Important Questions Class 12 Physics Chapter 3

Important Questions for CBSE Class 12 Physics Chapter 3 – Current Electricity

These Class 12 Physics Chapter 3 Important Questions will teach students about current electricity. After studying these important questions, students should understand the concept of electric charges and fields better. These Class 12 Physics Chapter 3 Important Questions will prepare students for their exams and help them solve many CBSE sample papers.

These important question notes contain important formulas and CBSE extra questions that can help students test their understanding. In addition, these notes are crafted as per the prescribed CBSE syllabus.

These important questions are as per the  NCERT books. After going through them, students will know all about electric charges and fields and will gain confidence to solve CBSE past years’ question papers.

CBSE Class 12 Physics Chapter 3 Important Questions

Study Important Questions for Class 12 Physics Chapter 3 – Current Electricity

In Class 12 Physics, chapter 3, which is on current electricity, is both crucial and fascinating. As a result, it is essential for the students to have important questions about current electricity. This will enable students to fully comprehend the chapter with clarity and precision. These Class 12 Physics Chapter 3 Important Questions are made by filtering the entire content and comparing it to the standards and patterns set by the CBSE board. By using the study guides students can succeed in this chapter and earn better grades. The students will also benefit from creating notes from the chapter based on all the relevant questions.

Here are some questions that may come up in your physics exams from this topic.

Short Answer Questions

Q.1. How would the drift velocity of the electron vary if the potential difference V that is applied across a conductor is raised to 2V?

Ans. The electron’s drift velocity doubles when the potential difference along a conductor doubles. It can be represented by the following equation.

Vd = eEm = eE1m

Q.2. The resistance is the same for two equal-length wires made of copper and manganin. What kind of wire is thicker?

Ans. As we know that R = plA       ∴ A = plR

Both wires’ R and l are the same, and p copper < p manganin.

∴ A copper < A manganin

The manganin wire will be thicker as opposed to the copper wire.

Q.3 As depicted in the image, a 5 V battery with a very low internal resistance is connected to a 200 V battery with a resistance of 39. Determine the current’s value.

Ans. The value of the current will be

i = 200-539 = 5A

Q.4. Name one requirement for the cell to draw its maximum current.

Ans. We know that

I = ER+r

Internal resistance must be zero in order to calculate the maximum current.

Q.5. The resistivity of copper is 1.710-8m, silver is 1.010-8m, and manganin is 4410-8. Which among these will be the best conductor?

Ans. When the cross-sectional area and length are held constant, the resistance is directly proportional to the specific resistance (resistivity).

R = plR

As a result, silver is the best conductor because of its low specific resistance.

Q.6. If you stretch a wire till it doubles in length, what is its new resistivity?

Ans. Since resistivity is solely dependent on the composition of the material, there won’t be any change in resistivity.

Q.7. Give the name of any substance with a low-temperature coefficient of resistance. Write one way this material can be used.

Ans. Nichrome is a material with a low-temperature coefficient of resistance. It is an alloy that is used to create typical resistance coils.

Q.8. Parallel connections are made between two identical cells, each with an emf of E and negligible internal resistance, by means of an external resistance R. How much current is passing via this resistance?

Ans. ∴ I = ER

EMF does not alter when cells are connected in parallel.

Q.9. A and B, two students were instructed to choose a 15 k resistor from a selection of carbon resistors. While B selected a resistor with black, green, and red bands, A selected one with brown, green, and orange bands. Who made the right resistor choice?

Ans. Student A has selected the right resistor of 15 k.

Short Answers 

Q.1. There are two bulbs, i.e., A, which is marked with 220V, 40W, and B, which is marked 220V, 60V. Determine which one has the higher resistance. 

Ans. We know that

R = V2P

When it comes to bulb A

R1 = (220)240 = 1210

When it comes to bulb B

R2 = (220)260 = 806

Hence, the resistance in bulb A is higher than the resistance in bulb B.

Q.2. Give an explanation of the concept of “drift velocity” for charge carriers in a conductor and describe how it relates to the current that is flowing through it.

Ans. The speed at which a free electron in a conductor drifts when an external electric field is introduced is known as the drift velocity.

∴Vd = eEm = 1neA

Here, refers to the average relaxation time.

n refers to the number of free electrons per unit volume in the conductor.

m refers to the mass of an electron.

E refers to the electric field.

Q.3. How does applying a potential difference across a conductor’s ends affect the free electrons’ random motion?

Ans. Free electrons move at random and are drawn to a spot with a higher potential.

Q.4. Give the potentiometer’s guiding principle.

