वह कंचे लेना चाहता था लेकिन स्कूल की घंटी सुनते ही दौड़ पड़ा। स्कूल में देरी से आने पर वह सबसे पीछे बैठा। उसके सहपाठी रामन, मल्लिका, अम्मू आदि आगे बैठे थे। जार्ज जो उसका सहपाठी था, आज बुखार होने के कारण स्कूल नहीं आया था। वह उसके बारे में सोचने लगा क्योंकि वह कंचे का अच्छा खिलाड़ी था। मास्टर साहब उस समय रेलगाड़ी के बारे में पढ़ा रहा था परंतु अप्पू का ध्यान पढ़ाई में नहीं था। वह अभी भी कंचे के बारे में सोच रहा था। इतने में ही उसे एक चॉक का टुकड़ा आ लगा और वह खड़ा हो गया। मास्टर जी उसके पास आकर डाँटने लगे। मास्टर जी उसका चेहरा देखकर समझ गए कि इसका ध्यान कहीं और था। उन्होंने अप्पू से प्रश्न पूछा जिसका जवाब वह नहीं दे पाया। मास्टर जी ने उसे बेंच पर खड़ा कर दिया। सभी बच्चे उसकी हँसी उड़ा रहे थे। वह रोने लगा। बेंच पर खड़ा अप्पू अभी भी कंचों के बारे में ही सोच रहा था। वह सोच रहा था कि जॉर्ज के आने पर कंचे खेलेगा। जॉर्ज के साथ वह दुकानदार के पास जाएगा।

शाम को वह इधर-उधर घूमता रहा। मोड़ पर उसी दुकान पर पहुँचकर वह शीशे के जार में रखे कंचे देखने लगा। उसने अपनी फ़ीस के एक रुपया पचास पैसे के उस दुकानदार से कंचे खरीद लिए। जब वह कंचे लेकर घर आ रहा था तो रास्ते में उसे देखने के लिए जैसे ही कागज़ की पुड़िया खोला तो सारे कंचे बिखर गए। अब वह उन्हें चुनने लगा। अपनी किताबें बाहर निकाल वह कंचे बस्ते में डालने लगा। वह उसे चुनने लगा तभी एक गाड़ी आई और वहाँ रुक गई। गाड़ी की ड्राइवर को अप्पू पर बहुत गुस्सा आया पर उसे खुश देख वह मुसकराकर चला गया। जब अप्पू घर पहुँचा और माँ को कंचा दिखाया तो माँ इतने सारे कंचे देखकर हैरान हो गई| अप्पू ने बताया कि वह फ़ीस के पैसों से ये कंचे खरीद लाया है। माँ ने कहा कि अब खेलोगे किसके साथ? यह कहकर माँ रोने लगी क्योंकि उसकी एक बहन थी मुन्नी, जो अब दुनिया में नहीं रही थी। तब अप्पू ने माँ से पूछा कि आपको कंचे अच्छे नहीं लगे। माँ उसकी भावनाओं को समझ गई और हँसकर बोली बहुत अच्छे हैं।

Course A: This course has two textbooks, Kshitiz and Kritika. The compositions of Kshitiz textbook give literary sensations in the students while at the same time also combine them with various issues of life. The textbook comprises the compilations of Bhakti Kal. The prose are of famous writers such as Premchand, Rahul Sankrityayan, Mahadevi Verma and Hazariprasad Diwedi etc. On the other side Kritika, supplementary reader contains five chapters which are lengthy and informative in nature.

Course B: This course has also two textbooks, Sparsh and Sanchyan. Sparsh textbook is divided into two parts-prose section and poetic section. In the prose section, the sequence of lessons is done according to grammatical points of the composition texts. In the poetic section the poets have been kept in their chronological order. Ramdhari Singh Dinkar and Harivansh Rai Bachchan are contemporary poets. So here the Dinkar has been placed before and 'Bachchan' after him. Sanchayan textbook has total of six chapters which includes story, memoir, autobiography and travelogue. Translated compositions have been also given place.

History textbook has five chapters of which you have to study four chapters. In the first chapter, we will learn about the French Revolution which led to the end of monarchy in France. This revolution gave the ideas of of liberty, freedom and equality which is basic fundamentals of democracy. In the second chapter, you will read about the coming of socialism in Europe, and the dramatic

events that forced the ruling monarch, Tsar Nicholas II, to give up power.

CBSE has limited the course of English and now only English language and literature course is in the syllabus. This course has two textbooks First Flight and Footprints without Feet. Here you can find the NCERT Solutions for Class 10th English of both textbook. The latter textbook is supplementary reader and two short answer questions and one long answer type questions will be asked from this book in the examination. We have provided the detailed and accurate answers of these textbooks through which you easily understand the theme and concepts of chapters. Select your desired textbook from the list.

• NCERT Solutions for Class 10th First Flight English
• NCERT Solutions for Class 10th Footprints without Feet English

NCERT Solutions are one of the most important things which you need practice to ace in the examination. English NCERT Solutions will also build your language and thinking skills that will help you in future examinations also.

NCERT Solutions are one of the most basic and sought thing but finding NCERT Solutions are that are reliable, accurate and detailed is not an easy task. But NCERT Solutions prepared by our experts try to provide all round clarity of questions. Thus, you can trust on us as you're in good company. You can also take help of these solutions in formulating your own answers that will improve your thinking skills and build your language skills at the same time.

CBSE has removed Course Communicative and students are only left with English Language and Literature course. This course has been designed to help the children to learn to communicate in English with confidence and accuracy. NCERT Solutions for Class 9 English of both textbook, Beehive and Moments. These NCERT Solutions are very beneficial for the purpose of the examination. You can select your desired textbook from the list to get well prepared solutions.

• NCERT Solutions for Class 9 Beehive English
• NCERT Solutions for Class 9 Moments English

Class 9 English contains rich variety of reading material has been provided to include the literary,
cultural and sociological dimensions of texts. It is a core subject thus students need to prepare well in order to get more marks in the examination.

Beehive NCERT Solutions for Class 9 English

The book contains texts of variety of genres such as story, biography and autobiography, science fiction, humour, travelogue, and the one-act play. The number of poems are increased so that students can understand and enjoy the music of words. The poems are also of various types such as the lyric, the ballad and the humorous poem. Number of total chapters are eleven and each chapter has a poem. Before the beginning of each unit, activity based questions are provided in 'Before you read' Section. The section ‘Thinking about the Text’ contains such questions which are useful for the purpose of the exams.

Moments NCERT Solutions for Class 9 English

It is a supplementary reader which is meant for extensive reading. The pieces in the book has stories of mystery, adventure, courage, growing up, romance, wit and humour. There are total ten chapters. The few questions given under ‘Think about It’ are designed to help learners develop their ability of intelligent and imaginative reading. The ‘Talk about It’ section has discussion topics. Only the first section is important for the purpose of the examination.

