# NCERT Solutions Class 10 Science Chapter 12

## NCERT Solutions for Class 10 Science Chapter 12: electricity

Class 10 Science Chapter 12 is about electricity. Electricity is one of the most fundamental needs  of our society. Electricity has shaped our civilization ever since the dawn of the industrial revolution, powering entire industries and businesses. Today, we can’t think of life without electricity,  it’s an indispensable resource.

NCERT Solutions for Class 10 Science Chapter 12, provided by Extramarks, explains the essential concepts of electricity, electric current and circuit, circuit diagram, how electricity works at the molecular level, applications of electricity, etc. These solutions strive to impart maximum informational value and answers to all the textbook questions. Referring and practising thoroughly to our  Class 10 Science NCERT Solutions Chapter 12 can help students score high in class tests and in the CBSE  board examination.

Sign up at Extramarks’ website to get  access to our NCERT Solutions for Class 10 Science Chapter 12 and explore a repository of  study materials available for students of all classes from 1 to 12.

### Key Topics Covered In NCERT Solutions for Class 10 Science Chapter 12

Our NCERT Solutions for Chapter 12 Science Class 10 discusses electricity related topics in great detail.

Following are the key topics covered in this chapter:

Electricity

Definition

electricity is the amount of the flow of charges in a conductor and is part of electromagnetism. A few common phenomena associated with electricity include magnetism, lightning, static electricity, electric discharges and electric heating.

Any object with moving electric charges, either positive or negative, creates both electric and magnetic fields giving rise to electric current. A magnitude force will act on that object when a charge is placed in a non-zero electric field. Coulomb's law explains the magnitude of force. Hence, the electric field will work on the electric charge when the charges move.

Best electricity Conductors

Metals like Aluminium, Gold, Silver, Copper, Steel and Brass are considered the best electricity conductors. Students can learn more about good conductors and bad conductors from our NCERT Solutions for Class 10 Science Chapter 12.

Sources of electricity

The two types of electricity are:

1. Static electricity: Static electricity is generated by rubbing two or more objects to build friction. The first observations of this electrical phenomenon were made in ancient Greece by Thales of Miletus when amber bars were rubbed against tanned skin, which produced an attractive characteristic.
2. Current electricity: This type of electricity is produced by the flow of electric charges in a conductor. This is one of the accurate sources of electricity. Examples of generating current electricity include:
• The pressure applied by underground water flow is used as alternative energy in large ships to the central system.
• electricity in dams is produced by releasing a controlled flow of high-pressure water through a forced conduit.
• The water drives turbines that move the generators, producing an electric current. This low-voltage of high current passes through a voltage booster, transforming it into electricity.

The topic and difference between static electricity and current electricity has been elaborated by our Science subject matter experts  in our NCERT Solutions for Class 10 Science Chapter 12. Students can register at  Extramarks’ website and gain access to our  study solutions that can help them understand the important points easily and make them revise quickly

Electric Current and Circuit

As defined in our Class 10 Science Chapter 12 NCERT Solutions, Electric Current is the rate at which the electrons flow in a conductor. Electric current is measured in Ampere. Electrons are tiny particles that are within the molecular structure of a substance. Sometimes, these electrons are tightly held, and sometimes loosely. When the nucleus loosely holds electrons, they can travel freely within the body's limits.

Electrons are the negatively charged particles. When they move, several other particles move and this movement of electrons is known as electric current. We should note that the number of electrons that can move controls the capacity of a substance to produce electricity.

Some materials let current move better than others.

Based on the power of the material to conduct electricity, they are classified into two categories:

1.Conductors

The materials allow the free flow of electrons from one particle to another. The flow of electrons inside the conductor generates an electric current. The force required to drive the current flow through the conductor is voltage.

Examples of conductors include the human body and metals like iron, silver, gold, brass and copper.

2. Insulators

Insulators are materials that restrict the free flow of electrons from one particle to another. Eventually, the charge is seldom distributed evenly across the surface of an insulator.

Examples of insulators are plastic, glass and wood.

Further in our NCERT Solutions for Class 10 Science Chapter 12, the prerequisites for the electric current to flow in a conductor are discussed and understood through a circuit. The circuit comprises an energy source like a battery that produces voltage. Voltage creates pressure, also known as the electromotive force, on the electrons, channelling them to flow in a single direction.

