NCERT Solutions for Class 12 Physics Chapter 14 - Semiconductor Electronics Materials Devices And Simple Circuit
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This Story also Contains
- Semiconductor Electronics Materials Devices And Simple Circuit NCERT Solutions: Download PDF
- Semiconductor Electronics Materials Devices And Simple Circuit NCERT Solutions: Exercise Questions
- Semiconductor Electronics Materials Devices And Simple Circuit NCERT Solutions: Additional Questions
- Class 12 Physics Chapter 14 - Semiconductor Electronics Materials Devices And Simple Circuit: Higher Order Thinking Skills (HOTS) Questions
- Class 12 Physics Chapter 14 - Semiconductor Electronics Materials Devices And Simple Circuit: Topics
- Class 12 Physics Chapter 14 - Semiconductor Electronics Materials Devices And Simple Circuit: Notes
- Approach to Solve Questions of Semiconductor Electronics: Materials, Devices and Simple Circuits
- How Can NCERT Solutions for Class 12 Physics Chapter 14 Help in Exam Preparation?
- What Extra Should Students Study Beyond NCERT for JEE/NEET?
- NCERT solutions for class 12 physics chapter-wise
The NCERT Solutions for Class 12 Physics Chapter 14 - Semiconductor Electronics Materials Devices And Simple Circuit have been prepared by experts in the subject and would give step-by-step answers to all the questions in the exercises, hence helping the students to grasp the more complicated topics such as semiconductor materials, p-n junction diodes, rectifiers, transistors and logic gates. These NCERT solutions are not just an answer to the textbook exercises, but also a boost to the analytical and problem-solving skills necessary to be able to succeed in exams. These NCERT Solutions for Class 12 Physics Chapter 14 - Semiconductor Electronics Materials Devices And Simple Circuit will allow students to develop a solid conceptual foundation, revise well on solved examples and gain the confidence required to not only score high on CBSE exams but also to score well in JEE and NEET.
Also read :
Semiconductor Electronics Materials Devices And Simple Circuit NCERT Solutions: Download PDF
The Class 12 Physics Chapter 14 - Semiconductor Electronics Materials Devices And Simple Circuit question answers offer detailed answers to each of the textbook questions, so that complicated concepts become easy to understand. These Class 12 physics chapter 14 Semiconductor Electronics Materials Devices And Simple Circuit questions answers are formatted according to the current CBSE syllabus and are very useful during revisions, examinations at the board level and even competitive applications like JEE/NEET examinations. They can also print the PDF to study offline and get last-minute reading practice.
Semiconductor Electronics Materials Devices And Simple Circuit NCERT Solutions: Exercise Questions
The Class 12 Physics Chapter 14 - Semiconductor Electronics Materials Devices And Simple Circuit question answers (Exercise questions) offer step-by-step solutions to all questions in the textbook, and that is why it becomes easy to learn and apply the semiconductor concepts among students. Such Class 12 physics chapter 14 Semiconductor Electronics Materials Devices And Simple Circuit questions answers are accurate, clear, and they are used to prepare board exams as well as competitive exams such as JEE/NEET.
Q. 14.1 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.
Answer:
An N-type semiconductor has electrons as majority carriers and holes as minority carriers. It is formed when we dopea pentavalent impurity into a Silicon atom. Some pentavalent dopants are phosphorus, arsenic, and bismuth.
Hence, the correct option is C.
Q. 14.2 Which of the statements given in Exercise 14.1 is true for p-type semiconductors.
(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
Answer:
In a p-type semiconductor, holes are the majority carrier and electrons are the minority carrier. It is formed when a trivalent atom-like aluminium is doped in a silicon atom. Hence correct option for a p-type conductor would be (d).
(a) $(E_{g})_{Si} < (E_{g})_{Ge}< (E_{g})_{C}$
(b) $(E_{g})_{C} < (E_{g})_{Ge}> (E_{g})_{Si}$
(c) $(E_{g})_{C} > (E_{g})_{Si}> (E_{g})_{Ge}$
(d) $(E_{g})_{C} = (E_{g})_{Si}= (E_{g})_{Ge}$
Answer:
Since carbon is a non-metal, its energy band gap would be the highest, and the energy band gap of Ge would be the least, as it is a metalloid.