Hindi subject is divided into two courses in Class 10th, Course A and B. A student only has to study one of these courses. NCERT Solutions for Class 10 Hindi for both courses are provided here. Each course has two textbooks. These solutions are prepared by the experts of StudyRankers which are detailed and accurate so you don't need to waste your time in finding exact answers. Select your desired textbook from the list to start reading NCERT Solutions.

• NCERT Solutions for Class 10 Kshitiz Hindi
• NCERT Solutions for Class 10 Kritika Hindi
• NCERT Solutions for Class 10 Sparsh Hindi
• NCERT Solutions for Class 10 Sanchyan Hindi

Class 9th Hindi Language subject textbook contains various pieces of writings that will enable learners to analyse the moral values and understand literary arts in well manner. Hindi is one of the main language subject which have mass students and learners. You also need to study Grammar and Writing section in order to score more marks in the examination.

Course A

Kshtiiz and Kritika are the textbooks of this course. Kritika is supplementary reader while Kshitiz is the main textbook. The main textbook consist of seventeen chapters, of which eight are prose works
and nine are poems. The poems are sequenced according to chronological order so that students can understand the literary writings of Bhakti Kal, Ritikal and modern times and become familiar with the development of the Hindi poetry. The prose part includes story, memoir, drawing, autobiography, etc. with thoughtful and sarcastic essays.

On the other hand, Kritika supplementary textbook has been created according to the age, interest and merit of the students. Five compositions have been compiled in the textbook. These compositions are specific in its story, craft and presentation.

Course B

Sparsh and Sanchyan are textbook of Course B. The main textbook Sparsh contains different types of prose such as essays, memorable, drama, folklore etc. This will help learner in understanding the difficult terms and words of the Hindi language and build their vocabulary.

Sanchayan textbook has three chapters. The chapter will introduce the students to cultural heritage in the field of folk tradition, literature, art, science, society etc. and develop positive attitude towards them. Thus, careful selection of the classical text has been done.

Hindi language subject is one of the core subject of masses. CBSE has divided Class 9th Hindi into two course, Course A and B. Student only have to opt for a single course. There are two textbook in both courses. We have provided the NCERT Solutions for Class 9 Hindi of all four textbooks. These solutions are helpful in getting maximum marks in the examination. You can get solutions of your desired textbook by clicking on the link provided below.

• NCERT Solutions for Class 9 Kshitiz Hindi
• NCERT Solutions for Class 9 Kritika Hindi
• NCERT Solutions for Class 9 Sparsh Hindi
• NCERT Solutions for Class 9 Sanchyan Hindi

Course A

Kshitiz and Kritka are the textbooks of Course A. There are total 17 chapters given in the main textbook that create literary sensations in the students. The poem is taken from the medieval era also known as Bhakti Kal in Hindi Literature such as compositions of Kabir, Raskhan and ladlad. In order to give importance to freedom movement, describing nature and human dignity, the poems of nine poets representing different periods of Hindi poetry have been compiled in kshitiz. These poems are given in order of easy to hard. The prose parts contains different style of prose such as story, travelogue, essay, diary, satire, autobiography etc. which ensures the form of pluralistic and secular society should came forward.

Kritika textbook contains total five prose compositions. Leaving story, all other four compositions are long which will help student in developing habit of patient reading. At the end of every chapter, some questions have been given which will give students the opportunity to engage deeply with the chapter.

Course B

Sparsh and Sanchyan are the textbooks of Course B. Sparsh book is divided into two parts-prose section and poetic section. The sequence of lessons in the prose section has been done keeping in mind the grammatical difficulty of the composition texts. In the poetic section, the poems have been kept in their chronological order.

Sanchyan textbook contains six chapters which introduce students their cultural heritage. The texts includes story, memoir, autobiography, travelogue and translated compositions which will help students to get acquainted with the stylistic characteristics of these different disciplines.

2) An ammeter has a high resistance.
3) A voltmeter has a low resistance.A. 1, 2, 3
B. 1, 2
C. 2, 3
D. 1

Answer

→ An ammeter has to measure to current flowing through the circuit. Resistance offers an obstruction to the current flow. So, if the resistance of an ammeter is large, the current measured by the ammeter will be quite less as compared to the actual amount of current flowing through the circuit which is undesirable. If ammeter has zero resistance, then it will give the exact value of current. But this is not practically possible because every material has some value of internal resistance which we can't control. For this reason, ammeter must have small resistance.

Multiple Choice Questions (MCQs)-Pg-27 


25. The resistance of a wire of length 300 m and cross-section area 1.0 mm2 made of material of resistivity 1.0 x 10-7 Ω m is :
A. 2 Ω
B. 3 Ω
C. 20 Ω
D. 30 Ω


Answer



26. When the diameter of a wire is doubled, its resistance becomes:
A. double
B. four times
C. one-half
D. one-fourth

Answer

→ When the diameter of a wire is doubled, its resistance becomes one-fourth.

27. If the resistance of a certain copper wire is 1Ω, then the resistance of a similar nichrome wire will be about:
A. 25 Ω
B. 30 Ω
C. 60 Ω
D. 45 Ω

Answer

→ Alloy of nickel, chromium, manganese and iron having a resistivity of about 60 times more than that of copper.

28. If the diameter of a resistance wire is halved, then its resistance becomes:
A. four times
B. half
C. one-fourth
D. two times

Answer

→ If the diameter of a resistance wire is halved, then its resistance becomes four times.

29. The resistivity of a certain material is 0.6 Ωm. The material is most likely to be:
A. an insulator
B. a superconductor
C. a conductor
D. a semiconductor

Answer

→ The resistivity of semiconductor like silicon and germanium is in between those of conductor and insulator.

30. When the area of cross-section of a conductor is doubled, its resistance becomes:
A. double
B. half
C. four times
D. one-fourth

Answer

→ When the area of cross-section of a conductor is doubled, its resistance becomes.

31. The resistivity of copper metal depends on only one of the following factors. This factor is :
A. length
B. thickness
C. temperature
D. area of cross-section

Answer

→ The resistivity of copper metal depends on temperature.

32. If the area of cross-section of a resistance wire is halved, then its resistance becomes :
A. one-half
B. 2 times
C. one-fourth
D. 4 times

Answer

→ If the area of cross-section of a resistance wire is halved, then its resistance becomes twice.

Questions Based on High Order Thinking Skills (HOTS)-Pg-27

33. A piece of wire of resistance 20 Ω is drawn out so that its length is increased to twice its original length. Calculate the resistance of the wire in the new situation.

Answer

→ We known that

Now if we stretch the same wire in such a way that its length get double. At the same time its area of cross-section decrease to half. So that

34. The electrical resistivities of three materials P, Q and R are given below :
P 2.3 x 10³ Ω m
Q 2.63 x 10^-8 Ω m
R 1.0 x 10¹⁵ Ω m
Which material will you use making (a) electric wires (b) handle for soldering iron, and (c) solar cells? Give reasons for your choices.

