NCERT Class 12 Physics Chapter 3 Notes Current Electricity - Download PDF

NCERT Class 12 Physics Chapter 3 Notes Current Electricity - Download PDF

Edited By Vishal kumar | Updated on Jan 23, 2024 11:18 AM IST

You've come to the right place if you're looking for class 12 physics chapter 3 notes to help you improve your academic journey and achieve high marks in board and state exams, including those in fields like engineering and medicine. On this page dedicated to NCERT notes, Careers360 provides detailed notes for quick revision.

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Candidates can have a look at the JEE Main 2025 physics syllabus below.

JEE Main 2025 physics topics

Number of questions

Marks

Modern Physics

5

20

Heat and Thermodynamics

3

12

Optics

3

12

Current Electricity

3

12

Electrostatics

3

12

Magnetics

2

8

Unit, Dimension and Vector

1

4

Kinematics

1

4

Laws of Motion

1

4

Work, Power and Energy

1

4

Centre of Mass, Impulse, and Momentum

1

4

Rotation

1

4

Gravitation

1

4

Simple Harmonic Motion

1

4

Solids and Fluids

1

4

Waves

1

4

Electromagnetics Induction; AC

1

4

Read More

CBSE Class 12 Physics chapter 3 notes moves around the concepts of Ohms's law and Kirchoff's laws also. The NCERT Class 12 Physics chapter 3 Notes are prepared by expert faculties and mainly they can be used for revising the concepts. These Current Electricity class 12 notes have been carefully created by Careers360 physics experts and are presented in simple language with comprehensive details. Furthermore, these CBSE class 12 physics ch 3 notes are available in PDF format for easy download and use at any time and from any location.

Also, students can refer,

NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity

NCERT Exemplar Class 12 Physics Chapter 3 Current Electricity

NCERT Class 12 Physics Chapter 3 Notes

Electric Current: The flow of charge through a conductor per unit of time is defined as electric current. It is measured in amperes (A) and is essential for understanding electrical circuits and electromagnetism.

\small \small i=\frac{q}{t}1625645340959

Where, i is the current, q is the charge and t is the time.

  • If the rate of flow of charge is variable, the current at any time is\small \small i=\frac{dq}{dt}1625645341129

Current Density: The amount of electric current flowing through a material per unit cross-sectional area is referred to as its current density. It is a vector quantity denoted by vecJ and can be written as,

  • If the cross-sectional area is not perpendicular to the current but forms an angle θ with the current direction, then
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  • Relation between current density and electric field

\small \vec{J}=\sigma \vec{E} =\frac{\vec{E}}{\rho }1625645341771

Where, σ is conductivity and ρ is resistivity or specific resistance of the substance

Mobility: Mobility for electrons is defined as the drift velocity per unit electric field.

\small \small \mu=\frac{v_d}{E}1625645343394

Where, μ is mobility and vd is drift velocity

Ohm’s Law:

Ohm's Law states that in a conductor, under constant external conditions such as temperature and pressure, the current flowing through the conductor is directly proportional to the potential difference across its two ends.

\small V\propto I1625645343925

\small V= IR1625645344104

R- Electric Resistance

\small R=\rho\frac{l}{A}1625645344272

where ρ is resistivity / specific resistance, l is the length of conductor and A is the area of the cross-section of the conductor

  • Ohmic Substance: An Ohmic substance is a substance that obeys Ohm's Law. It has a linear I-V graph, and the slope gives the conductance, which is the reciprocal of resistance.
  • Non-ohmic Substances: Non-ohmic or non-linear conductors are substances that do not obey Ohm's Law, such as gases and crystal rectifiers.
  • Superconductor: Superconductors are materials that have zero resistivity below a critical temperature. Electrical resistance is zero in this state
  • In Superconductor resistivity is zero

,16256453792021625645344625

Resistivity:

\small \rho=\frac{m}{ne^2\tau}1625645344803

Where, m is the mass, n is the number of electrons per unit volume, e is the charge of the electron, and τ is the relaxation time.