The circuit is a closed loop through which electrons flow. A circuit is said to be closed when a switch is turned on.

Electric Potential and Potential Difference

Electric potential is a point in an electric field which is the work done in moving a unit of electric charge from infinity to that point opposite of electric forces.

On the other hand, a potential difference between two points in an electric field is the work done in moving a unit of electric charge from one point to another point against the electric force.

Potential difference= work done/charges i.e, V=W/Q .

Circuit Diagram

A circuit diagram is an electrical diagram of the components of an electrical circuit. In other words it’s a graphical representation of an electrical circuit which shows actual electrical connections. It shows the respective positions of all the components and their link to one another. Given below is an example of a simple circuit diagram.

Components of Circuit Diagram:

1. Electric cell: It is the source of current. The large terminal is positive, and the small terminal is negative.
2. Battery: The utility of this is equal to a cell.
3. Switch: Used to start or stop the current flow upon being pressed.
4. Wire Joint: Used to connect the devices, represented by drawing 'blobs' at the connection point.
5. Wires crossing without joining: In this, the cables do not touch each other and are shown without ‘blobs.’
6. Electric bulb: Devices used to glow using electricity.
7. Resistor (R): It controls the amount of current flow in the circuit.
8. Variable resistance: Regulating the amount of current flow by increasing or decreasing the resistance is done using variable resistance. It is also known as a rheostat.
9. Ammeter: The ammeter measures the current passing at a particular point.
10. Voltmeter measures the voltage between two points in a circuit.

Circuit diagrams are very important for understanding a lot of future concepts and theories in Physics. Our subject matter experts  have explained the circuit diagram with a step-by-step explanation  and illustrated with examples for students to comprehend this easily. We recommend students to  sign up at Extramarks’ website and check out what we have in store for you. Besides NCERT Solutions for Class 10 Science Chapter 12, every student registered at Extramarks can access study materials on all subjects and boost their academic performance.

Ohm’S Law

Georg Simon Ohm was the first to verify Ohm's law experimentally. The law explains the relationship between electric current and potential difference. The current that flows through the conductors is directly proportional to its voltage, provided the physical condition and the temperature remain constant.

The law equation is V=IR, where V is the voltage, I is the current flowing through the conductor, and R is the resistance.

Applications of Ohm's law:

• To ascertain the voltage, resistance or current of an electric circuit.
• To maintain the required voltage drop across the electronic devices.
• Used in DC ammeter and DC shunts to reroute the current.

Constraints of Ohm’s Law:

• Electrical elements like diodes and transistors only allow the current to flow through in one direction. Hence Ohm's law cannot be applied here.
• The voltage and current are not constant with time for non-linear electrical elements with parameters like capacitance, resistance etc., making it challenging to apply Ohm's law.

Ohm’s Law will be extensively used in higher grades in Science stream  and we recommend students to check their understanding by solving sample papers, Long and short questions , MCQs to strengthen their  understanding  from  our NCERT Solutions for Class 10 Science Chapter 12.  Solutions cover various concepts  of the chapter and are considered ideal study material for examinations and score high marks. No wonder students have complete faith and trust in Extramarks

Factors affecting the Resistance of a Conductor

Resistance is the polarity to the flow of electrical current through a conductor. Resistance = Potential difference/ Current. The resistance of the conductor depends on factors like temperature of the conductor, length of the conductor, type of material of the conductor and the cross-sectional area.

Resistance of A System of Resistors

A resistor is a non-resistant two-terminal electrical component that works as an electrical resistance in the circuit. Resistors decrease the electric charges and voltage levels within circuits. Almost all the circuits often have more than one resistor.

The two most direct connections of resistors are:

• Series.
• Parallels.

1. Resistors in series

When an exact amount of current flows through the resistor, but the voltage across each resistor is different, the circuit is connected in series, as shown below.

The total resistance is calculated as Rtotal=R1+R2+…..+Rn. In this type of connection, if the resistor or any other component is broken, the entire circuit is turned off.