Answer

→ (a) Material Q with resistivity 2.63×10^-8 ohm-m can be used for making electric wires because it has very low resistivity.
(b) Material R with resistivity 1.0 ×10¹⁵ ohm-m can be used for making handle of soldering iron because it has very high resistivity.
(c) Material P with resistivity 2.3 × 10³ ohm-m can be used for making solar cell because it is a semiconductor.

35. The electrical resisitivities of four materials A, B, C and D are given below :
A) 110 x 10^-8 Ω m
B) 1.10 x 10¹⁰ Ω m
C) 01.0 x 10^-8 Ω m
D) 2.3 x 10³ Ω m
Which material is : (a) good conductor (b) resistor (c) insulator, and (d) semi-conductor ?

Answer

→ (a) Good conductor = C (10×10^-8 ohm-m)
(b) Resistor= A (110×10^-8 ohm-m)
(c) Insulator= B (1×10¹⁰ ohm-m)
(d) Semiconductor= D (2.3 ×10³ ohm-m)

36. The electrical resistivities of five substances A, B, C, D and E are given below :
A) 5.20 x 10^-8 Ω m
B) 2.60 x 10^-8 Ω m
C) 10.0 x 10^-8 Ω m
D) 1.70 x 10^-8 Ω m
(a) Which substance is the best conductor of electricity? Why?
(b) Which one is better conductor: A or C? Why?
(c) Which substance would you advice to be used for making heating elements of electric irons? Why?
(d) Which two substances should be used for making electric wires? Why?

Answer

→ (a) E is the best conductor of electricity due to its least electrical resistivity.
(b) C, because its resistivity is lesser than that of A.
(c) B, because it has highest electrical resistivity
(d) C and E, because of their low electrical resistivities.

Very Short Answer Type Questions-Pg-37

1. Give the law of combination of resistances in series.

Answer


→ When resistors are joined from end to end, it is called in series. In this case, the total resistance of the system is equal to the sum of the resistance of all the resistors in the system.

2. If five resistances, each of value 0.2 ohm, are connected in series, what will be the resultant resistance?

Answer

→ According to the law of combination of series connection
We know that


3. State the law of combination of resistances in parallel.

Answer

→ When resistors are joined in parallel, the reciprocal of total resistance of the system is equal to the sum of reciprocal of the resistance of resistors.

4. If 3 resistances of 3 ohm each are connected in parallel, what will be their total resistance?

Answer

→ We know that in parallel combination, the reciprocal of total resistance is


5. How should the two resistances of 2 ohms each be connected so as to produce an equivalent resistance of 1 ohm?

Answer

→ According to the question the resultant resistance is less than the individual resistances, so the resistances should be connected in parallel not in series.

6. Two resistances X and Y are connected turn by turn : (i) in parallel, and (ii) in series. In which case the resultant resistance will be less than either of the individual resistance?

Answer

→ The resultant resistance will be less than either of the individual resistance in parallel connection.

7. What possible values of resultant resistance one can get by combining two resistances, one of value 2 ohm and the other 6 ohm?

Answer

→ Here, R₁=2 ohm, R₂=6 ohm
When the resistors are arranged in parallel combination we get,

Again, when the resistors are arranged in series combination we get
R = R₁ + R₂ = 2 + 6 = 8ohm

8. Show how you would connect two 4 ohm resistors to produce a combined resistance of (a) 2 ohms (b) 8 ohms.

Answer

→ When two 4 ohm resistors are connected in parallel then the equivalent resistance will be
1/R_equivalent = 1/4 + 1/4 = 2/4 = 1/2
Or R = 2 ohms
When two 4 ohm resistors are connected in series then the equivalent resistance will be
Requivalent = 4 ohm + 4 ohm = 8 ohm

9. Which of the following resistor arrangement, A or B, has the lower combined resistance?


Answer

→ Equivalent resistance in arrangement A is 10 ohm as this is in series.
And in arrangement B both 10 ohm and 1000 ohm are in parallel.
So, combined resistance of arrangement B is caculated as follows:
1/R = 1/10 + 1/1000 = (100+1)/1000
R = 1000/101 = 9.9 ohm
So, arrangement in B has lower combined resistance.

10. A wire that has resistance R is cut into two equal piece. The two parts are joined in parallel. What is the resistance of the combination?

Answer

→ As the wire is cut into two equal pieces. So resistance of each part is R/2.
The two parts are joined in parallel. So the equivalent resistance will be
1/R_equivalent = 2/R + 2/R
1/R_equivalent= 4/R
R_equivalent = R/4

11. Calculate the combined resistance in each case:


Answer

→ (a) Here both resistance i.e. R1 = 500 ohm, R2 = 1000 ohm are connected in series
So equivalent resistance R = R1 + R2 = 500 + 1000 = 1500ohm.
(b) Here both resistance i.e. R1=2ohm, R2=2ohm are connected in parallel
So equivalent resistance
1/R=1/R1+1/R2
1/R=1/2+1/2
R=1ohm
(c) Here resistance R1=4ohm and R2=4ohm are in parallel and R3=3ohm are in series
According to the given figure,
1/R=1/R1+1/R2
1/R=1/4+1/4
R=2ohm
Total resistance of the given circuit =R+R3
=2+3=5ohm.

12. Find the current in each resistor in the circuit shown below:


Answer

→ Let R1=6Ὠ, R2=4Ὠ and V=24V
As the two resistances are connected in parallel. So the current across R1 will be

And the Current across R2 will be


13. Explain with diagrams what is meant by the “series combination” and “parallel combination” of resistances. In which case the resultant resistance is: (i) less, and (ii) more, than either of the individual resistances?

Answer

→ In a circuit, resistors are connected end-to-end are said to be in series, if the same current
exists in all of them through a single path. When resistances are connected in series, their equivalent resistance is equal to the sum of the individual resistances.

R = R₁ + R₂ + R₃ + - - - - -

When resistances are connected in parallel, the reciprocal of their equivalent resistance is equal to the sum of the reciprocals of the individual resistances.
1/R = 1/R₁ + 1/R₂ + 1/ R₃ + - - - - -

The resultant resistance is less than either of the individual resistances.

14. 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. How much current would flow through the 12 Ω resistor?

Answer

→ Here a battery of 9 V is connected in series with resistors of R₁=0.2ohm, R₂=0.4ohm, R₃=0.3ohm, R₄=0.5ohm, R₅=12ohm,
So the resultant resistance = R₁ + R₂ + R₃ + R₄ + R₅
R = 0.2+0.4+0.3+0.5+12=13.4ohm
As we know that
V= IR
Thus the current flow through 12ohm resistance will be equal to the current flowing across the whole circuit as in case of Series Connection same current flows through each resistance
∴ I = V/R
I = 9/13.4
I = 0.67amp.

15. An electric bulb of resistance 20Ω and a resistance wire of 4 Ω are connected in series with a 6 V battery. Draw the circuit diagram and calculate:
(a) total resistance of the circuit
(b) current through the circuit.
(c) potential difference across the electric bulb.
(d) potential difference across the resistance wire.

