NCERT Solutions For Class 10 Science Chapter 11 Electricity

NCERT Solutions For Class 10 Science Chapter 11 Electricity

Vishal kumarUpdated on 16 Sep 2025, 07:58 PM IST

Electricity has become one of the most important aspects of modern life, and it serves to power up not only household appliances such as lights, fans and refrigerators but also devices such as mobile phones, laptops and even heavy machinery in industries and agriculture. It is extensively used in homes, schools, hospitals, factories and businesses. We are so used to it that even a power outage of a few hours is enough to shake the whole of our daily routine. Class 10 Science Chapter 11 - Electricity helps the students to learn more about how electric circuits operate and discusses some of the basic ideas (including electric current, potential difference (voltage), resistance, Ohm laws Law, heating effect of current, and electric power). The concepts are essential in solving theoretical as well as numerical problems.

This Story also Contains

  1. Electricity NCERT Solutions: Download Solution PDF and Formulae Handbook
  2. Class 10 Science Chapter 11 - Electricity: Intext Questions Solution
  3. Class 10 Science Chapter 11 - Electricity Question Answers - Exercise
  4. Class 10 Physics NCERT Chapter 11: Higher Order Thinking Skills (HOTS) Questions
  5. NCERT Solutions for Class 10 Science Chapter 11 - Electricity: Important Topics
  6. Class 10 Science Chapter 11 - Electricity Question Answers: Important Formulae
  7. Approach to Solve Questions of Class 10 Science Chapter 11 - Electricity
  8. Benefits of Class 10 Science Chapter 11 - Electricity Question Answers
  9. NCERT Solutions for Class 10 Science: Chapter Wise
NCERT Solutions For Class 10 Science Chapter 11 Electricity
NCERT Solutions for Class 10 Science Chapter Electricity(Image: Shutterstock)

NCERT Solutions for Class 10 Science Chapter 11 - Electricity are given in step-by-step explanations to the in-text and exercise questions, and thus, students will find it easy to master concepts and numericals. By using these NCERT solutions, students become assured of the ability to solve problems involving resistance in series and parallel, calculation of current, and consumption of power in electrical appliances. The NCERT Solutions for Class 10 Science Chapter 11 - Electricity are designed according to the current CBSE syllabus, so as to prepare students not only to take up board exams, but also to prepare a solid ground for higher classes and competitive exams. It is also provided with a downloadable PDF format, important formulas, extra HOTS (Higher Order Thinking Skills) questions, and real-life uses of electricity, making it fully prepared.

Electricity NCERT Solutions: Download Solution PDF and Formulae Handbook

The Class 10 Science Chapter 11 - Electricity question answers provide step-wise as well as clear explanations to all the textbook questions, and thus make the concepts of electricity fundamental to the students easy to grasp. All these solutions are provided in PDF format so that one can download all solutions and read them conveniently offline.

Download Solution PDF

Download the chapter-wise formula of NCERT class 10 science by clicking on the link given in the box.

Download Formulae Handbook

Class 10 Science Chapter 11 - Electricity: Intext Questions Solution

Electricity Class 10 question answers give a clear as well as detailed explanation of the answers to all in-text questions present in the NCERT textbook. Through these solutions, the students will be able to grasp some important concepts without prior knowledge of electric currents, potential differences, resistance, and Ohm’s law in a sequential manner, guaranteeing efficient board preparation.

Electricity Class 10 question answers - Topic:11.1

Q.1 What does an electric circuit mean?

Answer:

A closed and continuous path of electric current is known as an electric circuit. It consists of electric circuit elements like batteries, resistors, etc, and electric devices like a switch and measuring devices like ammeters, etc.

Q.2 Define the unit of currcurrent ent.

Answer:

Ampere(A) is a unit of current. 1A is the flow of 1C of charge through a wire in 1s of time.

current I=qt

1A=1C1sec

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

Answer:

Given: Q=1C

We know that the charge of an electron e=1.6×1019C

Q=nen=Qen=11.6×1019=6.25×1018


Thus, the number of electrons constituting one coulomb of charge is 6×1018 electrons.

Electricity Class 10 question answers - Topic:11.2

Q.1 Name a device that helps to maintain a potential difference across a conductor.

