CBSE Class 12th Exam Date:17 Feb' 26 - 17 Feb' 26
Have you ever wondered how simple electronic circuits work or how semiconductors are used in devices like diodes and transistors? Chapter 14: Semiconductor Electronics Materials Devices and Simple Circuitss in Class 12 Physics explains the basics of how these devices work. In this chapter you will learn about semiconductors, diodes, transistors (NPN and PNP) and logic gates (Like - OR gate, AND gate etc) . These topics are very important for understanding how modern electronic gadgets are built. The NCERT Exemplar Solutions for class 12 are prepared by experienced teachers as per the CBSE syllabus.
The NCERT Exemplar Solutions for Class 12 Physics Chapter 14 by Careers360 provide, step-by-step answers to all the exemplar problems. The solutions cover MCQs, Very short answer (VSA) short answer(SA) and long answer(LA) questions. solving these NCERT exemplar questions will help you understand electronics better, improve your application skills, and prepare well for exams like CBSE Board Exams, JEE, and NEET.
The JEE Mains 2026 total marks for paper 1 are given below.
Subjects | Section A | Section B | Marks |
Maths | 20 | 5 | 100 |
Physics | 20 | 5 | 100 |
Chemistry | 20 | 5 | 100 |
Total Marks | 75 | 300 |
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Question:1
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.
Answer:
The answer is the option (d)Question:2
In Fig. 14.1, Vo is the potential barrier across a p-n junction, when no battery is connected across the junction
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.
Answer:
Answer: The answer is the option (b)Question:3
In Fig. 14.2, assuming the diodes to be ideal,
A. $D_1$ is forward biased and $D_2$ is reverse biased and hence current flows from A to B
B. D2 is forward biased and $D_1$ is reverse biased and hence no current flows from B to A and vice versa.
C. $D_1$ and $D_2$ are both forward biased and hence current flows from A to B.
D. $D_1$ and $D_2$ are both reverse biased and hence no current flows from A to B and vice versa.
Answer:
The answer is the option (b)Question:4
A 220 V A.C. supply is connected between points A and B (Fig. 14.3). What will be the potential difference V across the resistor?
A. 220V
B. 110V
C. 0V
D. $220 \sqrt{2}V$
Answer:
As p-n junction diode will conduct during positive half cycle only during the negative half cycle diode is reverse biases. during this diode will not give any output so potential difference across capacitor C-peak voltage of the given AC voltage
$V_{0}=V_{rms}\sqrt{2}=220 \sqrt{2}V$
Question:5
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.
Answer:
The answer is the option (b)Question:6
The output of the given circuit in Fig. 14.4.
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.
Answer:
The answer is the option (c)
would be like a half wave rectifier with negative cycles in output
Question:7
In the circuit shown in Fig. 14.5, if the diode forward voltage drop is 0.3 V, the voltage difference between A and B is
A. 1.3 V
B. 2.3 V
C. 0
D. 0.5 V
Answer:
The answer is the option (b)Question:8
Truth table for the given circuit (Fig. 14.6) is
A.
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C
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1
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0
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Answer:
The answer is the option (c)
A
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B
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$\bar{A}$
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C=A.B
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d=$\bar{A}$.B
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E=(C+D)
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0
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Question:9
When an electric field is applied across a semiconductor
A. electrons move from lower energy level to higher energy level in the conduction band.
B. electrons move from higher energy level to lower energy level in the conduction band.
C. holes in the valence band move from higher energy level to lower energy level.
D. holes in the valence band move from lower energy level to higher energy level.
Answer:
The answer is the option (a, c)Question:10
Consider an npn transistor with its base-emitter junction forward biased and collector base junction reverse biased. Which of the following statements are true?.
A. Electrons crossover from emitter to collector.
B. Holes move from base to collector.
C. Electrons move from emitter to base.
D. Electrons from emitter move out of base without going to the collector.
Answer:
The answer is the option (a, c)Question:11
Figure 14.7 shows the transfer characteristics of a base biased CE transistor. Which of the following statements are true?