Answer


(a) Here resistance are connected in series
Total resistance of the circuit = R₁+R₂=20+4=24Ω(ohm)
(b) According to ohm’s law.
V=IR
Therefore,
6V=I x 24Ω(ohm)
I=6 V /24 ohm = 0.25A(Ampere)
(c) Potential difference across the electric bulb
V1= IR₁=0.25 X 20=5V
(d) Potential difference across the resistance wire
V2= IR₂=0.25X4=1V

16. Three resistors are connected as shown in the diagram.

Through the resistor 5 ohm, a current of 1 ampere is flowing
(i) What is the current through the other two resistors?
(ii) What is the p.d. across AB and across AC?
(iii) What is the total resistance?

Answer

→ (i) According to the figure.

R₁ and R₂ are connected in parallel.
Let I be the current following in the circuit, which is equal to 1A. After passing through the resistance R₃
Current I is divided into two part say I₁ and I₂

(ii) Potential difference across AB = IR₃ = 1 x 5 = 5V
As R₁ and R₂ are connected in parallel
So the equivalent resistance 1/Requivalent = 1/R₁ +1/R₂
1/R_equivalent= 1/10 + 1/15
R_equivalent= 6 ohm
Total resistance in the circuit
Rtotal= = 5+6 = 11 ohm
Potential difference across AC = IR = 1x11 = 11V

17. For the circuit shown in the diagram below:
What is the value of:

(i) current through 6 Ω resistor?
(ii) potential difference across 12 Ω resistor?

Answer

→ The resistance of 6 ohm and 3 ohm are connected in series. Therefore, their net resistance can be calculated as
R= R₁+R₂
Here, R₁= 6ohm
R₂= 3ohm
so R= 6ohm +3ohm= 9 ohm
The current through 6 ohm resistor= current through line 1 = I= V/R =4/9=0.44V
(ii) The resistance of 12ohm and 3 ohm are connected in series. Therefore, their net resistance can be calculated as
R= R₁+R₂
Here, R₁= 12ohm
R₂= 3ohm
so R= 12ohm +3ohm= 15 ohm
The current through them = I= V/R =4/15
Potential difference across 12 ohm resistor= 4/15 ×12 =3.2

18. Two resistors, with resistances 5 Ω and 10 Ω respectively are to be connected to a battery of emf 6 V so as to obtain:
(i) minimum current flowing
(ii) maximum current flowing
(a) How will you connect the resistances in each case?
(b) Calculate the strength of the total current in the circuit in the two cases.

Answer

→ (i) For obtaining minimum current, the two resistors should be connected in parallel.
(ii) For obtaining maximum current, the two resistors should be connected in series.
(a) R₁= 5 ohm, R₂= 10 ohm and V = 6 V
For parallel combination

R= 10/3 ohm
Total current in the circuit

For series combination
R=R₁+R₂
R=5+10=15 ohm
Total current in the circuit.

(b) So the current in each case is 1.8 A and 0.4A in parallel and series circuit respectively.

19. The circuit diagram given below shows the combination of three resistors R₁, R₂ and R₃ :

Find:
(i) total resistance of the circuit.
(ii) total current flowing in the circuit.
(iii) the potential difference across R₁.

Answer

→ (i) As shown in the figure, the resistor R₂ and R₃ are connected in parallel. Their total resistance is given by

Here R₂= 3 ohm and R₃=6 ohm
So

R =2 ohm
The resistance R₁ is in series with the equivalent resistance which are parallel.
Total resistance of the circuit = 2+4 ohms = 6 ohms
(ii) Total current flowing through the circuit=Potential difference/total resistance
Here p.d =12V, and total resistance = 6 ohms
I = 12/6 = 2amps
(iii) The potential difference across R₁.
V = IR₁=2×4 =8V

20. In the circuit diagram given below, the current flowing across 5 ohm resistor is 1 amp. Find the current flowing through the other two resistors.


Answer

→ Given I= 1 A (Across 5 ohm)
R= 5 ohm
The potential drop across AB, V = I × R
or V= 5 ohm × 1A = 5V
In a parallel circuit, the potential difference across the ends of all resistors remains the same. Therefore the potential difference across 4 ohm and 10 ohm will be 5 V
The current flowing through the 4 ohm resistance
= V/R = 5/4 = 1.25 A
Current through 10 ohm resistor = V/R = 5/10 = 0.5 A

21. A resistor has a resistance of 176 ohms. How many of these resistors should be connected in parallel so that their combination draws a current of 5 amperes from a 220 volt supply line?

Answer

→ We Know according to Ohms Law that:-
V= IR, Where V= Voltage/Potential Difference          I= Current          R= Resistance According to Question :-V= 220V                                   I= 5A∴ Total Resistance of the Combination is :- 
Suppose x resistors should be connected in parallel to draw a current of 5A
for parallel combination:-
1/R = 1/R₁ +1/R₂+ 1/R₃........1/R_X(Where Resistance of each Resistor is 176Ω)
According to the question,
R_Combination=176/x
⇒44=176/x( We know from above the total resistance of combination)
x=176/44=4 resistor
Hence 4 resistors each of 176Ω should be connected in parallel so as to draw the current of 5A from a 220 volt supply line.

22. An electric heater which is connected to 220 V supply line has two resistance coils A and B of 24 Ω resistance each. These coils can be used separately (one at a time), in series or in parallel. Calculate the current drawn when:
(a) only one coil A is used.
(b) coils A and B are used in series
(c) coils A and B are used in parallel.

Answer

→ (a) When only one coil, A is used
V= IR
220= 24I
I=9.2
(b) When coils A and B are used in series.
Total resistance R= R_A +R_B = 24+24 = 48ohms
I = V/R = 220/48
=4.58amps
(c) When coils A and B are used in parallel.

R=12ohms
I = V/R = 220/12
=18.33amps

23. In the circuit diagram given below five resistances of 10 Ω, 40 Ω, 30 Ω, 20 Ω and 60 Ω are connected as shown to a 12 V battery.

Calculate:
(a) total resistance in the circuit.
(b) total current flowing in the circuit.

Answer

→ (a) when three resistors are connected in parallel, the net resistance can be obtained as followed

The resistance of 30 ohm, 20 ohm and 60 ohm are connected in parallel. therefore, the net resistance R will be
 ---------------------- (1)
The resistance of 10 ohm and 60 ohm are connected in parallel. Therefore, the net resistance R will be

Now R₁ and R₂ are connected in series
So total resistance in the circuit is R = 10+8 =18 ohm
(b) Total current flowing in the circuit


24. In the circuit diagram given below, three resistors, R₁, R₂ and R₃ of 5 Ω, 10 Ω and 30 Ω respectively are connected as shown.

Calculate :
(a) current through each resistor.
(b) total current in the circuit.
(c) total resistance in the circuit.

