doping, the depletion layer becomes very thin and electric field, across the junction, becomes very high even for a small reverse bias voltage. 1

Commonly Made Error

 Many students able to defined the working of zener diode but couldn't write the reason of heavily doping of p and n sides of the junction.

Q. 15. Give reasons for the following : (i) A photodiode, when used as a detector of optical signals is operated under reverse bias. (ii) The band gap of the semiconductor used for fabrication of visible LED’s must at least be 1.8 eV. A&E [CBSE SQP 2014] Ans. (i) When operated under reverse bias, the photodiode can detect the changes in current with the changes in light intensity more easily. 1½ (ii) The photon energy, of visible light photons varies from about 1.8 eV to 3 eV. Hence for visible LED’s, the semiconductor must have a band gap of 1.8 eV. [CBSE Marking Scheme 2014] 1½

Long Answer Type Questions Q. 1. (i) Explain with the help of a diagram, how a depletion layer and barrier potential are formed in a junction diode. [Delhi & Delhi Comptt. I, II, III 2014]

(5 marks each)

(ii) Draw a circuit diagram of full-wave rectifier. Explain its working and draw input and output waveforms. U [Delhi Comptt. I, II, III 2014; O.D. I, II, III 2013; Delhi I, II, III 2012; O.D. I, II, III, 2011]

[ 389

SEMICONDUCTOR ELECTRONIC : MATERIALs, DEVICEs and simple circuits



Electron diffusion

Electron drift

Hole diffusion

Biasing The V-I characteristics are obtained by connecting the battery, to the diode, through a potentiometer (or rheostat). The applied voltage to the diode is changed. The values of current, for different values of voltage, are noted and a graph between V and I is plotted. The V-I characteristics of a diode, have the form shown here. 1

Depletion region

{

Ans. (i)

½

Hole drift

gr

te le

Q. 3. (i) Describe briefly, with the help of a diagram, the role of the two important processes involved in the formation of a p-n junction. (ii) Name the device which is used as a voltage regulator. Draw the necessary circuit diagram and U R [OD I, II, III 2012] explain its working.

Voltmeter(V)

applied electric field ( E ) which results in drifting →

of holes along E and of electrons opposite to that →

p n Milliammeter (mA) Switch







Reverse biasing

+ –

Ans. (i) For diagram see Q. 1(a) Long Answer Type Questions. 1 The two processes are (a) Diffusion ½ (b) Drift ½ Diffusion : Holes diffuse from p-side to n-side (p ® n) and electrons diffuse from n-side to p-side ( n ® p). Drift : The motion of charge carriers, due to the →

An i

Ans. (i) Forward biasing

ke

tS ha _o ffi

ci

Q. 2. (i) Draw the circuit arrangement for studying the I – V characteristics of a p-n junction diode in (a) forward and (b) reverse bias. Briefly explain how the typical I – V characteristics of a diode are obtained and draw these characteristics. [Delhi I, II, III, O.D. Comptt. I, II, III 2014] (ii) With the help of necessary circuit diagram explain the working of a photodiode used for detecting optical signals. U [O.D. Comptt. I, II, III 2014, 2013]

½ + ½ (ii) Try yourself, Similar to Q. 12, Short Answer Type Questions-II. 2 [CBSE Marking Scheme 2014, 13]

al

Due to the diffusion of electrons and holes, from their majority zone to minority zone, a layer of positive and negative space charge region on either side on the junction is formed. This is called the depletion region. ½ The loss of electrons, from n-region and the gain of electrons by the p-region, cause a difference of potential across the junction. This tends to prevent the movement of charge carriers across the junction and is, therefore, termed as barrier potential. 1 (ii) Try yourself, Similar to Q. 1 (ii), Short Answer Type Question-II. 3 [CBSE Marking Scheme, 2014, 13, 12, 11]

am



½



½ of electric field ( E ). (ii) Name of device : Zener Diode 1 [Note : for diagram see Q. 4 (ii) (SATQ-II)] 1 Working : Any increase / decrease in the input voltage results in an increase / decrease of the voltage drop across Rs without any change in voltage across the zener diode. Thus Zener diode acts as a voltage regulator. [CBSE Marking Scheme, 2012] ½