  • Resistivity is a material's intrinsic property, and its value tends to increase with the presence of impurities and mechanical stress.
  • The reciprocal of resistivity is called conductivity.
  • The reciprocal of resistance is termed conductance, and its SI unit is either Ω-1 or Siemens.

Temperature-Dependent Resistivity:

\small \rho=\rho_0(1+\alpha(T-T_0))1625645345335

ρ: Resistivity at temperature T

ρo: Resistivity at temperature To

Temperature-Dependent Resistance

  • Temperature Coefficient of Resistance

\small R_T=R_{T_0}[1+\alpha(T-T_0)]1625645346123

RT: Resistance at the temperature T1625645346515

Ro: Resistance at the temperature T_{o}1625645346857

α: Temperature coefficient of resistance

\small \alpha=\frac{R_T-R_0}{R_0(T-T_0)}1625645347206

Where the value of α is different at different temperatures

Colour Coding of a Resistance:

The carbon resistance typically consists of four coloured ring bands labelled as A, B, C, and D.

16256453827071625645347912

  • Colour bands A and B indicate the significant digits of the resistance value.
  • Colour band C represents the decimal multiplier.
  • Colour band D indicates the tolerance, expressed as a percentage, around the specified resistance value.

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Tolerance of Gold is \small \pm5\%1625645348090

Tolerance of Silver is \small \pm 101625645348265%

Tolerance if no colour \small \pm 201625645348444%

16256453867961625645348618

Grouping of Resistance

Series Grouping of resistance

\small R_{eq}=R_1+R_2+.........+R_n1625645348797

Parallel Grouping of Resistance:

\small \frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+............+\frac{1}{R_n}1625645348977

Heat developed in a resistor: When a constant current flows through a resistance R for t seconds, the loss in electric potential energy appears itself as increased thermal energy (heat H) in the resistor. This relationship is expressed mathematically as,

\small H=i^2Rt1625645349160

\small \text{ The power developed =}1625645349408\small \frac{energy}{time}=i^2R=iR=\frac{V^2}{R}1625645349777 \small \text{ (from Ohm's law) }1625645349967

  • The unit of heat is the joule (J), and the unit of power is the watt (W).

Cell: The device which converts Chemical energy into electrical energy is known as an electric cell.

Internal resistance: In the case of a cell the opposition of electrolyte to the flow of current through it. It is shown by r.

  • The internal resistance of a cell depends on the distance between electrodes.

\small r\propto d1625645350328

  • The internal resistance of a cell depends on the area of the electrodes

\small r\propto A1625645350543

  • The internal resistance of a cell concentration of electrolyte

\small r\propto c1625645350742

  • The internal resistance of a cell temperature of the electrolyte

\small r\propto \frac{1}{temp}1625645350944

Emf of a cell: The electromotive force (emf) of a cell is defined as the work done or energy carried by unit charge when it completes one full cycle within the circuit.

Potential difference: The potential difference, also known as voltage, is the electrical pressure across the terminals of a cell when it is actively supplying current to an external resistance in the circuit.

Equation of cell

  1. When supplying the current: \small E=V+iR1625645351148
  2. When the cell is being charged: \small E=V- iR1625645351324

Current supplied by the cell:

16256453851581625645351588

  • Cell supplies a constant current in the circuit.

\small i=\frac{E}{R+r}1625645351795 ,

Where, R- External resistance and r- internal resistance

  • Potential drop inside the cell=ir
  • The internal resistance of the cell \small r=(\frac{E}{V}-1)R1625645351941
  • The power dissipated in external resistance\small P=(\frac{E}{R+r})^2R1625645352111
  • Maximum power is obtained when the resistance value of the load is equal in value to that of the voltage source's internal resistance.