A. At Vi = 0.4V, transistor is in active state.
B. At Vi = 1V, it can be used as an amplifier.
C. At Vi = 0.5V, it can be used as a switch turned off.
D. At Vi = 2.5V, it can be used as a switch turned on.
Answer:
The answer is the option (b, c, d)Question:12
In a npn transistor circuit, the collector current is 10mA. If 95 per cent of the electrons emitted reach the collector, which of the following statements are true?
A. The emitter current will be 8 mA.
B. The emitter current will be 10.53 mA.
C. The base current will be 0.53 mA.
D. The base current will be 2 mA.
Answer:
The answer is the option (b, c)Question:13
In the depletion region of a diode
A. there are no mobile charges
B. equal number of holes and electrons exist, making the region neutral.
C. recombination of holes and electrons has taken place.
D. immobile charged ions exist.
Answer:
The answer is the option (a, b, d)Question:14
What happens during regulation action of a Zener diode?
A. The current in and voltage across the Zenor remains fixed.
B. The current through the series Resistance (Rs) changes.
C. The Zener resistance is constant.
D. The resistance offered by the Zener changes.
Answer:
The answer is the option (b,d)Question:15
To reduce the ripples in a rectifier circuit with capacitor filter
A. RL should be increased.
B. input frequency should be decreased.
C. input frequency should be increased.
D. capacitors with high capacitance should be used.
Answer:
The answer is the option (a, c, d)Question:16
The breakdown in a reverse biased p–n junction diode is more likely to occur due to
A. large velocity of the minority charge carriers if the doping concentration is small.
B. large velocity of the minority charge carriers if the doping concentration is large.
C. strong electric field in a depletion region if the doping concentration is small.
D. strong electric field in the depletion region if the doping concentration is large.
Answer:
The answer is the option (a, d)
Reverse biasing is when the positive terminal of the battery is connected to the N-crystal and negative terminal is of the battery is connected to P-crystal.
In reverse biasing, ionization takes place because the minority charge carriers get accelerated due to reverse biasing. They strike with the electrons which in turn increase the number of charge carriers. And when the doping region is large, there will be a large number of ions in the depletion region. This will give rise to a strong electric field.
Question:17
Why are elemental dopants for Silicon or Germanium usually chosen from group XIII or group XV?
Answer:
Silicon and germanium are chosen as a dopants because their size is compatible with the gaps in semiconductors. And they even are capable of forming covalent bonds.Question:18
Answer:
Sn is a conductor as it has an energy gap of 0 EV while C has an energy gap of 5.4 eV therefore, it is an insulator. Si and Ge have an energy gap of 1.1 eV and 0.7 eV, which makes them a semiconductor. The gaps in energy are related to their individual atomic size, which is responsible for making one insulator, conductor and semiconductor.Question:19
Answer:
We cannot measure the potential barrier of a PN-junction by connecting a sensitive voltmeter across its terminals because in the depletion region, there are no free electrons and holes and in the absence of forward biasing, PN- junction offers infinite resistance.Question:20
Draw the output waveform across the resistor (Fig.14.8).
Answer:
The diode act as a half wave rectifier, it offers low resistance when forward biased and high resistance when reverse biased.Question:21
The amplifiers X, Y and Z are connected in series. If the voltage gains of X, Y and Z are 10, 20 and 30, respectively and the input signal is 1 mV peak value, then what is the output signal voltage (peak value)
(i) if dc supply voltage is 10V?
(ii) if dc supply voltage is 5V?
Answer:
The ratio of output signal voltage to the input signal voltage is known as the total voltage amplification.Question:22
Answer:
$(i) dc \: current\: gain: \beta_{dc}=\frac{i_{c}}{i_{b}}$Question:23
(a)
(b)
(i) Name the type of a diode whose characteristics are shown in Fig. 14.9 (A) and Fig. 14.9(B).