½

Q. 4. (a) Why are photodiodes preferably operated under reverse bias when the current in the forward bias is known to be more than that in reverse bias? (b) The two optoelectronic devices: Photodiode and solar cell, have the same working principle but differ in terms of their process of operation. Explain the difference between the two devices in terms of (i) biasing, (ii) junction area and (iii) I-V characteristics. [SQP 2018]

390 ]

Oswaal CBSE Chapterwise & Topicwise Question Bank, PHYSICS, Class – XII

Q. 5. Explain with the help of suitable diagram, the two processes which occur during the formations of a p-n junction diode. Hence define the terms (i) depletion region and (ii) potential barrier. (b) Draw a circuit diagram of a p-n junction diode under forward bias and explain its working. [CBSE Comptt. 2018]

Ans. (a) The fractional change in majority charge carriers is very less compared to the fractional change in minority charge carriers on illumination. 1 (b) The difference in the working of two devices : Photodiode

Solar cell

(i) Biasing

Used in reverse biasing

No external biasing is given

(ii) J u n c t i o n Area

Small

Large for solar radiation to be incident on it.

Ans. (a) Explaining the two processes 1+1 Defining the two terms ½+½ (b) Circuit diagram 1 Working 1 (a) The two important processes are diffusion and drift. Due to concentration gradient, the electrons diffuse from the n side to the p side and holes diffuse the p side to the n side. ½ Electron diffusion Electron drift

(iii) I-V characteristics

am

mA

I1 I2 I3 I4

{

gr

Reverse bias

te le

Hole diffusion

ci

al

A I

tS ha _o ffi

I1>I2>I3>I4

Voc (open circuit voltage) V

ke

ISC



Depletion region

Hole drift Due to the diffusion, an electric field develops across the junction. Due to the field, an electron moves from the p-side. The flow of the charge carriers due to the electric field, is called drift. ½ (i) Depletion region : It is the space charge region on either side of the junction, that gets depleted of free charges. is known as the depletion region. ½ (ii) Potential Barrier : The potential difference, that gets developed across the junction and opposes the diffusion of charge carries and brings about a condition of equilibrium, is known as the barrier potential. ½ (b) Try yourself similar to Q. 5 SAT Question-II [CBSE Marking Scheme 2018]

Short circuit current

An i

[CBSE Marking Scheme 2018]

oswaal learning tools For Suggested Online Videos Visit : https://qrgo.page.link/v6rh6

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





Other Important Chapter/Topics for Competitive Examinations Preparation. Unit – x

communication Systems

CHAPTER

15

Communication Systems

Chapter Analysis

Topic-1

D

OD

–

1Q (2 marks) 1Q (3 marks)

1Q (2 marks)

1Q (3 marks)

–

1Q (1 mark) 1Q (3 marks)

1Q (1 mark)

1Q (3 marks)

1Q (3 marks)

1Q (2 marks) 1Q (3 marks)

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OD

D/OD

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Modulation

2018

D

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Elements of Communication System

2017

am

2016

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List of Topics

Revision Notes

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Elements of Communication System

TOPIC - 1

An i

¾¾ Communication is the method of transmitting and receiving information Elements of Communication System of data. .... P. 391 ¾¾ There may be two types of communication : • Point to point communication : In this type of communication Topic - 2 .... P. 399 transmitter and receiver are single point. Example : Telephonic Modulation communication. • Broadcasting communication : In this type of communication, there is one transmitter and many receivers. Example : transmission of radio or television. ¾¾ When input message (any form of signal variation) is combined with some intelligence input (encoding and compatibility in machine language), it becomes information. ¾¾ Generalised communication system : Communication System

Transmitted Received Information Message User of Transmitter Receiver Channel Message Information Signal Signal Signal Source Signal

Noise

Block diagram of a generalised Communication System

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Oswaal CBSE Chapterwise & Topicwise Question Bank, PHYSICS, Class – XII

