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NCERT Solutions Class 12 Physics Chapter 3 Current Electricity

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NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity - FREE PDF Download

Chapter 3 Current Electricity class 12 NCERT solutions by Vedantu, explains the principles and applications of electric current in conductors. This chapter explores the behaviour of electric charges in motion within conductors and is pivotal for understanding how electrical circuits operate, which is essential for both theoretical knowledge and practical applications in technology and engineering. This chapter also explores the properties of materials, understanding how different substances influence the flow of current through their electrical resistivity and conductivity. By practising with these Class 12 Physics NCERT Solutions, students can build confidence in their understanding and excel in their studies.

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Table of Content
1. NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity - FREE PDF Download
2. Glance on Physics Chapter 3 Class 12 - Current Electricity
3. Access NCERT Solutions for Class 12 Physics Chapter 3 – Current Electricity
4. Current Electricity Chapter Summary - Class 12 NCERT Solutions
5. Overview of Deleted Syllabus for CBSE Class 12 Physics Current Electricity
6. Other Study Material for CBSE Class 12 Physics Chapter 3
7. Chapter-Specific NCERT Solutions for Class 12 Physics
FAQs


Glance on Physics Chapter 3 Class 12 - Current Electricity

  • Chapter 3 of Class 12 Physics, "Current Electricity," is an essential topic that explores the behaviour and properties of electric currents in conductors. It forms the foundation for understanding electric circuits and their applications in various technologies and devices.

  • It introduces the concept of electric current, defined as the rate of flow of electric charge. Discussions cover closed and open circuits, voltage sources (like batteries), and resistors. Understanding the flow of current in a circuit is essential.

  • In Current Electricity class 12 exercise solutions, Different arrangements of resistors in circuits are discussed, along with how they affect the total resistance and current flow.

  • Lso in current electricity class 12 solutions, Kirchhoff's laws are introduced to analyse more complex circuits. Formulas to calculate electrical energy (E) and power (P) are discussed. Various arrangements of cells in a circuit are discussed, such as series and parallel combinations, and their effects on the voltage and current.

  • This article contains chapter notes, important questions, exemplar solutions, exercises, and video links for Chapter 3 - Current Electricity, which you can download as PDFs.

  • There are 9 fully solved questions in class 12th physics chapter 3 Current Electricity.

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Access NCERT Solutions for Class 12 Physics Chapter 3 – Current Electricity

1. The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is \[0.4\Omega \] , what is the maximum current that can be drawn from the battery?

Ans: In the above question it is given that:

Emf of the battery, $E=12V$

Internal resistance of the battery, $r=0.4\Omega $

Consider the maximum current drawn from the battery to be $I$.

Therefore, using Ohm’s law,

$E=Ir$

$\Rightarrow I=\frac{E}{r}$

$\Rightarrow I=\frac{12}{0.4}$

$\Rightarrow I=30A$

Clearly, the maximum current drawn from the given battery is $30A$ .


2. A battery of emf 10 V and internal resistance $3\Omega $ is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?

Ans: In the above question it is given that:

Emf of the battery, E = 10 V

Internal resistance of the battery, $r=3\Omega $

Current in the circuit, $I=0.5A$ 

Consider the resistance of the resistor to be $R$.

Therefore, using Ohm’s law,

$I=\frac{E}{R+r}$

$R+r=\frac{E}{I}$

$\Rightarrow R+r=\frac{10}{0.5}$

$\Rightarrow R+r=20$

$\Rightarrow R=20-3=17\Omega $

Let the terminal voltage of the resistor be $V$.

Using the Ohm’s law,

$V=IR$

$\Rightarrow V=0.5\times 17=8.5V$

Thus, the resistance of the resistor is $17\Omega $ and the terminal voltage is $8.5V$ .


3. At room temperature ${{27.0}^{\circ }}C$, the resistance of a heating element is $100\Omega $. What is the temperature of the element if the resistance is found to be $117\Omega $, given that the temperature coefficient of the material of the resistor is $1.70\times {{10}^{-4}}^{\circ }{{C}^{-1}}$ ?

Ans: In the above question it is given that at room temperature $(T={{27.0}^{\circ }}C)$, the resistance of the heating element is $100\Omega $ (say R).