1. Resistors in Parallel
• The voltage remains the same, and the electric current is branched out and recombined when the branches meet at the same point. If a resistor or other component appears broken, it can be disconnected easily without affecting other elements in a parallel circuit. The total resistance is calculated as 1/Rtotal=1/R1+1/R2+…..+1/Rn.

Heating Effect Of Electric Current

The heating effect of Electric Current is experienced; by passing an electric current through a conductor, the conductor becomes hot and generates heat. This application is widely used across devices like electric coils, bulbs, iron, heater, fuse, etc.

Electric Power

Chapter 12 Science Class 10 defines electric power as the rate of energy transformed into an electrical circuit. Simply put, it measures how much energy is used in a period. The equation of electric power is P=VI, where P is the power, V is the potential difference, and I is the electric current.

### NCERT Solutions for Class 10 Science Chapter 12: Exercises &  Solutions

NCERT Solutions for Class 10 Science Chapter 12, provided by Extramarks, covers all essential topics from the examination point of view. Our NCERT Solutions comprises chapter notes, various questions such as MCQs, short and long answer questions, and their solutions, revision notes, etc. Students may refer to and practise different types of questions consolidated in NCERT Solutions for Class 10 Science Chapter 12 by clicking on the link given below.

NCERT Solutions - MCQ's

Furthermore, in addition to NCERT Solutions for Class 10 Science Chapter 12, students may access NCERT Solutions for all other chapters. Students may click on the respective links to refer to NCERT Solutions  for other classes from 1 to 12. Students need to make use of these solutions to excel in academics and stay ahead of the pack.

NCERT Solutions Class 12

NCERT Solutions Class 11

NCERT Solutions Class 10

NCERT Solutions Class 9

NCERT Solutions Class 8

NCERT Solutions Class 7

NCERT Solutions Class 6

NCERT Solutions Class 5

NCERT Solutions Class 4

NCERT Solutions Class 3

NCERT Solutions Class 2

NCERT Solutions Class 1

### NCERT Solutions Class 10 Science Exemplar

By practising these advanced level of NCERT Exemplar questions, students can have better clarity on the concepts like Electric Current and Circuit, Factors affecting the Resistance of a Conductor, etc., and also develop better problem-solving skills to be confident in facing all types of questions in the examination.

The Exemplar book is a collection of a variety of questions for students to practice and become strong   in  their respective subjects. Students can enjoy a repository  of questions covered in Exemplars - MCQs, short answer questions, long format questions, etc.

While solving the questions from Exemplar, students will also get to revise key topics of the chapter. They will understand where they lack or need little more practice. . Students can clear their doubts with the help of detailed and authentic solutions and become more confident.

Students can first study the chapter from NCERT textbook and refer to our NCERT Solutions for Class 10 Science Chapter 12. Then they can solve questions given in NCERT Exemplar. This way students will have a comprehensive understanding of the electricity  chapter. Our study notes also contain solutions for all the chapters  covered in NCERT textbooks

Furthermore, students can refer to the chapter-wise Exemplar questions and solutions for Class 10 Science, available on the Extramarks’ website.

• NCERT Solutions Class 10 Science - Exemplar Questions and Answers

### Key features of NCERT Solutions for Class 10 Science Chapter 12

Extramarks is one of  the  best online learning platforms for a wide range of subjects for lakhs of students from Classes 1 to  12. Our academic research team has prepared the study materials based on the latest CBSE syllabus and as per the NCERT guidelines. Students can confidently rely on our learning solutions for their studies and exam preparation.

### Benefits of NCERT Solutions for Class 10 Science Chapter 12 include:

• Our NCERT Solutions are structured to present all the Chapter 12 topics in an easy to understand language and provide authentic, reliable and to the point study material. .
• NCERT Solutions provides detailed Solutions to the chapter questions that help students to fetch more marks in the examinations. It helps them to continue their learning experience with Extramarks which tries to do away with rote learning and supplements their studies  with experiential learning and other innovative educational materials.
• The notes are prepared by highly qualified and experienced faculty with decades of experience to make you understand the concepts easily and effectively.
• Our NCERT Solutions for Class 10 Science Chapter 12 provides well-explained answers to all the questions  in the textbook. Also, the tricky in text questions wherein   apply the concepts learned in the chapter.
• The content on NCERT Solutions provided by Extramarks is regularly updated as per the CBSE Board and the NCERT syllabus.