Answer:

A battery, cell or power supply source helps to maintain a potential difference across a conductor.

Q.2 What is meant by saying that the potential difference between two points is 1 V?

Answer:

The potential difference between two points is 1 V means 1 J of work is required to move a charge of 1C from one point to another.

Q.3 How much energy is given to each coulomb of charge passing through a 6 V battery?

Answer:

Given: potential difference = 6V and charge =1C.

Potential difference = work done charge 6= work done 1 work done =6×1=6 J


Thus, 6J energy is given to each coulomb of charge passing through a 6 V battery.

Electricity Class 10 question answers - Topic 11.5

Q.1 On what factors does the resistance of a conductor depend?

Answer:

The resistance of a conductor depends on :

1. Cross-section area of the conductor.

2. Length of conductor

3. The temperature of the conductor.

4. Nature of material of the conductor.

Q.2 Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?

Answer:

We know that resistance is given as

R=ρlAremain

R= resistance
ρ= resistivity
l=length of wire
A= area of cross-section
Resistance is inversely proportional to the area of the cross-section.
The thicker the wire, the more cross-sectional area, resulting in less resistance and thus more current flows.

Q.3 Let the resistance of an electrical component remains constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?

Answer:

By Ohm's law,

V=IRI=VR

V= potential difference
I =current
R= resistance
Now, the potential difference is reduced to half i.e.

V=V2

R=R= resistance
I' =current

I=V2R=V2R=I2

Q.4 Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?

Answer:

The resistivity of an alloy is higher than pure metal, so the alloy does not melt at high temperatures. Thus, coils of electric toasters and electric irons are made of an alloy rather than a pure metal.

Q.5 Use the data in Table to answer the following –

(a) Which among iron and mercury is a better conductor?
(b) Which material is the best conductor?

1644209855219

Answer:

The lower the value of resistivity, the lower will be the resistance for a material of given area and length. If resistance is low, current flow will be high when a potential difference is applied across the conductor.

(a) Resistivity of iron =10.0×108Ω m,
the resistivity of mercury =94×108Ω m
The resistivity of mercury is more than iron, so iron is a better conductor than mercury.

(b) From the table, we can observethat silver has the lowest resistivity, so it is the best conductor.

Electricity Class 10 question answers - Topic 11.6

Q.1 Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5Ω resistor, an 8Ω resistor, a 12Ω resistor, and a plug key, all connected in series.

Answer:

The schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5Ω resistor, an 8Ω resistor, and a 12Ω resistor, and a plug key, all connected in series, is as shown below :

1644209887262

Q.2 Redraw the circuit of Question 1, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12Ω resistor. What would be the readings in the ammeter and the voltmeter?

Answer:

The diagram is as shown :

1644210295069

Resistance of circuit, R=5+8+12=25

Potential = 6V

V=IRI=VR=625=0.24 A


Now, for a 12 ohm resistor, current =0.24 A.
By Ohm's law,

V=IR=0.24×12=2.88 V


The reading of the ammeter is 0.24 A and the voltmeter is 2.88 V.


Electricity Class 10 question answers - Topic 11.7

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

Answer:
(a) 1Ω and 106Ω
R= Equivalent resistance

1R=1R1+1R21R=11+11061R=106+1106R=106106+11Ω

(b) 1Ω and 103Ω, and 106Ω,
R= Equivalent resistance

1R=1R1+1R2+1R3

1R=11+1103+11061R=106+103+1106R=106106+103+1=0.999Ω1Ω

Q.2 An electric lamp of 100Ω , a toaster of resistance 50Ω and a water filter of resistance 500Ω 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 all three appliances, and what is the current through it?

Answer:

Given : R1=100Ω,R2=50Ω,R3=500Ω
R= Equivalent resistance

1R=1R1+1R2+1R31R=1100+150+15001R=5+10+1500=16500R=50016=31.25Ω


By Ohm's law,

I=VR=22031.25=7.04A


Hence, the resistance of the electric iron is 31.25, and the current through it is 7.04 A.

Q.3 What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?