Also, the heating element’s temperature coefficient is given to be $\alpha =1.70\times {{10}^{-4}}{}^\circ {{C}^{-1}}$.

Now, it is said that the resistance of the heating element at an increased temperature (say ${{T}_{1}}$) is $117\Omega $ (say ${{R}_{1}}$). To compute this unknown increased temperature ${{T}_{1}}$, the formula for temperature coefficient of a material can be used. It is known that temperature co-efficient of a material provides information on the nature of that material with respect to its change in resistance with temperature. Mathematically,

$\alpha =\frac{{{R}_{1}}-R}{R\left( {{T}_{1}}-T \right)}$

$\Rightarrow {{T}_{1}}-T=\frac{{{R}_{1}}-R}{R\alpha }$

Substituting the given values,

$\Rightarrow {{T}_{1}}-27=\frac{117-100}{100\times 1.70\times {{10}^{-4}}}$

$\Rightarrow {{T}_{1}}-27=1000$

$\Rightarrow {{T}_{1}}={{1027}^{\circ }}C$

Clearly, it is at ${{1027}^{\circ }}C$ when the resistance of the element is $117\Omega $.


4. A negligibly small current is passed through a wire of length 15 m and uniform cross-section $6.0\times {{10}^{-7}}{{m}^{2}}$ , and its resistance is measured to be $5.0\Omega $ . What is the resistivity of the material at the temperature of the experiment?

Ans: In the above question it is given that:

Length of the wire, $l=15m$

Area of cross-section of the wire, $a=6.0\times {{10}^{-7}}{{m}^{2}}$

Resistance of the material of the wire, $R=5.0\Omega $

Let resistivity of the material of the wire be $\rho $

It is known that, resistance is related with the resistivity as:

$R=\rho \frac{l}{A}$

$\Rightarrow \rho =\frac{RA}{l}$

$\Rightarrow \rho =\frac{5\times 6.0\times {{10}^{-7}}}{15}$

$\Rightarrow \rho =2\times {{10}^{-7}}{{m}^{2}}$

Therefore, the resistivity of the material is $2\times {{10}^{-7}}{{m}^{2}}$ .


5. A silver wire has a resistance of \[2.1\Omega \] at ${{27.5}^{\circ }}C$ , and a resistance of $2.7\Omega $ at ${{100}^{\circ }}C$. Determine the temperature coefficient of resistivity of silver.

Ans: In the above question it is given that:

Temperature, ${{T}_{1}}={{27.5}^{\circ }}C$.

Resistance of the silver wire at ${{T}_{1}}$ is ${{R}_{1}}=2.1\Omega $ .

Temperature, ${{T}_{2}}={{100}^{\circ }}C$ .

Resistance of the silver wire at ${{T}_{2}}$ is ${{R}_{2}}=2.7\Omega $ .

Let the temperature coefficient of silver be $\alpha $ . It is known that temperature co-efficient of a material provides information on the nature of that material with respect to its change in resistance with temperature. Mathematically, it is related with temperature and resistance by the formula:

$\alpha =\frac{{{R}_{2}}-{{R}_{1}}}{{{R}_{1}}\left( {{T}_{2}}-{{T}_{1}} \right)}$

$\Rightarrow \alpha =\frac{2.7-2.1}{2.1\left( 100-27.5 \right)}={{0.0039}^{\circ }}{{C}^{-1}}$

Clearly, the temperature coefficient of silver is ${{0.0039}^{\circ }}{{C}^{-1}}$.


6. A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2A which settles after a few seconds to a steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is ${{27}^{\circ }}C$ ? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is $1.70\times {{10}^{-4}}^{\circ }{{C}^{-1}}$ .

Ans: In the above question it is given that:

Supply voltage is $V=230V$

Initial current drawn is ${{I}_{1}}=3.2A$.

Let the initial resistance be ${{R}_{1}}$ .

Therefore, using Ohm’s law,

${{R}_{1}}=\frac{V}{{{I}_{1}}}$

$\Rightarrow {{R}_{1}}=\frac{230}{3.2}=71.87\Omega $

Steady state value of the current is ${{I}_{2}}=2.8A$.

Let the resistance of the steady state be ${{R}_{2}}$ .

Therefore, using Ohm’s law.

