NCERT Solutions Class 10 Science Chapter 11 Electricity

NCERT Solutions Class 10 Science Chapter 11 Electricity is one of the important chapters. Understanding electricity prepares students for the CBSE board exams and lays a solid foundation for future competitive exams, such as engineering entrances.

NCERT Solution class 10 Science chapter 11 is one of the major resources for practising this chapter. These papers help you cover the entire syllabus and provide an in-depth analysis. These NCERT Solutions are crafted by experienced teachers that cover theoretical aspects and include numerical and practical exercises, making it easier for students to handle assignments and excel in competitive exams like Engineering. 

We want you to strengthen your grasp of this. So, in this article, we will provide essential insights and resources to help students master the concepts of electricity, a crucial topic in physics.

1.0Download Class 10 Science Chapter 11 Electricity NCERT Solutions: Free PDF

Be prepared to practice and perfect your concepts of chapter 11 with the PDF of NCERT Solution class 10 Science! ALLEN's expert educators have created these solutions to perfectly match the CBSE Class 10 Science syllabus.

2.0NCERT Solutions Class 10 Science Chapter 11 Electricity : Overview

NCERT Solutions Chapter 11, "Electricity," in Class 10 Science enhances students' understanding of the flow of electric current and its effects. The topics covered in the NCERT Solutions for Class 10 Science Chapter 11: Electricity include power, resistance, and current calculations, as well as numerical problems pertaining to series and parallel circuits. The NCERT solutions include a variety of questions, all designed to enhance students' understanding of the concepts related to electricity and its applications.

Subtopics of Class 10 Science Chapter 11 Electricity 

The concepts and fundamental laws you will learn in Chapter 11, Electricity, are mentioned below. You must know about these subtopics before getting in-depth details about the NCERT Solutions for Class 10 Science.

• Electric Current and Circuit
• Electric Potential and Potential Difference
• Circuit Diagram 
• Ohm’s Law
• Factors on Which the Resistance of a Conductor Depends 
• Resistance of a System Of Resistors
• Heating Effect of Electric Current
• Electric Power

3.0Electric Current and Circuit

It is the flow of charge through a conductor. The electric current I is referred to as the charge Q passing through a point in a circuit per unit time.

$I=\frac{Q}{t}$. Ampere is the SI Unit of current(A). Current is a scalar quantity.

Static Electricity: Concerns stationary charges.

Current Electricity: Involves moving charges.

Properties of Electric Charge

• Like charges repel; unlike charges attract.
• Charge is conserved and additive.
• Charge is quantized, meaning it exists in discrete amounts.

Electric Potential (V): The work done per unit positive charge to move it from infinity to a point in an electric field: $V=\frac{W}{Q_0}$

Ohm’s Law: Ohm’s Law states that the current through a conductor is directly proportional to the potential difference across it, provided the temperature remains constant: $\frac{V}{I}=R$

R=Resistance of the conductor which is a constant

4.0Factors Affecting Resistance

Resistance R of a conductor depends on:

• Material: Different materials have different resistivities.
• Length (l): Resistance is directly proportional to length.
• Cross-Sectional Area (A): Resistance is inversely proportional to the area.
• Temperature: Resistance typically increases with temperature.

Resistivity (ρ) is a material property and is measured in Ohm-Meters (Ω·m). It is independent of the physical dimensions of the conductor.

Also Read Electric Resistance and Ohm's Law

5.0Heating Effect of Electric Current

• If an electric circuit is purely resistive, it consists of resistors only connected to a battery, the energy of the source is dissipated in the form of heat.This effect is utilized in devices such as electric geysers, electric heaters, electric irons etc.

$H=I^{2}Rt\to \text{Joule Heating Effect of current}$

6.0Electric Power

• The rate at which electric energy is used or dissipated in an electric circuit .

P=VI

$P=I^{2}R=\frac{V^2}{R}$

• Electric Power measured in Watt(w)

$\text{Watt}=1\text{Volt}\times 1\text{Ampere}=1VA$

$\text{KWh}=1000\text{Watt}\times 3600s=3.6\times 10^6\text{Watt Second}=3.6\times 10^6\text{Joule}(J)$

7.0Sample NCERT Solutions For Class 10 Science Chapter 11 Electricity

1. What does an electric circuit mean?

Solution:

An electric circuit is a closed and continuous path consisting of many devices like resistors, electric bulbs, etc., through which an electric current can flow.