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Important terminology and their application in communication system ¾¾ Information source : The information source is the generator of information which we want to communicate. It may be audio, video or data. ¾¾ Electric Transducer : Electric transducer convert physical variable into electrical signal variable. ¾¾ Signal: Information converted in electrical form and suitable for transmission is called signal. ¾¾ Transmitter : Process of the incoming message signal and making it suitable for transmission through particular channel. ¾¾ Noise : Unwanted signal which interferes with the information signal and disturbs the information. ¾¾ Channel : It is the medium through which signal from transmitter propagates to the receiver. For example : optical fibre, coaxial cables etc. ¾¾ Receiver : Collects the message from the channel and extracts signal. ¾¾ Attenuation : Signal loses energy during propagation through channel. This is called attenuation. ¾¾ Amplifier : It is the device which increases the strength by increasing its amplitude. ¾¾ Range : Maximum distance between transmitter and receiver at which signal can be recovered is called the range of communicating system. ¾¾ Bandwidth : It is the frequency range over which an equipment operates or range of frequencies a signal has. ¾¾ Modulation : Mixing of signal with carrier frequency is known as modulation. ¾¾ Demodulation : Extracting of signal from carrier frequency is known as demodulation. ¾¾ Repeater : It receives the signal, reconditions it and then retransmits it. Signal may be classified in two categories : ¾¾ Analog Signal : Continuous variation of signal with respect to time is known as analog signal. For example : telephonic signal, video signal etc. In modern technology we can convert analog signal to digital signal for communicating and convert back to analog signal at receiver. ¾¾ Digital Signal : Discrete value of signal variation with respect to time is known as digital signal. For example : computer etc. • Coding helps in sending digital signal with much more accuracy. There are several coding techniques. For example in computer data, we employ suitable combinations of number systems such as the binary coded decimal (BCD), American Standard Code for Information Interchange (ASCII) ¾¾ Operational advantages of digital communication system over analog communication systems are. • An improved form of sending messages securely. • Increased immunity to noise and external interference. • A common format for encoding different kinds of message signals for the purpose of transmission. • Flexibility in configuration of digital communication system. ¾¾ Hence in modern technology, analog signals are transmitted through digital communication. In the final stage they are converted back to analog signals. ¾¾ Different types of message signals have different range of frequencies. • Audio signal – 20 Hz to 20kHz • Video signal – 4.2 MHz • TV signal – 6 MHz (audio + video) ¾¾ Large bandwidth is required to accommodate complete information of wave. ¾¾ Frequency bands of some important wireless communications : Service

Frequency bands

Comments

Standard AM broadcast

540-1600 kHz

FM broadcast

88-108 MHz

Television

54-72 MHz

VHF (very high frequencies)

76-88 MHz

TV

174-216 MHz

UHF (ultra high frequencies)

420-890 MHz

TV

896-901 MHz

Mobile to base station

840-935 MHz

Base station to mobile

5.925-6.425 GHz

Uplink

3.7-4.2 GHz

Downlink

Cellular Mobile Radio Satellite Communication

[ 393

Communication systems

Propagation of Electromagnetic wave : ¾¾ Earth’s atmosphere plays a vital role in propagation of electromagnetic wave. There are three ways of communication through electromagnetic wave. ¾¾ Ground Wave : (i) The radio waves which travel through atmosphere following the surface of the earth are called ground waves or surface waves and their propagation is known as ground wave propagation or surface wave propagation. (ii) The ground waves have vertical orientation and travel parallel to the ground. (iii) The ground wave propagation is suitable for low and medium frequency, i.e., from few hundred kHz to 2 MHz only. (iv) Its power is less as they operate in low frequency. λ (v) It can bend round the corners of the object on the earth, hence can jump the restriction. ( θ = , low a