${{R}_{2}}=\frac{V}{{{I}_{2}}}$

$\Rightarrow {{R}_{2}}=\frac{230}{2.8}=82.14\Omega $

Temperature co-efficient of nichrome is $\alpha =1.70\times {{10}^{-4}}^{\circ }{{C}^{-1}}$ .

Initial temperature of nichrome is ${{T}_{1}}={{27}^{\circ }}C$.

Let steady state temperature reached by nichrome be ${{T}_{2}}$ .

Now, it is known that temperature co-efficient of a material provides information on the nature of that material with respect to its change in resistance with temperature. Mathematically, it is given by

$\alpha =\frac{{{R}_{2}}-{{R}_{1}}}{{{R}_{1}}\left( {{T}_{2}}-{{T}_{1}} \right)}$

$\Rightarrow \left( {{T}_{2}}-{{T}_{1}} \right)=\frac{{{R}_{2}}-{{R}_{1}}}{{{R}_{1}}\alpha }$

Substituting the given values,

$\Rightarrow \left( {{T}_{2}}-27 \right)=\frac{82.14-71.87}{71.87\times 1.70\times {{10}^{-4}}}$

$\Rightarrow {{T}_{2}}-27=840.5$

$\Rightarrow {{T}_{2}}={{867.5}^{\circ }}C$

Clearly, the steady temperature of the heating element is ${{867.5}^{\circ }}C$.


7. Determine the current in each branch of the network shown in figure:


Circuit diagram

Ans: Current flowing through various branches of the circuit is represented in the given figure.


Circuit diagram to calculate the current

Consider

${{I}_{1}}=$Current flowing through the outer circuit

${{I}_{2}}=$Current flowing through branch AB

${{I}_{3}}=$Current flowing through branch AD

${{I}_{2}}-{{I}_{4}}=$Current flowing through branch BC

${{I}_{3}}+{{I}_{4}}=$Current flowing through branch CD

${{I}_{4}}=$Current flowing through branch BD

For the closed circuit ABDA, potential is zero i.e.,

$10{{I}_{2}}+5{{I}_{4}}-5{{I}_{3}}=0$

$2{{I}_{2}}+{{I}_{4}}-{{I}_{3}}=0$

${{I}_{3}}=2{{I}_{2}}+{{I}_{4}}$ …… (1)

For the closed circuit BCDB, potential is zero i.e.,

$5\left( {{I}_{2}}-{{I}_{4}} \right)-10\left( {{I}_{3}}+{{I}_{4}} \right)-5{{I}_{4}}=0$

$5{{I}_{2}}+5{{I}_{4}}-10{{I}_{3}}-10{{I}_{4}}-5{{I}_{4}}=0$

$5{{I}_{2}}-10{{I}_{3}}-20{{I}_{4}}=0$

${{I}_{2}}=2{{I}_{3}}+4{{I}_{4}}$ …… (2)

For the closed circuit ABCFEA, potential is zero i.e.,

$-10+10\left( {{I}_{1}} \right)+10\left( {{I}_{2}} \right)+5\left( {{I}_{2}}-{{I}_{4}} \right)=0$

$10=15{{I}_{2}}+10{{I}_{1}}-5{{I}_{4}}$

$3{{I}_{3}}+2{{I}_{1}}-{{I}_{4}}=2$ …… (3)

From equations (1) and (2), we obtain

${{I}_{3}}=2\left( 2{{I}_{3}}+4{{I}_{4}} \right)+{{I}_{4}}$

${{I}_{3}}=4{{I}_{3}}+8{{I}_{4}}+{{I}_{4}}$

$-3{{I}_{3}}=9{{I}_{4}}$

$-3{{I}_{4}}=+{{I}_{3}}$ …… (4)

Putting equation (4) in equation (1), we obtain

${{I}_{3}}=2{{I}_{2}}+{{I}_{4}}$

$-4{{I}_{4}}=2{{I}_{2}}$ …… (5)

It is evident from the given figure that,

${{I}_{1}}={{I}_{3}}+{{I}_{2}}$ ……. (6)

Putting equation (6) in equation (1), we obtain

$3{{I}_{2}}+2\left( {{I}_{3}}+{{I}_{2}} \right)-{{I}_{4}}=2$

$5{{I}_{2}}+2{{I}_{3}}-{{I}_{4}}=2$ …… (7)