2. Define the SI unit of current.

Solution:

The SI unit of current is ampere (A). Current flowing through a conductor is said to be 1 ampere if 1 coulomb of charge flows through it in 1 second.

3. Calculate the number of electrons constituting one coulomb of charge.

Solution:

Number of electrons constituting 1 coulomb is given by,

n = Q / e

Where, Q = 1 C and e = charge of a single electron = 1.6 × 10⁻¹⁹ C

or n = 1 C / (1.6 × 10⁻¹⁹ C)

= 6.25 × 10¹⁸ ≈ 6 × 10¹⁸ electrons.

4. Name a device that helps to maintain a potential difference across a conductor.

Solution:

A battery can be used to maintain a potential difference across a conductor.

5. What is meant by saying that the potential difference between two points is 1 V?

Solution:

Potential difference between two points is 1 volt if 1 joule of work is done to carry a charge of 1 coulomb from one point to the other.

6. How much energy is given to each coulomb of charge passing through a 6 V battery?

Solution:

Work done, W = QV

where Q = 1 C; V = 6 V

W = 1 C × 6 V = 6 J

7. On what factors does the resistance of a conductor depend?

Solution:

The resistance (R) of a conductor depends upon:

(1) Its length (ℓ): R ∝ ℓ

(2) Its cross-sectional area (A): R ∝ 1/A

(3) Nature of material i.e., resistivity (ρ) of its material: R ∝ ρ

(4) Temperature: The resistance increases with an increase in temperature.

8. Does current flow more easily through a thick wire or a thin wire of the same material when connected to the same source? Why?

Solution:

The current flows more easily through a thick wire than through a thin wire. This is because the resistance R of a thick wire (large area of cross-section) is less than that of a thin wire (small area of cross-section) as R ∝ 1/A.

9. Let the resistance of an electrical component remain constant while the potential difference across the two ends of the component decrease to half its former value. What change will occur in the current through it?

Solution:

We know that I = V/R. When the potential difference becomes V/2, and the resistance remains constant, then the current becomes (V/2)/R = (1/2) * (V/R) = 1/2 of its former value.

10. Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?

Solution:

This is because:

(i) The resistivity of an alloy is generally higher than that of pure metals.

(ii) An alloy does not oxidize easily at high temperatures.

11. (a) Which among iron and mercury is a better conductor?

Given, ρ iron = 10.0 × 10⁻⁸ Ω m and ρ mercury = 94.0 × 10⁻⁸ Ω m.

(b) Which material is the best conductor?

Solution:

(a) Iron is a better conductor than mercury as the resistivity (ρ) for iron is less than that for mercury. A lower resistivity indicates better conductivity.

(b) Silver is the best conductor because its resistivity (ρ) is the least among common materials.

12. Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 ohm resistor, an 8 ohm resistor, and a 12 ohm resistor, and a plug key, all connected in series.

Solution:

13. Redraw the circuit of Q.12, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the voltage across the 12 -ohm resistor. What would be the readings in the ammeter and the voltmeter?

Solution

Since all the three resistances are in series, total resistance in the circuit,

R = 5 + 8 + 12 = 25 Ω

Current in the circuit,

I = V/R = (2 + 2 + 2) / 25 = 6 / 25 = 0.24 A,

Thus, the ammeter will read 0.24 A.

Potential difference across 12-ohm resistor V = I × R = 0.24 × 12 = 2.88 V

14. Judge the equivalent resistance when the following are connected in parallel (a) 1 Ω and 10⁶ Ω (b) 1 Ω, 10³ Ω and 10⁶ Ω.

Solution

(a) Approx. 1 Ω (slightly less than 1 Ω) as the other one (10⁶ Ω) is very large compared to 1 Ω. In a parallel combination of resistors, the equivalent resistance is lesser than the least resistance (in this case, 1 Ω).