$(E_{g})_{C} > (E_{g})_{Si}> (E_{g})_{Ge}$
Hence correct option would be (c)
Q14.4 In an unbiased p-n junction, holes diffuse from the p-region to n-region beca
(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
Answer:
Charge flows from the higher concentration to the lower concentration in a junction. In this case, holes are diffusing from the p-region to the n-region, and hence the concentration of holes is greater in the p-region.
and hence the correct option would be (c)
Q. 14.5 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) lowers the potential barrier.
(d) none of the above.
Answer:
When a p-n junction is forward biased, the negative voltage repels the electrons toward the junction and gives them the energy to cross the junction and combine with the hole, which is also being pushed by a positive voltage. This leads to a reduction in the depletion layer, which means a reduction in the potential barrier across the junction.
Hence correct option would be (c)
Answer:
As we know :
output frequency for a half-wave rectifier = input frequency, and hence output frequency in the half-wave rectifier will be 50Hz.
also, output frequency for full-wave rectifier = 2*(input frequency) and Hence output frequency in full-wave rectifier will be 2*50 = 100 Hz.
Semiconductor Electronics Materials Devices And Simple Circuit NCERT Solutions: Additional Questions
Class 12 Physics Chapter 14 - Semiconductor Electronics Materials Devices And Simple Circuit additional questions give students extra practice on concepts like p-n junctions, diodes, transistors, and logic gates. These questions go beyond the NCERT exercise, helping in deeper understanding and exam-focused preparation for boards as well as JEE/NEET.
Answer:
Given
the energy band gap of photodiode is 2.8eV.
wavelength = $\lambda$ = 6000nm = $6000*10^{-9}$
The energy of signal will be $\frac{hc}{\lambda }$
where c is speed of light(300000000m/s) , h is planks constant ( = $6.626 * 10^{-34}Js$ )
putting the corresponding value
The energy of signal = $\frac{(6.626 * 10^{-34} * 3*10^8)}{6000*10^{-9}}$
= $3.313*10^{-20}J$
= $0.207eV (since 1.6*10^{-20}= 1eV)$
The energy of the signal is 0.207eV, which is less than 2.8eV ( the energy and gap of the photodiode). Hence signal can not be detected by the photodiode.
Answer:
Given:
number of Silicon atoms per $m^{3}$ = $5\times 10^{28}.$
number of Arsenic atoms per $m^{3}$ = $5\times 10^{22}.$
number of Indium atoms per $m^{3}$ = $5\times 10^{20}$
number of thermally generated electrons $n_{i}=1.5\times 10^{16}\; m^{-3}.$
Now,
Number of electrons
$n_e =$ $5 * 10 ^{22}-1.5*10^{16}$ = $4.99*10^{22}(approx)$
number of holes is $n_h$
in thermal equilibrium
$n_h*n_e=n_i^2$
$n_h=n_i^2/n_e$
$n_h= (1.5*10^{16})^2/4.99*10^{22}$
$n_h= 4.51 * 10^9$
Now, since the number of electrons is higher than the number of holes, it is an n-type semiconductor.
Q.3 In an intrinsic semiconductor the energy gap $E_{g}$ is $1.2\; eV.$ Its hole mobility is much smaller than electron mobility and independent of temperature. What is the ratio between conductivity at $600K$ and that at $300K$ Assume that the temperature dependence of intrinsic carrier concentration $n_{i}$ is given by $n_{i}=n_{0}\; exp\left [ -\frac{E_{g}}{2K_{B}T} \right ]$ Where, $n_{0}$ is constant.
Answer:
Energy gap of given intrinsic semiconductor = E g = 1.2eV
temperature dependence of intrinsic carrier concentration $n_{i}$ is given by
$n_{i}=n_{0}\; exp\left [ -\frac{E_{g}}{2K_{B}T} \right ]$
Where is constant, $K_B$ is Boltzmann constant = $8.862 * 10^{-5}eV/K$ ,
T is temperature
Initial temperature = T1 = 300K
the intrinsic carrier concentration at this temperature :
$n_{i1} = n_0exp[\frac{-E_g}{2K_B*300}]$
Final temperature = T2 = 600K
the intrinsic carrier concentration at this temperature :
$n_{i2} = n_0exp[\frac{-E_g}{2K_B*600}]$
The ratio between the conductivities at 300K and at 600K is equal to the ratio of their intrinsic carrier concentration at these temperatures
$\frac{n_{i2}}{n_{i2}} = \frac{n_0exp[\frac{-E_g}{2K_B*600}]}{n_0exp[\frac{-E_g}{2K_B*300}]}$
$= exp\frac{E_g}{2K_B}[\frac{1}{300}-\frac{1}{600}]=exp[\frac{1.2}{2*8.62*10^{-5}}* \frac{2-1}{600}]$
$= exp[11.6] = 1.09 * 10^{5}$
Therefore, the ratio between the conductivities is $1.09 * 10^{5}$.