Answer:

In parallel, there is no division of voltage among the appliances so the potential difference across all appliances is equal and the total effective resistance of a circuit can be reduced. Hence, connecting electrical devices in parallel with the battery is beneficial instead of connecting them in series.

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

Answer:
(a) R1=3,R2=6,R3=2
R= Equivalent resistance

1R12=1R1+1R21R12=13+161R12=2+16R12=63=2R=R12+R3=2+2=4

1644210349354

(b) 1 Ohm

R=Equivalent resistance

Connect all three resistors in parallel

1R=1R1+1R2+1R31R=12+13+161R=6+3+16R=166=1Ω

Q.5 (a) What is the highest total resistance that can be secured by combinations of four coils of resistance 4Ω,8Ω,12Ω,24Ω?

Answer:

The highest total resistance is obtained when all the resistors are connected in a series

4+8+12+24=48Ω

Q.5 (b) What is the lowest total resistance that can be secured by combinations of four coils of resistance 4Ω,8Ω,12Ω,24Ω?

Answer:

The lowest total resistance is R is obtained when all the resistors are connected in parallel.

1R=14+18+112+1241R=6+3+2+124=1224R=2412=2

Electricity Class 10 question answers - Topic 11.6

Q. 1. Why does the cord of an electric heater not glow while the heating element does?

Answer:

The heating element of an electric heater is a resistor.

The amount of heat production is given as, H=I2Rt.

The resistance of elements (alloys) of an electric heater is high. As current flows through this element, it becomes hot and glows red.

The resistance of the cord (metal like Cu or Al)of an electric heater is low. As current flows through this element, it does not become hot and does not glow red.

Q. 2. Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.

Answer:

Given: potential difference = 50 V.

Charge = 9600 C

time = 1 hr=3600 s

I= charge time ( seconds )=96003600=803 A So, H= VIt H=50×803×3600H=4800000 J=4.8×106 J

Q. 3 An electric iron of resistance 20Ω takes a current of 5 A. Calculate the heat developed in 30 s.

Answer:

Given : resistance =R=20Ω, Time =30 s, current =I=5 A

V=IRV=5×20=100 VH=VItH=100×5×30H=15000 J=1.5×104 J

Electricity Class 10 question answers - Topic 11.8

Q.1 What determines the rate at which energy is delivered by a current?

Answer:

The rate at which energy is delivered by a current is power.

Power P=I2 R, where I is the current and R is the resistance of the circuit/ appliance. Power depends on the current drawn by the appliance and the resistance of the appliance.

Q.2 An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.

Answer:

Given: I= 5 A and V= 220 V.

Power=P=VI

P=220×5=1100 W
Time =2hr=2×60×60=7200s
The energy consumed = power × time=1100×7200 J=7920000 J
Power of motor =1100 W
Energy consumed by motor =7920000 J

Class 10 Science Chapter 11 - Electricity Question Answers - Exercise

NCERT Solutions for Class 10 Science Chapter 11 - Electricity comprise the well-organised answer to all the exercises presented in the textbook, including key topics such as resistance, resistivity, electrical power, heating effects of current etc. The solutions facilitate the solving of problems and enhance better exam preparations with proper explanations.

Q. 1. 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 –

(a) 1/25 (b) 1/5 (c) 5 (d) 25

Answer:

Given: A piece of wire of resistance R is cut into five equal parts.

Resistance of each part is R5.

1R=5R+5R+5R+5R+5R


1R=25R

R=R25

Hence, RR=RR25=25

Thus, option d is correct.

Q. 2. Which of the following terms does not represent electrical power in a circuit?

(a) I2R(b)IR2(c)VI(d)V2/R

Answer:
We know that power =P=VI-------------- (1)

Put, V=IR in equation 1

P=I2R


Pur, I=VR in equation 1,

P=V×VR=V2R

P= power, V= potencial difference, I= current, R= resistance
P cannot be IR2.

Thus, option b is correct.

Q. 3. An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be

(a) 100 W (b) 75 W (c) 50 W (d) 25 W

Answer:

Given : V=220 V,P=100 W

P=V2R


The resistance of the bulb

R=V2P=(220)2100=484Ω


If the bulb is operated on 110 V and the resistance is the same, the power consumed will be P

P=V2R=(110)2484=25W

Hence, option d is correct.