Ans. The potential difference across any length of a wire with uniform cross-section and uniform composition while a constant current flows through it is directly proportional to its length, i.e.,

V1 l

Q.5. Explain the relationship between conductivity and temperature using the mathematical expression for the conductivity of a material. Explain how the temperature of conductivity of material caries from

(i) semiconductor

(ii) good conductors

Ans. Conductivity = ne2m

(i) Semiconductors: The conductivity of semiconductors rises with temperature. It results from an increase in V. It outweighs the impact of the decline in “x”.

(ii) Good conductors: The conductivity of good conductors decreases as temperature rises. It is brought on by a decline in the value of downtime. An increasing value of V has little impact.

Q.6. Define a galvanometer’s current sensitivity and voltage sensitivity. Explain why the galvanometer’s voltage sensitivity may not always rise as its current sensitivity does.

Ans. The deflection that the galvanometer experiences when a unit current is supplied via its coil is referred to as current sensitivity.

IS = 1 = nBAk   radian/ampere or division A-1.

Here, n is the number of turns in the galvanometer.

k is the torsional constant or the restoring couple per unit twist.

Voltage sensitivity is the deflection that the galvanometer experiences when a unit voltage is placed across its coil.

VS = V = nBRkR radian/volt or div. V-1

Increased current sensitivity may not always result in increased voltage sensitivity since VS =IsR. The resistance applied may have an impact on it.

Q.7. A wire with a 15 resistance is stretched so that its length can be doubled. Then, it is divided into two equal pieces. The 3.0-volt battery is then used to connect these components in parallel. Find the amount of current drawn from the battery.

Ans. R is given as 15 .

When it stretched, the resistance becomes R1 = 60 because the volume is constant and R l2.

The two separated parts will have 30 resistance each because they are parallelly connected.

Req = 302 =  15

That current that will be drawn from the supply will be

I = VReq,

i.e., I = 315 = 15 = 0.2 A

Q.8. A 20 resistance wire is slowly stretched until it has doubled in length. Then, it is divided into two equal pieces. Then, these components are coupled across a 4.0-volt battery in parallel. Calculate the battery’s current usage.

Ans. The resistance of the wire will increase four times the original resistance when stretched, i.e., 80 as the volume is constant and R l2.

So the separated parts will have 40 resistance each.

The 20 will be the equivalent resistance when they are connected parallel.

∴ Current drawn = VReq = 420 = 15 =  0.2 A

Q.9. A cell’s emf is consistently higher than its terminal voltage. Explain why.

Ans. A cell’s internal resistance causes a decrease in potential when current flows through it. We refer to this as the “lost volt.” As a result, the terminal voltage is lower than the cell’s emf.

Q.10. An automobile’s storage battery has an emf of 12 volts. What is the highest current that a battery can support if its internal resistance is 0.4?

Ans. The information given is as follows.

Emf of the battery is 12 V.

The battery’s internal resistance is 0.4.

We know that

V = E-iR

In order to get the maximum current from the battery,

E-iR = 0

E = iR

i = ER = 120.4

i =  30 A

Hence, the maximum current that can be drawn from the battery will be 30 A.

Long answer question

Q.1. What do you mean by drift velocity? How can you calculate the electrons’ drift velocity in a good conductor in terms of their relaxation time?

Ans. The average speed at which free electrons drift away from an electric field is referred to as the drift velocity.

Let m represent the electron’s mass and e represent its charge.

The acceleration that the electron acquires when electric field E is applied

a = eEm

From the first motion equation, i.e., v = u + at

The average initial velocity will be

u = OV = vd

The relaxation time, t =

vD = a

vD = eEm

Here, e is the charge present on the electron, E is the intensity of the electric field intensity, represents the relaxation time, and m represent the mass of the electron.

Hence, the expression vD = eEm can be used to express the drift velocity of the electron passing through a good conductor regarding the relaxation time of electrons.

Q.2. The balance or null point of the potentiometer circuit is located at X. List the reasons behind the adjustment in the balancing point when

(i) all other parameters remain the same as resistance R is increased.

(ii) S is increased while maintaining the same resistance R.

(iii) another cell that has a lower emf than cell Q replaces cell P.

Ans. (i) The current flowing through potentiometer wire AB will drop if the resistance R is raised. As a result, the potential difference between A and B will be smaller, and the equilibrium point will move toward B.

(ii) The battery’s terminal potential difference will decrease as resistance S is raised. As a result, the balancing point will be found at a shorter length, shifting it towards A.

(iii) The potential difference across AB will be less than that of emf Q if cell P is replaced by a cell whose emf is lower than that of cell Q.

As a result, the balance point won’t be reached.

Q.3. In terms of drift velocity, express the current in a conductor having a cross-sectional area denoted by A.