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frequency means l is more so more bending) (vi) Attenuation is high for ground wave transmission and increases with increase in frequency. This is because more absorption of ground waves (near earth) takes place at higher frequency during propagation through atmosphere. (vii) Length of antenna is directly proportional to the wavelength of EM wave. Hence, for ground wave large antenna is required. (viii) The ground wave propagation is generally used for local band broadcasting and is commonly known as medium wave. Local transmitter, police walkie talkie , AM transmitter are some of its examples. Sky wave propagation : (i) The sky waves are the radio waves of frequency between few MHz to 40 MHz. (ii) These radio waves can propagate in atmosphere and are reflected back by the ionosphere of earth’s atmosphere. (iii) The sky waves travel from transmitter antenna to receiver antenna, through sky they reflect back from ionosphere. Hence, their propagation is called sky wave propagation. (iv) Critical frequency (fc) is that highest frequency of radio waves, which when sent straight (i.e., normally) towards the layer of ionosphere gets reflected and returns to the earth. If the frequency of radio waves is more than the critical frequency, it will not be reflected by the ionosphere. (v) The value of C.F. is found to be 4 MHz, 5 MHz and 6 to 8 MHz for D (part of stratosphere), E (part of stratosphere), F1 (part of mesosphere) and F2 (Thermosphere) layers of ionosphere which are at heights about 110 km, 180 km and 300 to 350 km respectively from the surface of earth. (vi) Its range is very large as compared to range of ground wave. Range can be targeted and can be increased by multiple transmitters. • Limitations : 3 MHz to 30 MHz is very small bandwidth of frequency for present application. Higher frequencies penetrate the ionosphere and can’t be reflected. 1.5 (c) voltage (b) (d) 1

(a)

0.5

0 0

0.2

0.4

0.6

0.8

time 1

–0.5

–1 –1.5

Space wave propagation : (i) It is used for very high frequency (> 40 MHz). These can penetrate ionosphere more efficiently. (ii) Due to high frequency, wavelength is very small and energy is very high. (iii) Television broadcast, microwave links and satellite communication are some examples of communication systems that use space wave mode of propagation.

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Oswaal CBSE Chapterwise & Topicwise Question Bank, PHYSICS, Class – XII

Line of sight communication by space wave : ¾¾ We also use this space wave in ground transmission. It is known as line of sight transmission. ¾¾ These are (space wave) high frequency hence they travel nearly in a line. Mobile transmission or microwave links are based upon this. ¾¾ Earth’s curvature restrict the range of line of sight transmission. There is limited space between two antennas. ¾¾ If h is the height of transmitting antenna then its signal range is d = 2hR ¾¾ T he range of communication dM between the transmitting antenna of height hT and the receiving antenna of height hR is given by dM = 2 hT R + 2 hR R

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where, R is the radius of the earth. Area covered through one tower = pd2 = p × 2hR Population covered = population density × area covered. Satellite Communication : ¾¾ The satellite communication is a mode of communication of signal between a transmitter and a receiver through a satellite. ¾¾ The satellite communication is like a line of sight microwave communication. ¾¾ Since, the satellite communication is through space hence, it is also part of space communication. ¾¾ A communication satellite is a space craft, provided with microwave receiver and transmitter. It is placed in an orbit around the earth. ¾¾ In satellite communication, a beam of modulated microwave from the transmitter is sent directly towards the communication satellite, which receives the coming signal, amplifies it and returns it to the earth. Transmitting frequency (uplink) and receiving frequency (downlink) are different to avoid interference between the uplink and the downlink. ¾¾ A satellite communication is possible through geostationary satellites. ¾¾ A single geostationary satellite cannot cover the whole part of the earth for microwave communication. It is so because, the large part of the earth is out of sight due to the curvature of the earth. One satellite roughly covers one third of earth. ¾¾ In order to have global transmission, at least three geostationary satellites are required, which are at particular distance from each other. ¾¾ Global positioning system is also based upon satellite communication.

Very Short Answer Type Questions

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Q. 1. Which mode of propagation is used by short wave broadcast service ? R Ans. Sky wave propagation is used in short wave broadcast service. Its frequency range is from few MHz upto 40 MHz. 1 Q. 2. Name the essential components of a communication system. R [O.D. I, II, III 2016] Ans. Transmitter, Medium or Channel and Receiver. 1 Q. 3. Why are micro waves considered suitable for radar systems used in aircraft navigation ?