(b) Again, the resistance is approx. 1 Ω (slightly less than 1 Ω).

15. An electric lamp of 100 Ω, a toaster of resistance 50 Ω and a water filter of resistance 500 Ω are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as in three appliances and what is the current through it?

Solution

Resistance of the electric lamp, R₁ = 100 Ω; resistance of toaster, R₂ = 50 Ω; resistance of water filter, R₃ = 500 Ω

Since R₁, R₂ and R₃ are connected in parallel, their equivalent resistance (Rp) is given by:

1/Rp = 1/R₁ + 1/R₂ + 1/R₃ = 1/100 + 1/50 + 1/500

= (5 + 10 + 1) / 500 = 16 / 500 = 4 / 125

Rp = 125 / 4 Ω

Current through the three appliances, i.e., I = V/Rp = 220 / (125 / 4) = 7.04 A

Since the electric iron draws the same current when connected to the same source (220 V), its resistance must be equal to Rp.

Thus, resistance of the electric iron = 125 / 4 = **31.25 Ω**

Current through the electric iron, I = 7.04 A

16. What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?

Solution

(a) In the case of devices in parallel, if one device gets damaged (or open), all others will work as usual as the whole circuit does not break. This is not so with devices connected in series because when one device fails, the circuit breaks and all devices stop working.

(b) Since the potential difference across all devices is the same in a parallel circuit, they will draw the required current according to their resistances. This is not so in a series circuit where the same current flows through all the devices, irrespective of their resistances.

17. How can three resistors of resistances 2 Ω, 3 Ω, and 6 Ω be connected to give a total resistance of (a) 4 Ω (b) 1 Ω?

Solution

(a) To get a total resistance of 4 Ω from resistors of resistances 2 Ω, 3 Ω and 6 Ω, the resistors are joined as shown below.

The resistors having resistances 3 Ω and 6 Ω are connected in parallel. This combination is connected in series with the resistor of resistance 2 Ω. Let us check it mathematically, the equivalent resistance of the 3 Ω and 6 Ω resistors is:

R₁ = (3 × 6) / (3 + 6) = (3 × 6) / 9 = 18 / 9 = 2 Ω

Now, R₁ (which is 2 Ω) and the 2 Ω resistor are in series, their equivalent resistance is:

(b) To get a resistance of 1 Ω from three given resistors of resistances 2 Ω, 3 Ω, and 6 Ω, they are joined as shown below 

They are all connected in parallel. Their equivalent resistance is given by:

1/R = 1/2 + 1/3 + 1/6 = (3 + 2 + 1)/6 = 6/6 = 1

Therefore, R = 1 Ω

18. What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistances 4 Ω, 8 Ω, 12 Ω, 24 Ω?

Solution

(a) The highest resistance is secured when all the resistors are connected in series. The equivalent resistance is given by:

Re = 4Ω + 8Ω + 12 Ω + 24 Ω = 48 Ω.

(b) The lowest resistance is secured when all four coils are connected in parallel. The equivalent resistance is given by:

1/Re = 1/4 + 1/8 + 1/12 + 1/24

= (6 + 3 + 2 + 1)/24 = 12/24 = 1/2

or Re = 2 Ω

19. Why does the cord of an electric heater not glow while the heating element does?

Solution

The cord of an electric heater is made of thick copper wire and has much lower resistance than the heating element. For the same current (I) flowing through the cord and the element, the heat produced (H = I²R) in the element is much more than that produced in the cord because of the higher resistance (R) of the element. As a result, the element becomes very hot and glows, whereas the cord does not become hot enough to glow.

20. Compute the heat generated while transferring 96000 coulombs of charge in one hour through a potential difference of 50 V.

Solution

Here, charge (Q) = 96000 C; time (t) = 1 hour; potential difference (V) = 50 V.

Heat produced (H) = VIt

We know that charge (Q) = Current (I) × time (t), so I = Q/t.

Therefore, H = V × (Q/t) × t = V × Q

H = 50 V × 96000 C = 4,800,000 J = 4.8 × 10⁶ J.

21. An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 s.

Solution

Here, resistance (R) = 20 Ω, current (I) = 5 A, time (t) = 30 s.