Ans. The drift velocity means the speed at which free electrons move toward the positive terminal when influenced by an external electric field. The electron’s drift velocity is on the order of 10-5 metres per second. Current can be expressed in terms of drift velocity as follows.

I = Anevd

Consider a conductor with a uniform cross-section area denoted with A and length dented with I.

∴  Volume of the conductor = Al

Total free electrons in the conductor, where n is the number of conductors, will be Aln.

If e represents the charge on each electron, then q = Alne represents the total charge on all of the free electrons in the conductor.

The potential difference V conductor’s conductor’s electric field is given by,

E = Vl

I = Anevd

The free electrons in the wire will start to drift towards the battery’s positive terminal due to this field, with a speed of vd.

∴ The time that will be taken by the free electrons to cross the conductor will be

t = lvd

Therefore, current I =  lvdqt = Almelvd I = Anevd

Since A, n, and e are constants

Hence, I ∝ vd

Hence, the current flowing and the drift velocity will be directly proportional.

CBSE Class 12 Physics Chapter 3 Important Questions

Class 12 Physics Chapter 3 Important Questions – What is Ohm’s Law?

Ohm’s Law is an important law that students should know. The link between electric current and the potential difference is established by this law. The applied voltage has a directly proportional relationship with the current that flows through the conductor. Additionally, it stipulated that the temperature and all other environmental factors must remain constant. This is crucial because certain components experience an increase in temperature when electricity is routed through them. The filament of a light bulb is a well-known illustration of this violation of Ohm’s Law. Georg Simon Ohm, a German physicist, validated this law; as a result, the law bears his name.

The equation of Ohm’s law is

V = IR


V is the voltage across the conductor.

I is the current that is passing through the conductor.

R is the resistance that is being offered by the conductor towards the flow of current.

State the Factors Affecting Resistivity in Class 12 Physics Chapter 3 Important Questions.

Certain materials allow an object to pass through an electric charge quite effortlessly. The amount of restriction placed on the flow of electric charge within the circuit is determined by the electric resistance. These are a few of the elements that have an impact on electric resistance. The conductor’s electrical resistance is determined by the following variables.

  • The conducting device’s length
  • The conducting device’s cross-sectional area
  • The material of which the conducting device is made of
  • The conducting material’s temperature

Here, the electrical resistance is inversely proportional to the cross-sectional area of the conductor and directly proportional to the conducting device’s length. This relation can be represented by the following equation.

R = pLA

Here, P is the material’s resistivity which is usually measured in ohm meter, m.

Explain Kirchhoff’s Law in Context to Class 12 Physics Chapter 3 Important Questions.

Kirchhoff’s law is a key concept in Class 12 about current electricity. German physicist Gustav Kirchhoff created a few principles in 1845 that govern the exchange of energy and current within an electrical circuit. Kirchhoff’s current and voltage laws are the names given to both of these laws. Let’s examine each of these two laws in greater detail.

The first law of Kirchhoff states that no charge is lost and that the total current entering the node or junction equals the charge exiting the node or junction. According to this, every current entering and leaving the node has an algebraic sum of zero.

The second law of Kirchhoff states that for every closed network, the sum of all voltage drops is equal to the voltage around the loop in the same loop and also equals zero. The conservation of energy dictates that the algebraic total of each voltage in the loop must be zero.

CBSE Class 12 Physics Chapter 3 Important Questions – What are the Two Types of Resistance?

Resistors come in essentially two different categories and are connected either in series or in parallel. The difference between each resistor, when used in parallel, remains constant. Additionally, the applied difference and this difference are equal. When there is a change, the same amount of current flows through each resistor when the resistors are connected in series.

FAQs (Frequently Asked Questions)

1. In the potentiometer experiment, the driver cell's emf should be higher than the cell's expected emf. Why?

The potentiometer wire won’t reach a null point if the driver cell’s emf is lower.

2. An internal resistance R and an emf cell are connected by a resistance R. The potential difference between the cell's terminals is now measured as V using a potentiometer. In terms of V and R, create the expression for the letter "r".

r = V-1R

3. How may the sensitivity of a given four-wire potentiometer be increased?

The sensitivity can be increased if the primary circuit’s potentiometer wire is connected in series with resistance to lessen the potential drop across the wire.

4. Why aren't copper wires used to make potentiometers?

Copper is not recommended due to its high-temperature coefficient of resistance.

5. When does a cell's terminal voltage (i) exceed its emf (ii) become less than its emf?

(i) The terminal potential difference exceeds emf when the cell is being charged; V = E+Ir.

(ii) The terminal potential is lower than emf when the cell is discharged; V = E-Ir.