Answer

→ (a) Let I₁, I₂ and I₃ be the current flowing through the resistors of 5 ohm, 100 ohm and 30 ohm, respectively.
According to ohm's law V=IR
Here V= 12 V and R = 5 ohm
I = V/R
Current through R₁ = V/R₁ = 12/5 = 2.4 A
Current through R₂ = V/R₂ = 12/10 = 1.2 A
Current through R₃ = V/R₂ = 12/30 = 0.4 A
(b) Total current in the circuit =Current through R₁ +Current through R₂ +Current through R₃
= 2.4 + 1.2 + 0.4 = 4 A
(c) Total resistance in the circuit = R

R= 3 ohm

25. A p.d. of 4 V is applied to two resistors of 6 Ω and 2 Ω and 6 Ω connected in parallel. Calculate:
(a) the combined resistance
(b) the current flowing in the main circuit
(c) the current flowing in the 6Ω resistor.

Answer

→ (a) When two resistors are connected in series, their resultant resistance is given by
R= R₁+R₂
Here R₁ = 6 ohm
R₂= 2 ohm
Combined resistance, R = R₁ + R₂ = 6+2 = 8 ohm
(b) We know that the current, I= V/R
Here R = 8 ohm
V= 4V
I = 4/8 = 0.5 amp
(c) Potential difference across 6ohm resistor V= I x R₁ = 0.5 x 6 = 3 V

26. A p.d. of 6 V is applied to two resistors of 3Ω and 6Ω connected in parallel. Calculate:
(a) the combined resistance
(b) the current flowing in the main circuit
(c) the current following in the 3Ω resistor

Answer

→ (a) ∵ the 3 Ω and 6 Ω resistance are in parallel, hence the eq. resistance =

(b) I = V/Req = 4/2 = 2A
(c) Current Flowing in 3 Ω resistor =


27. In the circuit shown below, the voltmeter reads 10 V.

(a) What is the combined resistance?
(b) What current flows?
(c) What is the p.d. across 2 Ω resistor?
(d) What is the p.d. across 3 Ω resistor?

Answer

→ When two resistors are connected in series, their resultant resistance is given by
R= R₁+R₂
Here R₁ = 2 ohm
R₂= 3 ohm
Combined resistance, R = R₁ + R₂ = 2+3 = 5 ohm
(b) We know that the current, I= V/R
Here R = 5 ohm
V= 10V
I = 10/5 = 2 amp
(c) Potential difference across 2 ohm resistor V= I x R₁= 2×2=4V
(d) Potential difference across 3 ohm resistor V= I x R₂= 2×3=6V

28. In the circuit given below:

(a) What is the combined resistance?
(b) What is the p.d. across the combined resistance?
(c) What is the p.d. across the 3 Ω resistor?
(d) What is the current in the 3Ω resistor?
(e) What is the current in the 6 Ω resistor?

Answer

→ (a)The resistors of 6 ohm and 3 ohm are connected in parallel. Therefore, their combined resistance can be calculated as

R= 2 ohms
(b) The p.d. across the combined resistance, V= IR
Here I = 6A and combined resistance = 2 ohms.
So, V= 6 x 2 = 12 V
(c) In parallel connection potential difference remains constant, so Potential difference across 3 ohm resistor =12 V
(d) The current in the 3Ω resistor
I= V/R1
I= 12/3 =4A
(e) Current flowing through the 6 ohm resistor = V/R₂ = 12/6 = 2 A

29. (a) 5 V battery is connected to two 20 Ω resistors which are joined together in series.
(a) Draw a circuit diagram to represent this. Add an arrow to indicate the direction of conventional current flow in the circuit.
(b) What is the effective resistance of the two resistors?
(c) Calculate the current that flows from the battery.
(d) What is the p.d. across each resistor?

Answer

→ (a)

(b) Since two resistors of 20 ohm are connected in series, the effective resistance will be
R= R₁ +R₂
Here, R₁ = 20 ohm
R₂ = 20 ohm
Therefore the effective resistance of the two resistors = 20+20 =40 ohm
(c) Current flowing through the circuit = I = V/R = 5/40 = 0.125 amps
(d) p.d. across each resistance = I x R = 0.125 x 20 = 2.5 V

30. The figure given below shows an electric circuit in which current flows from a 6V battery through two resistors.

(a) Are the resistors connected in series with each other or in parallel?
(b) For each resistor, state the p.d. across it.
(c) The current flowing from the battery is shared between the two resistors. Which resistor will have bigger share of the current
(d) Calculate the effective resistance of the two resistors.
(e) Calculate the current that flows from the battery.

Answer

→ (a) In the given figure resistors are connected in parallel.
(b) In a parallel arrangement, the voltage remains the same across each resistor i.e 6V
(c) Due to lower resistance 2 ohm have bigger share of current.
(d) The resistors of 2 ohm and 3 ohm are connected in parallel. Therefore, their combined resistance can be calculated as

R= 1.2 ohm
(e) Current flowing through battery, I=V/R=6/1.2 = 5amps

31. A 4Ω coil and a 2 Ω coil are connected in parallel. What is their combined resistance? A total current of 3A passes through the coils. What current passes through the 2Ω coil?

Answer

→ The coils of the resistors 4 ohm and 2 ohm are connected in parallel. Therefore

R=4/3 ohm
Total current =3 A (given)
Potential difference V= 3×4/3=4V
Now current through 2Ω coil= V/2 =4/2 = 2A

Long Answer Type Questions-Pg-41

32A. With the help of a circuit diagram, deduce the equivalent resistance of two resistances connected in series.

Answer


Let the current in the circuit be I amperes and the battery be of strength V volts. Let the combined resistance of the three resistors be R ohms.
Therefore, according to Ohm's law, we have
V=IR ---- (i)
We know that when resistors are connected in series, the current with a battery of V volts.
V₁=I×R₁ ------- (ii)
V₂ =I×R₂ ------ (iii)
Let the potential difference across R₁ is V₁ and the potential difference across R₂ is V₂
V=V₁+V₂--------- (iv)
From equation (i),(ii), (iii) and (iv) we get.
IR= I×R₁+ I×R₂
R= R₁+ R₂

32B. Two resistances are connected in series as shown in the diagram:

(i) What is the current through the 5 ohm resistance?
(ii) What is the current through R?
(iii) What is the value of R?
(iv) What is the value of V?

Answer

→ (i) Current through 5 ohm resistor
I= V/R
I = 10/5 = 2A
(ii) Two resistances are connected in series. So same current flow through the circuit,
So current flowing through R =2A
(iii) As we know that according to ohm's law
V=IR
So R= V/I = 6/2 = 3 ohm
(iv) Total V in the given circuit = V₁+V₂ = 10+6= 16 V

33A. With the help of a diagram, derive the formula for the resultant resistance of three resistors connected in series.

