NCERT Class 12 Physics Chapter 3 Notes, Current Electricity Class 12 Chapter 3 Notes

Think of your phone charger- that moment when you switch it on, and there is an electric flow to charge your phone, that is a real-life example of the concepts in Class 12 Physics Chapter 3: Current Electricity. We explore the underlying concepts of electric currents, resistances, circuits, and flow of charge in conductive materials a very important concept in the way electronic devices operate in this chapter of NCERT Notes.

This chapter is an essential part of the CBSE Class 12 Physics curriculum, and at the same time, essential to JEE, NEET, and other competitive exam candidates because it reinforces the conceptual understanding of concepts such as Ohm Law, Kirchoff Law, internal resistance of cells, Wheatstone bridge, meter bridge, and resistivity and conductivity. Being good at these concepts will lay a foundation in solving theoretical and numerical problems in exams. In order to facilitate students in revising the topics well, Careers360 provides well-organised NCERT Notes of Class 12 Chapter 3 Current Electricity, developed through the efforts of experienced Physics teachers. This revision notes include summaries of the main concepts with fast learning, an extended list of significant formulas, neatly labelled diagrams to support visual understanding, solved sample questions to improve problem-solving ability

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: Download PDF

Download the PDF version of Current Electricity NCERT Notes to view them offline. Using these notes, students can revise anytime without the internet and is most suitable for fast and easy access.

Download PDF

NCERT Class 12 Physics Chapter 3 Notes

The NCERT Class 12 Physics Chapter 3 Notes – Current Electricity provide a clear and concise summary of essential concepts like Ohm’s Law, Kirchhoff’s Laws, and circuit analysis. These notes are ideal for quick revision and exam preparation for CBSE, JEE, and NEET.

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.

$I=\frac{q}{t}$

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 i=dqdt
• 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 J and can be written as,

$\overline{j}=\frac{\mathrm{\Delta }i}{\mathrm{\Delta }A}$

• If the cross-sectional area is not perpendicular to the current but forms an angle θ with the current direction, then

$\begin{array}{rl}{J}_{av}& =\frac{di}{dA\cos \theta }\\ di& =JdA\cos \theta =\overrightarrow{J}\cdot d\overrightarrow{A}\end{array}$

• Relation between current density and electric field

$\overrightarrow{J}=\sigma \overrightarrow{E}$

$\overrightarrow{J}=\frac{\overrightarrow{E}}{\rho }$

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.

$\mu =\frac{v_d}{E}$

Where, μ is mobility and v_d 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.

$\begin{array}{c}V\propto I\\ V=IR\end{array}$

R- Electric Resistance

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

where ρ is resistivity / specific resistance, l is the length of the 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

,

Resistivity:

$\rho =\frac{m}{n{e}^2\tau }$

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:

ρ=ρ_o(1+α(T−T_o))

ρ: Resistivity at temperature T

ρ_o: Resistivity at temperature T_o

• Temperature Coefficient of Resistance

$R_T=R_0[1+\alpha [T-T_0]]$

$R_T$ - Resistance at temperature $T$
$R_0$ - Resistance at temperature $T_o$
$\alpha$ - temperature coefficient of resistance

$\alpha =\frac{R_T-R_o}{R_o(T-T_o)}$

Where the value of $\alpha$ is different at different temperatures

Colour Coding of Resistance:

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

• 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.
• May be remembered as BBROY Great Britain Very Good Wife.

Tolerance of Gold is ±5%
Tolerance of Silver is ±10%
Tolerance if no colour ±20%

Grouping of Resistance:

Series Grouping of resistance

$R_{eq}=R_1+R_2+R_3+\cdots +R_n$

$R_{eq}=\text{Equivalent Resistance }$
For n-identical resistance: $R_{eq}=nR$

$V^{\prime }=\frac{V}{n}$

Parallel Grouping of Resistance:

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\cdots +\frac{1}{R_n}$

If two resistances are in Parallel:

$R_{eq}=\frac{R_1{R}_2}{R_1+R_2}$

Heat developed in a resistor:

• Heat developed in a resistor: When a steady current flows through a resistance R for time $t$, the loss in electric potential energy appears as increased thermal energy(Heat H) of resistor and $H=i^{2}Rt$

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

• Unit of heat is the joule (J)

• Unit of power is 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.

r∝d

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

r∝A

• The internal resistance of a cell concentration of electrolyte

r∝c

• The internal resistance of a cell temperature of the electrolyte

r∝1/temp

• 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: E=V+iR
2. When the cell is being charged: E=V−iR

• Current supplied by the cell:

Cell supplies a constant current in the circuit.