R [2016 Delhi All set-3] Ans. Due to their short wavelengths, they are suitable for radar system used in aircraft navigation. 1 Q. 4. What is the meaning of the term ‘attenuation’ used in communication system ? R [O.D. Comptt. I, II, III 2014] Ans. Attenuation is the loss of strength of a signal, while propagating through a medium. 1 [CBSE Marking Scheme, 2014] Q. 5. Give one example of point-to-point communication mode. R [O.D. Comptt. I, II, III 2014] Ans. Telephony service is based upon point to point communication. 1

(1 mark each)

Q. 6. How does the effective power radiated from a linear antenna depend on the wavelength of the signal to be transmitted ? U [Delhi Comptt. I, II, III 2014]





Ans. Effective power ∝

1 λ2

Alternatively, The effective power which is radiated decreases with an increase in wavelength. 1 [CBSE Marking Scheme, 2014]

Q. 7. Draw a block diagram of a generalized communication system. R [Delhi Comptt. I, II, III 2014] Ans. Refer Block diagram (Topic-I) Revision Notes Alternatively, Also accept if the student gives only the following diagram : Message Signal





Transmitter

Receiver

User

[CBSE Marking Scheme, 2014] 1

[ 395

Communication systems

Short Answer Type Questions-I

Q. 7. State the concept of mobile telephony and explain its working. R [2016-OD, east] Ans. Concept of mobile telephony is to divide the service area into a suitable number of cells centered on an office MTSO (Mobile Telephone Switching Office). Mobile telephony means that you can talk to any person from anywhere. 1 Explanation : (i) Entire service area is divided into smaller parts called cells. (ii) Each cell has a base station to receive and send signals to mobiles. (iii) Each base station is linked to MTSO. MTSO coordinates between base station and TCO (Telephone Control Office) 1 Q. 8. Distinguish between any two types of propagation of electromagnetic waves with respect to (i) frequency range over which they are applicable, (ii) communication systems in which they are U [CBSE SQP 2016] used. Ans.

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Ans. Broadcast/point to point, mode of communication ½ Space wave propagation ½ Above 40 MHz ½ Because EM waves, of frequency above 40 MHz, are not reflected back by the ionosphere / penetrate through the ionosphere. ½ [CBSE Marking Scheme, 2017]

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[CBSE Marking Scheme, 2017]



Q. 6 A device X used in communication system can convert one form of energy into another. Name the device X. Explain the function of a repeater in a communication system. R [2016- Foreign Sets-I, II, III] Or Write the function of a (i) transducer and (ii) repeater in a communication system. [2016 -OD; south] Ans. ‘X’ is a transducer It is a device which converts one form of energy to another. 1 A repeater picks up the signal from the transmitter, amplifies and transmits it to the receiver sometimes with a change in carrier frequency. Repeaters are used to extend / increase the range of a communication system. 1

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Q. 2. Which basic mode of communication is used in satellite communication ? What type of wave propagation is used in this mode ? Write, giving reason, the frequency range used in this mode of propagation. R [2017 Delhi set-1]

Example : Point-to-point : telephone (any other) ½ Broadcast : T.V. Radio (any other) ½ [CBSE Marking Scheme, 2017]

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Ans. (i) Communication, using waves which travel in straight line from transmitting antenna to receiving antenna is called line of sight communication. 1 (ii) Because T.V. signal waves are not reflected back by the ionosphere ½ d = 2hR ½



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Q. 1. (i) What is line of sight communication ? (ii) Why is it not possible to use sky waves for transmission of TV signals? Upto what distance can a signal be transmitted using an antenna of height ‘h’? R [2017 Delhi set-3]

(2 marks each)

Q. 3. Distinguish between a transducer and a repeater. R [2017 Delhi set-2]

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Ans. Transducer : A device which converts one form of energy into another. 1 Repeater : A combination of receiver and transmitter. It picks signals from a transmitter; amplifies and retransmits them. 1 [CBSE Marking Scheme, 2017]

Commonly Made Error

 Many students couldn’t define the tansducer. Q. 4. A TV transmission tower antenna is at a height of 20 m. How much range can it cover if the receiving antenna is at a height of 25 m? R [CBSE-SQP -2018] Ans. Range d =