Q. 4. Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be

(a) 1:2 (b) 2:1 (c) 1:4 (d) 4:1

Answer:

If resistors are connected in parallel, the net resistance is given as

1Rp=1R1+1R21Rp=1R+1R1Rp=2RRp=R2


If resistors are connected in series, the net resistance is given as

Rs=R1+R2=R+R=2R


Heat produced =H=V2Rt

HsHp=V2t2RV2tR2HsHp=14Hs:Hp=1:4

Thus, option c is correct.

Q. 5. How is a voltmeter connected in the circuit to measure the potential difference between two points?

Answer:

A voltmeter should be connected in parallel to measure the potential difference between two points.

Q. 6. A copper wire has a diameter of 0.5 mm and resistivity of 1.6×108Ω m. What will be the length of this wire to make its resistance 10Ω ? How much does the resistance change if the diameter is doubled?

Answer:

Given : diameter =d=0.5 mm and resistivity =ρ=1.6×108Ω m, resistance =R=10Ω
Area =A

A=πd24=3.14×0.5×0.54A=0.000000019625 m2


We know

R=ρlAl=RAρ=10×0.0000000196251.6×108=122.72 m


If the diameter is doubled.

d=1 mm


Area =A

A=πd24=3.14×1×14A=0.000000785 m2


We know

R=ρlAR=1.6×108×122.720.000000785R=2.5ΩRR=2.510=14


Hence, the new resistance is 14 of the original resistance.

Q.9. A battery of 9 V is connected in series with resistors of 02Ω , 0.3Ω , 0.4Ω , 0.5Ω and 12Ω respectively. How much current would flow through the 12Ω resistor?

Answer: Total resistance =R

R=0.2+0.3+0.4+0.5+12=13.4Ω


V=9 VI=VR=913.4=0.67 A


Hence, All resistors are in series, so 0.67 A current would flow through the 12Ω resistor.

Q.10. How many 176Ω resistors (in parallel) are required to carry 5 A on a 220 V line?

Answer:

Given: V=220V and I =5A

R=VI=2205=44Ω


Let x number of resistors be connected in parallel to obtain a 44 Ohm equivalent resistance.
Thus,

1R=1176+1176+............ x times 144=x176x=17644=4


Hence, 4 resistors of 176 Ohms are connected in parallel to obtain 44 Ohms.

Q.11. Show how you would connect three resistors, each of resistance 6Ω , so that the combination has a resistance of (i) 92Ω , (ii) 4Ω .

Answer:

(i) R1=R2=R3=6
R= Equivalent resistance

1R12=1R1+1R21R12=16+161R12=1+16R12=62=3R=R12+R3=3+6=9Ω

1644210424046

(ii)

R1=R2=R3=6

R= Equivalent resistance

R12=R1+R2=6+6=12Ω1R=1R12+1R31R=112+161R=1+212R=123=4

Q.12. 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?

Answer:

Given: V=220 V and P=10 W

R=V2P=220210=4840Ω


Let x be the number of bulbs.

I=5 A and V=220 VR=VI=2205=44Ω


For x bulbs of resistance 4680 Ohms, are connected in parallel to obtain 44 Ohms equivalent resistance.

Thus,

1R=14840+14840+... to x times 144=x4840x=484044=110

Hence, 110 bulbs of 4840 Ohms are connected in parallel to obtain 44 Ohms.

Q.13. A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24Ω resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?

Answer:

Given : V=220V and Resistance of each coil=R =24A

When the coil is used separately, the current in the coil is

I=VR=22024=9.16A


When two coils are connected in series, the net resistance is

R=R1+R2=24+24=48Ω

current in coil is I'

I=VR=22048=4.58 A


When two coils are connected in parallel, the net resistance is

1R=124+124=224R=12Ω

current in the coil is I"

I=VR=22012=18.33 A

Q.14. Compare the power used in the 2Ω resistor in each of the following circuits:

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

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

Answer:

i) Given: V=6V

R=1+2=30hm

I=VR=63=2A


In, a series current is constant.
So, power =P

P=I2R=22×2=8W

ii) Given: V=6 V,R=2 ohm

In, parallel combination voltage in the circuit is constant.
So, power =P

P=V2R=422=8W

Q.15. Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to an electric mains supply. What current is drawn from the line if the supply voltage is 220 V?

