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NCERT Solutions For Class 10 Science Chapter 12 Electricity

NCERT Solutions For Class 10 Science Chapter 12 Electricity

Edited By Vishal kumar | Updated on Jul 21, 2023 12:39 PM IST | #CBSE Class 10th

NCERT Solutions for Class 10 Science Electricity – CBSE Free PDF Download

NCERT Solutions for Class 10 Science Chapter 12: Students preparing for CBSE class 10 board exams, must know that NCERT chapter 12 class 10 science is one of the most important chapters of NCERT syllabus. In this article on electricity class 10 NCERT solutions, you will get the complete question and answer of the electricity chapter topic-wise in a very easy and systemic manner.

Since CBSE class 10 is a board class it is extremely important to understand topics carefully. The Electricity class 10 solutions will help you score well in the CBSE board exam. Students are advised to go through Class 10 science chapter 12 question answer to score good marks in the Board examination. Complete the electricity class 10 NCERT syllabus at the earliest to revise it in a proper way. Refer to the NCERT Class 10th Science Notes and cover all the topics for scoring well in the exams.

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Electricity class 10 questions and answers are created by highly qualified subject matter experts in very easy-to-understand and step-by-step which ensure a student scores well in exam. The main aim of NCERT solutions is to provide comprehensive guidance to students. By presenting clear explanations, solving examples, and offering valuable insights, these class 10 science ch 12 ncert solutions assists students in strengthening their understanding of the subject.

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NCERT Solutions for Class 10 Science: Important Formulas and Diagrams + eBook link

Electricity class 10 ncert solutions include important formulas, diagrams, and a helpful eBook link, providing essential resources for comprehensive exam preparation and a better understanding of the subject. Click the eBook link to access valuable study materials and excel in your studies.

  • Current:

Current (I)= Charge(Q)/ Time taken(t)

  • Voltage:

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Voltage(v)= Work Done(W) / Charge(Q)

  • Resistance:

R= ρl / A= V/I

Where:

ρ = Resistivity, l = length, A = Area

  • Ohm's Law:

V = I × R

  • Resistors in Series:

Rt = R1 + R2 + R3 + ...

  • Resistors in Parallel:

1/Rt = 1/R1 + 1/R2 + 1/R3 + ...

Download Ebook - NCERT Class 10 Science: Chapterwise Important Formulas, Diagrams, And Points

This ebook comprises crucial formulas, diagrams, and key points for each chapter, curated by experts. Utilizing this valuable resource, students can score well in less time and use it for quick revision during weekly tests and exams, ensuring better academic performance.

Class 10 electricity question answer - Important Topics

As science class 10 chapter 12 is crucial, all topics hold significance. However, here are some very important topics that will aid in scoring good marks in the exam:

  • Electric current potential

  • Electric resistance

  • Circuit diagrams

  • Ohm's law

  • The heating effect of electric current and power.

  • The electric current can be classified as direct current (DC) and alternating current ac (AC).

Also students can refer,

Download Class 10 electricity exercise solutions Electricity for free.

This chapter has been renumbered as Chapter 11 in accordance with the CBSE Syllabus 2023–24.

NCERT Class 10 Science Chapter 12 Question Answer - Electricity

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Access Answer to NCERT Class 10 Science Chapter 12: Electricity

Topic:12.1

Q.1 What does an electric circuit mean?

Answer:

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

Q.2 Define the unit of current.

Answer:

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

\\current\ I=\frac{q}{t}\\1A=\frac{1 \ C}{1\ sec}

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\times 10^{-19}C

Q=ne

\Rightarrow n=\frac{Q}{e}

\Rightarrow n=\frac{1}{1.6\times 10^{-19}}=6.25\times 10^{18}

Thus, the number of electrons constituting one coulomb of charge is 6\times 10^{18} electrons.

Class 10 science ch 12 ncert solutions - topic 12.2

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

Answer:

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 amount 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 carge =1C.