2 hR + 2 hR R





d=

2 × 20 × 6·4 × 10 6 + 2 × 25 × 6·4 × 10 6





d = 33.9 km 1 [CBSE Marking Scheme, 2018]



1

Q. 5. Distinguish between broadcast mode and pointto-point mode of communication and give one example for each. R [2017-Foreign Sets-I, II, III] Ans. In point-to-point communication mode, communication takes place over a link between a single transmitter and a single receiver. ½ In the broadcast mode, there are large number of receivers corresponding to a single transmitter. ½

Type of EM wave propagation

Frequency range

Ground wave

500–1500 KHz

Space wave

Above 40 MHz

Q. 9. Write the functions of communication systems : (i) Transmitter (ii) Modulator

Use

Standard AM broadcast

Television ½+½+½+½ the following in

R [Delhi 2014]

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Oswaal CBSE Chapterwise & Topicwise Question Bank, PHYSICS, Class – XII

Ans. (i) Device used for communication which transmits the signal by amplifying it.

Ans. S.No

(ii) Device which merges the high frequency carrier wave and low frequency base-band signal.

Q. 10. Write

the

functions

of

the

following

1+1



Restricted up to a Can take place (even) few MHz frequency beyond 40 MHz fre(3 to 40 MHz). quency.

(ii)

Waves are reflected Space waves travel in back from iono- a straight line, either sphere. directly from transmitting antenna to receiving antenna or, through satellite.

(i) Transducer R [O.D. I, II, III 2014]

(ii) Repeater

Ans. (i) Transducer : Any device that converts one

transmitter, amplifies and retransmits it to the receiver.

[CBSE Marking Scheme, 2014] 1

Ans. Analog signals are continuous variations of voltage or current. 1 Digital signals are those which can take only discrete (step wise) values of current or voltage. 1 [CBSE Marking Scheme, 2014]

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Q. 11. Distinguish between ‘sky wave’ and ‘space wave’ modes of propagation in a communication system.  R [Delhi Comptt., I, II, III, 2016]

[CBSE Marking Scheme, 2016] 2 Q. 12. Distinguish between ‘Analog and Digital signals’. U [Delhi I, II, III 2014]

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(ii) Repeater : A repeater accepts the signal from the

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1

form of energy to another.

(3 marks each)

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Short Answer Type Questions-II

Ans. Propagation of waves, along a straight path from the transmitting antenna to receiving antenna, using line of sight (LOS) communication is called space wave propagation. 1

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Q. 1. What is sky wave propagation ? Which frequency range is suitable for sky wave propagation and why ? Over which range of frequencies can communication through free space using radio U [2017, Foreign Set-II] waves take place ?

Space Wave

(i)

in

communication systems :

Sky Wave



Relevant system of communication :



Television broadcast, microwave links and satellite communication (any one) ½



‘Radio horizon‘ equals the distance between the transmitting antenna and the point on the earth where the direct waves get blocked due to the curvature of the earth.



[Also accept d =

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Ans. In sky wave propagation, long distance communication is achieved by ionospheric reflection of radio waves back towards the earth. 1 The frequency range is from a few Mega hertz to 30/40 Mega hertz. The ionospheric layers can act as a reflector over this frequency range (3 MHz to 30/40 MHz). Higher frequencies penetrate through it. 1 The frequency range of radio frequencies is a few hundred kHz to a few GHz. (waves having frequency beyond 40 MHz) 1 [CBSE Marking Scheme, 2017]



Commonly Made Error  Many students confused ‘sky wave propogation’ with’ space wave propagation.‘



2hR ; h = height of transmitting

antenna, R = Radius of the earth.] At frequencies above 40 MHz, relatively smaller antennas are needed and communication is essentially limited to line of sight paths. 1 [Alternatively, At frequencies (more than 40 MHz), EM waves do not get bent or reflected by ionosphere. Therefore space wave propagation has to be used for frequencies above 40 MHz.] 1 [CBSE Marking Scheme, 2017]

Q. 2. What is space wave propagation ? Which systems of communication use space waves ? What is ‘radio horizon’ of a transmitting antenna of height h ? Why is space wave propagation suitable for frequencies above 40 MHz ? U [2017 Foreign Set-III] Q. 3. Draw a block diagram of a generalized communication system. Write the functions of each of the following : (i) Transmitter (ii) Channel U [2017, OD Set-III] (ii) Receiver

[ 397

Communication systems

Ans.