Answer:

Given :

For lamp one: Power =P1=100 W and V=220 V

I1=P1V=100220=0.455 A


For lamp two: Power =P2=60 W and V=220 V

I2=P2V=60220=0.273 A


Thus, the net current drawn from the supply is 0.455+0.273=0.728 A

Q.16. Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?

Answer:

For TV set :

Given : Power =250 W and time =1hr=3600 seconds
Energy consumed =H=PI

H=250×3600=900000J


For toaster :
Given : Power =1200 W and time =10 minutes =600 seconds
Energy consumed =H=PI

H=1200×600=720000J

Thus, the TV set uses more energy than a toaster.

Q.17. An electric heater of resistance 8Ω draws 15 A from the service mains 2 hours. Calculate the rate at which heat is developed in the heater.

Answer:

Given: R=8 Ohm ,I=15A and t= 2hr

The heat developed in the heater is H.

The heat developed in the heater is H .

H=I2Rt


The rate at which heat is developed is given as

I2Rtt=I2R=152×8=1800 J/s

Q.18(a). Explain the following: Why is the tungsten used almost exclusively for filament of electric lamps?

Answer:

Tungsten is used almost exclusively for the filament of electric lamps because the melting point of tungsten is very high. The bulb glows at a very high temperature. So tungsten filament does not melt when the bulb glows.

Q.18.(b) Explain the following: Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal?

Answer:

The conductors of electric heating devices, such as bread toasters and electric irons, are made of an alloy rather than a pure metal because the resistivity of the alloy is greater than a pure metal. So the resistance will be high and the heating effect will be high.

Q.18.(c) Explain the following: Why is the series arrangement not used for domestic circuits?

Answer:

If any of the elements in the circuit get damaged, the entire circuit will be affected. If an element breaks, there will not be any current flow through the circuit. So the series circuit is not preferred in the domestic circuit.

Q.18.(d) Explain the following: How does the resistance of a wire vary with its area of cross-section?

Answer:

We know that

R=ρlARα1A

Thus, the resistance of a wire is inversely proportional to its area of cross-section.

Q.18.(e) Explain the following: Why are copper and aluminium wires usually employed for electricity transmission?

Answer:

Copper and aluminium wires are usually employed for electricity transmission because they have low resistivity and are good conductors of electricity.

Class 10 Physics NCERT Chapter 11: Higher Order Thinking Skills (HOTS) Questions

The NCERT Class 10 Physics Chapter 11 Electricity HOTS (Higher Order Thinking Skills) questions aim at challenging high-order conceptual insight in topics such as Ohm's Law, series and parallel circuits, and power consumption. These complicated questions lead to increased analytical thinking and problem-solving abilities, which also help the students in all types of competitive exams.

Q1:

If the electronic charge is 1.6×1019C, then the number of electrons passing through a section of wire per second when the wire carries a current of 2 A, is

Answer:

Given : e=1.6×1o19C,I=2A,t=1 s
Using I= ne t
Number of electrons n=Ite=2×11.6×1019=1.25×1019 electrons


Q2:

When two bulbs of power 60 W and 40 W are connected in series, then the power of their combination will be:

Answer:

Given:- Power of two bulbs P1=60 and P2=40 W

To find the Power of their combination,
We know, Power P=V2R
where V denotes voltage across bulb and R denotes resistance of bulb.
So, Resistance of first bulb R1=V2p1=V260....(1)

Similarly,
Resistance of and bulb, R2=V2P2=V240......(2)
Now, R1 and R2 are Joined in Series
then,
equivalent resistance Req=R1+R2

V2Peq=V260+V2401Peq=160+140Peq=60×4060+40=24 WPeq=24 W


Q3:

At a constant potential difference, the resistance of any electric circuit is halved. The value of heat produced will be:

Answer:

We know,
The heat produced =i2Rt....(1)
where i is current, R is resistance and is tee.