\small P_{max}=\frac{E^{2}}{4r}1625645352295

Series grouping of cells:

In series grouping anode of one cell is connected to the cathode of other cells

16256453907241625645352446

n- identical cells which are connected in series, then

  • Equivalent e.m.f of combination is

\small E_{eq}=nE1625645352595

  • Equivalent internal resistance

\small r_{eq}=nr1625645352799

  • Main current/current from each cell

\small i=\frac{nE}{R+nr}1625645352996

  • The power dissipated in the external circuit is

\small (\frac{nE}{R+nr})^2.R1625645353224

  • Condition for Maximum Power is R=nr
  • \small P_{max}=n(\frac{E^{2}}{4r})1625645353374\small when1625645353558\small nr<<R1625645353815

Parallel grouping of cells:

In parallel grouping, all anodes are connected to one point and all cathode together at other points

16256453926181625645353981

For n cells connected in parallel then,

  • Equivalent e.m.f

\small E_{eq}=E1625645354126

  • Equivalent internal resistance

\small R_{eq}=\frac{r}{n}1625645354278

  • The main current is

i\small =\frac{E}{R+r/n}1625645354433

  • The potential difference across an external resistance \small V=iR1625645354637
  • Current from each cell \small i'=i/n1625645354888
  • The power dissipated in the circuit

\small P=(\frac{E}{R+r/n})^2 R1625645355117

  • Condition for Maximum Power

\small R= \frac{r}{n}1625645355326

\small P_{max}=n \left ( \frac{E^{2}}{4r} \right )1625645355480 when \small r>>nR1625645355682

Kirchoff's first law: In a circuit, at any junction, the sum of the currents entering the junction must equal the sum of the currents leaving the junction. This law is also known as Junction rule or Kirchoff's current law (KCL)

\small \sum i=01625645355932

16256453948371625645356109

\small i_1+i_3=i_2+i_41625645356280

This law is simply based on the conservation of charge.

Kirchoff's second law: Algebraic sum of all the potential across a closed loop is zero. This law is also known as Kirchhoff's Voltage law (KVL)

This law is based on the conservation of energy.

\small \sum V=01625645356460

In closed-loop

16256453955511625645356652

\small -i_1R_1+i_2R_{2}-E_1-i_3R_3+E_2+E_3-i_4R_4=01625645356844

Potentiometer: A potentiometer is a device that measures potential difference without drawing current from the circuit. It is commonly employed to measure the electromotive force (e.m.f) of a cell accurately and to compare e.m.f values of different cells. Additionally, it is utilized for determining the internal resistance of a given cell.

16256453952531625645357023

The potentiometer consists of wires of length 5 to 10 meters arranged on a wooden block as parallel strips of wires with 1-meter length each and ends of wires are joined by thick coppers. The wire has a uniform cross-section and is made up of the same material. A driver circuit that contains a rheostat, key, and a voltage source with internal resistance r. The driver circuit sends a constant current (I) through the wire. The potential across the wire

V=IR

R is proportional to l since area and resistivity are constant. Therefore V is proportional to length.

\small V\propto L1625645357202

The secondary circuit contains cell/resistors whose potential is to be measured. Whose one end is connected to a galvanometer and the other end of the galvanometer is connected to a jockey which is moved along the wire to obtain a point where there is no current through the galvanometer. So that So the potential of the secondary circuit is proportional to the length at which there is no current through the galvanometer. This is how the potential of a circuit is measured using the potentiometer

Calibration of potentiometer

16256453961011625645357378

In the potentiometer a battery of known emf E. A constant current I is flowing through AB from the driver circuit (that is the circuit above AB). The jockey is slid on potentiometer wire AB to obtain null deflection in the galvanometer. Let l be the length at which the galvanometer shows null deflection. The potential of wire AB (V) is proportional to the length AB(L).

Now

\small \\\frac{V}{E}=\frac{L}{l}\\V=E\frac{L}{l}1625645357571

Thus we obtained the potential of wire AB when a constant current is passing through it. This is known as calibration.