Potential\, \, difference=\frac{work\, \, done}{charge}

\Rightarrow 6=\frac{work\, \, done}{1}

\Rightarrow work\, \, done=6\times 1=6 J

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

NCERT solutions for class 10 science chapter 12 Electricity topic 12.5

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=\frac{\rho l}{A}

R=resistance

\rho =resistivity

l=length of wire

A=area of cross section

Resistance is inversely proportional to the area of cross-section.

Thicker the wire more is cross-sectional area resulting in less resistance resulting in more current flow.

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=IR

I=\frac{V}{R}

V= potential difference

I =current

R = resistance

Now, the potential difference is reduced to half i.e.

V'=\frac{V}{2}

R' = R=resistance

I' =current

I'=\frac{\frac{V}{2}}{R}=\frac{V}{2R}=\frac{I}{2}

Current flowing is reduced to half.

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 alloy does not melt at high temperature. Thus, coils of electric toasters and electric irons made of an alloy rather than a pure metal.

Q.5 Use the data in Table 12.2 to answer the following –

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

1644209855219

Answer:

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

(a).resistivity of iron = 10.0\times 10^{-8} \Omega m

the resistivity of mercury = 94\times 10^{-8} \Omega m

the resistivity of mercury is more than iron so iron is a better conductor than mercury.

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

Class 10 electricity exercise solutions topic 12.6

Q.1 Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 \Omega resistor, an 8 \Omega resistor, and a 12 \Omega 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 5 \Omega resistor, an 8 \Omega resistor, and a 12 \Omega 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 12\; \Omega 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=IR

\Rightarrow I=\frac{V}{R}=\frac{6}{25}=0.24A

Now, for 12 ohm resistor, current = 0.24 A.

By Ohm's law,

V=IR=0.24\times 12=2.88V

The reading of the ammeter is o.24A and the voltmeter is 2.88V.

NCERT solutions for class 10 science chapter 12 Electricity topic 12.6.2

Q.1 Judge the equivalent resistance when the following are connected in parallel –

(a)\; 1\; \Omega \; and \; 10^{6}\Omega (b)\; 1\; \Omega \; and \; 10^{3}\Omega,\; and \; 10^{6}\Omega,

Answer:

(a)\; 1\; \Omega \; and \; 10^{6}\Omega

R=Equivalent resistance

\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}

\Rightarrow \frac{1}{R}=\frac{1}{1}+\frac{1}{10^{6}}

\Rightarrow \frac{1}{R}=\frac{10^6+1}{10^{6}}

\Rightarrow R=\frac{10^6}{10^{6}+1}\approx 1\Omega

(b)\; 1\; \Omega \; and \; 10^{3}\Omega,\; and \; 10^{6}\Omega,

R=Equivalent resistance

\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}

\Rightarrow \frac{1}{R}=\frac{1}{1}+\frac{1}{10^3}+\frac{1}{10^{6}}

\Rightarrow \frac{1}{R}=\frac{10^6+10^3+1}{10^{6}}

\Rightarrow R=\frac{10^6}{10^{6}+10^3+1}=0.999\Omega \approx 1\Omega


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 appliance 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 2\Omega ,3\Omega , \; and \; 6\Omega be connected to give a total resistance of

(a)4\Omega , (b)1\Omega ?

Answer:

(a) R_1=3,R_2=6,R_3=2

R=Equivalent resistance

\frac{1}{R_1_2}=\frac{1}{R_1}+\frac{1}{R_2}

\Rightarrow \frac{1}{R_1_2}=\frac{1}{3}+\frac{1}{6}

\Rightarrow \frac{1}{R_1_2}=\frac{2+1}{6}

\Rightarrow R_1_2=\frac{6}{3}=2

R=R_1_2+R_3=2+2=4

1644210349354

(b).1 Ohm

R=Equivalent resistance

Connect all the three resistors in parallel

\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}

\Rightarrow \frac{1}{R}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}

\Rightarrow \frac{1}{R}=\frac{6+3+1}{6}

\Rightarrow R=\frac{16}{6}=1\Omega

Q.5 (a) What is (a) the highest total resistance that can be secured by combinations of four coils of resistance

  • 4\; \Omega , 8\; \Omega ,12\; \Omega ,24\; \Omega ?