Putting equations (4) and (5) in equation (7), we obtain

$5\left( -2{{I}_{4}} \right)+2\left( -3{{I}_{4}} \right)-{{I}_{4}}=2$

$-10{{I}_{4}}-6{{I}_{4}}-{{I}_{4}}=2$

$17{{I}_{4}}=-2$

${{I}_{4}}=-\frac{2}{17}A$

Equation (4) reduces to

${{I}_{3}}=-3\left( {{I}_{4}} \right)$

${{I}_{3}}=-3\left( -\frac{2}{17} \right)=\frac{6}{17}A$

${{I}_{2}}=-2\left( {{I}_{4}} \right)$

${{I}_{2}}=-2\left( -\frac{2}{17} \right)=\frac{4}{17}A$

${{I}_{2}}-{{I}_{4}}=\frac{4}{17}-\left( -\frac{2}{17} \right)=\frac{6}{17}$

${{I}_{3}}+{{I}_{4}}=\frac{6}{17}+\left( \frac{-2}{17} \right)=\frac{4}{17}A$

${{I}_{1}}={{I}_{3}}+{{I}_{2}}$

$\therefore {{I}_{1}}=\frac{6}{17}+\frac{4}{17}=\frac{10}{17}A$

Therefore, current in branch AB $=\frac{4}{17}A$

Current in branch BC $=\frac{6}{17}A$

Current in branch CD $=\frac{-4}{17}A$

Current in branch AD $=\frac{6}{17}A$

Current in branch BD $=\left( \frac{-2}{17} \right)A$

Total current $=\frac{4}{17}+\frac{6}{17}+\frac{-4}{17}+\frac{6}{17}+\frac{-2}{17}=\frac{10}{17}A$ .


8. A storage battery of emf 8.0 V and internal resistance $0.5\Omega $ is being charged by a 120 V DC supply using a series resistor of $15.5\Omega $. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?

Ans: In the above question it is given that:

Emf of the storage battery is $E=0.8V$.

Internal resistance of the battery is $r=0.5\Omega $ .

DC supply voltage is $V=120V$

Resistance of the resistor is $R=15.5\Omega $.

Consider effective voltage in the circuit to be $V'$, which would be the difference in the supply voltage and the emf of the battery.

$V'=V-E$

$\Rightarrow V'=120-8=112V$

Now, current flowing in the circuit is $I$ and the resistance $R$ is connected in series to the storage battery. 

Therefore, using Ohm’s law,

$I=\frac{V'}{R+r}$

$\Rightarrow I=\frac{112}{15.5+0.5}=7A$

Thus, voltage across resistor $R$would be:

$IR=7\times 15.5=108.5V$

DC supply voltage = Terminal voltage of battery + Voltage drop across $R$

Terminal voltage of battery $=120-108.5=11.5V$

A series resistor in a charging circuit takes the responsibility for controlling the current drawn from the external source. Excluding this series resistor is dangerous as the current flow would be extremely high if so.


9. The number density of free electrons in a copper conductor estimated in Example 3.1 is $8.5\times {{10}^{28}}{{m}^{-3}}$ . How long does an electron take to drift from one end of a wire 3.0 m long to its other end? The area of cross-section of the wire is $2.0\times {{10}^{-6}}{{m}^{2}}$ and it is carrying a current of 3.0 A.

Ans: In the above question it is given that:

Number density of free electrons in a copper conductor is $n=8.5\times {{10}^{28}}{{m}^{-3}}$.

Length of the copper wire is $l=3.0m$.

Area of cross-section of the wire is $A=2.0\times {{10}^{-6}}{{m}^{2}}$.

Current carried by the wire is $I=3.0A$.

Now we know that:

$I=nAe{{V}_{d}}$

Where,

$e$ is the electric charge of magnitude $1.6\times {{10}^{-19}}C$.

${{V}_{d}}$ is the drift velocity and

\[Drift\text{ }velocity=\frac{\text{Length of the wire }\left( \text{l} \right)}{\text{Time taken to cover }\left( \text{t} \right)}\]

$I=nAe\frac{l}{t}$

$\Rightarrow t=\frac{nAel}{I}$

$\Rightarrow t=\frac{3\times 8.5\times {{10}^{28}}\times 2\times {{10}^{-6}}\times 1.6\times {{10}^{-19}}}{3.0}$

$\therefore t=2.7\times {{10}^{4}}s$ .