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

$\frac{\mathrm{R/}}{\mathrm{R}‘}$

is

Ans-

$\begin{array}{l}\left(\mathrm{a}\right)\text{}\frac{1}{25}\end{array}$ $\begin{array}{l}\left(\mathrm{b}\right)\text{}\frac{1}{5}\end{array}$ $\begin{array}{l}\left(\mathrm{c}\right)\text{5}\end{array}$ $\begin{array}{l}\text{(d) 25}\end{array}$

The correct option is (d).

$\begin{array}{l}\text{The resistance of a piece of wire is}\mathrm{R}\text{and it is proportional}\\ \text{to the length of wire. As the wire is cut into five equal parts}\\ \text{h}\mathrm{ence},\text{resistance of each part =}\frac{\mathrm{R}}{5}\\ \text{Let the equivalent resistance of these piece of wire in}\\ \text{parallel is R’ then,}\\ \frac{1}{\mathrm{R}‘}=\frac{5}{\mathrm{R}}+\frac{5}{\mathrm{R}}+\frac{5}{\mathrm{R}}+\frac{5}{\mathrm{R}}+\frac{5}{\mathrm{R}}=\frac{25}{\mathrm{R}}\\ \mathrm{or},\text{}\frac{\mathrm{R}}{\mathrm{R}‘}\text{= 25}\end{array}$

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

(a) I2R
(b) IR2
(c) VI

Ans-

$\left(\mathrm{d}\right)\frac{{\mathrm{V}}^{2}}{\mathrm{R}}$ $\begin{array}{l}\text{Electrical power is given by,}\\ \text{P= VI}\dots \text{(1)}\\ \text{According to ohm’s law,}\\ \text{V = IR}\dots \text{(2)}\\ \text{Where, V = potential difference}\\ \text{I = current}\\ \text{R = resistance}\\ \therefore \text{P = VI}\\ \text{From equation (1), P = (IR)×I}\\ \therefore {\text{P = I}}^{\text{2}}\text{R}\\ \text{From equation (2),}\\ \text{I =}\frac{\text{V}}{\text{R}}\\ \therefore \text{P =}\frac{{\text{V}}^{\text{2}}}{\text{R}}\text{}\\ {\text{Hence, P = VI = I}}^{\text{2}}\text{R =}\frac{{\text{V}}^{\text{2}}}{\text{R}}\\ {\text{Thus, IR}}^{\text{2}}\text{does not represent electrical power}\text{.}\end{array}$

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

Ans-

$\begin{array}{l}{\text{Given, V}}_{\text{1}}{\text{= 220 V, V}}_{\text{2}}=110\text{V}\\ {\text{P}}_{\text{1}}=\text{100 W}\\ \mathrm{On}\text{using the relation,}\\ {\text{P}}_{\text{1}}\text{=}\frac{{\text{V}}_{1}^{2}}{\mathrm{R}}\\ \text{or, R =}\frac{{\mathrm{V}}_{1}^{2}}{{\mathrm{P}}_{1}}\text{=}\frac{{\left(220\text{V}\right)}^{2}}{\left(100\text{W}\right)}=484\text{}\mathrm{\Omega }\\ \text{If the bulb is operated on 110 V, then the energy}\\ \text{consumed,}\\ {\text{P}}_{\text{2}}=\frac{{\text{V}}_{2}^{2}}{\mathrm{R}}=\text{}\frac{{\left(110\text{V}\right)}^{2}}{484\text{\hspace{0.17em}}\mathrm{\Omega }}=\text{25 W}\\ \text{Therefore, the power consumed will be 25 W.}\end{array}$

