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NCERT Solutions for Class 10 Physics Chapter 12 : Electricity

NCERT Solutions for Class 10 Science Chapter 12 are made by our team of subject experts. Class 10 Electricity NCERT Solutions is framed strictly in accordance with the CBSE Curriculum and the exam pattern. These are the best resources designed after proper research and study to assist the students in scoring good marks. As Class 10 exams are Board exams, NCERT Solutions provided by Toppr are the best study material to excel in the exams. These solutions will not only help the students in preparing for the board exams but also for the Olympiads. With the help of Science Chapter 12 Electricity Class 10 NCERT Solutions you can also analyze your shortcomings and work on them before the exams.

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Introduction

Access NCERT Solutions for Class 10 Physics Chapter 12 : Electricity

In Text 12.1

Question 1

What does an electric circuit mean?

Medium

Solution

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$Explanation:$

$∙$ The closed-loop or path of electrical components in which electrons are able to flow. This path consists of electrical components source, like a battery, resistance, wire, bulb, etc.

$∙$ Electric circuits are useful in many ways like all of our electrical appliances run by a particular type of circuit.

Question 2

Define electric current. What is its

S I

unit?

Easy

Solution

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Electric current is basically the flow or charge of the electric charge in motion in a conductor. It is said to exist when there is a net flow of charge through the region.

the current is given as the flow of charge per second:

I = \frac{c h a r g e}{t i m e}

The SI unit of current is Ampere.

Question 3

Calculate the number of electrons constituting one coulomb of charge.

Medium

Solution

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we know that the magnitude of charge on one electron

= 1.6 \times 1 0^{- 19} C

So, the number of electrons constituting one coulomb of charge

= \frac{1}{1 . 6 \times 1 0 ^{- 19}}

= 6.25 \times 1 0^{18}

Question 4

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

1:2

2:1

1:4

4:1

Medium

Solution

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Since the two wires have same material, length and diameter, they have the same resistance. Let the resistance of each wire be R.

Let applied potential difference be

V

R_{s e r i e s} = R_{1} + R_{2}

= R + R = 2 R

P_{s e r i e s} = V^{2} / R_{s e r i e s}

= V^{2} / (2 R) . . . . . . . . . . (i)

\frac{1}{R _{p a r a l l e l}} = \frac{1}{R _{1}} + \frac{1}{R _{2}}

R_{p a r a l l e l} = R / 2

P_{p a r a l l e l} = V^{2} / R_{p a r a l l e l}

= 2 V^{2} / R . . . . . . . . . . . . (i i)

Dividing (ii) from (i),

\frac{P _{s e r i e s}}{P _{p a r a l l e l}} = 1 / 4

Question 5

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

Easy

Solution

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Explanation

∙

To measure the potential difference between two points, the voltmeter is connected to parallel to those points.

Question 6

A copper wire has diameter 0.5mm and resistivity of

1.6 \times 1 0^{- 8} Ω m

. 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?

Medium

Solution

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Given, diameter,

d = 0.5 m m

resistivity,

ρ = 1.6 \times 1 0^{- 8} Ω m

Resistance,

R = 10 Ω

Let the length of wire be

l

A = π d^{2} / 4

R = ρ l / A

= \frac{ρ l}{π d ^{2} / 4}

⟹ l = \frac{R π d ^{2}}{4 ρ}

l = \frac{1 0 \times 3 . 1 4 \times ( 0 . 5 \times 1 0 ^{- 3} ) ^{2}}{4 \times 1 . 6 \times 1 0 ^{- 8}}

l = 122.7 m

R \propto 1 / d^{2}

If the diameter is doubled, resistance will be one-fourth.

Hence, new resistance

= 2.5 Ω

Question 7

The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below -

I(amperes)	0.5	1	2	3	4
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.

Medium

Solution

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Plot of

I

V

is shown in the figure.

Resistance = \frac{1}{Slope}

From the graph,

Slope = \frac{1 - 0 . 5}{3 . 4 - 1 . 6} = 0.278 Ω^{- 1}

R = 1 / 0.278 = 3.6 Ω

Question 8

When a 12V 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.

Medium

Solution

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Given:

V = 12 V

I = 2.5 m A

Let the resistance be

R

By Ohm's Law,

V = I R

12 = 2.5 \times 1 0^{- 3} R

R = 4.8 \times 1 0^{3} Ω

Question 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 will flow through the

12 Ω

resistor?

Medium

Solution

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Hint:

Use the Ohm's law and the equivalence resistance in series combination.

Step 1 Finding the equivalence resistance

For resistors in series,

R_{e q} = R_{1} + R_{2} + R_{3} + R_{4} + R_{5}

= 0.2 + 0.3 + 0.4 + 0.5 + 12

= 13.4 Ω

Step 2: Finding the current.

Using the ohm's Law:

V = I R_{e q}

9 = 13.4 I

I = 0.67 A

When resistors are connected in series, the current is same in all the resistors. Hence, current in

12 Ω

resistor

= 0.67 A

Question 10

How many

176 Ω

resistors (in parallel) are required to carry 5A on a 220V line?

