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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Ohm’s Law	Combination of Resistors	Cells, emf, and internal resistance
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CBSE CLASS XII Related Questions

1.
Write the mathematical forms of three postulates of Bohr’s theory of the hydrogen atom. Using them prove that, for an electron revolving in the $ n $-th orbit,
(a) the radius of the orbit is proportional to $ n^2 $, and
(b) the total energy of the atom is proportional to $ \frac{1}{n^2} $.
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2.
A vertically held bar magnet is dropped along the axis of a copper ring having a cut as shown in the diagram. The acceleration of the falling magnet is:
- zero
- less than $ g $
- $ g $
- greater than $ g $
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3.
A system of two conductors is placed in air and they have net charge of $ +80 \, \mu C $ and $ -80 \, \mu C $ which causes a potential difference of 16 V between them.
(1) Find the capacitance of the system.
(2) If the air between the capacitor is replaced by a dielectric medium of dielectric constant 3, what will be the potential difference between the two conductors?
(3) If the charges on two conductors are changed to +160µC and −160µC, will the capacitance of the system change? Give reason for your answer.
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4.
The resistance of a wire at 25°C is 10.0 $ \Omega $. When heated to 125°C, its resistance becomes 10.5 $ \Omega $. Find (i) the temperature coefficient of resistance of the wire, and (ii) the resistance of the wire at 425°C.
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5.
A rectangular glass slab ABCD (refractive index 1.5) is surrounded by a transparent liquid (refractive index 1.25) as shown in the figure. A ray of light is incident on face AB at an angle $i$ such that it is refracted out grazing the face AD. Find the value of angle $i$.
$A rectangular glass slab ABCD (refractive index 1.5)$
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6.
(a) Consider the so-called ‘D-T reaction’ (Deuterium-Tritium reaction).
In a thermonuclear fusion reactor, the following nuclear reaction occurs: \[ \ ^{2}_1 \text{H} + \ ^{3}_1 \text{H} \longrightarrow \ ^{4}_2 \text{He} + \ ^{1}_0 \text{n} + Q \] Find the amount of energy released in the reaction.
% Given data Given:
$ m\left(^{2}_1 \text{H}\right) = 2.014102 \, \text{u} $
$ m\left(^{3}_1 \text{H}\right) = 3.016049 \, \text{u} $
$ m\left(^{4}_2 \text{He}\right) = 4.002603 \, \text{u} $
$ m\left(^{1}_0 \text{n}\right) = 1.008665 \, \text{u} $
$ 1 \, \text{u} = 931 \, \text{MeV}/c^2 $
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