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

Ans-

$\begin{array}{l}\text{The correct option is}\left(\text{c}\right).\\ \text{Suppose, R = resistance of each wire}\\ \therefore \text{The equivalent resistance in series}\\ {\text{R}}_{\text{s}}\text{= R+R = 2R}\\ \text{If V is the applied potential difference, then it is the}\\ \text{voltage across the equivalent resistance.}\\ {\text{V = I}}_{\text{s}}×{\text{2R (I}}_{\text{s}}\text{= current in series combination)}\\ \\ ⇒{\text{I}}_{\text{s}}\text{=}\frac{\mathrm{V}}{2\mathrm{R}}\\ \mathrm{The}\text{heat dissipated in time t is,}\\ {\text{H}}_{\text{s}}={\text{I}}_{\mathrm{s}}^{2}×2\mathrm{R}×\mathrm{t}={\left(\frac{\mathrm{V}}{2\mathrm{R}}\right)}^{2}×2\mathrm{R}×\mathrm{t}\\ ⇒{\text{H}}_{\text{s}}=\frac{{\mathrm{V}}^{2}\mathrm{t}}{2\mathrm{R}}\text{}...\text{(i)}\\ \mathrm{The}\text{equivalent resistance of the parallel connection is}\\ {\text{R}}_{\text{p}}=\frac{1}{\frac{1}{\mathrm{R}}+\frac{1}{\mathrm{R}}}=\frac{\mathrm{R}}{2}\\ \mathrm{V}{\text{is the applied potential difference across R}}_{\text{p}}.\\ \therefore \text{}\mathrm{V}={\text{I}}_{\mathrm{p}}×\frac{\mathrm{R}}{2}\end{array}$

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

Ans-

A voltmeter need to be connected in parallel in a circuit to measure the potential difference between two points.

Q.6 A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10−8 Ωm. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?

Ans-

$\begin{array}{l}\text{Given,}\rho =1.6×{10}^{-8}\text{}\Omega \text{m}\\ \text{A =}\pi {\text{r}}^{\text{2}}\\ r=\frac{0.5}{2}mm=\frac{0.0005}{2}m\\ \text{R = 10}\Omega \\ Hence,\text{length of the wire}\\ \text{l =}\frac{RA}{\rho }=\frac{10×3.14{\left(\frac{0.0005}{2}\right)}^{2}}{1.6×{10}^{-8}}=122.72\text{m}\\ \text{If the diameter of the wire is doubled then, new diameter}\\ \text{= 2}×\text{0}\text{.5 mm =1 mm =0}\text{.001m}\\ \text{Hence, Resistance R’ =}\rho \frac{l}{A}=\frac{1.6×{10}^{-8}×122.72}{\pi ×{\left(\frac{1}{2}×{10}^{-3}\right)}^{2}}\\ \text{= 2}\text{.5}\Omega \end{array}$

Q.7 The value of current (I) flowing in a given resistor for the corresponding values of potential difference (V) across the resistor are given below-

 I (amperes) 0.5 1 2 3 4.0 V (volts) 1.6 3.4 6.7 10.2

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

When the graph is plotted between voltage and current then, it is called V-I characteristics. Here, the voltage is plotted on X axis and current is plotted on Y axis.

$\begin{array}{l}\mathrm{The}\text{slop of the line gives the value of resistance (R)}\\ \text{as,}\\ \text{Slope, =}\frac{1}{\mathrm{R}}=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{2}{6.8}\\ \therefore \mathrm{R}=\text{}\frac{6.8}{2}=3.4\text{}\mathrm{\Omega }\text{}.\end{array}$

Q.8 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.

$\begin{array}{l}\mathrm{According}\text{to ohm’s law,}\\ \text{V = IR}\\ \text{or, R =}\frac{\mathrm{V}}{\mathrm{I}}\\ \text{Here, potential difference, V = 12 volts}\\ \text{i = 2.5 mA}\\ \text{=2.5}×{\text{10}}^{\text{-3}}\text{A}\\ \therefore \text{R =}\frac{12}{2.5×{\text{10}}^{\text{-3}}}\text{=4.8}×{\text{10}}^{\text{-3}}\mathrm{\Omega }.\text{}\end{array}$

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

Ans-

$\begin{array}{l}\text{The current would be same throughout the series circuit.}\\ \text{Equivalent resistance of the circuit,}\\ {\text{R}}_{\text{eq}}=\left(0.2+0.3+0.4+0.5+12\right)\text{}\mathrm{\Omega }=13.4\text{}\mathrm{\Omega }\text{}\\ \mathrm{V}=\text{9 V}\\ \text{By using ohm’s law,}\\ \text{V =IR}\\ \mathrm{or},\text{I =}\frac{\mathrm{V}}{\mathrm{R}}\\ \therefore \text{I =}\frac{9}{13.4\text{ }\mathrm{\Omega }}\text{= 0.671 A}\\ \text{Therefore, the current that would flow through}\\ \text{12}\mathrm{\Omega }\text{resistor is 0.671 A.}\end{array}$