Communication System

Transmitted Received Information Message User of Transmitter Receiver Channel Message Information Signal Signal Signal Source Signal

Noise



[Also accept the following diagram

1½



am

Information Communication Receiver of → → source channel Information (i) Transmitter : A transmitter processes the incoming message signal so as to make it suitable for transmission through a channel and subsequent reception. ½ (ii) Channel : It carries the message signal from a transmitter to a receiver. ½ (iii) Receiver : A receiver extracts the desired message signals from the received signals at the channel output. ½ [CBSE Marking Scheme, 2017]

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Commonly Made Error  Few students couldn’t write the function of channel.

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(ii) The service area is divided into a suitable number of hexagonal cells centered on MTSO (Mobile Telephone Switching Office). Each cell contains a low-power transmitter called a base station and caters to a large number of mobile receivers / cell phones. When a mobile receiver crosses one base station it is handed over to another base station . It is called handover or handoff. 1½ Q. 6. Mention three applications of the internet. Explain U [2016-OD, south] one of these in detail.



An i

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Q. 4. (i) Which mode of propagation is used by shortwave broadcast services having frequency range from a few MHz up to 30 MHz ? Explain diagrammatically how long distance communication can be achieved by this mode. (ii) Why is there an upper limit to frequency of waves used in this mode ? U [O.D. I, II, III, 2016] Ans. (i) Sky wave propagation. ½

1

Long distance communication can be achieved by reflection of radio waves by the ionosphere, back towards the Earth. This ionosphere layer acts as a reflector only for a certain range of frequencies. (few MHz to 30 MHz) ½ (ii) Electromagnetic waves of frequencies higher than 30 MHz, penetrate the ionosphere and escape whereas the waves less than 3 MHz are reflected back to the earth by the ionosphere. 1 Q. 5. (i) Distinguish between point to point and broadcast modes of communication. Give an example of each. (ii) Explain the basic concept of mobile telephone. U [2016 Foreign Set-1] Ans. (i) In point to point communication mode, communication takes place over a link between a single transmitter and a receiver. In broadcast mode , there are a large number of receivers corresponding to a single transmitter. Examples; Point to point : telephony Broadcast : radio / Television. 1½

Ans. Application of internet–e-mail, social networking sites, e-commerce, mobile telephony, GPS. [Any three] ½+½+½ Explanation of any one. 1½ [CBSE Marking Scheme, 2016] Detailed Answer : One of the main use of internet is instant global connectivity. It is very useful in e-learning sector. Nowadays students from remote area can access quality of education from different educational website and e-lectures through internet. Interactive videos available in internet help them also. Q. 7. What is global positioning system ? Explain its U [2016 OD, east] working in brief. Ans. Global Positioning System (GPS) is method of identifying location or position of any point or a person on earth using a system of 24 satellites, which are continuously orbiting, observing, monitoring and mapping the earth. 1 Working Principle : (i) The unique location of GPS user is determined by measuring its distance from at least three GPS satellites. 1 (ii) Using these values of distances, obtained from three satellites, a microprocessor, fitted in GPS device, determines the exact location. 1

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Oswaal CBSE Chapterwise & Topicwise Question Bank, PHYSICS, Class – XII

Q. 8. What is space wave propagation ? State the factors which limit its range of propagation. Derive an expression for the maximum line of sight distance between two antennas for space wave propagation. U [2016; OD, North Set] Ans. Space Wave Propagation : The mode of propagation in which radio waves travel, along a straight line, from the transmitting to the receiving antenna. 1 Limiting Factors : (i) Curvature of the earth (ii) Insufficient height of the receiving antenna (Award this ½ mark if the student writes any one of these two factors) 1 Derivation : W h Transmitting Antenna

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R



From the figure, we have (R + h)2 = R2 + d2 Or 2Rh @ d2 (as h2