V=iRi=V/R

putt in (1)

H=(VR)2Rt=V2Rt

So, V is constant
and R is now R/2 given.
and t remains same

H=V2R/2t=2V2Rt

H=2V2Rt

Hence, the heating produced will be doubled


Q4:

A wire of resistance 9Ω is bent to form an equilateral triangle. Then the equivalent resistance across any two vertices will be ____ ohm.

Answer:

9Ω is the resistance of whole wire
resistance of each wire =3Ω.
Equivalent resistance =2Ω


Q5:

A wire of resistance R is bent into an equilateral triangle and an identical wire is bent into a square. The ratio of resistance between the two endpoints of an edge of the triangle to that of the square is

Answer:

R=ρA

So, R
Side length of triangle is 1/3 of total length.

(Req)1=2r/3×r/32r/3+r/3=2r/9

(Req)2=3r/4×r/43r/4+r/4=3r/16

(Req)1(Req)2=2r/93r/16=3227


NCERT Solutions for Class 10 Science Chapter 11 - Electricity: Important Topics

Electricity NCERT Solutions cover all the important topics needed to understand how electric circuits work and how electrical quantities like current, voltage, resistance, and power are calculated. These topics are essential for both board exam preparation and real-life applications.

11.1 Electric Current And Circuit
11.2 Electric Potential And Potential Difference
11.3 Circuit Diagram
11.4 Ohm’s Law
11.5 Factors On Which The Resistance of A Conductor Depends
11.6 Resistance of A System Of Resistors
11.6.1 Resistors In Series
11.6.2 Resistors In Parallel
11.7 Heating Effect Of Electric Current
11.7.1 Practical Applications Of Heating Effect Of Electric Current
11.8 Electric Power

Also, students can refer,

Class 10 Science Chapter 11 - Electricity Question Answers: Important Formulae

The Electricity Class 10 question answers provide all the important formulae related to current, voltage, resistance, power, and energy. These formulas make problem-solving easier, help in quick revision, and are essential for scoring well in board as well as competitive exams.

1. Ohm’s Law

V=IR
Where:
V= potential difference (volts)
I= current (amperes)
R= resistance (ohms, Ω)

2. Resistance of a Conductor

R=ρlA
Where:
R= resistance
ρ= resistivity of the material
l= length of the conductor
A= cross-sectional area

3. Resistance in Series

Rtotal =R1+R2+R3+

4. Resistance in Parallel

1Rtotal =1R1+1R2+1R3+

5. Electric Power

P=VI=I2R=V2R
Where:
P= power (watts)

6. Electrical Energy

E=Pt=VIt=I2Rt=V2tR
Where:
E= electrical energy (in joules)

Approach to Solve Questions of Class 10 Science Chapter 11 - Electricity

Class 10 Science Chapter 11 - Electricity concentrates on significant concepts such as electric current, potential difference, resistance, Ohm's law, heating effect, and power. The formulas, circuit rules, and techniques of solving numerical problems need to be clearly understood to allow students to solve the questions in this chapter. Through the right approach, it becomes much easier to address both theoretical and numerical questions.

  • Carefully read the question to determine what you are given and what to find.

  • Write down the known variables, which can be current, voltage, resistance or power or time.

  • Get familiar with the nature of the circuit, either series or parallel.

  • Use the proper idea relative to the type of circuits and provided values.

  • Use proper units and convert as appropriate to standard SI units.

  • Carry out calculations in steps with the concept needed.

  • Check your last answer against being correct and related to the question.

Benefits of Class 10 Science Chapter 11 - Electricity Question Answers

There are various advantages of the Class 10 Science Chapter 11 - Electricity question answers, which offer answers to the textbook questions in a step-by-step manner and are correct. Such solutions not only simplify concepts such as current, resistance, Ohm's law, and electric power, but they are also useful in improving numerical solving. They are used to revise students swiftly, instil confidence in them, and achieve better scores in CBSE board as well as competitive exams with clear explanations.