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

$\begin{array}{l}Given,\text{V = 220 volt}\\ \text{I = 5 A}\\ \text{Equivalent resistance (R) of the combination is}\\ \text{given as}\\ \text{}\frac{1}{R}\text{= x}×\frac{1}{176}\\ \text{or, R =}\frac{176}{x}\\ \text{From ohm’s law,}\\ \text{}\frac{V}{I}=\frac{176}{x}\\ or,\text{x =}\frac{176×5}{220}=4\\ Hence,\text{4 resistors of 176}\Omega \text{are required to draw the}\\ \text{given amount of current}\text{.}\end{array}$

Q.11 Show how would you connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of (i) 9 Ω (ii) 4 Ω.

(i)

$\begin{array}{l}\text{When two resistors are connected in parallel and}\\ \text{the third one is connected in series with these}\\ \text{two then, Two 6}\mathrm{\Omega }\text{resistors are connected in parallel.}\\ \text{Their equivalent resistance (R’) will be}\\ \text{}\frac{1}{\mathrm{R}‘}=\text{}\frac{1}{6\text{ }\mathrm{\Omega }}+\frac{1}{6\text{ }\mathrm{\Omega }}=\frac{2}{6\text{ }\mathrm{\Omega }}=\frac{1}{3\text{ }\mathrm{\Omega }}\\ \\ \text{or, R’ = 3}\mathrm{\Omega }\\ \text{The third 6}\mathrm{\Omega }\text{resistor is connected in series with 3}\mathrm{\Omega }.\\ {\text{Hence, the equivalent resistance R}}_{\text{1}}\text{= (6+3)}\mathrm{\Omega }\text{= 9}\mathrm{\Omega }\end{array}$

(ii)

$\begin{array}{l}\text{Two resistors of 6 Ω are in series. Their equivalent}\\ \text{resistance will be = (6}+6\text{) Ω = 12 Ω}\\ \text{The third 6 Ω resistor is connected in parallel with 12 Ω.}\\ {\text{Hence, equivalent resistance (R}}_{\text{2}}\text{)}\\ \frac{\text{1}}{{\text{R}}_{\text{2}}}\text{=}\frac{\text{1}}{\text{12\hspace{0.17em}Ω}}\text{+}\frac{\text{1}}{\text{6 Ω}}\text{=}\frac{\text{3}}{\text{12 Ω}}\text{=}\frac{\text{1}}{\text{4 Ω}}\\ {\text{or, R}}_{\text{2}}\text{= 4 Ω}\end{array}$

Question

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?

Ans-

$\begin{array}{l}\mathrm{Given},\text{V = 220 V}\\ \text{i = 5 A}\\ {\text{P}}_{\text{1}}\text{= 10 W}\\ \text{On using the relation,}\\ \text{P =}\frac{{\mathrm{V}}^{2}}{{\mathrm{R}}_{1}}\\ {\text{or, R}}_{\text{1}}\text{=}\frac{{\mathrm{V}}^{2}}{{\mathrm{P}}_{1}}\\ \therefore \text{=}\frac{{\left(220\right)}^{2}}{10}\text{= 4840}\mathrm{\Omega }\\ \text{As per ohm’s law,}\\ \text{V= IR}\\ \therefore \text{R =}\frac{\mathrm{V}}{\mathrm{I}}\text{=}\frac{220}{5}=44\text{}\mathrm{\Omega }\\ \mathrm{Resistance}\text{of each electric bulb,}\\ {\text{R}}_{\text{1}}=4840\text{}\mathrm{\Omega }\\ \therefore \text{}\frac{1}{\mathrm{R}}=\frac{1}{{\mathrm{R}}_{1}}+\frac{1}{{\mathrm{R}}_{1}}+.\dots +\mathrm{upto}\text{x times}\\ \text{or, x =}\frac{{\mathrm{R}}_{1}}{\mathrm{R}}=\frac{4840}{44}=110\\ \mathrm{Thus},\text{110 electric bulbs are connected in parallel.}\end{array}$

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 resistances which may be used separately, in series or in parallel. What are the currents in the three cases?