Answer


Let the current be I amperes and the battery be of strength V volts. Let the combined resistance of the three resistors be R ohms.
Therefore, according to Ohm's law, we have
V= IR -----(i)
We know that when the resistors are connected in series, the current is the same in all the resistors. Therefore
V= V₁ +V₂+ V₃ ------------------ (ii)
Let the current flowing through the whole circuit is I, and the equivalent resistance be R.
According to ohm's law
V= IR
V₁= IR₁ -------- (iii)
V₂= IR₂-------- (iv)
V₃= IR₃ ------ (v)
By using eq (i),(ii),(iii) ,(iv) and (v) we get
IR= IR₁+ IR₂+ IR₃
R= R₁+ R₂+ R₃

33B. For the circuit shown in the diagram given below

Calculate:
(i) the value of current through each resistor.
(ii) the total current in the circuit.
(iii) the total effective resistance of the circuit.

Answer

→ (i) Here R₁= 5 ohm, R₂= 10 ohm and R₃= 30 ohm
The value of current through each resistor is different as they are arranged in parallel connection.
So the current through 5 ohm resistor, I₁ = V/R = 6/5 = 1.2A
So the current through 10 ohm resistor, I₂ = V/R = 6/10 = 0.6A
So the current through 30 ohm resistor, I₃ = V/R = 6/30 = 0.2A
(ii) The total current in the circuit = I= I₁+ I₂+ I₃
1.2A + 0.6A + 0.2A = 2A
(iii) Total resistance in the circuit = R, As the circuit is in parallel connection, so


R= 3 ohm

34A. With the help of a circuit diagram, obtain the relation for the equivalent resistance of two resistances connected in parallel.

Answer


Let the individual resistance of the two resistors be R₁ and R₂ and their combined resistance be R. Let the total current flowing in the circuit be I and strength of the battery be V volts. Then, from Ohm's law, we have: V= IR... (1)
We know that when resistors are connected in parallel, the potential drop across each resistance is the same. Therefore: =I=I₁ + I₂ = V/R₁ + V/R₂
I = V/(1 /R₁ + 1/R₂) ... (2)
From the equations (1) and (2) we have:
1/R = 1 /R₁ + 1/R₂

34B. In the circuit diagram shown below, find:
(i) Total resistance.
(ii) Current shown by the ammeter A


Answer

→ (i) Let the total resistance of the circuit =R
As the connections are parallel. So the total resistance will be
1/R=1/R₁+1/R₂
R₂=3+2=5ohms
R₁=5 ohms
1/R=1/5+1/5
1/R=2/5
R=2.5ohms
(ii) Current shown by the ammeter A will be equal to the current following in the circuit.
I=V/R=4/(2.5)
=1.6amps

35A. Explain with the help of a labelled circuit diagram, how you will find the resistance of a combination of three resistors of resistances R₁, R₂ and R₃ joined in parallel.

Answer

Let the resistance of the three resistors be R₁, R₂ and R₃, respectively. Let their combined resistance be R.
Let the total current flowing in the circuit be I and the strength of the battery be V. Then from Ohm's law, we have: V= IR------- (1)
We know that when the resistors are connected in parallel, the potential drop across each resistance is the same. Therefore:
I = I₁+I₂+I₃
I = V/R₁+ V/R₂+ V/R₃
1/R=1/R₁+1/R₂+ 1/R₃
If two resistances are connected in parallel, then the resultant resistance will be
1/R=1/R₁+1/R₂+ 1/R₃

35B. In the diagram shown below, the cell and the ammeter both have negligible resistance. The resistors are identical

With the switch K open, the ammeter reads 0.6 A. What will the ammeter reading when the switch is closed?

Answer 

→ When switch is open, the upper two resistances are connected in parallel in the circuit.
Effective resistance is 1/R_eq=1/R+1/R=2/R
R_eq = R/2
So the current=I=V/(R/2)=0.6A (given)
V/R = 0.3 A
When the switch closes, the third resistance also comes in the circuit.
The effective resistance of the circuit becomes R/3
Hence, Current I = V/(R/3) = 3 (V/R) = 3 x 0.3 = 0.9 A

36. The figure given below shows three resistors
Their combined resistance is:

A. 7.5 Ω
B. 14 Ω
C. 10 Ω
D. 15 Ω

Answer

→ Here two resistor 6 ohm and 2 ohm are connected in parallel

Now R ohm and 6 ohm are in series

So the combined resistance


37. If two resistors of 35 Ω and 15 Ω are joined together in series and then placed in parallel with a 40 Ω resistor, the effective resistance of the combination is :
A. 0.1 Ω
B. 10 Ω
C. 20 Ω
D. 40 Ω

Answer

→ When the resistors are in series, then the combined resistance = 35 Ω+15 Ω = 50Ω
Now 50Ω and 40 Ω are connected in parallel, so the combined resistance


38. The diagram below shows part of a circuit:

If this arrangement of three resistors was to be placed by a single resistor, its resistance should be :
A. 9 Ω
B. 4 Ω
C. 6 Ω
D. 18 Ω

Answer

→ The two 6 Ω resistors are in parallel. Hence, the Equivalent Resistor =

Now, 3 Ω and 6 Ω resistors are in series. Hence the Total Equivalent Resistance = 3 + 6 = 9 Ω

39. In the circuit shown below :

The potential difference across the 3 Ω resistor is:
A. 5V 
B. 4V
C. 1V
D. 2V

Answer

→ Here all the resistance are connected in series. Therefore net resistance = 2Ω+3Ω+1Ω =6Ω
So current in the circuit =

Potential across 3Ω resistor =


40. A battery and three lamps are connected as shown:

Which of the following statements about the currents at X, Y and Z is correct?
A. The current at Z is greater than that at Y.
B. The current at Y is greater than that at Z.
C. The current at X equal the current at Y.
D. The current at X equals the current at Z.

Answer

→ The current at Y is greater than that at Z due to the arrangement of the circuit.

41. V₁, V₂ and V₃ are the p.d. across the 1 Ω, 2 Ω and 3 Ω resistors in the following diagram, and the current is 5A.

Which one of the columns (a) to (d) shows the correct values of V₁, V₂ and V₃ measured in volts?
A. V₁- 1.0, V₂- 2.0, V₃- 3.0
B. V₁- 5.0, V₂-10.0, V₃ -15.0
C. V1- 5.0, V2-2.5, V3- 1.6
D. V₁- 4.0, V₂-3.0, V₃- 2.0

Answer

→ As V = IR

42. A wire of resistance R₁ is cut into five equal pieces. These five pieces of wire are then connected in parallel. If the resultant resistance of this combination be R₂, then the ratio is:
A. 20:21
B. 25:1
C. 5:19
D. 25:7

Answer

→ When the wire is cut into 5 pieces, resistance of each part is R/5
So, equivalent resistance of the 5 pieces when connected in parallel is

R₁ = R/25
So required ratio R: R₁=25:1

Questions Based on High Order Thinking Skills (HOTS)-Pg-43

43. Show with the help of diagrams, how you would connect three resistors each of resistance 6 Ω so that the combination has resistance of (i) 9 Ω (ii) 4 Ω

Answer

→ (i)

 To obtain 9Ω resistance two resistance are connected in parallel,
Let the resultant resistance for parallel circuit=R
1/R=1/6+1/6
1/R=2/6
R=3
Effective resistance=6+3 = 9 ohms
(ii) To obtain 4Ω resistance all resistance are connected in parallel

So the effective resistance will be R for each circuit

Or R= 2 Ω
Total resistance = 2 Ω+2 Ω=4Ω

44. To resistances when connected in parallel give resultant value of 2 ohm; when connected in series the value becomes 9 ohm. Calculate the value of each resistance.