Medium

Solution

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Step 1 : Equivalent resistance

Let

n

be the number of resistors to be connected in parallel

Then,

R_{1} = R_{2} = . . . . . . = R_{n} = R = 176 Ω

Now,

\frac{1}{R _{e q}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + . . . . . \frac{1}{R _{n}}

\Rightarrow \frac{1}{R _{e q}} = \frac{n}{R _{1}} = \frac{n}{1 7 6}

\Rightarrow R_{e q} = 176 / n

. . . . (1)

Step 2 : Applying Ohm’s law

V = I R_{e q}

\Rightarrow 220 = 5 R_{e q}

\Rightarrow 220 = 5 \times 176 / n

(

Using Equation

(1))

\Rightarrow n = 4

Hence, four resistors are required.

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NCERT Solutions for Class 10 Physics Chapter 12 : Electricity

NCERT Science Chapter 12 - Electricity Class 10 Solutions

Class 10 Chapter 12 – Electricity explains how electricity works at the molecular level along with its applications. Electricity is one of the most important sources of energy today that drives and powers entire industries and businesses. Today, one cannot imagine a life without electricity. NCERT solutions for class 10 also discuss the important concepts such as electric current and charge, potential and potential difference, Circuit diagrams, Ohm’s law, Resistivity and Resistance, Factors that affect the Resistance of a Conductor, Parallel and Series Combination of Resistors and their applications, Heating Effect of Electric Current and its Applications, Electric Power, etc.

Key Features of for Class 10 Electricity NCERT Solutions

Class 10 NCERT Solutions provide a better understanding of the subject and concepts.
These are curated by the experts after thorough research.
They are the best means to evaluate your preparations and overcome your shortcomings.
The Class 10 NCERT Solutions will help the students in board exams as well as Olympiads.
These are absolutely free to download.

Related Chapters

Chapter 1 : Chemical reactions

Chapter 2 : Acids, Bases and Salts

Chapter 3 : Metals and Non metals

Chapter 4 : Carbon and its compounds

Chapter 5 : Periodic Classification of Elements

Chapter 6 : Life Processes

Chapter 7 : Control and Co-ordination in Plants and Animals

Chapter 8 : Reproduction

Chapter 9 : Heredity and Evolution

Chapter 10 : Light- Reflection and Refaction

Chapter 11 : The Human Eye and colourful World

Chapter 13 : Magnetic Effect of Electric Current

Chapter 14 : Sources of Energy

Chapter 15 : Our Environment

Chapter 16 : Management of Natural Resources

Frequently Asked Questions on NCERT Class 10 Physics Chapter 12 : Electricity

Question 1. Define Electric Power.

Answer. Electric power refers to the rate at which work is done or electrical energy is transformed into an electrical circuit per unit of time. It is commonly denoted by P and its SI unit is watt or one joule per second. Electric power is a Scaler quantity and is commonly supplied by electric batteries and produced by electric generators. The formula for electric power is: P = VI Where, P = electric power V = potential difference in the circuit I = electric current

Question 2. List the factors that influence the resistance of a conductor.

Answer. Resistance refers to the opposition to the flow of electrical current through a conductor. The resistance of an electric circuit is numerically measurable. Also, conductivity and resistivity are inversely proportional. Thus, more conductivity implies less resistance. Resistance = Potential difference/ Current The factors that influence the resistance of the conductor are as follows: The nature of the material of which the conductor is made up. The length of the conductor. The temperature of the conductor The cross-sectional area of the conductor.

Question 3. State the difference between Resistance and Resistivity.

Answer. Resistance refers to the property of the conductor that opposes the flow of electric current. Alternatively, it can also be defined as the ratio of the voltage applied to the electric current flowing through it. The resistance of a conductor depends on the nature of the material of which the conductor is made up, the length of the conductor, the area of cross-section, and the temperature of the conductor. Also, the resistance is directly proportional to the length of the conductor while it is inversely proportional to the area of the cross-section of the conductor. Resistivity refers to the resistance offered by the material per unit length for the unit cross-section of the conductor. Its SI unit is Ohms-meter. It is only proportional to the nature and temperature of the particular material. Also, it increases linearly with temperature. Moreover, the resistivity of conductors is low as compared to the resistivity of the insulators. Therefore, we can say that Resistivity of conductors < Resistivity of alloys < Resistivity of insulators.

Related Chapters

Chapter 1 : Chemical reactions

Chapter 2 : Acids, Bases and Salts

Chapter 3 : Metals and Non metals

Chapter 4 : Carbon and its compounds

Chapter 5 : Periodic Classification of Elements

Chapter 6 : Life Processes

Chapter 7 : Control and Co-ordination in Plants and Animals

Chapter 8 : Reproduction

Chapter 9 : Heredity and Evolution

Chapter 10 : Light- Reflection and Refaction

Chapter 11 : The Human Eye and colourful World