Answer:

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

4+8+12+24=48\Omega

Q.5 (b) What is (b) the lowest total resistance that can be secured by combinations of four coils of resistance

  • 4\; \Omega , 8\; \Omega ,12\; \Omega ,24\; \Omega ?

Answer :

(b) the lowest total resistance is R is obtained when all the resistors are connected in parallel.

\frac{1}{R}=\frac{1}{4}+\frac{1}{8}+\frac{1}{12}+\frac{1}{24}

\frac{1}{R}=\frac{6+3+2+1}{24}=\frac{12}{24}

R=\frac{24}{12}=2


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=I^2Rt .

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

The resistance of cord (metal like Cu or Al)of an electric heater is low. As current flow through this element, it does not becomes 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=\frac{charge}{time(seconds)}=\frac{9600}{3600}=\frac{80}{3}A

So, H=VIt

\Rightarrow H=50\times \frac{80}{3}\times 3600

\Rightarrow H=4800000J=4.8\times 10^6J

NCERT solutions for class 10 science chapter 12 Electricity topic 12.8

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=I 2 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\times 5=1100W

Time = 2hr= 2\times 60\times 60=7200s

The energy consumed = power\times time

=1100\times 7200J=7920000J

Power of motor = 1100W

Energy consumed by motor = 7920000J

Class 10 electricity question answer - Exercise Solution

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 \frac{R}{5} .

\frac{1}{R'}=\frac{5}{R}+\frac{5}{R}+\frac{5}{R}+\frac{5}{R}+\frac{5}{R}

\frac{1}{R'}=\frac{25}{R}

R'=\frac{R}{25}

Hence, \frac{R}{R'}=\frac{R}{\frac{R}{25}}=25

Thus, option D is correct.

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

(a)\; I^{2}R (b)\; IR^{2} (c)\; VI (d)\; V^{2}/R

Answer:

We know that power = P=VI....................................1

Put , V=IR in equation 1

P=I^2R

Pur , I=\frac{V}{R} in equation 1,

P=V\times \frac{V}{R}=\frac{V^2}{R}

P= power, V=potencial difference, I= current,R=resistance

P cannot be IR^{2} .

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=220V,P=100W

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

The resistance of the bulb

\Rightarrow R=\frac{V^2}{P}=\frac{(220)^2}{100}=484\Omega

If bulb is operated on 110 Vand resistance is the same , the power consumed will be P'

P'=\frac{V^2}{R}=\frac{(110)^2}{484}=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

\frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}

\Rightarrow \frac{1}{R_p}=\frac{1}{R}+\frac{1}{R}

\Rightarrow \frac{1}{R_p}=\frac{2}{R}

\Rightarrow R_p=\frac{R}{2}

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

\Rightarrow R_s=R_1+R_2=R+R=2R

Heat produced = H = \frac{V^2}{R}t

\frac{H_s}{H_p}=\frac{\frac{V^2t}{2R}}{\frac{V^2t}{\frac{R}{2}}}

\Rightarrow \frac{H_s}{H_p}=\frac{1}{4}

\Rightarrow H_s:H_p=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\times 10^{-8}\; \Omega m. What will be the length of this wire to make its resistance 10\; \Omega ? How much does the resistance change if the diameter is doubled?

Answer:

Given : diameter=d= 0.5 mm and resistivity = \rho = 1.6\times 10^{-8}\; \Omega m.,resistance =R=10

Area =A

A=\frac{\pi d^2}{4}=\frac{3.14\times 0.5\times 0.5}{4}

\Rightarrow A=0.000000019625 m^2

We know

R=\frac{\rho l}{A}

\Rightarrow l=\frac{RA}{\rho }=\frac{10\times 0.000000019625}{1.6\times 10^-^8}

=122.72m

If the diameter is doubled.

d=1 mm

Area =A'

A'=\frac{\pi d^2}{4}=\frac{3.14\times 1\times 1}{4}

\Rightarrow A=0.000000785 m^2

We know

R'=\frac{\rho l}{A'}

\Rightarrow R'=\frac{1.6\times 10^{-8}\times 122.72}{0.000000785}

\Rightarrow R'=2.5\Omega

\frac{R'}{R}=\frac{2.5}{10}=\frac{1}{4}

Hence, new resistance is \frac{1}{4} of original resistance.