Hence the time taken by an electron to drift from one end of the wire to the other is $2.7\times {{10}^{4}}s$.


Current Electricity Chapter Summary - Class 12 NCERT Solutions

  • Current through a given area of a conductor is the net charge passing per unit time through the area.

  • Motion of conduction electrons in electric field E is the sum of (i) motion due to random collisions and (ii) that due to E. The motion due to random collisions averages to zero and does not contribute to vd.

  • Current is a scalar, although we represent current with an arrow. Currents do not obey the law of vector addition. That current is a scalar also follows from its definition. The current I through an area of cross-section is given by the scalar product of two vectors:

I = j. ΔS, where j and ΔS are vectors. 

  • The resistance R of a conductor depends on its length l and cross-sectional area A through the relation,

$R=\frac{\rho \l }{A}$ 

Where ρ, called resistivity, is a property of the material and depends on temperature and pressure.

  • Electrical resistivity of substances varies over a very wide range. Metals have low resistivity, in the range of 10–8 Ω m to 10–6 Ω m. Insulators like glass and rubber have 1022 to 1024 times greater resistivity. Semiconductors like Si and Ge lie roughly in the middle range of resistivity on a logarithmic scale.

  • In most substances, the carriers of current are electrons; in some cases, for example, ionic crystals and electrolytic liquids, positive and negative ions carry the electric current.

  • Current density j gives the amount of charge flowing per second per unit area normal to the flow, j = nq vd where n is the number density (number per unit volume) of charge carriers each of charge q, and vd is the drift velocity of the charge carriers. For electrons q = – e. If j is normal to a cross-sectional area A and is constant over the area, the magnitude of the current I through the area is nevd A.

  • Using E = V/l, I = nevd A, and Ohm’s law, one obtains $\frac{eE}{m}=\rho\frac{ne^2}{m}v_d$

The proportionality between the force eE on the electrons in a metal due to the external field E and the drift velocity vd (not acceleration) can be understood, if we assume that the electrons suffer collisions with ions in the metal, which deflect them randomly. If such collisions occur on an average at a time interval τ, 

vd = aτ = eEτ/m

where a is the acceleration of the electron. This gives 

$\rho=\frac{m}{ne^2\tau }$

  • When a source of emf ε is connected to an external resistance R, the voltage Vext across R is given by

$V_{ext}=IR=\frac{\varepsilon }{R+r}R$

  • (a) Total resistance R of n resistors connected in series is given by R = R1 + R2 +..... + Rn

(b) Total resistance R of n resistors connected in parallel is given by  $\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+.....\frac{1}{R_n}$

  • Kirchhoff’s Rules

(a) Junction Rule: At any junction of circuit elements, the sum of currents entering the junction must equal the sum of currents leaving it. Kirchhoff’s junction rule is based on conservation of charge.

(b) Loop Rule: The algebraic sum of changes in potential around any closed loop must be zero.

  • The Wheatstone bridge is an arrangement of four resistances – R1, R2, R3, R4 as shown in the text. The null-point condition is given by

$\frac{R_1}{R_2}=\frac{R_3}{R_4}$ using which the value of one resistance can be determined, knowing the other three resistances.

  • The measurement of resistance by Wheatstone bridge is not affected by the internal resistance of the cell.

  • If a skeleton cube is made with 12 equal resistance each having resistance R then the net resistance across.

(a) The longest diagonal (EC or AG) $=\frac{5}{6}R$

(b) The diagonal of face (e.g. AC, ED, ....) $=\frac{3}{4}R$

(c) A side (e.g. AB, BC.....) $=\frac{7}{12}R$

  • The potentiometer is a device to compare potential differences. Since the method involves a condition of no current flow, the device can be used to measure potential difference; internal resistance of a cell and compare emf’s of two sources.


