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NCERT Solutions for 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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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

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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.


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FAQs on NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity

1. What key topics are covered in NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity as per the CBSE 2025–26 syllabus?

NCERT Solutions for Chapter 3 Current Electricity comprehensively address:

  • Definition and calculation of electric current and its direction
  • Ohm’s Law and its applications in circuits
  • Concepts of resistivity and conductivity
  • Calculation of drift velocity
  • Methods of combining resistors in series and parallel
  • Understanding internal resistance of cells
  • Working with Kirchhoff’s Rules for complex networks
  • Wheatstone bridge and potentiometer principle and applications

Each topic is approached with stepwise, CBSE-approved methodology for the Class 12 Board exam.

2. How should students approach solving NCERT Class 12 Physics Chapter 3 questions for maximum marks?

To maximise marks:

  • Read the question carefully, identifying all given values
  • State the relevant formula before solving (e.g., Ohm’s Law: V = IR)
  • Write every calculation step, showing substitution of values and units
  • Justify steps with correct physical reasoning
  • Add clear, neatly-labelled diagrams if required
  • Highlight or box the final answer

This method, aligned with the CBSE marking scheme, increases accuracy and scoring potential.

3. Why is Ohm’s Law essential for solving NCERT Class 12 Physics Chapter 3 exercises?

Ohm’s Law (V = IR) is a fundamental principle that relates voltage, current, and resistance in a circuit. It enables students to:

  • Determine unknown quantities in simple and complex circuits
  • Understand how changes in one parameter affect the others
  • Solve both theoretical and numerical problems in Chapter 3

Mastery of Ohm’s Law is required for scoring in both short and long answer questions as per the latest Physics NCERT Solutions.

4. What are common pitfalls to avoid when solving numericals from NCERT Solutions for Current Electricity?

Students should avoid these typical mistakes:

  • Omitting units or using incorrect units (e.g., confusion between ohms and kilo-ohms)
  • Failure to convert temperatures when necessary (e.g., Celsius to Kelvin)
  • Ignoring the internal resistance of cells in calculations
  • Applying Ohm’s Law to non-ohmic conductors where it does not hold
  • Not drawing or mislabelling required circuit diagrams

Awareness of these pitfalls leads to more accurate and CBSE-compliant answers.

5. How do Kirchhoff’s Laws enable solving complex circuit problems in Chapter 3 NCERT Solutions?

Kirchhoff’s Laws—the Current Law (KCL) and Voltage Law (KVL)—allow students to set up equations for electrical quantities in multi-loop and bridge circuits where simple series/parallel rules do not apply. By applying these laws, students can:

  • Formulate simultaneous equations
  • Calculate unknown currents and voltages
  • Analyze circuits featured in advanced NCERT and board exam questions

6. What is the role of drift velocity in the flow of current according to the NCERT Solutions?

Drift velocity (vd) is the average velocity attained by free electrons in a conductor due to an electric field. It determines the current using the formula: I = nAe vd, where n = carrier density, A = area, and e = charge of an electron. Calculating drift velocity is crucial for connecting theory with practical current flow in Class 12 Physics numericals.

7. Why does resistance of a metallic conductor increase with temperature as explained in Chapter 3?

As temperature rises, metal ions vibrate more intensely, causing free electrons to undergo more frequent and stronger collisions with the lattice. This increased scattering raises the resistance. The temperature coefficient of resistance quantifies this relationship—a commonly tested concept in NCERT Physics and CBSE exams.

8. How can the Wheatstone Bridge principle be applied to find unknown resistances as per NCERT Solutions?

The Wheatstone Bridge is balanced when the ratio of two known resistors equals the ratio of the unknown to the standard resistor (R1/R2 = R3/R4). At this balanced point, with zero galvanometer current, students can accurately calculate unknown resistance. This technique is often applied in board-style NCERT questions.

9. What are the limitations of Ohm’s Law discussed in Class 12 Physics NCERT Solutions?

Ohm’s Law only applies to ohmic materials—those exhibiting a linear relationship between current and voltage and a constant resistance (such as most metals at constant temperature). It does not apply to semiconductors, diodes, or materials with variable resistance (affected by temperature, voltage, or light). Recognizing these limitations is necessary to avoid misconceptions on MCQs and HOTS questions.

10. If two resistors with different temperature coefficients are placed in series, how is the overall temperature coefficient determined?

For resistors R1 and R2 with temperature coefficients α1 and α2, the overall temperature coefficient α is calculated using:
α = (α1R1 + α2R2) / (R1 + R2). This formula ensures precise analysis in application-based circuit problems in the NCERT curriculum.

11. What is the difference between resistance and resistivity based on the NCERT Solutions for Current Electricity?

Resistance (R) depends on size and geometry: R = ρ(l/A), where ρ is resistivity, l is length, and A is the cross-sectional area. Resistivity (ρ) is a material property, independent of the conductor’s dimensions, characterizing how strongly a material opposes current flow. It remains constant for a material at a given temperature.

12. How should students revise Chapter 3 Current Electricity for the CBSE 2025–26 board exam?

Effective revision strategy includes:

  • Reviewing all solved NCERT exercise questions using correct stepwise method
  • Memorising important formulas and their application
  • Drawing and interpreting circuit diagrams
  • Practising previous years’ questions for pattern understanding
  • Clarifying concepts such as drift velocity, resistance, and combination of resistors

This preparation method ensures coverage of both conceptual and numerical aspects for board exams.

13. Why is understanding internal resistance crucial in Chapter 3 NCERT Solutions?

Internal resistance determines the terminal voltage and actual current delivered by a cell or battery, not just the theoretical emf value. It is vital when calculating real-world circuit outputs. All CBSE-compliant NCERT Solutions for Chapter 3 require incorporating internal resistance in stepwise calculations for accurate answers.

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

The SI unit of electric current is ampere (A). One ampere is defined as one coulomb of charge passing through a section of conductor per second. This unit is essential for all calculations in Current Electricity as per the CBSE 2025–26 syllabus.

15. How do stepwise NCERT Solutions for Current Electricity help improve exam performance?

Using stepwise, syllabus-based solutions:

  • Enhances conceptual clarity and problem-solving skills
  • Encourages the correct use of formulas and units
  • Reduces common errors seen in board exams
  • Boosts confidence in tackling both numerical and theory questions
  • Improves answer presentation, which leads to higher scores as per CBSE marking guidelines