Answer

R₁+R₂=9
R₂= 9-R₁

R²₁-9R₁+18=0
(R₁-3)(R₁-6)=0
R₁=3,6


45. A resistor of 8 ohm. Calculated in parallel with another resistor X. The resultant resistance of the combination is 4.8 ohms. What is the value of the resistor X?

Answer

→ Parallel combinations of the resistors will be written as
1/x + 1/8 = 1/4.8
On solving the above relation,
x = 12 ohms

46. You are given three resistances of 1, 2 and 3 ohms. Show by diagrams, how with the help of these resistances you can get :
(i) 6 Ω (ii) C (iii) 1.5 Ω

Answer

→ (i) 6 ohm can be obtained by connecting the 1,2 and 3 ohm in series.

(ii) 6/11 ohm can be obtained by connecting the 1,2 and 3 ohm in parallel connection.

(iii) 1.5 ohm can be obtained by connecting 1 and 2 ohm in series and then by connecting 3 ohm to the resultant resistance i.e 3 ohm in parallel connection.


47 How will you connect three resistors of 2Ω, 3Ω and 5Ω respectively so as to obtain a resultant resistance of 2.5 Ω? Draw the diagram to show the arrangement.

Answer

→ Resistor 2 and 3 ohm are connected in series, 5 ohm resistance is connected parallel to 2 and 3 ohm.
Thus the resultant resistance will be 2.5 ohm


48. How will you connect three resistors of resistance 2Ω, 3 Ω and 6 Ω to obtain a total resistance of: (a) 4Ω, and (b) 1 Ω?

Answer

→ (a) For total Resistance to be 4
we will connect 2 ohm resistance in series with parallel combination of 3 and 6 ohm resistance. When 3 and 6 ohm are in parallel combination then 1/R=1/3+1/6
or R=2
Now this R connected in series with 2 ohm resistance will give equivalent resistance = 2+ 2 = 4 ohm.
For total Resistance to be 1 ohm
All the given resistance should be connected in parallel combination 1/R’= 1/2+1/3+1/6
Or R'=1 ohm

49. What is (a) highest, and (b) lowest, resistance which can be obtained by combining four resistors having the following resistances?
4Ω, 8Ω, 12Ω, 24Ω

Answer

→ (a) To obtain the highest resistance we must connect the given resistances in series.
Highest resistance R = 4 + 8 + 12 + 24 = 48ohms
(b) To obtain the lowest resistance we must connect the given resistances in parallel.
1/R=1/4+1/8+1/12+1/24
On solving we get, R=2ohms.

50. What is the resistance between A and B in the figure given below?


Answer

→ According to the figure, 20 ohm, 10 ohm and 20 ohm are connected in series so the resultant resistance in the extreme side = 20+10+20 = 50 ohms
Now this 50 ohm is in parallel with 30 ohm, so the resultant resistance become
On solving we get R = 18.75 ohm
Now 18.75 ohm, 10 ohm and 10 ohm are connected in series. So the resultant resistance become

R= 18.75+10+10 = 38.75 ohm.

51. You are given one hundred 1 Ω resistors. What is the smallest and largest resistance you can make in the circuit using these?

Answer

→ The greatest possible resistance is the series combination of the resistors.
In series combination of hundred 1 ohm resistors equivalent resistance will be 100 ohm
The least possible resistance is the parallel combination of the resistors.
When resistances are connected in parallel then
1/R =100
R= 1/100 =0.01 ohm

52. You are supplied with a number of 100 Ω resistors. How could you combine some of these resistors to make a 250 Ω resistor?

Answer

→ To make equivalent resistance of 250 ohms one may connect:
1) 2 resistors in series Thus, the resultant would be:
100+100= 200 ohms
2) 2 resistors in parallel Thus, the resultant of 2 parallel combination would be:

3) Now connect the combination of the two (series and parallel) in series: Thus, we may obtain a net resistance of: 200+50 = 250 ohms.

53. The resistors R1, R2, R3 and R4 in the figure given below are all equal in value.

What would you expect the voltmeters A, B and C to read assuming that the connecting wires in the circuit have negligible resistance?

Answer



54. Four resistances of 16 ohms each are connected in parallel. Four such combinations are connected in series. What is the total resistance?

Answer

→ When four 16 ohm resistances is connected in parallel combination then the equivalent resistance will be
1/R = 1/16 + 1/16 + 1/16 + 1/16 = 4/16
R = 4 ohm
When four such combinations are connected in series, than the total resistance = 4+4+4+4 = 16 ohm.

55. If the lamps are both the same in the figure given below and if A₁ reads 0.50 A, what do A₂, A₃, A₄ and A₅ read?

Answer

→ It is given that the two lamps are identical. So, they will have equal resistance.
The ammeter A₁ reads 0.5 A current. This current will now get distributed equally in the two branches as resistances are equal.
Hence, the ammeters A₂ and A₄ will read 0.25 A each.
Now, A₃ and A₅ are connected in series with A₂ and A₄, so they will also read 0.25 A.

Very Short Answer Type Questions-Pg-47

1. Are the lights in your house wired in series?

Answer

→ No, When appliances are connected in a parallel arrangement, each of them can be put on and off independently. This is a feature that is essential in a house's wiring.

2. What happens to the other bulbs in a series circuit if one bulb blows off?

Answer

→ When bulbs are connected in a series arrangement, if one of them can be put off all bulbs are switched off.

3. What happens to the other bulbs in a parallel circuit if one bulb blows off?

Answer 

→ When bulbs are connected in a parallel arrangement, each of them can be put on and off independently.

4. Which type of circuit, series of parallel, is preferred while connecting a large number of bulbs:
(a) for decorating a hotel building from outside?
(b) for lighting inside the rooms of the hotel?

Answer

→ (a) For decorating a hotel building from outside we use series connection.
(b) For lighting inside the rooms of the hotel we use parallel connection.

5. Draw a circuit diagram to show two 4 V electric lamps can be lit brightly from two 2 V cells.

Answer


Short Answer Type Questions-Pg-47

6. Why is series arrangement not used for connecting domestic electrical appliances in a circuit?

Answer

→ In series arrangement if one electrical appliance stops working due to some defect, then all other appliance also stop working as the whole circuit is broken.So we should not used series arrangement for connecting domestic electrical appliances in a circuit.


