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.
Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.
View Solution
2.
Assertion (A) : All atoms have a net magnetic moment. Reason (R) : A current loop does not always behave as a magnetic dipole.
- Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
- Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
- Assertion (A) is true, but Reason (R) is false.
- Both Assertion (A) and Reason (R) are false.
View Solution
3.
Assertion (A) : The mass of a nucleus is less than the sum of the masses of the constituent nucleons. Reason (R) : Energy is absorbed when the nucleons are bound together to form a nucleus.
- Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
- Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
- Assertion (A) is true, but Reason (R) is false.
- Both Assertion (A) and Reason (R) are false.
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4.
If both the number of protons and the neutrons are conserved in each nuclear reaction, in what way is mass converted into energy (or vice versa) in a nuclear reaction? Explain.
View Solution
5.
A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of 60° in 0.2 s. The value of emf induced in the loop will be:
- 5 V
- 3.5 V
- 2.5 V
- Zero V
View Solution
6.
Write any two features of nuclear forces.
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NCERT Solutions For Class 12 Physics Chapter 3: Current Electricity

NCERT Solutions for Class 12 Physics Chapter 3

Chapter 3 Physics Class 12 Important Topics

CBSE CLASS XII Related Questions

1.
Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.

2.
Assertion (A) : All atoms have a net magnetic moment. Reason (R) : A current loop does not always behave as a magnetic dipole.

3.
Assertion (A) : The mass of a nucleus is less than the sum of the masses of the constituent nucleons. Reason (R) : Energy is absorbed when the nucleons are bound together to form a nucleus.

4.
If both the number of protons and the neutrons are conserved in each nuclear reaction, in what way is mass converted into energy (or vice versa) in a nuclear reaction? Explain.

5.
A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of 60° in 0.2 s. The value of emf induced in the loop will be:

6.
Write any two features of nuclear forces.

Similar Physics Concepts

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NCERT Solutions For Class 12 Physics Chapter 3: Current Electricity

NCERT Solutions for Class 12 Physics Chapter 3

Chapter 3 Physics Class 12 Important Topics

CBSE CLASS XII Related Questions

1.Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.

2.Assertion (A) : All atoms have a net magnetic moment. Reason (R) : A current loop does not always behave as a magnetic dipole.

3.Assertion (A) : The mass of a nucleus is less than the sum of the masses of the constituent nucleons. Reason (R) : Energy is absorbed when the nucleons are bound together to form a nucleus.

4. If both the number of protons and the neutrons are conserved in each nuclear reaction, in what way is mass converted into energy (or vice versa) in a nuclear reaction? Explain.

5.A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of 60° in 0.2 s. The value of emf induced in the loop will be:

6. Write any two features of nuclear forces.

Similar Physics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.

2.
Assertion (A) : All atoms have a net magnetic moment. Reason (R) : A current loop does not always behave as a magnetic dipole.

3.
Assertion (A) : The mass of a nucleus is less than the sum of the masses of the constituent nucleons. Reason (R) : Energy is absorbed when the nucleons are bound together to form a nucleus.

4.
If both the number of protons and the neutrons are conserved in each nuclear reaction, in what way is mass converted into energy (or vice versa) in a nuclear reaction? Explain.

5.
A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of 60° in 0.2 s. The value of emf induced in the loop will be:

6.
Write any two features of nuclear forces.