Q. 7. The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given be low –

I (amperes) 0.5 1.0 2.0 3.0 4.0
V (volts) 1.6 3.4 6.7 10.2 13.2

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

Answer:

The plot between voltage and current is as shown :

1644210395023

The slope of the line gives resistance (R)

R=\frac{6.8}{2}=3.4\Omega


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

Answer:

Total resistance =R

R=0.2+0.3+0.4+0.5+12=13.4\Omega

V = 9V

I=\frac{V}{R}=\frac{9}{13.4}=0.67 A

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

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

Answer:

Given: V=220V and I =5A

R=\frac{V}{I}=\frac{220}{5}=44\Omega

Let x number of resistors are connected in parallel to obtain 44 Ohm equivalent resistance.

Thus,

\frac{1}{R}=\frac{1}{176}+\frac{1}{176}+.......................to\, \, x\, \, times

\Rightarrow \frac{1}{44}=\frac{x}{176}

\Rightarrow x=\frac{176}{44}=4

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

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=220V and P=10W

R=\frac{V^2}{P}=\frac{220^2}{10}=4840\Omega

Let x be the number of bulbs.

I = 5A and V=220V

R=\frac{V}{I}=\frac{220}{5}=44\Omega

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

Thus,

\frac{1}{R}=\frac{1}{4840}+\frac{1}{4840}+.......................to\, \, x\, \, times

\Rightarrow \frac{1}{44}=\frac{x}{4840}

\Rightarrow x=\frac{4840}{44}=110

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

Q.13. A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24\Omega 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 coil is used separately,current in coil is

I=\frac{V}{R}=\frac{220}{24}=9.16A

When two coils are connected in series, net resistance is

R=R_1+R_2=24+24=48\Omega

current in coil is I'

I'=\frac{V}{R}=\frac{220}{48}=4.58A

When two coils are connected in parallel, net resistance is

\frac{1}{R}=\frac{1}{24}+\frac{1}{24}=\frac{2}{24}

\Rightarrow R=12\Omega

current in the coil is I''

I''=\frac{V}{R}=\frac{220}{12}=18.33A

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

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

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

Answer:

i) Given: V=6V

R=1+2=3Ohm

I=\frac{V}{R}=\frac{6}{3}=2A

In, a series current is constant.

So, power =P

P=I^2R=2^2\times 2=8W

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

In, parallel combination voltage in the circuit is constant.

So, power =P

P=\frac{V^2}{R}=\frac{4^2}{2}=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=100W and V = 220V

I_1=\frac{P_1}{V}=\frac{100}{220}=0.455A

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

I_2=\frac{P_2}{V}=\frac{60}{220}=0.273A

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

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 = 250W and time = 1 hr = 3600 seconds

Energy consumed = H=PI

H=250\times 3600=900000J

For toaster :

Given : Power = 1200W and time = 10 minutes = 600 seconds

Energy consumed = H=PI

H=1200\times 600=720000J

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

Q.17. An electric heater of resistance 8\Omega 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.

H=I^2Rt

The rate at which heat is developed is given as

\frac{I^2Rt}{t}=I^2R=15^2\times 8=1800J/s

Q.18(a). Explain the following.

(a) Why is the tungsten used almost exclusively for filament of electric lamps?

Answer:

The tungsten used almost exclusively for 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 bulb glows.

Q.18.(b) Explain the following.

(b) 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, made of an alloy rather than a pure metal because the resistivity of the alloy is more than a pure metal. So the resistance will be hight and the heating effect will be high.

Q.18.(c) Explain the following.

(c) 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 brake 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.