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

Where, R- External resistance and r- internal resistance

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

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

• Series grouping of cells:

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

$n=$ identical cells which are connected in series, then
- Equivalent e.m.f of combination is $E_{eq}=nE$
- Equivalent internal resistance $r_{eq}=nr$
- Main current / current from each cell $i=\frac{nE}{R+nr}$
- Power dissipated in the external circuit is ${\left(\frac{nE}{R+nr}\right)}^2\cdot R$
- Conditions for Maximum Power is $R=nr$
- $P_{\text{max }}=n\left(\frac{E^2}{4r}\right)$ when $nr<R$

• Parallel grouping of cells:

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

For n cells connected in parallel
Equivalent e.m.f $E_{eq}=E$
Equivalent internal resistance $R_{eq}=\frac{r}{n}$
The main current is

$i=\frac{E}{R+\frac{r}{n}}$
The potential difference across the external resistance

$V=iR$
Current from each cell

$i^{\prime }=\frac{i}{n}$

The power dissipated in the circuit

$P={\left(\frac{E}{R+\frac{r}{n}}\right)}^2\cdot R$

Condition for Maximum Power

$\begin{array}{rl}& R=\frac{r}{n}\\ & P_{max}=n\left(\frac{E^2}{4r}\right)\text{when }r>>nR\end{array}$

Kirchhoff'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)

∑i=0

i₁+i₃=i₂+i₄

This law is simply based on the conservation of charge.

Kirchhoff's second law:

• The algebraic sum of all the potentials 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.

∑V=0

In closed-loop

$-i_1{R}_1+i_2{R}_2-E_1-i_3{R}_3+E_2+E_3-i_4{R}_4=0$

Wheatstone's Bridge:

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

$\begin{array}{rl}& \frac{P}{Q}=\frac{R}{S}\\ & V_B=V_D\end{array}$

(Balanced condition)
No current will flow through the galvanometer
unbalanced condition: $V_B>V_D$

$(V_A-V_B)<(V_A-V_D)$

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.

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

Current Electricity Previous year Question and Answer

Q1: n identical cells, each of e.m.f. E and internal resistance r, are connected in series. Later on it was found out that two cells ' X ' and ' Y ' are connected in reverse polarities. Calculate the potential difference across the cell ' X '.

Answer:

Two cells are connected in reverse polarity
So total emf $=(n-2)E-2E$

$\begin{array}{rl}& =(n-2-2)E\\ & =(n-4)E\end{array}$

$\begin{array}{rl}& \text{ Req }=nr\\ & I=\frac{E_{net}}{R_{eq}}=\frac{(n-4)E}{nr}\end{array}$

potential difference across cell $X=E+Ir$

$\begin{array}{rl}& =E+\frac{(n-4)E}{nr}\times r\\ & =\frac{nE+(n-4)E}{n}\\ & =\frac{nE+(n-4)E}{n}\\ & =\frac{E(n+n+4)}{n}=\frac{2E(n+2)}{n}\end{array}$

Q2: Two batteries of emf's 3 V and 6 V and internal resistances 0.2Ω and 0.4Ω are connected in parallel. This combination is connected to a 4Ω resistor. Find the equivalent internal resistance of the combination

Answer:

Given:
Two batteries are connected in parallel, and then connected to an external resistor of $R=4\mathrm{\Omega }$.
Battery $1:E_1=3\mathrm{V},r_1=0.2\mathrm{\Omega }$
Battery $2:E_2=6\mathrm{V},r_2=0.4\mathrm{\Omega }$

The equivalent internal resistance of the combination is calculated using the formula:

$r_{eq}=\frac{r_1\cdot r_2}{r_1+r_2}$

Substitute the values into the formula:

$r_{eq}=\frac{0.2\cdot 0.4}{0.2+0.4}=\frac{0.08}{0.6}=\frac{2}{15}\mathrm{\Omega }\approx 0.133\mathrm{\Omega }$

Hence, the equivalent internal resistance of the combination is 0.133 ohm.

Q3: Two wires P and Q are made of the same material. The wire Q has twice the diameter and half the length as that of wire P. If the resistance of wire P is R, the resistance of the wire Q will be:

Answer:

$\begin{array}{c}R=\frac{\rho l}{A}=\frac{\rho l}{\pi r^2}\\ \text{ Resistance of }Q=R\text{ ' }=\frac{\rho \left(\frac{l}{2}\right)}{\pi (2r{)}^2}\\ R^{\prime }=\frac{\rho l}{\pi r^2\cdot 4\times 2}=\frac{\rho l}{8\cdot \pi r_2}=\frac{R}{8}\end{array}$

NCERT Class 12 Notes Chapterwise

Subject Wise NCERT Exemplar Solutions

• NCERT Exemplar Class 12 Solutions
• NCERT Exemplar Class 12 Maths
• NCERT Exemplar Class 12 Physics
• NCERT Exemplar Class 12 Chemistry
• NCERT Exemplar Class 12 Biology

Subject Wise NCERT Solutions

• NCERT Solutions for Class 12 Mathematics
• NCERT Solutions for Class 12 Chemistry
• NCERT Solutions for Class 12 Physics
• NCERT Solutions for Class 12 Biology

NCERT Books and Syllabus