Q.4 In a p-n junction diode, the current I can be expressed as $I=I_{0}\; [exp \frac{eV}{K_{B}T}-1 ]$ where $I_{0}$ is called the reverse saturation current, V is the voltage across the diode and is positive for forward bias and negative for reverse bias, and $I$ is the current through the diode, $k_{B}$ is the Boltzmann constant $(8.6\times 10^{-5}eV/K)$ and $T$ is the absolute temperature. If for a given diode $I_{0}=5\times 10^{-12}A$ and $T=300\; K,$ then what will be the forward current at a forward voltage of $0.6\; V\; ?$
Answer:
As we have
$I=I_{0}\; [exp \frac{eV}{K_{B}T}-1 ]$
Here, $I_{0}=5\times 10^{-12}A$ , $T=300\; K,$ and , $k_{B}$ = Boltzmann constant = $(8.6\times 10^{-5}eV/K)$ $=(1.376*10^{-23}J/K)$
When the forward voltage is 0.6V:
$I=5*10^{-12}\;[ exp \frac{1.6*10^{-19}*0.6}{1.376*10^{-23}*300}-1 ]=0.0625A$
Hence, the forward current is 0.0625A
Q.5 In a p-n junction diode, the current I can be expressed as $I=I_{0}\; [exp \frac{eV}{K_{B}T}-1 ]$ where I0 is called the reverse saturation current, $V$ is the voltage across the diode and is positive for forward bias and negative for reverse bias, and $I$ is the current through the diode, $k_{B}$ is the Boltzmann constant $(8.6\times 10^{-5}\; eV/K)$ and $T$ is the absolute temperature. If for a given diode $I_{0}=5\times 10^{12}A$ and $T=300\; K,$ then what will be the increase in the current if the voltage across the diode is increased to $0.7 \; V?$
Answer:
As we have
$I=I_{0}\; [exp \frac{eV}{K_{B}T}-1 ]$
Here, $I_{0}=5\times 10^{-12}A$ , $T=300\; K,$ and , $k_{B}$ = Boltzmann constant = $(8.6\times 10^{-5}eV/K)$ $=(1.376*10^{-23}J/K)$
When the forward voltage is 0.7V:
$I=5*10^{-12}\;[ exp \frac{1.6*10^{-19}*0.7}{1.376*10^{-23}*300}-1 ]=3.029A$
When the forward voltage is 0.6V:
$I=5*10^{-12}\;[ exp \frac{1.6*10^{-19}*0.6}{1.376*10^{-23}*300}-1 ]=0.0625A$
Hence the increase in the forward current is
$I(whenv=0.7) - I(whenv=.6)$ $= 3.029- 0.0625 = 2.967A$
Q.6 In a p-n junction diode, the current I can be expressed as $I=I_{0}\; [exp \frac{eV}{K_{B}T}-1 ]$ where $I_{0}$ is called the reverse saturation current, $V$ is the voltage across the diode and is positive for forward bias and negative for reverse bias, and $I$ is the current through the diode, $k_{B}$ is the Boltzmann constant $(8.6\times 10^{-10}\; eV/K)$ and $T$ is the absolute temperature. If for a given diode $I_{0}=5\times 10^{-12}A$ and $T=300\; K,$ then what is the dynamic resistance?
Answer:
Dynamic Resistance = $\frac{voltage-change}{ current-change}$
Resistance change = 0.7 - 0.6 = 0.1
Current change = 2.967(calculated in prev question)
Therefore
, $Dynamic Resistance = \frac{0.1}{2.967} = 0.0337\Omega$
Answer:
As we have
$I=I_{0}\; [exp \frac{eV}{K_{B}T}-1 ]$
Here, $I_{0}=5\times 10^{-12}A$ , $T=300\; K,$ and , $k_{B}$ = Boltzmann constant = $(8.6\times 10^{-5}eV/K)$ $=(1.376*10^{-23}J/K)$
When reverse voltage is 1V, V = -1
$I=5*10^{-12}\;[ exp \frac{1.6*10^{-19}*(-1)}{1.376*10^{-23}*300}-1 ]\approx5\times10^{-12}$
When the reverse voltage is -2V:
$I=5*10^{-12}\;[ exp \frac{1.6*10^{-19}*(-2)}{1.376*10^{-23}*300}-1 ]\approx5\times10^{-12}$
In both cases, the current is very small and approximately equal to the reverse saturation current; hence, their difference is negligible, which causes the dynamic resistance of infinity.