Answer:
i) Figure a) it represents the characteristics of Zener diode and figure b) represents a solar cell.Question:24
Answer:
Question:25
Answer:
IbR1 + Vbe = Vbb
Base current = Ib = Vbb – Vbe/R1
Ib is inversely proportional to R1
Hence, if R1 is increased, then Ib gets reduced.
Question:26
Answer:
Here, the OR gate is used to explain the situation:
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Y=A+B
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Question:28
Explain why elemental semiconductor cannot be used to make visible LEDs.
Answer:
In an elemental semiconductor, the bandgap is such that the emission is in the infrared region and not in the visible region.Question:29
Write the truth table for the circuit shown in Fig.14.11. Name the gate that the circuit resembles.
Answer:
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B
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V0=A.B
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Question:30
Answer:
According to the problem Power=1 WattQuestion:31
Answer:
According to the problem forward biases resistance = $25 \Omega$ and reverse biases resistance=$\infty$Question:32
Answer:
According to the problem V1 =10V, Resistance RB=400k$\Omega$, VBE=0, VCE=0 and Rc=3k$\Omega$Question:33
Draw the output signals C1 and C2 in the given combination of gates (Fig. 14.15).
Answer:
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Question:34
Consider the circuit arrangement shown in Fig 14.16 (a) for studying input and output characteristics of n-p-n transistor in CE configuration.
Select the values of RB and RC for a transistor whose VBE = 0.7 V, so that the transistor is operating at point Q as shown in the characteristics shown in Fig. 14.16 (b).
Given that the input impedance of the transistor is very small and VCC = VBB = 16 V, also find the voltage gain and power gain of circuit making appropriate assumptions.
Answer:
According to the problem at point Q, from graph $V_{BE}=0.7V,V_{CC}=V_{BB}=16V$ and $V_{CE}=8V$Question:35
Answer:
The waveform obtained from the circuit will be a sine wave with a little dip in the input wave.Question:36
Answer:
$n_{e}=N_{D}=10^{-6}\times 5 \times 10^{28}atoms/m^{3}\\ =5 \times 10^{22}/m^{3}$Question:37
An X-OR gate has following truth table:
A
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B
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Y
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0
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0
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0
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0
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1
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1
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Answer:
XOR can be obtained by combining two NOT gates, two AND gates and one OR gate. The logic relation for the given table is as follows:Question:38
Consider a box with three terminals on top of it as shown in Fig.14.18 (a):
Three components namely, two germanium diodes and one resistor are connected across these three terminals in some arrangement.
A student performs an experiment in which any two of these three terminals are connected in the circuit shown in Fig. 14.18 (b).
The student obtains graphs of current-voltage characteristics for unknown combination of components between the two terminals connected in the circuit.
The graphs are
(i) when A is positive and B is negative
(ii) when A is negative and B is positive
(iii) When B is negative and C is positive
(iv) When B is positive and C is negative
(v) When A is positive and C is negative
(vi) When A is negative and C is positive
From these graphs of current – voltage characteristic shown in Fig. 14.18 (c) to (h), determine the arrangement of components between A, B and C.
Answer:
a) n-side of the PN junction is connected to the A terminals while B is connected to the top-side of PN junction.Question:39
Answer:
As we know the base current is very small So,
$
\begin{aligned}
& \quad I_C=I_E \\
& R_C=7.8 \mathrm{k} \Omega \\
& \text { from th fig. } I_C\left(R_c+R_E\right)+V_{C E}=12 \\
& \left(R_E+R_C\right) \times 1 \times 10^{-3}+3=12 \\
& \left(R_E+R_C\right)=9 \times 10^3=9 \mathrm{k} \Omega \\
& R_E=9-7.8=1.2 \mathrm{k} \Omega \\
& V_E=I_E \times R_E \\
& =1 \times 10^{-3} \times 1.2 \times 10^3=1.2 \mathrm{~V} \\
& \text { VoltageV } B=V_E+V_{B E}=1.2+0.5=1.7 \mathrm{~V} \\
& \text { Current } I=\frac{V_B}{20 \times 10^3}=\frac{1.7}{20 \times 10^3}=0.085 \mathrm{~mA} \\
& \text { Resistance } R_B=\frac{12-1.7}{\frac{I_C}{\beta}+0.085} \times 10^3=\frac{10.3}{0.01+0.085} \quad \text { [Given } \beta=100 \text { ] } \\
& =108 \mathrm{k} \Omega
\end{aligned}
$
Question:40
In the circuit shown in Fig.14.20, find the value of RC.