Heat produced (H) = I²Rt

H = (5 A)² × 20 Ω × 30 s

H = 25 × 20 × 30

H = 500 × 30

H = 15000 J = 1.5 × 10⁴ J.

22. What determines the rate at which energy is delivered by a current?

Solution:

Electric power determines the rate at which energy is delivered by a current.

23. An electric motor takes 5 A from a 220 V supply. Determine the power and energy consumed in 2 hr.

Solution:

Here, current, I = 5 A; potential difference, V = 220 V; time, t = 2 hr

= 2 × 60 × 60 = 7200 s.

Power, P = V × I = 220 × 5 = 1100 W

Energy consumed, H = V × I × t

= 220 × 5 × 7200

= 7.92 × 10⁶ J.

24. A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R', then the ratio R / R' is:

(1) 1/25

(2) 1/5

(3) 5

(4) 25

Solution:

Resistance of each one of the five parts = R/5

Resistance of five parts connected in parallel is given by:

1/R' = 1/(R/5) + 1/(R/5) + 1/(R/5) + 1/(R/5) + 1/(R/5)

or 1/R' = 5/R + 5/R + 5/R + 5/R + 5/R = 25/R

or R/R' = 25

Thus, (4) is the correct answer.

25. Which of the following terms does not represent electrical power in a circuit?

(1) I²R

(2) IR²

(3) VI

(4) V²/R

Solution:

Electrical power, P = VI = (IR)I = I²R

P = VI = V(V/R) = V²/R

Obviously, IR² does not represent electrical power in a circuit.

Thus, (2) is the correct answer.

26. An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be:

(1) 100 W

(2) 75 W

(3) 50 W

(4) 25 W

Solution:

Resistance of the electric bulb,

R = V²/P (P = V²/R) or R = (220)²/100 = 484 Ω

Power consumed by the bulb when it is operated at 110 V is given by:

P' = V'²/R = (110)²/484 = (110 × 110)/484 = 25 W

Thus, (4) is the correct answer.

27. Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then in parallel in an electric circuit. The ratio of the heat produced in series and parallel combinations would be:

(1) 1: 2

(2) 2: 1

(3) 1: 4

(4) 4: 1

Solution:

Since both the wires are made of the same material and have equal lengths and equal diameters, they have the same resistance. Let it be R.

When connected in series, their equivalent resistance is given by:

Rs = R + R = 2R

When connected in parallel, their equivalent resistance is given by:

1/Rp = 1/R + 1/R = 2/R or Rp = R/2

Further, electrical power is given by:

P = V²/R

Power (or heat produced) in series,

Ps = V²/Rs

Power (or heat produced) in parallel,

Pp = V²/Rp

Thus, Ps/Pp = (V²/Rs) / (V²/Rp) = Rp/Rs = (R/2) / (2R) = 1/4

or Ps : Pp = 1 : 4

Thus, (3) is the correct answer.

28. How is a voltmeter connected in the circuit to measure the potential difference between two points?

Solution:

A voltmeter is always connected in parallel across the points between which the potential difference is to be determined.

29. A copper wire has a diameter of 0.5 mm and a resistivity of 1.6 × 10⁻⁶ ohm cm. How much of this wire would be required to make a 10 ohm coil? How much does the resistance change if the diameter is doubled?

Solution:

We are given that, diameter of the wire,

D = 0.5 mm = 0.5 × 10⁻³ m

Radius r = D/2 = (0.5 × 10⁻³)/2 = 2.5 × 10⁻⁴ m

Resistivity of copper, ρ = 1.6 × 10⁻⁶ ohm cm

= 1.6 × 10⁻⁸ ohm m

Required resistance, R = 10 ohm

As, R = ρl/A, l = RA/ρ = R(πr²)/ρ [A = πr²]

or l = (3.14 × 10 × (2.5 × 10⁻⁴)²) / (1.6 × 10⁻⁸) = 122.65 m

Since,

R = ρl/A = ρl/(πr²) = ρl/(π(D/2)²) = 4ρl/(πD²), R ∝ 1/D²

When D is doubled, R becomes 1/4 times.