Q.8(a) You are given the two circuits as shown in Fig. Show that circuit acts as OR gate while the circuit
Answer:
Here, THE Input = A and B
Output = Y
The left part of the figure acts as a NOR, and the right part acts as a NOT Gate.
The output of NOR gate = $\overline{A+B}$
The output of the NOR gate would be the input of the NOT Gate, and hence
$\mathrm{Y}=\overline{\overline{A+B}}=\mathrm{A}+\mathrm{B}$
Hence, the figure functions like an OR Gate.
or compare the truth table by giving different inputs and observing the output
INPUTS | OUTPUT |
|---|---|
| A B | Y |
| 0 0 | 0 |
| 0 1 | 1 |
| 1 0 | 1 |
| 1 1 | 1 |
Q. 8(b) You are given the two circuits as shown in Fig. Show that circuit acts as AND gate.
Answer:
The output of the NOT gate ( left part of the circuit) is the input of the NOR gate
Hence the output of total circuit Y = $\over(\overline A + \overline B)$
= $\overline{\overline A}.\overline{\overline B}$ $\overline{A+B}=\overline A. \overline B$
= $A*B$
Hence, the circuit functions as an AND gate.
or give the inputs 00,01,10,11, and observe the truth table
INPUTS | OUTPUT |
|---|---|
| A B | Y |
| 0 0 | 0 |
| 0 1 | 0 |
| 1 0 | 0 |
| 1 1 | 1 |
The truth table is the same as that of the AND gate
Q.9 Write the truth table for a NAND gate connected as given in the figure.
Hence, identify the exact logic operation carried out by this circuit.
Answer:
Here, A is both the input of the NAND gate and hence, Output Y will be
$Y = \overline {A*A}$
$Y = \overline {A} + \overline A$
$Y = \overline {A}$
Hence circuit functions as a NOT gate.
The truth table for the given figure:
| Input | Output |
| A | Y |
| 0 | 1 |
| 1 | 0 |
Answer:
a)
A and B are inputs of a NAND gate, and the output of this gate is the input of another NAND gate, so,
Y = $\over(\overline {A.B})(\overline {A.B})$
Y= $\over(\overline {A.B})$ $+$ $\over(\overline {A.B})$
Y= $AB$
Hence, this circuit functions asan AND gate.
b)
A is input to the NAND gate whose output goes to the rightmost NAND gate. Also, B is input to the NAND gate whose output goes to the rightmost NAND gate.
Y = $\over \overline A .\overline B$
Y = $\over\overline A .$ + $\over\overline B.$
Y = A + B
Hence, the circuit functions as an OR gate.
Alternative method
fig. a
Construct the truth table by giving various inputs and observing the output
| INPUT | INTERMEDIATE OUTPUT | OUTPUT |
| 00 | 1 | 0 |
| 01 | 1 | 0 |
| 10 | 1 | 0 |
| 11 | 0 | 1 |
The above truth table is the same as that of an AND gate
fig. b
| INPUTS | OUTPUT |
| 00 | 0 |
| 01 | 1 |
| 10 | 1 |
| 11 | 1 |
The above truth table is the same as that of an OR gate
Answer:
A and B are the inputs of a NOR gate, and the Output of this NOR gate is the Input of another NOR gate whose Output is Y. Hence,
Y = $\over(\overline{A+B} + \overline{A+B})$
Y = $\over\overline {A+B}$ . $\over\overline {A+B}$
Y = A + B
Hence Circuit behaves as an OR gate.
Truth table
| INPUTS | OUTPUT |
| 00 | 0 |
| 01 | 1 |
| 10 | 1 |
| 11 | 1 |
Answer:
a)
A are the two inputs of the NOR gate, and hence Output Y is:
Y = $\overline {A+A}$
Y = $\overline {A}$
Hence circuit functions as a NOT gate.
