Have you ever observed how your laptop turns on as soon as you plug in the charger, requiring electricity to flow through wires in order to function? NCERT Class 12 Physics Chapter 3 Notes Current Electricity is perfectly represented by that everyday phenomenon. This chapter examines how current is affected by resistance, how charges flow through conductors, and how circuits serve as the foundation for current electronics.
The NCERT Class 12 Physics Chapter 3 Notes Current Electricity are a crucial part of the CBSE curriculum. Current electricity is also important for preparing students for competitive tests like JEE and NEET. Key concepts like Ohm's Law, Kirchhoff's Laws, resistivity, conductivity, internal cell resistance, Wheatstone bridge, and meter bridge are covered in this chapter. A solid understanding of these concepts improves theoretical comprehension and helps in developing numerical problem-solving abilities. Careers360's carefully curated NCERT Class 12 Physics Chapter 3 Notes Current Electricity offers concise definitions, labelled diagrams that help visualise concepts, well-structured summaries, and crucial formulas for fast revision to streamline your preparation. Furthermore, practice questions and solved examples improve exam readiness by enabling students to confidently answer board and entrance exam questions. Mastering these NCERT Class 12 Physics Chapter 3 Notes is the key to excelling in exams, strengthening fundamentals in electronics, and gaining a deeper appreciation for how electricity powers everyday life.
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NCERT Class 12 Physics Chapter 3 Notes are a concise and clear summary of the Current Electricity chapter, which includes important definitions, key concepts, formulas, solved examples, and diagrams to make revision easier. Students are able to download the PDF version for easy study, last-minute preparation, and perform better in both Board and Competitive exams like JEE & NEET.
NCERT Class 12 Physics Chapter 3 Notes discuss the chapter in a simple and organised way. NCERT Class 12 Physics Chapter 3 Notes include many key concepts like Ohm’s Law, resistivity, Kirchhoff’s laws, electrical circuits, etc, in order to understand these topics in a simple way. These notes are meant to facilitate faster revision by students and enhance problem-solving skills for examinations.
$
I=\frac{q}{t}
$
Where, i is the current, q is the charge and t is the time.
$\bar{j}=\frac{\Delta i}{\Delta A}$
$
\begin{aligned}
J_{a v} & =\frac{d i}{d A \cos \theta} \\
d i & =J d A \cos \theta=\vec{J} \cdot d \vec{A}
\end{aligned}
$
$\vec{J}=\sigma \vec{E}$
$\vec{J}=\frac{\vec{E}}{\rho}$
Where, σ is conductivity and ρ is resistivity or specific resistance of the substance
$
\mu=\frac{v_d}{E}
$
Where, μ is mobility and vd is drift velocity
$
\begin{gathered}
V \propto I \\
V=I R
\end{gathered}
$
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
,
$\rho=\frac{m}{n e^2 \tau}$
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.
ρ=ρo(1+α(T−To))
ρ: Resistivity at temperature T
ρo: Resistivity at temperature To
Temperature Coefficient of Resistance
$
R_T=R_{0}\left[1+\alpha\left[T-T_0\right]\right]
$
$R_T$ - Resistance at temperature $T$
$R_0$ - Resistance at temperature $T_o$
$\alpha$ - temperature coefficient of resistance
$
\alpha=\frac{R_T-R_o}{R_o\left(T-T_o\right)}
$
Where the value of $\alpha$ is different at different temperatures
The carbon resistance typically consists of four coloured ring bands labelled as A, B, C, and D.
Tolerance of Gold is ±5%
Tolerance of Silver is ±10%
Tolerance if no colour ±20%
Series Grouping of resistance
$
R_{e q}=R_1+R_2+R_3+\cdots+R_n
$
$R_{e q}{ } = {\text {Equivalent Resistance }}$
For n-identical resistance: $R_{e q}=n R$
$
V^{\prime}=\frac{V}{n}
$
Parallel Grouping of Resistance:
$
\frac{1}{R_{e q}}=\frac{1}{R_1}+\frac{1}{R_2}+\cdots+\frac{1}{R_n}
$
If two resistances are in Parallel:
$
R_{e q}=\frac{R_1 R_2}{R_1+R_2}
$
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 $H=i^2 R t$
The power developed $=\frac{\text { energy }}{\text { time }}=i^2 R=i R=\frac{V^2}{R} \quad$ (from Ohm's law)
Unit of heat is the joule (J)
Unit of power is Watt (W)
r∝d
r∝A
r∝c
r∝1/temp
Cell supplies a constant current in the circuit.