Ans-

$\begin{array}{l}\mathrm{Given},\text{Supply voltage, V = 220 V}\\ \text{Resistance, R = 24}\mathrm{\Omega }\text{}\\ \text{(i) When coils are used separately:}\\ \text{By using the relation,}\\ {\text{V = I}}_{\text{1}}{\text{R}}_{\text{1}}\\ {\text{Where, I}}_{\text{1}}\text{= current flowing through the coil}\\ \text{}\therefore {\text{I}}_{\text{1}}\text{=}\frac{\text{V}}{{\text{R}}_{\text{1}}}\text{=}\frac{\text{220}}{\text{24}}\text{= 9.166 A.}\\ \\ \text{(ii) When coils are connceted in series:}\\ \text{total resistance of the two coils}\\ {\text{R}}_{\text{2}}\text{= (24+24)}\mathrm{\Omega }\text{= 48}\mathrm{\Omega }\text{}\\ \text{By using the relation,}\\ {\text{V = I}}_{2}{\text{R}}_{2}\\ \therefore {\text{I}}_{\text{2}}\text{=}\frac{\text{V}}{{\text{R}}_{\text{2}}}\text{=}\frac{\text{220}}{\text{48}}\text{= 4.58 A.}\\ \text{(iii) When coils are connceted in parallel:}\\ {\text{Total resistance (R}}_{\text{3}}\right)\text{would be}\\ \text{}\frac{1}{{\text{R}}_{\text{3}}}=\frac{1}{24}+\frac{1}{24}=\frac{2}{24}\\ \mathrm{or},{\text{R}}_{\text{3}}=12\mathrm{\Omega }\\ \text{On using the relation}\\ {\text{V = I}}_{3}{\text{R}}_{3}\\ \therefore {\text{I}}_{3}=\text{}\frac{\mathrm{V}}{{\text{R}}_{3}}=\text{}\frac{220}{12}=18.33\text{A.}\end{array}$

Q.14 Compare the power used in the 2 Ω 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.

Ans-

$\begin{array}{l}\left(\mathrm{i}\right)\text{Given, Potential difference, V = 6 V}\\ {\text{R}}_{\text{1}}{\text{= 1 Ω, R}}_{\text{2}}\text{= 2 Ω}\\ \text{}\therefore {\text{Equivalent resistance in series, R = R}}_{\text{1}}{\text{+ R}}_{\text{2}}\\ \text{}=\text{(1+2) Ω = 3 Ω}\\ \text{On using the relation,}\\ \text{V = IR (I = current through the circuit)}\\ \\ \text{}\therefore \text{I =}\frac{\text{V}}{\text{R}}\text{=}\frac{\text{6}}{\text{3}}\text{= 2 A}\\ \text{Now, power in the circuit}\\ {\text{P = I}}^{\text{2}}{\text{R = (2)}}^{\text{2}}\text{×2=8 W.}\\ \text{(ii) Given, potential difference, V = 4 V}\\ {\text{R}}_{\text{1}}=\text{12}\mathrm{\Omega },{\text{R}}_{2}=2\text{}\mathrm{\Omega }\\ \text{The voltage across each component of a}\\ \text{parallel circuit remains the same. Hence, voltage across}\\ \text{}2\text{}\mathrm{\Omega }\text{resistor will be 4 V.}\\ \text{Power consumed by 2}\mathrm{\Omega }\text{is given by}\\ \text{P =}\frac{{\mathrm{V}}^{2}}{\mathrm{R}}=\frac{{4}^{2}}{2}=\text{8 W.}\end{array}$