Overview of Deleted Syllabus for CBSE Class 12 Physics Current Electricity

Chapter

Dropped Topics

Current Electricity

3.7 Resistivity of Various Materials (delete Tables 3.1 and 3.2 and Carbon resistors, Colour code for carbon resistor)

3.10 Combinations of Resistors – Series and Parallel

Example 3.5

3.15 Meter Bridge

3.16 Potentiometer

Exercises 3.3, 3.4, 3.10, 3.12, 3.14–3.23



Conclusion

NCERT Class 12 Physics Chapter 3 Solutions on Current Electricity provided by Vedantu explains the fundamental concepts underlying the flow of electric charge in conductors. This chapter comprehensively introduces students to how electric current behaves in various materials and circuits. This knowledge is critical for both academic success and practical applications in numerous fields. These concepts are essential for mastering the topic and are often tested in exams. From previous year's question papers, typically around 5–6 questions are asked from this chapter. These questions test students' understanding of theoretical concepts as well as their problem-solving skills.


Other Study Material for CBSE Class 12 Physics Chapter 3



Chapter-Specific NCERT Solutions for Class 12 Physics

Given below are the chapter-wise NCERT Solutions for Class 12 Physics. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.




Related Links for NCERT Class 12 Physics in Hindi

Discover relevant links for NCERT Class 12 Physics in Hindi, offering comprehensive study materials, solutions, and resources to enhance understanding and aid in exam preparation.




Chapter-Specific NCERT Solutions for Class 12 Physics

Given below are the chapter-wise NCERT Solutions for Class 12 Physics. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.


FAQs on NCERT Solutions Class 12 Physics Chapter 3 Current Electricity

1. Which Topics are Covered in Current Electricity?

This chapter primarily deals with the first motion laws, second motion laws, Newton's third motion laws, pseudo forces, the horse and cart and inertia. Moreover, students will also become familiar with electric current, Ohm's law, electric current in conductors, the origin of resistivity, the drift of electrons, Ohm's law limitations, mobility, the resistivity of different materials, electrical power and energy and temperature vulnerability of resistivity.


By studying Current Electricity, students will also get to learn about the resistor combination in both parallel and series, cells, internal resistance, emf, Kirchhoff’s rule, Wheatstone bridge, potentiometer and meter bridge. Some other topics included in this chapter are non-conservation and conservation force field, work-energy theorem, kinetic energy, types of power and equilibrium, the law of conservation of mechanical energy, etc.

2. What is Meant by a Wheatstone Bridge?

This device is utilised to observe mechanical and electrical quantities. In other words, Wheatstone Bridge is a tool for light measurement with the help of a photo resistive instrument. A light-dependent resistor is a type of sensor which is passive-resistive. It is used to convert light levels to change in resistance, and eventually to voltage.


The circuit of the Wheatstone bridge has many applications today. Along with the current functional amplifiers, the Wheatstone Bridge can be used to interface different sensors and transducers into amplifier circuits.

3. How to Obtain Full Marks in Class 12 Physics?

In order to achieve the full score in the Class 12 Physics exam, the primary requirement is to study the NCERT thoroughly. Also, students must practise multiple choice questions from NCERT and recommended books.


Preparing a formula chart is vital for the learners to practise the numerical problems, and diagrams must also be practised with proper labelling. In the case of long answer questions, students must refrain from writing long paragraphs; instead, they must write in bullet points. Further, solving previous years question papers is also mandatory to score well.

4. How to prepare Class 12 Physics Chapter 3 well for exams?

Current Electricity in Physics for Class 12 can often confuse students due to the many laws and numerical in the chapter. To prepare this chapter well for exams, first, go through all the topics thoroughly. Clear the concepts and solve the numerical problems. Solve the exercises in the books and then re-check them with NCERT Solutions so you can get an idea of how to answer questions in the exams. Make notes so that the revision becomes easier.

5. How can I make a study plan for Class 12 Physics Chapter 3?

Class 12 Chapter 3 can be stressful for the students, so forming a good study plan is very important. To make a study plan for Physics, follow these steps:

  • Mark the topics you find difficult and give them more time.

  • Solve some numerical problems every day.

  • Study at odd hours of the day.

  • Take breaks in between to soak in all you have studied.

  • Make notes as you go through the chapters.

