Electric Current: Important Questions

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Electric current refers to the flow of electric charges. The study of electric current is called Current Electricity. This is an important concept from the CBSE Class 12 Physics Syllabus. Electric current is established due to the movement of negatively charged particles called electrons. This article covers some of the important questions related to electric current.

The SI unit of electric current is Amperes.
It is defined as one coulomb of charge flowing through the conductor in one second.
Metallic materials are rich in electrons and hence they allow the flow of electric current easily through them.
Such materials are called conductors. Insulators do not conduct electricity due to the absence of electrons in their subatomic structure.

Important Questions on Electric Current

Ques. Define Ampere. (1 Mark)

Ans. One Ampere is defined as the flow of one coulomb of charge in one second. It is denoted by ‘A’. AMpere is the SI unit of electric current.

Ques. What are the current carriers in gases? (1 Mark)

Ans. Positive ions and electrons are the current carriers in gases.

Ques. Define electric current. (1 Mark)

Ans. Electric current is defined as the amount of electric charges flowing through any cross-sectional area of a conductor.

Ques. Define Insulators. (2 Marks)

Ans. Materials that do not allow the flow of electrons from one place to another in an electric circuit are termed insulators. Examples are wood, paper, clay, etc.

Ques. What is an Electric Charge? (2 Marks)

Ans. The subatomic particles of matter experience a force when placed in an electric or magnetic field. This force is called an electric charge. For example, electrons are negatively charged particles.

Ques. How many electrons per second flow through the cross-section of a conductor so that the conductor carries a current of 1 A? (3 Marks)

Ans. We know, Current, I = Charge (Q)/ Time (t)

But Q = ne

Where n is the number of electrons

⇒ I = ne/t

⇒ n = Ie/t = (1 x 1.6 x 10^-19)/1 = 6.25 x 10¹⁸ electrons per second

Ques. State Coulomb’s Law. (2 Marks)

Ans. Coulomb’s Law states that the magnitude of the electrostatic force of attraction or repulsion between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them.

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Ques. What are the types of electric Currents? (2 Marks)

Ans. Electric current is divided into two types:

Direct Current - current flows only in one direction. Eg. computers, satellites, etc
Alternating Current- current changes its direction.

Ques. What is the conventional flow of current? (2 Marks)

Ans. By convention, current flows from the positive to the negative terminal and this indicates the direction in which positive charges flow. This is the conventional direction of current flow.

Ques. What is the number of electrons passing through a wire per minute if the current flowing through it is 500 mA? (2 Marks)

Ans. Number of electrons, n = It/e

I = 500 x10^-3 A

t =1 sec

Substituting we get,

n= 500 x10^-3 x 1/ 1.625 x 10^-19

= 1.875 x 10²⁰ Ω

Ques. How does electric current flow? (2 Marks)

Ans. The energy to the electric circuit is provided by the batteries. These give energy to the negatively charged particles or electrons to freely move through the circuit from the negative terminal to the positive terminal.

Ques. What are the applications of the heating effects of electric current? (3 Marks)

Ans. The most common applications of the heating effects of electric current are:

Glowing of the incandescent bulb when the filament inside the bulb heats up.
Electric stoves and heaters work on the heating effect
Soldering iron and cartridge heaters
Electric fuses and thermistors.

Ques. What is an emf of a cell? (2 Marks)

Ans. The amount of energy supplied by the cell to allow the flow of unit positive charge once around the entire circuit is emf or the electromotive force. It can also be termed as the potential difference between the two terminals of the cell. Emf is hence not a force but a source of energy.

Ques. A copper wire of radius 1.0 mm carries a current of 10 A. Calculate the drift velocity of electrons. Metallic copper has one conduction electron per atom. The atomic mass of copper is 64 u and the density of copper is 8900 kg/m³. Given 1 u = 1.66 x 10^-27 kg (5 Marks)

Ans. Let n = number of atoms per unit volume

As copper has one conduction electron per atom, therefore

n x mass of one atom of copper (M) = Mass of unit volume of copper = Density (ρ)

⇒ nM = ρ

⇒ n = ρ/M

Given, density, ρ = 8900 kg/m³ and M = 64 u = 64 x 1.66 x 10^-27 kg

⇒ n = 8900/(64 x 1.66 x 10^-27) = 8.38 x 10²⁸ m^-3

We know, drift velocity v_d = I/neA = I/(ne x πr²)

⇒ v_d = 10/(8.38 x 10²⁸ x 1.6 x 10^-19 x 3.14 x 10^-6)

⇒ v_d = 2.37 x 10^-4 m/s

Check Out:

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Class 12 Maths Book Pdf	NCERT Solutions for Class 12 Maths	Class 12 English Book Pdf

CBSE CLASS XII Related Questions

1.
Figure shows variation of Coulomb force (F) acting between two point charges with \( \frac{1}{r^2} \), \( r \) being the separation between the two charges \( (q_1, q_2) \) and \( (q_2, q_3) \). If \( q_2 \) is positive and least in magnitude, then the magnitudes of \( q_1, q_2 \), and \( q_3 \) are such that:

\( q_2<q_1<q_3 \)
\( q_3<q_1<q_2 \)
\( q_1<q_2<q_3 \)
\( q_2<q_3<q_1 \)

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2.
A parallel plate capacitor has plate area \( A \) and plate separation \( d \). Half of the space between the plates is filled with a material of dielectric constant \( K \) in two ways as shown in the figure. Find the values of the capacitance of the capacitors in the two cases.