30. The values of current, I, flowing in a given resistor for the corresponding values of potential difference, V, across the resistor are given below:

I (ampere) : 0.5   1.0   2.0    3.0    4.0

V (volt)   : 1.6   3.4   6.7   10.2   13.2

Plot a graph between V and I and calculate the resistance of the resistor.

Solution:

Here, the I-V graph is almost straight. The resistance of the resistor can be calculated as given below:

R = (V₂ - V₁) / (I₂ - I₁) = (13.2 - 1.6) / (4 - 0.5) = 11.6 / 3.5 = 3.3 Ω

31. When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.

Solution:

Here, V = 12 V, I = 2.5 mA = 2.5 × 10⁻³ A

Resistance of the resistor,

R = V / I = 12 / (2.5 × 10⁻³) = 4800 Ω = 4.8 kΩ

32. A battery of 9 V is connected with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω in series. How much current would flow through the 12 Ω resistor?

Solution:

Since all the resistors are in series, the equivalent resistance is:

R<0xE2><0x82><0x98> = 0.2 +

The current flowing through the circuit is:

I = V / R<0xE2><0x82><0x98> = 9 / 13.4 = 0.67 A

In a series circuit, the same current (I) flows through all the resistors. Thus, the current flowing through the 12 Ω resistor is 0.67 A.

33. How many 176 Ω resistors (in parallel) are required to carry 5 A in a 220 V line?

Solution:

Here, I = 5 A, V = 220 V.

The resistance required in the circuit is:

R = V / I = 220 V / 5 A = 44 Ω

The resistance of each individual resistor is, r = 176 Ω.

If n resistors, each of resistance r, are connected in parallel to get the required resistance R, then:

R = r / n

or

44 = 176 / n

or

n = 176 / 44 = 4

34. Show how you would connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of:

(i) 9 Ω

(ii) 2 Ω

Solution:

(i) In order to get a resistance of 9 Ω from three resistors, we connect two 6 Ω resistors in parallel and this parallel combination in series with the third 6 Ω resistor.

Two 6 Ω resistors are in parallel, their equivalent resistance is:

R₁ = (6 × 6) / (6 + 6) = (6 × 6) / 12 = 3 Ω

Now, R₁ and the third 6 Ω resistor are in series, their equivalent resistance is:

Re = R₁ + 6 = 3 + 6 = **9 Ω**

(ii) In order to get a resistance of 2 Ω, we connect all three resistors in parallel as shown in the figure. Their equivalent resistance is:

Rp = 6 Ω / 3 = 2 Ω.

34. Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line, if the maximum allowable current is 5 A ?

Solution:

Resistance of each bulb,

R = V²/P = (220)² / 10 = 4840 Ω

Total resistance in the circuit, Re = V / I = 220 / 5 = 44 Ω

Let n be the number of bulbs (each of resistance R) to be connected in parallel to obtain a resistance Re.

Clearly, Re = R / n or n = R / Re = 4840 / 44 = **110**

35. A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 Ω resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?

Solution:

Here, potential difference, V = 220 V

Resistance of each coil, R = 24 Ω

(i) When each of the coils A or B is connected separately, current through each coil, i.e.,

I = V / R = 220 / 24 = 9.2 A

(ii) When coils A and B are connected in series, equivalent resistance in the circuit, Rs = R + R = 2R = 48 Ω

Current through the series combination, i.e.,

Is = V / Rs = 220 / 48 = 4.6 A

(iii) When the coils A and B are connected in parallel, equivalent resistance in the circuit, Rp = R / 2 = 24 Ω / 2 = 12 Ω

Current through the parallel combination, i.e.,

Ip = V / Rp = 220 / 12 = 18.3 A

36. Compare the power used in the 2 Ω resistor in each of the following circuits:

(i) a 6 V battery in series with 1 Ω and 2 Ω resistors, and

(ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.

Solution:

(i) Since a 6 V battery is in series with 1 Ω and 2 Ω resistors, current in the circuit, I = 6 / (1 + 2) = 6 / 3 = 2 A

Power used in 2 Ω resistor,

P1 = I²R = (2)² × 2 = 8 W

(ii) Since a 4 V battery is in parallel with 12 Ω and 2 Ω resistors, potential difference across 2 Ω resistors, V = 4 V.

