NCERT Solutions For Class 12 Physics Chapter 3: Current Electricity

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NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity are given in this article. Current electricity is the electricity that powers our homes and electrical devices. Current electricity is named for the way electrons move. They “flow” in one direction- like a river current. The study of electrons in motion like this is called Electrodynamics.

The chapter along with the unit Electrostatics has a weightage of 16 marks in CBSE Class 12 Physics exams. The NCERT Solutions for Class 12 Physics Chapter 3 covers concepts of electric current, Ohm’s law, emf, cells in series and parallel, Kirchhoff’s Rules, etc.

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NCERT Solutions for Class 12 Physics Chapter 3

The NCERT Solutions for class 12 physics chapter 3: Current Electricity is as given below.

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Chapter 3 Physics Class 12 Important Topics

Current electricity is the flow of electrons from one section of the circuit to the another.

There are two types of Current Electricity –

Direct Current is the current electricity whose direction stays the same. It is the constant flow of electrons from a high electron density region to a region of low electron density.
Alternating Current is the current electricity that keeps changing the direction of the charge flow.

When 2 bodies at different potentials are linked with a wire, the free electrons move from Point 1 to Point 2, until both objects reach the same potential. The current stops flowing after that.

Electromotive Force: Electromotive force is the electric potential that is either produced by an electrochemical cell or produced by changing the magnetic field.
Voltage: Voltage is the electric potential difference between any two points.

Ohm’s Law states that the electric current flowing through a conductor is directly proportional to the potential difference (V) applied across its ends.

It can be represented as

V = IR

Electrical Resistance is the obstruction that is offered by a conductor in the path of the flow of current.

The formula of electrical resistance is R = V/l.

Electrical resistance of a conductor R = ρl/A

where l = length of the conductor,

A = cross-section area, and

ρ = resistivity of the material of a conductor.

The current electricity can be generated through various methods.
- Both alternating and direct current can be generated by moving a metal wire through a magnetic field.
- Direct Current can be generated by a battery through chemical reactions.

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CBSE CLASS XII Related Questions

1.
A charged particle $+q$ in an electric field $\vec{E}$ experiences a force in the direction of the electric field. As a result, its kinetic energy changes. Similarly, the charged particle also experiences a force when it moves in a magnetic field $\vec{B}$. But this magnetic force is perpendicular to both velocity $\vec{v}$ of the charged particle and the magnetic field $\vec{B}$, so it cannot change the kinetic energy of the charged particle. Consider two charged particles 1 and 2 of masses $m$ and $\frac{m}{2}$ having charges $-q$ and $+2q$ respectively. They are accelerated from rest through the same potential difference $V$ and acquire kinetic energy $K_1$ and $K_2$. Then they enter in a region of uniform magnetic field $\vec{B}$ perpendicular to their velocities.
2.
Consider the nuclear reaction $ X \to Y + Z $. Let $ M_x $, $ M_y $, and $ M_z $ be the masses of the three nuclei X, Y, and Z respectively. Then which of the following relations hold true?
- $ (M_x - M_z)<M_y $
- $ (M_x - M_y)<M_z $
- $ M_x>(M_y + M_z) $
- $ M_x<(M_y + M_z) $
View Solution
3.
Consider a cylindrical conductor of length $ l $ and area of cross-section $ A $. Current $ I $ is maintained in the conductor and electrons drift with velocity $ \vec{v}_d \, (|\vec{v}_d| = \frac{eE}{m} \tau) $, where symbols have their usual meanings. Show that the conductivity of the material of the conductor is given by \[ \sigma = \frac{n e^2 \tau}{m}. \]
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4.
The resistance of a metal wire at $ 20^\circ \text{C} $ is $ 1.05 \, \Omega $ and at $ 100^\circ \text{C} $ is $ 1.38 \, \Omega $. Determine the temperature coefficient of resistivity of this metal.
View Solution
5.
Derive an expression for the capacitance of a parallel plate capacitor of plate area A and plate separation d with air present between the plates.
View Solution
6.
An astronomical telescope consists of two converging lenses. One of them of large aperture and large focal length is called objective lens and the other one, of smaller focal length and smaller aperture is called the eyepiece. It is used to see distant objects which are not seen clearly with naked eyes. The image formed by the objective lens acts as an object for the eyepiece and the final image produced by the eyepiece is magnified.