TRUTH TABLE:
| INPUT | OUTPUT |
| 0 | 1 |
| 1 | 0 |
b) A is the two inputs of a NOR gate whose output(which is $\overline {A}$ ) is the one input of another NOR gate. B is the two inputs of the NOR gate whose output (which is $\overline {B}$ ) is the input of another NOR gate. Hence,
Y = $\over\overline {A} + \overline {B}$
Y = $\over\overline {A}$ . $\over\overline {B}$
Y = A.B
Hence, it functions as an AND gate.
TRUTH TABLE:
| INPUTS | OUTPUT |
| 00 | 0 |
| 01 | 0 |
| 10 | 0 |
| 11 | 1 |
Class 12 Physics Chapter 14 - Semiconductor Electronics Materials Devices And Simple Circuit: Higher Order Thinking Skills (HOTS) Questions
Class 12 Physics NCERT Chapter 14 HOTS (Higher Order Thinking Skills) Questions challenge students to apply semiconductor concepts like diode characteristics, transistor action, rectifiers, and logic circuits in tricky problem-solving. These questions enhance critical thinking and analytical skills, making students better prepared for competitive exams and advanced applications.
Q1:
For an amplifier (NPN), VBC = 0, $\beta=50$, IC = 2.475 mA. Then the value of IB = ?
Answer:
As we learn
Relation between emitter current, Base current, and collector current -
$
I_E=I_B+I_C
$
- wherein
$I_E=$ Emitter Current
$I_B=$ Base Current
$I_C=$ Collector Current
$
I_B=\frac{I_C}{\beta}=\frac{2.475}{50}=49.5 \mathrm{~mA}
$
Q2:
A Si diode has a saturation current of 10-7 A. Calculate the junction current for a forward bias of 0.7 V and 300 k (y = 2 for Si and V = 26 mV)
Answer:
As we learn,
Relation between current I & Voltage V -
$I=I_0\left(e^{\frac{c v}{K T}}-1\right)$
- wherein
K = Boltzmann constant
I0 = reverse saturation current
In forward bias
$e^{\frac{r v}{K T}}>>1$
Then, the forward biasing current is
$I=I_0 \cdot e^{\frac{c v}{K T}}$
$I=I_0\left(e^{\frac{\mathrm{cv}}{K T}}-1\right)=I_0\left(e^{\frac{v}{y+V_T}}-1\right)=10^{-7}\left(\frac{0.7}{2 \times 26}-1\right)=70 \mathrm{~mA}$
Q3:
Find the current through the circuit for Si diode`
Given that -
Knee voltage for Ge is 0.3 V
Knee voltage for Si is 0.7 V
Answer:
Knee voltage of P-N junction -
It is defined as that forward voltage at which the current through the junction starts rising rapidly with an increase in voltage.
Knee voltage for Ge is 0.3 V
Knee voltage for Si is 0.7 V
$I=\frac{3.4-0.7}{600}=\frac{2.7}{600}=4.5 \mathrm{~mA}$
Q4:
The truth table for the above logic circuit is the same as that of :
Answer:
As we learn
NOR Gate -
NOT + OR Gate
- wherein
$Y=\overline{A+B}$
A and B are input
Y is output
The output (y) of two input (A, B) NOR gates is :
$Y=\overline{A+B}$
Q5:
For the circuit shown current through 1.5K$\Omega$ is:
Answer:
As we learn
Zener diode can operate continuously without being damaged in the region of reverse bias
- wherein
1) It acts as a voltage regulator
2) In forward biasing, it acts as an ordinary diode.
$I_L=\frac{V_Z}{1.5 \mathrm{~K}}=\frac{6}{1.5 \times 10^3}=4 \mathrm{~mA}$
CBSE Class 12th Syllabus: Subjects & Chapters
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Class 12 Physics Chapter 14 - Semiconductor Electronics Materials Devices And Simple Circuit: Topics
Class 12 Physics Chapter 14 - Semiconductor Electronics Materials Devices And Simple Circuit introduces the working principles of semiconductors, diodes, transistors, rectifiers, and logic gates that form the backbone of modern electronics. Understanding these topics helps students connect theory with real-world applications in devices like mobiles, LEDs, and computers.
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Download EBookClass 12 Physics Chapter 14 - Semiconductor Electronics Materials Devices And Simple Circuit: Notes
Class 12 Physics Chapter 14 - Semiconductor Electronics Materials Devices And Simple Circuit Notes offer a neatly organised introduction to semiconductor electronics with all essential definitions, formulas and concepts being simplified. Such notes enable the students to revise and improve their basic knowledge quickly to succeed in the CBSE board exams, JEE, and NEET.