Answer:
Let us consider the ciruit diagram to solve this problemThe main subtopics of NCERT Exemplar Class 12 Physics Solutions Chapter 14 cover the basics of semiconductor materials, how devices like diodes and transistors work, and their use in simple electronic circuits. These topics help students connect physics concepts to real-world technology used in everyday electronics.
Frequently Asked Questions (FAQs)
Semiconductors are materials that have properties between conductors and insulators. They're super important because they power most modern electronic gadgets.
You’ll study diodes, transistors, logic gates, LEDs, and their working, applications, and how they’re used in electronic circuits.
Yes! The concepts you learn here are directly linked to real-world devices like smartphones, computers, calculators, and even traffic lights.
This chapter covers everything related to semiconductors like types, properties, junction transistors, diodes, logic gates, etc.
In NEET exam 2 to 3 questions are asked every year from the chapter Semiconductor Electronics. Mostly questions from both analog and digital electronics are included.
Students can get a better understanding of the concepts discussed in the chapter using NCERT Exemplar Solutions For Class 12 Physics Chapter 14, which in turn will be helpful for exams
On Question asked by student community
Hello,
The date of 12 exam is depends on which board you belongs to . You should check the exact date of your exam by visiting the official website of your respective board.
Hope this information is useful to you.
Hello,
Class 12 biology questions papers 2023-2025 are available on cbseacademic.nic.in , and other educational website. You can download PDFs of questions papers with solution for practice. For state boards, visit the official board site or trusted education portal.
Hope this information is useful to you.
Hello Pruthvi,
Taking a drop year to reappear for the Karnataka Common Entrance Test (KCET) is a well-defined process. As a repeater, you are fully eligible to take the exam again to improve your score and secure a better rank for admissions.
The main procedure involves submitting a new application for the KCET through the official Karnataka Examinations Authority (KEA) website when registrations open for the next academic session. You must pay the required application fee and complete all formalities just like any other candidate. A significant advantage for you is that you do not need to retake your 12th board exams. Your previously secured board marks in the qualifying subjects will be used again. Your new KCET rank will be calculated by combining these existing board marks with your new score from the KCET exam. Therefore, your entire focus during this year should be on preparing thoroughly for the KCET to achieve a higher score.
For more details about the KCET Exam preparation,
CLICK HERE.
I hope this answer helps you. If you have more queries, feel free to share your questions with us, and we will be happy to assist you.
Thank you, and I wish you all the best in your bright future.
Yes, you can switch from Science in Karnataka State Board to Commerce in CBSE for 12th. You will need a Transfer Certificate from your current school and meet the CBSE school’s admission requirements. Since you haven’t studied Commerce subjects like Accountancy, Economics, and Business Studies, you may need to catch up before or during 12th. Not all CBSE schools accept direct admission to 12th from another board, so some may ask you to join Class 11 first. Make sure to check the school’s rules and plan your subject preparation.
Hello
For the 12th CBSE Hindi Medium board exam, important questions usually come from core chapters like “Madhushala”, “Jhansi ki Rani”, and “Bharat ki Khoj”.
Questions often include essay writing, letter writing, and comprehension passages. Grammar topics like Tenses, Voice Change, and Direct-Indirect Speech are frequently asked.
Students should practice poetry questions on themes and meanings. Important questions also cover summary writing and translation from Hindi to English or vice versa.
Previous years’ question papers help identify commonly asked questions.
Focus on writing practice to improve handwriting and presentation. Time management during exams is key to answering all questions effectively.
This ebook serves as a valuable study guide for NEET 2025 exam.
This e-book offers NEET PYQ and serves as an indispensable NEET study material.
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