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3.
The electric field at a point in a region is given by \( \vec{E} = \alpha \frac{\hat{r}}{r^3} \), where \( \alpha \) is a constant and \( r \) is the distance of the point from the origin. The magnitude of potential of the point is:

\( \frac{\alpha}{r} \)
\( \frac{\alpha r^2}{2} \)
\( \frac{\alpha}{2r^2} \)
\( -\frac{\alpha}{r} \)

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4.
(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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5.
Two point charges \( q_1 = 16 \, \mu C \) and \( q_2 = 1 \, \mu C \) are placed at points \( \vec{r}_1 = (3 \, \text{m}) \hat{i}\) and \( \vec{r}_2 = (4 \, \text{m}) \hat{j} \). Find the net electric field \( \vec{E} \) at point \( \vec{r} = (3 \, \text{m}) \hat{i} + (4 \, \text{m}) \hat{j} \).

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6.
Answer the following giving reason:
(a) All the photoelectrons do not eject with the same kinetic energy when monochromatic light is incident on a metal surface.
(b) The saturation current in case (a) is different for different intensity.
(c) If one goes on increasing the wavelength of light incident on a metal sur face, keeping its intensity constant, emission of photoelectrons stops at a certain wavelength for this metal.

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Electric Current: Important Questions

Important Questions on Electric Current

CBSE CLASS XII Related Questions

2.
A parallel plate capacitor has plate area \( A \) and plate separation \( d \). Half of the space between the plates is filled with a material of dielectric constant \( K \) in two ways as shown in the figure. Find the values of the capacitance of the capacitors in the two cases.

3.
The electric field at a point in a region is given by \( \vec{E} = \alpha \frac{\hat{r}}{r^3} \), where \( \alpha \) is a constant and \( r \) is the distance of the point from the origin. The magnitude of potential of the point is:

5.
Two point charges \( q_1 = 16 \, \mu C \) and \( q_2 = 1 \, \mu C \) are placed at points \( \vec{r}_1 = (3 \, \text{m}) \hat{i}\) and \( \vec{r}_2 = (4 \, \text{m}) \hat{j} \). Find the net electric field \( \vec{E} \) at point \( \vec{r} = (3 \, \text{m}) \hat{i} + (4 \, \text{m}) \hat{j} \).

Similar Physics Concepts

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Electric Current: Important Questions

Important Questions on Electric Current

CBSE CLASS XII Related Questions

2.A parallel plate capacitor has plate area \( A \) and plate separation \( d \). Half of the space between the plates is filled with a material of dielectric constant \( K \) in two ways as shown in the figure. Find the values of the capacitance of the capacitors in the two cases.

3.The electric field at a point in a region is given by \( \vec{E} = \alpha \frac{\hat{r}}{r^3} \), where \( \alpha \) is a constant and \( r \) is the distance of the point from the origin. The magnitude of potential of the point is:

5.Two point charges \( q_1 = 16 \, \mu C \) and \( q_2 = 1 \, \mu C \) are placed at points \( \vec{r}_1 = (3 \, \text{m}) \hat{i}\) and \( \vec{r}_2 = (4 \, \text{m}) \hat{j} \). Find the net electric field \( \vec{E} \) at point \( \vec{r} = (3 \, \text{m}) \hat{i} + (4 \, \text{m}) \hat{j} \).

Similar Physics Concepts

Comments

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2.
A parallel plate capacitor has plate area \( A \) and plate separation \( d \). Half of the space between the plates is filled with a material of dielectric constant \( K \) in two ways as shown in the figure. Find the values of the capacitance of the capacitors in the two cases.

3.
The electric field at a point in a region is given by \( \vec{E} = \alpha \frac{\hat{r}}{r^3} \), where \( \alpha \) is a constant and \( r \) is the distance of the point from the origin. The magnitude of potential of the point is:

5.
Two point charges \( q_1 = 16 \, \mu C \) and \( q_2 = 1 \, \mu C \) are placed at points \( \vec{r}_1 = (3 \, \text{m}) \hat{i}\) and \( \vec{r}_2 = (4 \, \text{m}) \hat{j} \). Find the net electric field \( \vec{E} \) at point \( \vec{r} = (3 \, \text{m}) \hat{i} + (4 \, \text{m}) \hat{j} \).