6. Where can I download Physics NCERT Solutions for Class 12 Chapter 3 PDF?

You can download the PDF for Physics NCERT Solutions for Class 12: Chapter 3 using the official website of Vedantu or the Vedantu app. Vedantu provides all the study material free of cost. Visit NCERT Solutions chapter 3 Current and Electricity and you will land on the page with chapter-wise solutions. Choose Chapter 3: Current Electricity. Thereafter, you will be able to download the NCERT Solution and other study material for the chapter. These solutions can be a student's best study buddy as they help in providing the best guidance and make learning fun.

7. Why are the Class 12 Physics Chapter 3 NCERT Solutions important?

NCERT Solutions for Chapter 3 “Current Electricity” in Class 12 Physics is essential for the exams and can often confuse the students. Using the solutions will help you clear all the concepts and guide you in solving the numerical problems. They also give you a fair idea about the type of questions in exams and how to answer them. It is essential to be well-prepared for the exams and NCERT Solutions are the best guidance you can find online.

8. Do I need to practice all the questions provided in Class 12 Physics NCERT Solutions of Chapter 3?

Solving all the questions provided in Chapter 3 of Class 12 Physics NCERT Solutions will help you comprehend the concepts in the chapter intensively and clear your doubts. The answers to the questions from the NCERT textbook in the solutions are designed following the CBSE board guidelines by experts in each subject. Therefore, solving and practising all of them regularly will bring you a step closer to achieving good marks in the exams.

9. Why is current electricity important mentioned in Chapter 3 Physics Class 12?

Current electricity is essential because it forms the foundation of modern electrical and electronic systems. It is crucial for the functioning of household appliances, industrial machinery, communication devices, and medical equipment. Understanding current electricity helps in designing circuits, troubleshooting electrical issues, and developing new technologies.

10. What initiates current in Current Electricity Class 12 NCERT Solutions?

Electric current is initiated when there is a potential difference (voltage) across a conductor, causing the free electrons in the conductor to move. This movement of electrons constitutes an electric current. The potential difference can be created using a power source such as a battery or a generator.

11. How is current electricity produced that is mentioned in Current Electricity Class 12 Exercise Solutions?

Current electricity is produced when electrons move through a conductor due to a potential difference. This can be achieved by connecting a conductor to a power source (like a battery) that provides the necessary voltage to drive the electrons from one point to another, creating a flow of electric charge.

12. What is the SI unit of current according to Class 12 Physics Chapter 3 NCERT Solutions?

In Class 12 Physics Chapter 3 NCERT Solutions, the SI unit of electric current is the ampere (A). One ampere is defined as the flow of one coulomb of charge per second.

13. What is the theory of Current Electricity in Class 12 Solutions?

The theory of Current Electricity involves the study of how electric charges move through a conductor and the laws governing this movement. Key concepts include Ohm’s Law, Kirchhoff’s laws, and the relationship between voltage, current, resistance, and power. It also covers the behaviour of materials under the influence of an electric field and the properties of conductors, insulators, and semiconductors.

14. What is the symbol for the current in Class 12 Physics Ch 3 NCERT Solutions?

In physics chapter 3 class 12 exercise solutions, the symbol for electric current is I. It is typically represented in equations and circuit diagrams to denote the flow of electric charge.

15. How does current electricity work as described in Current Electricity NCERT Solutions?

Current electricity works by the movement of electrons through a conductor. When a potential difference is applied across a conductor, it creates an electric field that exerts a force on the free electrons, causing them to move. This flow of electrons constitutes an electric current. The current flows from the negative terminal to the positive terminal of the power source, although conventionally, it is considered to flow from positive to negative.

16. What is the formula for current electricity?

The basic formula for electric current is: I= $\frac{Q}{t}$

​where 

I is the current, 

Q is the charge, and 

t is the time.

Another important formula is Ohm’s Law:

V=IR

where 

V is the voltage, 

I is the current, and 

R is the resistance.

17. What is the gist of Current Electricity in Chapter 3 Physics Class 12 NCERT Solutions?

The gist of ch 3 physics class 12 current electricity is the study of how electric charges flow through conductors and the principles that govern this flow. It involves understanding the relationship between voltage, current, and resistance, and applying laws such as Ohm's Law and Kirchhoff's Laws to analyse electric circuits. This knowledge is fundamental for designing, operating, and troubleshooting electrical and electronic systems.