Comparison of emf:

16256453975291625645357753

\small \frac{E_{1}}{E_{2}}=\frac{l_{1}}{l_{2}}1625645357936

Determine the internal resistance of a cell

16256453989981625645358112

\small r=(\frac{l_{1}-l_{2}}{l_{2}})R1625645358328

\small r=(\frac{E}{V}-1)R1625645358510

\small \frac{E}{V}=\frac{l_{1}}{l_{2}}1625645358695

Comparison of resistances:

16256453998461625645358875

\small \frac{R_{2}}{R_{1}}=\frac{l_{2}-l_{1}}{l_{1}}1625645359037

Wheatstone's Bridge:

16256453994911625645359212

It is an arrangement of four resistances that can be used to measure one of them in terms of rest

\small \frac{P}{Q}= \frac{R}{S}1625645359397

\small V_{B} = V_{D}1625645359571

( Balanced condition )

No current will flow through the galvanometer

unbalanced condition:\small V_{B}>V_{D}1625645359747

\small (V_{A}-V_{B})<(V_{A}-V_{D})1625645359924

Current will flow from A to B

Meter bridge: The meter bridge is used to find the resistance of a wire, enabling the calculation of its specific resistance. Operating on Wheatstone's bridge principle, it provides a precise method for measuring resistance by balancing known and unknown resistances.

16256454000261625645360122

\small \frac{P}{Q}=\frac{R}{S}\Rightarrow S= \frac{(100-l)}{l}R1625645360312

After reading the above Current Electricity class 12 notes, you will be well-prepared to answer questions and gain a solid understanding of the Current Electricity chapter in CBSE Class 12 Physics. The comprehensive coverage, ranging from fundamental concepts to advanced topics, ensures that you have the knowledge required to confidently solve problems and succeed in assessments. The provided CBSE class 12 physics ch 3 notes is an invaluable tool for integrating key ideas and fostering a deeper understanding of the chapter's complexities, ultimately improving your ability to score well.

Importance of NCERT Class 12 Physics Chapter 3 Notes

  • Current Electricity Notes Class 12 play a crucial role in preparing for CBSE and various state board exams.
  • This chapter covers approximately 10% of the Class 12 CBSE Board Exam Physics Paper, underlining its significance in board exams.
  • The notes align with the CBSE Physics Syllabus for Chapter 3, providing comprehensive coverage of the required topics.
  • Chapter 3 is essential for competitive exams like NEET and JEE Main, expanding its importance beyond regular board exams.

NCERT Class 12 Notes Chapterwise

Subject Wise NCERT Exemplar Solutions

Subject Wise NCERT Solutions

NCERT Books and Syllabus

Frequently Asked Questions (FAQs)

1. How important is the chapter current electricity for CBSE Class 12 board exam?

As far as Class 12 CBSE Board Exam Physics Papers are concerned the chapter current electricity holds a weightage of 10 to 15 percentage. 

2. What are the topics covered in the Current electricity Class 12 notes?

The main topics covered in the CBSEch 3 physics class 12 notes are listed below.

  • Electric current, drift velocity, and related equations.

  • Ohm's Law.

  • Resistance, resistivity, and their temperature dependence.

  • Cells and their combinations.

  • Kirchhoff’s laws.

  • Potentiometer, Wheatstone’s bridge, and meter bridge.

3. Which laws are discussed in the Class 12 NCERT books to solve electric circuits?

The basic laws to solve electrical circuits are Ohms's law and Kirchhoff’s laws. 

4. Is the topic potentiometer important for exams?

Yes, from the topic potentiometer you can expect at least one question in NEET and JEE Main exam and also in various boards exams across the country.

5. What are the applications of the potentiometer?

The potentiometer can be used to compare EMFs of cells and to calculate the internal resistance of a cell. These topics are mentioned in class 12 current electricity notes

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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

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

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

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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