NCERT Exemplar Class 12 Physics Chapter 3 Current Electricity
NCERT Exemplar for Class 12 Physics Chapter 3
NCERT Exemplar for Class 12 Physics will aid in deepening the knowledge of students related to the topic of Current Electricity. This provided exemplar by SimplyAcad allows learners to cover up all the sections presented in the chapter 3 of the physics textbook. NCERT Exemplar for Class 12 Physics Chapter 3 contains 8 very short, 10 short and 3 long answer type questions. It will benefit the students as they can prepare their own study and revision notes from them. Students can easily access this NCERT exemplar for class 12 physics in this article below to perform incredibly well in their upcoming 12th board examinations. Apart from these there are several NCERT exemplar for class 12 science of all the chapters provided in a detailed manner.
Access the NCERT Exemplar Class 12 Physics Chapter 3 Current Electricity
Very Short Answers Ncert exemplar for class 12 Physics
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. 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.4. 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.5. 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.6. 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.7. 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.8. 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 Ncert exemplar for class 12 Physics
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 Ncert exemplar for class 12 Physics
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. Explain why the balancing point adjusts in the following scenarios:
(i) When the resistance R is increased, keeping all other parameters the same.
(ii) When resistance S is increased while maintaining the same resistance R.
(iii) When cell P is replaced with another cell that has a lower emf than cell Q.
Ans.
(i) Increasing the resistance R reduces the current flowing through the potentiometer wire AB. Consequently, the potential difference between A and B decreases, causing the balance point to shift towards B.
(ii) Raising the resistance S decreases the terminal potential difference of the battery. This results in the balancing point being found at a shorter length, shifting it towards A.
(iii) If cell P is replaced by a cell with an emf lower than that of cell Q, the potential difference across AB will be less than the emf of Q. As a result, the balance point will not 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.
NCERT Exemplar For Class 12 Science
Students must practise these additional questions for their own benefits, the ncert exemplar are curated by the best subject-matter experts to boost your knowledge on the presented topic. Students can easily access the ncert exemplar for class 12 science by visiting our website SimplyAcad and solve all the questions listed to secure maximum marks.
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