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.
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
2.
Suppose a pure Si crystal has \( 5 \times 10^{28} \) atoms per \( \text{m}^3 \). It is doped with \( 5 \times 10^{22} \) atoms per \( \text{m}^3 \) of Arsenic. Calculate majority and minority carrier concentration in the doped silicon. (Given: \( n_i = 1.5 \times 10^{16} \, \text{m}^{-3} \))
View Solution
3.
Write the expression for the magnetic field due to a current element in vector form. Consider a 1 cm segment of a wire, centered at the origin, carrying a current of 10 A in positive x-direction. Calculate the magnetic field \( \mathbf{B} \) at a point \( (1 \, \text{m}, 1 \, \text{m}, 0) \).
View Solution
4.
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
5.
Two parallel plate capacitors X and Y are connected in series to a 6 V battery. They have the same plate area and same plate separation but capacitor X has air between its plates, whereas capacitor Y contains a material of dielectric constant 4. Calculate the capacitances of X and Y, if the equivalent capacitance of the combination of X and Y is \( 4 \, \mu\text{F} \). Calculate the potential difference across the plates of X and Y.
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6.
Draw the number of scattered particles versus the scattering angle graph for scattering of alpha particles by a thin foil. Write two important conclusions that can be drawn from this plot.
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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.
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.

2.
Suppose a pure Si crystal has \( 5 \times 10^{28} \) atoms per \( \text{m}^3 \). It is doped with \( 5 \times 10^{22} \) atoms per \( \text{m}^3 \) of Arsenic. Calculate majority and minority carrier concentration in the doped silicon. (Given: \( n_i = 1.5 \times 10^{16} \, \text{m}^{-3} \))

3.
Write the expression for the magnetic field due to a current element in vector form. Consider a 1 cm segment of a wire, centered at the origin, carrying a current of 10 A in positive x-direction. Calculate the magnetic field \( \mathbf{B} \) at a point \( (1 \, \text{m}, 1 \, \text{m}, 0) \).

4.
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.

6.
Draw the number of scattered particles versus the scattering angle graph for scattering of alpha particles by a thin foil. Write two important conclusions that can be drawn from this plot.

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. 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.

2.Suppose a pure Si crystal has \( 5 \times 10^{28} \) atoms per \( \text{m}^3 \). It is doped with \( 5 \times 10^{22} \) atoms per \( \text{m}^3 \) of Arsenic. Calculate majority and minority carrier concentration in the doped silicon. (Given: \( n_i = 1.5 \times 10^{16} \, \text{m}^{-3} \))

3.Write the expression for the magnetic field due to a current element in vector form. Consider a 1 cm segment of a wire, centered at the origin, carrying a current of 10 A in positive x-direction. Calculate the magnetic field \( \mathbf{B} \) at a point \( (1 \, \text{m}, 1 \, \text{m}, 0) \).

4.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.

6. Draw the number of scattered particles versus the scattering angle graph for scattering of alpha particles by a thin foil. Write two important conclusions that can be drawn from this plot.

Similar Physics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
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.

2.
Suppose a pure Si crystal has \( 5 \times 10^{28} \) atoms per \( \text{m}^3 \). It is doped with \( 5 \times 10^{22} \) atoms per \( \text{m}^3 \) of Arsenic. Calculate majority and minority carrier concentration in the doped silicon. (Given: \( n_i = 1.5 \times 10^{16} \, \text{m}^{-3} \))

3.
Write the expression for the magnetic field due to a current element in vector form. Consider a 1 cm segment of a wire, centered at the origin, carrying a current of 10 A in positive x-direction. Calculate the magnetic field \( \mathbf{B} \) at a point \( (1 \, \text{m}, 1 \, \text{m}, 0) \).

4.
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.

6.
Draw the number of scattered particles versus the scattering angle graph for scattering of alpha particles by a thin foil. Write two important conclusions that can be drawn from this plot.