$
i=\frac{E}{R+r}
$
Where, R is the external resistance and r is the internal resistance
$
P_{\max }=\frac{E^2}{4 r}
$
In series grouping anode of one cell is connected to the cathode of the other cells
$n=$ identical cells which are connected in series, then
- Equivalent e.m.f of combination is $E_{e q}=n E$
- Equivalent internal resistance $r_{e q}=n r$
- Main current / current from each cell $i=\frac{n E}{R+n r}$
- Power dissipated in the external circuit is $\left(\frac{n E}{R+n r}\right)^2 \cdot R$
- Conditions for Maximum Power is $R=n r$
- $P_{\text {max }}=n\left(\frac{E^2}{4 r}\right)$ when $n r<R$
In parallel grouping, all anodes are connected to one point and all cathodes together at other points
For n cells connected in parallel
Equivalent e.m.f $E_{e q}=E$
Equivalent internal resistance $R_{e q}=\frac{r}{n}$
The main current is
$
i=\frac{E}{R+\frac{r}{n}}
$
The potential difference across the external resistance
$
V=i R
$
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
$\begin{aligned} & R=\frac{r}{n} \\ & P_{\max }=n\left(\frac{E^2}{4 r}\right) {\text {when } r>>n R}\end{aligned}$
∑i=0
i1+i3=i2+i4
This law is simply based on the conservation of charge.
∑V=0
In closed-loop
$-i_1 R_1+i_2 R_2-E_1-i_3 R_3+E_2+E_3-i_4 R_4=0$
It is an arrangement of four resistances that can be used to measure one of them in terms of rest
$
\begin{aligned}
& \frac{P}{Q}=\frac{R}{S} \\
& V_B=V_D
\end{aligned}
$
(Balanced condition)
No current will flow through the galvanometer
unbalanced condition: $V_B>V_D$
$
\left(V_A-V_B\right)<\left(V_A-V_D\right)
$
Current will flow from $A$ to $B$
$\frac{P}{Q}=\frac{R}{S} \Rightarrow S=\frac{(100-l)}{l} R$
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 the 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 $=(n-2) E-2 E$
$
\begin{aligned}
& =(n-2-2) E \\
& =(n-4) E
\end{aligned}
$
$
\begin{aligned}
& \text { Req }=n r \\
& I=\frac{E_{n e t}}{R_{e q}}=\frac{(n-4) E}{nr}
\end{aligned}
$
potential difference across cell $X=E+I r$
$
\begin{aligned}
& = E+ \frac{(n-4) E}{n r} \times r \\
& =\frac{nE+(n-4) E}{n} \\
& =\frac{n E+(n-4) E}{n} \\
& =\frac{E(n+n+4)}{n}=\frac{2 E(n+2)}{n}
\end{aligned}
$
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\Omega$.
Battery $1: {E}_1=3 \mathrm{~V}, r_1=0.2 \Omega$
Battery $2: {E}_2=6 \mathrm{~V}, r_2=0.4 \Omega$
The equivalent internal resistance of the combination is calculated using the formula:
$
r_{e q}=\frac{r_1 \cdot r_2}{r_1+r_2}
$
Substitute the values into the formula:
$
r_{e q}=\frac{0.2 \cdot 0.4}{0.2+0.4}=\frac{0.08}{0.6}=\frac{2}{15} \Omega \approx 0.133 \Omega
$
Hence, the equivalent internal resistance of the combination is 0.133 ohms.
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 wire Q will be:
Answer:
$
\begin{gathered}
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(2 r)^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{gathered}
$
Foundation of Electronics
Key for Competitive Exams
Easy to Revise
Better Numerical Problem Solving
Learning Aid
Directly Relevant to the Board Exam
NCERT Class 12 Notes chapterwise provide a structured way for students to revise all Physics concepts with clarity and ease. These notes cover every chapter in a concise format, including key definitions, formulas, diagrams, and solved examples. With chapterwise PDF links, students can access them quickly for board exam prep and competitive exams like JEE and NEET.
NCERT Class 12 Physics Chapter 3 Notes |
Frequently Asked Questions (FAQs)
Current Electricity carries significant weight in the Physics board exam, accounting for around 10% of the total marks. It covers fundamental concepts that form the base for other chapters and practicals.
Yes, this chapter is very important for entrance exams such as JEE Main and NEET. Questions based on Ohm’s Law, circuits, and resistivity often appear in these exams.
Start with simple problems and gradually move to complex circuits. Practice daily, understand formulas, and use Kirchhoff’s and Ohm’s Laws effectively to solve questions step-by-step.
A Wheatstone Bridge is a network of four resistors used to precisely measure unknown resistance. It is widely used in sensor applications like strain gauges and temperature measurement.
In most conductors, resistance increases with an increase in temperature due to more frequent collisions between electrons and atoms. However, in semiconductors, resistance decreases with temperature.
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