  1. Concept Clarity: Uses simple yet step-by-step explanations of Ohm's laws, resistance, power, and electric energy, thereby making these concepts easier to understand.
  2. Exam-Oriented: Solutions are designed according to the CBSE syllabus and pattern, and this aids in students preparing to appear in board examinations.
  3. Quick Revision: The entire chapter can be revised within a short period of time using important formulae and solved examples.
  4. Enhances Problem-Resolving Ability: Practising questions enhances problem-solving and aids in resolving both numerical and theoretical problems without fear.
  5. Helps in Competitive Exams and tests: Creates an excellent background towards entrance exams such as NTSE, Olympiads, NEET, and IIT-JEE.

NCERT Solutions for Class 10 Science: Chapter Wise

Chapter-wise NCERT solutions can make it extremely easy to prepare notes for the Class 10 Science exams. These solutions include explanations for all in-text and exercise questions, and they assist students in developing firm concepts and get along with their problem-solving abilities. These links serve as a comprehensive guide to studying all board exams and competitive tests with step-by-step explanations, relevant formulas.

Also, check the NCERT Books and NCERT Syllabus here:

Frequently Asked Questions (FAQs)

Q: What are the most crucial topics of Electricity to be studied in exams?
A:

The key concepts are Ohm's Law, Series and Parallel resistors, heating of electric current, electric power and energy, and numerical problems on the same.

Q: How can I score full marks in Electricity numericals?
A:

In order to get full marks, write down the provided data in an understandable way, state the formula applied (such as V = IR or P = VI), replace values in the right way, and add the SI units to the final solution.

Q: What type of questions are usually asked in board exams in this chapter?
A:

Numerical problems (3-5 marks), conceptual questions (1-2 marks), and application-based HOTS questions, such as why use fuses, why parallel connection of bulbs, etc., can be expected.

Q: Why does the component of a heater feel very warm, and the corresponding connecting wire that carries the same amount of current feel cold?
A:

The resistance is dependent on the heating effect. The heater coil is composed of a high-resistance material, such as nichrome, thus giving it a higher output of heat when it carries current. The connecting wires consist of low-resistance substances such as copper or aluminium; thus, a minimal amount of heat is produced in the connecting wires

Q: Can I download Class 10 NCERT Solutions PDF for Electricity?
A:

Yes, NCERT Solutions of Chapter 11 of Class 10 Science, which is about Electricity, can be downloaded as a free PDF document. These are in-text answers, exercise answers, numerical answers, and HOTS questions; thus they are also optimal in revision and exam preparation.

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Questions related to CBSE Class 10th

On Question asked by student community

Have a question related to CBSE Class 10th ?

To get the previous year question papers you can visit the official website of your board and search under the section of previous year question papers. You can also search on Google for the papers or visit the educational platforms like Careers 360 for the papers. They also provide with the papers and answer key also.

You must be at least 14 years old by December 31st of the year 2027

Since your date of birth is 29 January 2013, you will turn 14 in January 2027, which is before the December 31st deadline for the 2027 exam.

Hence you are eligible..

Good luck!!

Hello,

If you want to get your 10th marksheet online, you just need to visit an official website like https://www.cbse.gov.in/ or https://results.cbse.nic.in/ for the CBSE board, and for the state board, you can check their website and provide your roll number, security PIN provided by the school, and school code to download the result.

I hope it will clear your query!!

Hello, if you are searching for Class 10 books for exam preparation, the right study material can make a big difference. Standard textbooks recommended by the board should be your first priority as they cover the syllabus completely. Along with that, reference books and guides can help in practicing extra questions and understanding concepts in detail. You can check the recommended books for exam preparation from the link I am sharing here.
https://school.careers360.com/ncert/ncert-books-for-class-10
https://school.careers360.com/boards/cbse/cbse-best-reference-books-for-cbse-class-10-exam

Hello Dinesh !

As per CBSE board guidelines for internal assessment for class 10th you will have to give a 80 marks board exam and 20 marks internal assessment. The internal assessment will be at the end of your year.

For knowing the definite structure of the internal assessment you will have to ask your teachers or your seniors in the school as CBSE has provided flexibility in choosing the methods of internal assessment to schools. For more details related to assessment scheme for class 10 given by CBSE you can visit: Assessment scheme (http://cbseacademic.nic.in/web_material/CurriculumMain2Sec/Curriculum_Sec_2021- 22.pdf)

I Hope you have understood it!