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

Answer:

We know that

R=\frac{\rho l}{A}

R\alpha \frac{1}{A}

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

Q.18.(e) Explain the following.

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

Answer:

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

Electricity Class 10 Exercise Solutions - Types of Questions Asked

Before moving on to the board exam, students must know the types of questions which may be asked from Electricity class 10 ncert solutions. The following types of questions are commonly found in the exam:

  • Very short answer type

  • Short answer type

  • Long answer type

  • Practical based questions

Also, check - NCERT Solutions for Class 10 Maths

NCERT Class 10 Science Solutions Chapter 12 - Weightage in Board Exams

Electricity class 10 solutions hold significant weightage in board exams. This chapter explores essential concepts related to electric circuits, Ohm's law, and electrical power, making it crucial for exam preparation and scoring well. Based on examination trends observed in previous years, class 10 science chapter 12 question answer is anticipated to carry a weightage of at least 8 marks in the Class 10 Science board exam. However, in the 2018 exam, the questions for this chapter totalled up to 7 marks. To strengthen your understanding of the key concepts in this chapter, make use of the ncert solutions for class 10 science chapter 12, ensuring a solid grasp of the subject matter.

How to Access class 10 electricity question answer?

To access electricity class 10 numericals with solutions - "Electricity", follow these steps:

  • Scroll down the electricity class 10 solutions on the website where you are looking for solutions.
  • Look for the section related to class 10 electricity ncert solutions.
  • Click on the link provided for electricity solutions class 10.
  • From there, navigate to the subject "Science" and then select Class 10.
  • Find Chapter 12 - "Electricity" from the list of chapters available for Class 10 Science.
  • Click on Chapter 12 to access the class 10 science ch 12 ncert solutions for "Electricity"

These ncert class 10 science chapter 12 exercise solutions include not just exercise solutions but also topic-specific questions, offering thorough coverage of the chapter. By using these solutions, you will be able to confidently face questions in the CBSE board exam for Science, taking a step closer to academic success.

About NCERT Class 10 Science chapter 12 Electricity

The ch 12 science class 10 is one of the most interesting chapters that will help in learning about electrical components and how they work. To give a better understanding of the chapter we have provided class 10 science chapter 12 question answer above. Students will learn about water, steam, and nuclear energy, etc. electricity class 10 questions and answers also include some other topics related to resistors. The main aim of NCERT solutions is to give a better knowledge of how to use the concept while answering the questions.

Some key points to remember from Class 10 electricity NCERT solutions:

  • Current flows from the positive terminal to the negative terminal of a circuit.

  • A voltmeter is always connected in parallel across the circuit whereas an ammeter is always connected in series with the circuit whose current is to be measured.

  • Reciprocal of the resistance is known as conductance whereas reciprocal resistivity is known as conductivity.

Key Features of NCERT Solutions for Class 10 Science Book Chapter 12 – Electricity

Comprehensive Coverage: The electricity class 10 solutions encompass answers to both in-text questions and exercise questions, providing a thorough understanding of the chapter.

Numerical and Theoretical Problems: NCERT solutions for class 10 science chapter 12 address both numerical and theoretical problems, aiding students in mastering various aspects of electricity.

Step-by-Step Solutions: Numerical problems are solved methodically, listing necessary equations and steps, making it easier for students to follow and learn.

18 Questions in Exercise: Class 10 electricity ncert solutions consists of 18 questions, offering ample practice opportunities to students.

Also, Check NCERT Books and NCERT Syllabus here:

NCERT Solutions for Class 10 Science - Chapter wise

NCERT Science Exemplar Solutions Class 10 - Chapter Wise

At Careers360, students have the advantage of accessing NCERT Solutions for all classes, ranging from Class 6 to 12, across all subjects. These solutions are thoughtfully curated by expert subject tutors, aligning them with the latest CBSE syllabus and guidelines. With Careers360 comprehensive NCERT Solutions, students can enhance their understanding, excel in exams, and achieve academic excellence across all levels of their education.