Metals, Conductors & Semiconductors
Why do some materials light up your circuits while others block the buzz? Metals conduct electricity well. Conductors let current pass easily. Semiconductors—like silicon—are in-between, and insulators stop the flow completely.
Intrinsic Semiconductor
Pure semiconductors like silicon have equal electrons and holes. They don’t conduct much—until heat gets them moving!
Extrinsic Semiconductor
- Add a pinch of impurity, and things change!
- n-type: Extra electrons (via elements like phosphorus).
- p-type: Extra holes (via elements like boron).
- Doping boosts conductivity.
P-N Junction
A magical meeting of p-type and n-type. It only lets current flow one way—thanks to the mysterious depletion region!
Semiconductor Diode
A p-n junction in action. It’s a one-way gate for current—perfect for letting power flow forward and blocking the backflow.
Junction Diode as Rectifier
- Want to turn AC into DC? Diodes do the trick!
- Half-wave: Uses one side of AC.
- Full-wave: Uses both sides.
- It’s how your gadgets get steady power.
Approach to Solve Questions of Semiconductor Electronics: Materials, Devices and Simple Circuits
To answer the questions on Semiconductor Electronics, it is essential to have a good understanding of such concepts as p-n junctions, diodes, transistors and logic gates. Using a well-defined step-by-step approach, students will be able to work both theoretically and numerically with a sense of confidence. This approach is useful in achieving good scores in board exams and also JEE/NEET, where precision and usage of concepts are important.
- Master the basic fundamentals of semiconductors.
- Know what conductors, insulators, and semiconductors are.
- There are two categories of semiconductors: intrinsic (pure) and extrinsic (doped).
- Know the types of doping
- n-type: Introduction of pentavalent impurity → Additional electrons
- p-type: Introduction of trivalent impurity → Increased holes
- Learn important parts:
- Diode: Recognize forward and reverse bias.
- Zener diode: Used for voltage control
- Transistors (n-p-n, p-n-p): Current amplification and switching
- Learn circuit behavior
- Explain how voltage and current act in forward/reverse-biased diodes
- Study rectifier circuits: Half-wave and full-wave.
- Apply proper formulas-
-
Current gain in a transistor: $\beta=\frac{I_C}{I_B}$
-
$
\left.I_E=I_B+I_C \text { (Emitter current }=\text { Base }+ \text { Collector }\right)
$
- Study I-V characteristics- Draw and label graphs for diodes and transistors.
- Learn logic gates (brief introduction)- AND, OR, NOT (if included) – Know the truth tables and how they apply.
- Practice circuit numericals- Recognise biasing, compute currents, and perceive the role of the power supply
- Know applications-
- Diodes: Rectification, switching
- Transistors: Logic circuits, amplifiers
- Study the theory and practice NCERT questions- Most questions are about numbers or ideas, so put both.
How Can NCERT Solutions for Class 12 Physics Chapter 14 Help in Exam Preparation?
Semiconductor Electronics Materials Devices And Simple Circuit class 12 question answers have proven to be very useful in exam preparation since they make one of the most practical and scoring chapters in Physics very easy. The chapter describes how diodes, transistors, logic gates, and other semiconductor devices work, which are the basis of modern electronics. These NCERT solutions provide step-by-step solutions to numerical and conceptual questions to enable the students to know how the semiconductor components work in an actual circuit. Through these Semiconductor Electronics Materials Devices And Simple Circuit class 12 question answers, students are able to understand the distinction between conductors, insulators and semiconductors, PN junction behaviour and learn to solve questions related to circuits that are frequently used in board and competitive exams. Also, the NCERT solutions indicate important formulas, schemes, and derivations, and hence, last-minute revision is fast and efficient. They make students relate theory to real-life practice and increase confidence and accuracy in examinations.
What Extra Should Students Study Beyond NCERT for JEE/NEET?
Although the basics of semiconductors, diodes, and transistors are taught in NCERT, students who are going to take on the JEE and NEET exams, must take it a step further. Competitive exams require a more in-depth problem-solving, application-based numerals, and exposure to the conceptual questions that are tricky. Comparison of NCERT vs. JEE/NEET requirements can be used to determine the additional topics and practice to achieve success in the exams.