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

Ans-

$\begin{array}{l}\mathrm{The}\text{}\mathrm{potential}\text{difference across each bulb will be 220}\\ \text{V because no division of voltage occurs in a parallel}\\ \text{circuit.}\\ \text{Current drawn by the bulb of rating 100 W is given by,}\\ {\text{P = VI}}_{\text{1}}\\ \therefore {\text{I}}_{\text{1}}\text{=}\frac{\mathrm{P}}{\mathrm{V}}=\frac{100W}{220V}\\ \mathrm{Similarly},\text{current drawn by the bulb of rating 60 W is}\\ \text{given by,}\\ {\text{I}}_{\text{2}}\text{=}\frac{\mathrm{P}}{\mathrm{V}}=\frac{60W}{220V}\\ \therefore \text{Current drawn from the line =}\frac{100W}{220V}+\frac{60W}{220V}=0.727\text{}\mathrm{A}\end{array}$

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

Ans-

$\begin{array}{l}\text{Energy consumed by a TV set of power 250 W in 1 h}\\ \text{is given by the expression}\\ \text{H = Pt = 250 W}×\text{3600 s = 9}×{\text{10}}^{\text{5}}\text{}\mathrm{J}\\ \mathrm{Similarly},\text{}\mathrm{energy}\text{consumed by a toaster of power 1200 W}\\ \text{in 10 minutes = 1200}×\text{600 = 7.2}×{\text{10}}^{\text{5}}\text{}\mathrm{J}\\ \text{Hence, the energy consumed by a 250 W TV in 1 h is more than the energy consumed by a toaster of power 1200 W in 10 minutes.}\end{array}$

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

Ans-

$\begin{array}{l}\mathrm{Given},\text{R = 8}\mathrm{\Omega }\\ \text{I = 15 A}\\ \mathrm{Rate}\text{of heat produced can be expressed as power,}\\ {\text{P = I}}^{\text{2}}\mathrm{R}\\ \mathrm{On}\text{putting the values,}\\ \text{P =}{\left(\text{15 A}\right)}^{2}\text{\hspace{0.17em}}×8=1800\text{W or}1800\text{}\frac{\mathrm{J}}{\mathrm{s}}.\end{array}$

Q.18 Explain the following.

(a) Why is the tungsten used almost exclusively for filament of electric lamps?
(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?
(c) Why is the series arrangement not used for domestic circuits?
(d) How does the resistance of a wire vary with its area of cross-section?
(e) Why are copper and aluminium wires usually employed for electricity transmission?

Ans-

(a) Due to high melting point and high resistivity of tungsten, it does not burn at a high temperature. As the electric lamps glow at very high temperatures hence, the tungsten is used as heating element of electric bulbs.
(b) Bread toasters and electric irons are made of alloy because resistivity of an alloy is more than that of pure metals. Due to high resistivity, they produce large amount of heat.
(c) In a series circuit, there is a voltage distribution. Each element of a series circuit receives a small voltage for a large supply voltage. Thus, the amount of current decreases and the device becomes hot. Therefore, series arrangement is not used in domestic circuits.
(d) Resistance (R) of a wire is inversely proportional to its area of cross-section (A).
(e) The wires made of copper or aluminium have low resistivity and they are good conductors of electricity. Therefore, they are frequently used for electricity transmission.

1. Why should students refer to the NCERT Solutions for Class 10 Science Chapter 12 on Extramarks?

The NCERT Solutions Class, 10 Science Chapter 12 on Extramarks, explains the concepts of electricity well with various diagrams and provides answers to all the questions present in the textbook. Students will also improve their time management skills which are important and necessary  from the examination point of view.

2. Which are the important questions from the electricity Chapter?

Some of the important  questions are:

• Define electric current? Name its SI unit.
• What is the electrical resistivity of a material? Mention its SI unit.

3. Is NCERT Solutions for Class 10 Science Chapter 12 enough for reference and examination practice?

By referring to NCERT Solutions for Class 10 Science Chapter 12 provided by Extramarks and studying from the NCERT textbooks, authentic and reliable notes and solutions provided by Extramarks will definitely prove useful. Students can ace the exam with excellent results.

4. Is Class 10 Science Chapter 12 difficult?

Class 10 Science is  comparatively difficult to understand as the subject  .In Science, it’s very important for students to understand the concepts and fundamentals of each topic. Once the fundamental concepts are clearly understood it becomes easy to understand these topics in higher classes. However if students are regular in their school lectures and  study regularly from the NCERT textbook, then it won’t be that difficult. .

Extramarks understands how important it is to give step-by-step explanations for making the concepts easy for students and even clarify their doubts via live classes.