Power used in 2 Ω resistor,

P2 = V² / R = (4)² / 2 = 16 / 2 = 8 W

Clearly, P1 / P2 = 8 / 8 = **1**.

37. Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to the electric mains supply. What current is drawn from the line if the supply voltage is 220 V ?

Solution:

Resistance of the first lamp,

R1 = V² / P = (220)² / 100

Resistance of the second lamp,

R2 = V² / P = (220)² / 60

Since the two lamps are connected in parallel, the equivalent resistance is given by:

1 / Rp = 1 / R1 + 1 / R2 = 100 / (220)² + 60 / (220)² = 160 / (220)²

or

Rp = (220)² / 160 = 48400 / 160 = 302.5 Ω

Current drawn from the line, i.e.,

I = V / Rp = 220 V / 302.5 Ω = 0.727 A

38. Which uses more energy, a 250 W TV set in 1 h, or a 1200 W toaster in 10 minutes?

Solution:

Energy used by 250 W TV set in 1 h = 250 × 1 = 250 Wh

Energy used by 1200 W toaster in 10 min (i.e., 1/6 h) = 1200 × (1/6) = 200 Wh

Thus, a 250 W TV set uses more energy in 1 h than a 1200 W toaster in 10 minutes.

39. An electric heater of resistance 8 Ω draws 15 A from the service mains for 2 hours. Calculate the rate at which heat is developed in the heater.

Solution:

Here, I = 15 A, R = 8 Ω, t = 2 h

Rate at which heat is developed, i.e., electric power,

P = I²R = (15)² × 8 = 225 × 8 = 1800 W

40. Explain the following:

(1) Why is tungsten used almost exclusively for filament of incandescent lamps?

(2) Why are the conductors of electric heating devices, such as toasters and electric irons, made of an alloy rather than a pure metal?

(3) Why is the series arrangement not used for domestic circuits?

(4) How does the resistance of a wire vary with its cross-sectional area?

(5) Why are copper and aluminium wires usually employed for electricity transmission?

Solution:

(1) Tungsten has a high melting point (3380 °C) and becomes incandescent (i.e., emits light) at a high temperature without melting.

(2) The resistivity of an alloy is generally higher than that of pure metals, and they do not oxidize readily at high temperatures. This makes them suitable for heating elements.

(3) In a series arrangement, if any one appliance fails or is switched off, the entire circuit breaks, and all other appliances stop working. Also, the voltage across each appliance in a series circuit is less than the supply voltage, and it varies depending on the resistance of the appliance.

(4) The resistance of a wire (R) varies inversely as its cross-sectional area (A). This means that if the cross-sectional area of a wire increases, its resistance decreases, and vice versa (R ∝ 1/A).

(5) Copper and aluminium wires possess low electrical resistivity, which minimizes power loss during electricity transmission over long distances. They are also relatively ductile and cost-effective.

8.0Benefits of Electricity

There are various reasons why the Class 10 Science NCERT Solutions for Chapter 11, "Electricity", are significant.

• Basic Concepts: The NCERT solutions help students thoroughly understand essential concepts such as electric current, circuits, Ohm’s law, resistance, and electrical energy and develop a clear understanding.
• Solving problems step-by-step: NCERT solutions provide detailed steps to solve numerical problems related to electricity, such as calculating current, voltage, resistance, and power, that help students learn the correct method for approaching and solving problems.
• NCERT Exam Focus: Since the majority of CBSE exam questions are based on the NCERT textbook, these solutions cover all important questions and ensure students are well-prepared for the exams.
• Improvements in problem-solving abilities: The solutions include a variety of problems that improve problem-solving skills, which are essential for higher-level science and mathematics courses.
• Revision Time Saver: These solutions act as a quick revision tool. Before exams, students can quickly review the material with well-explained answers and avoid wasting time on long explanations.
• Foundation for Higher Education: Understanding electricity in Class 10 serves as a foundation for more advanced topics in physics and engineering.