NCERT solutions for class 12 physics chapter-wise
NCERT Solutions of Class 12 Physics contain step-by-step explanations on each chapter that make the complicated theories simple. These solutions have been prepared according to the new CBSE syllabus and are quite beneficial in preparation for board exams, JEE and NEET. The following links lead to chapter-wise solutions that the students can download and revise to access knowledge.
Also Check NCERT Books and NCERT Syllabus here:
- NCERT Books Class 12 Physics
- NCERT Syllabus Class 12 Physics
- NCERT Books Class 12
- NCERT Syllabus Class 12
NCERT solutions subject wise
Frequently Asked Questions (FAQs)
Q: Why are NCERT Solutions Class 12 Physics necessary to students?
A:
They give step-by-step explanations to make the not-so-easy concepts easier and to solve questions in exercises in a straightforward manner.
Q: Are NCERT Solutions sufficient in achieving good marks in Class 12 board exams?
A:
Yes, NCERT Solutions is enough to pass boards yet however having practice and doing sample papers will increase your mark.
Q: Are these NCERT Solutions useful in JEE and NEET preparation?
A:
Absolutely! Most of the questions in JEE and NEET are directly and indirectly linked to the NCERT concepts; hence, these solutions assist in building a great foundation.
Q: Is it possible to download NCERT Solutions of Class 12 Physics chapter-wise in PDF?
A:
Yes, all NCERT Solutions in each Chapter can be downloaded as PDFs to study and revise.
Q: What is the best way to use NCERT Solutions Class 12 Physics?
A:
You should read the NCERT textbook, go over the NCERT Solutions to clear any confusion, then do previous year papers and mock tests.
Q: How can students use NCERT Solutions to increase their speed and accuracy when answering numerical problems about rectifiers, transistors, and diodes?
A:
Writing mathematical equations for p-n junctions or transistor circuits, identifying known variables, and changing values to the appropriate units are only a few of the detailed approaches that solutions provide. Frequent use of these simple methods results in quicker computations and fewer careless mistakes in time-bound examinations.
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No, it’s not true that GSEB (Gujarat Board) students get first preference in college admissions.
Your daughter can continue with CBSE, as all recognized boards CBSE, ICSE, and State Boards (like GSEB) which are equally accepted for college admissions across India.
However, state quota seats in Gujarat colleges (like medical or engineering) may give slight preference to GSEB students for state-level counselling, not for all courses.
So, keep her in CBSE unless she plans to apply only under Gujarat state quota. For national-level exams like JEE or NEET, CBSE is equally valid and widely preferred.
Hope it helps.
Hello,
The Central Board of Secondary Education (CBSE) releases the previous year's question papers for Class 12.
You can download these CBSE Class 12 previous year question papers from this link : CBSE Class 12 previous year question papers (http://CBSE%20Class%2012%20previous%20year%20question%20papers)
Hope it helps !
Hi dear candidate,
On our official website, you can download the class 12th practice question paper for all the commerce subjects (accountancy, economics, business studies and English) in PDF format with solutions as well.
Kindly refer to the link attached below to download:
CBSE Class 12 Accountancy Question Paper 2025
CBSE Class 12 Economics Sample Paper 2025-26 Out! Download 12th Economics SQP and MS PDF
CBSE Class 12 Business Studies Question Paper 2025
CBSE Class 12 English Sample Papers 2025-26 Out – Download PDF, Marking Scheme
BEST REGARDS
Hello,
Since you have passed 10th and 12th from Delhi and your residency is Delhi, but your domicile is UP, here’s how NEET counselling works:
1. Counselling Eligibility: For UP NEET counselling, your UP domicile makes you eligible, regardless of where your schooling was. You can participate in UP state counselling according to your NEET rank.
2. Delhi Counselling: For Delhi state quota, usually 10th/12th + residency matters. Since your school and residency are in Delhi, you might also be eligible for Delhi state quota, but it depends on specific state rules.
So, having a Delhi Aadhaar will not automatically reject you in UP counselling as long as you have a UP domicile certificate.
Hope you understand.
Hello,
You can access Free CBSE Mock tests from Careers360 app or website. You can get the mock test from this link : CBSE Class 12th Free Mock Tests
Hope it helps !
Popular CBSE Class 12th Questions
A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is
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An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range
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A particle is projected at 600 to the horizontal with a kinetic energy . The kinetic energy at the highest point
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In the reaction,
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How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?
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0.02 |
Option 2)
3.125 × 10-2 |
| Option 3)
1.25 × 10-2 |
Option 4)
2.5 × 10-2 |
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