Frequently Asked Question (FAQs)

1. What is the weightage of NCERT chapter 12 electricity for CBSE board exam?

Class 10 Science Chapter 12 Electricity carries 6 to 8 marks.

2. Is NCERT exercise enough to solve questions in the board exam?

Along with questions and concepts of class 10 chapter, practicing CBSE previous year question papers are mandatory for board exams.

3. Is NCERT Class 10 Science chapter 12 Electricity useful in class 11 and 12?

Yes, it helps in higher studies if students chooses the Science stream because class 12 has current electricity chapter in syllabus.

4. How many practice questions are included in chapter 12 class 10 science?

class 10 science chapter 12 exercise solutions contains the following number of exercises and questions:

Exercise 12.1 includes 3 questions.

Exercise 12.2 consists of 3 questions.

Exercise 12.3 has 5 questions.

Exercise 12.4 comprises 2 questions.

Exercise 12.5 has 5 questions.

Exercise 12.6 contains 3 questions.

Exercise 12.7 comprises 2 questions.

Exercise 12.8 has 18 questions.

5. Why is it beneficial for students to obtain ncert solutions for class 10 science electricity in PDF format?

The PDF format of class 10 electricity solutions Chapter 12 provides diagrams and solutions to all questions which are present in the textbook. The answers are designed with consideration for the comprehension level of students, and they strictly align with the CBSE syllabus and exam pattern, allowing students to approach exams with confidence.

6. What is Q in electricity chapter class 10?

"Q" refers to the electrical charge or quantity of charge.

7. What is the chapter number of Electricity class 10?

 As per the revised CBSE Syllabus 2023-24, Chapter 12 "Electricity" has been renumbered as Chapter 10.

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Questions related to CBSE Class 10th

Have a question related to CBSE Class 10th ?

Here are some options you can explore to get admission in a good school even though admissions might be closed for many:

  • Waitlist: Many schools maintain waitlists after their initial application rounds close.  If a student who secured a seat decides not to join, the school might reach out to students on the waitlist.  So, even if the application deadline has passed,  it might be worth inquiring with schools you're interested in if they have a waitlist and if they would consider adding you to it.

  • Schools with ongoing admissions: Some schools, due to transfers or other reasons, might still have seats available even after the main admission rush.  Reach out to the schools directly to see if they have any open seats in 10th grade.

  • Consider other good schools: There might be other schools in your area that have a good reputation but weren't on your initial list. Research these schools and see if they have any seats available.

  • Focus on excelling in your current school: If you can't find a new school this year, focus on doing well in your current school. Maintain good grades and get involved in extracurricular activities. This will strengthen your application for next year if you decide to try transferring again.


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Dear aspirant !

Hope you are doing well ! The class 10 Hindi mp board sample paper can be found on the given link below . Set your time and give your best in the sample paper it will lead to great results ,many students are already doing so .

https://www.google.com/url?sa=t&source=web&rct=j&opi=89978449&url=https://school.careers360.com/download/sample-papers/mp-board-10th-hindi-model-paper&ved=2ahUKEwjO3YvJu5KEAxWAR2wGHSLpAiQQFnoECBMQAQ&usg=AOvVaw2qFFjVeuiZZJsx0b35oL1x .

Hope you get it !

Thanking you

Hello aspirant,

The dates of the CBSE Class 10th and Class 12 exams are February 15–March 13, 2024 and February 15–April 2, 2024, respectively. You may obtain the CBSE exam date sheet 2024 PDF from the official CBSE website, cbse.gov.in.

To get the complete datesheet, you can visit our website by clicking on the link given below.

https://school.careers360.com/boards/cbse/cbse-date-sheet

Thank you

Hope this information helps you.

Hello aspirant,

The paper of class 7th is not issued by respective boards so I can not find it on the board's website. You should definitely try to search for it from the website of your school and can also take advise of your seniors for the same.

You don't need to worry. The class 7th paper will be simple and made by your own school teachers.

Thank you

Hope it helps you.

The eligibility age criteria for class 10th CBSE is 14 years of age. Since your son will be 15 years of age in 2024, he will be eligible to give the exam.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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