7. Give three reasons why different electrical appliances in a domestic circuit are connected in parallel.

Answer

→ Due to the following advantages different electrical appliances in a domestic circuit are connected in parallel:
(i) Each electrical appliance gets the same voltage as that of the power supply line.
(ii) When appliances are connected in a parallel arrangement, each of them can be put on and off independently.
(iii) If one electrical appliance stops working due to some defect, then all other appliances keep working properly.

8. Ten bulbs are connected in a series circuit to a power supply line. Ten identical bulbs are connected in a parallel circuit to an identical power supply line.
(a) Which circuit would have the highest voltage across each bulb?
(b) In which circuit would the bulbs be brighter?
(c) In which circuit, if one bulb blows out, all others will stop glowing?
(d) Which circuit would have less current in it?

Answer

→ (a) Parallel circuit would have the highest voltage across each bulb.
(b) In Parallel circuit the bulbs will be brighter.
(c) In series circuit, if one bulb blows out, all others will stop glowing
(d) Series circuit would have less current in it.

9. Consider the circuits given below:

(a) In which circuit are the lamps dimmest?
(b) In which circuit or circuits are the maps or equal brightness to the lamps in circuit (i)?
(c) Which circuit gives out the maximum light?

Answer

→ (a) In circuit (ii) the lamps are dimmest as it is connected in series arrangement.
(b) In circuit (iii) the lamps are equal brightness in circuit (i) as both are connected in parallel.
(c) Circuit (ii) give the maximum light.

10. If you were going to connect two light bulbs to one battery, would you use a series or a parallel arrangement? Why? Which arrangement takes more current from the battery?

Answer

→ If we were going to connect two light bulbs to one battery, we would use a parallel arrangement, so the current will flow through them equally. Series arrangement take more current from battery.

Long Answer Type Questions-Pg-48

11A. Which is the better way to connect light and other electric appliances in domestic wiring: series circuits or parallel circuits? why?

Answer

→ When appliances are connected in a parallel arrangement, each of them can be put on and off independently. This is a feature that is essential in a house's wiring.

11B. Christmas tree lamps are usually wired in series. What happens if one lamp breaks?

Answer

→ In series circuit, if one bulb blows out, all others will stop glowing.

11C. An electrician has wired a house in such a way that if a lamp gets fused in one room of the house, all the lamps in other rooms of the house stop working. What is the defect in the wiring?

Answer

→ The circuit arrangement is in series, so if one bulb blows out, all others will stop glowing.

11D. Draw a circuit diagram showing two electric lamps connected in parallel together with a cell and a switch that works both lamps. Mark an (A) on your diagram to show where an ammeter should be placed to measure the current.

Answer



Multiple Choice Questions (MCQs)-Pg-48

12. The lamps in a household circuit are connected in parallel because:
A. this way they require less current
B. if one lamp fails the others remain lit
C. this way they require less power
D. if one lamp fails the others also fail

Answer

→ The lamps in a household circuit are connected in parallel because in a parallel arrangement, each of them can be put on and off independently.

13. Using the circuit given below, state which of the following statement is correct?

A. When S₁and S₂ are closed, lamps A and B are lit.
B. With S₁open and S₂ closed, A is lit and B is not lit.
C. With S2 open and S1 closed A and B are lit.
D. With S₁ closed and S₂ open, lamp A remains lit even if lamp B gets fused.

Answer

→ When S₂ open and S₁ closed A and B are lit.

Questions Based on High Order Thinking Skills (HOTS)-Pg-48

14A. Draw a circuit diagram showing two lamps, one cell and a switch connected in series.

Answer



14B. How can you change the brightness of the lamps?

Answer


→ When lamps are connected in parallel the brightness of the lamps increases.

15. Consider the circuit given below where A, B and C are three identical light bulbs of constant resistance.

(a) List the bulbs in order of increasing brightness.
(b) If C burns out, what will be the brightness of A now compared with before?
(c) If B burns out instead, what will be the brightness of A and C compared with before?

Answer

→ (a) Here A and B are connected in series so the brightness of the A and B will same while C are connected in parallel so the brightness of the bulb C is maximum.
(b) Their will be no effect on A and B.
(c) A and B are connected in series so if B burns than A also not glow but their will be no effect on C, as C is connected in parallel.

16. How do you think the brightness of two lamps arranged in parallel compares with the brightness of two lamps arranged in series (both arrangements having one cell)?

Answer

→ The brightness of two lamps which are arranged in parallel is maximum than the the brightness of two lamps arranged in series.

17. If current flows through two lamps arranged:
(a) in series,
(b) in parallel,
and the filament of one lamps breaks, what happens to the other lamp? Explain your answer

Answer

→ (a) In series circuit, if one bulb blows out, all others will stop glowing.
(b) When lamps are connected in a parallel arrangement, each of them is independently.

18. The figure below shows a variable resistor in a dimmer switch.

How would you turn the switch to make the lights:
(a) brighter, and
(b) dimmer?
Explain your answer.

Answer

→ (a) To make the light brighter we turn the switch towards right side.
(b) To make the light dimmer we turn the switch towards left side.

Very Short Answer Type Questions-Pg-58

1. State two factors on which the electrical energy consumed by an electrical appliance depends.

Answer

→ Two factors on which the electrical energy consumed by an electrical appliance depends on.
1) Power rating of the appliance.
2) Time for which the appliance is used.

2. Which one has a higher electrical resistance: a 100-watt bulb or a 60 watt bulb?

Answer

→ As we know that

P is inversely proportional to resistance, so 60 watt has a higher electrical resistance.

3. Name the commercial unit of electric energy.

Answer

→ The commercial unit of electric energy is Kilowatt-hour.

4. An electric bulb is rated at 220 V, 100 W. What is its resistance?

Answer

→ Here, V = 220 V, P = 100W
R = ?
We know that

Thus
R = V²/P = (220)²/100 = 484 ohm.

5. What is the SI unit of (i) electric energy, and (ii) electric power?

Answer

→ (i) The SI unit of electric energy is joule.
(ii) The SI unit electric power is watt.

6. Name the quantity whose unit is (i) kilowatt, and (ii) kilowatt-hour.

Answer

→ (i) Kilowatt is the S.I unit of Electric power.
(ii) kilowatt-hour is the S.I unit of Electric energy.

7. Which quantity has the unit of watt?

Answer

→ Watt is the SI unit of Electric power.

8. What is the meaning of the symbol kWh? Which quantity does it represent?

Answer

→ kWh is the commercial unit of electrical energy. It full form is kilowatt-hour.

9. If the potential difference between the end of a wire of fixed resistance is doubled, by how much does the electric power increase?

Answer

→ As we know that

i.e. as power is directly proportional to the square of potential difference. So, the electric power becomes four times its previous value.


10. An electric lamp is labeled 12 V, 36 W. This indicates that it should be used with a 12 V supply. What other information does the label provide?

Answer

→ The electric lamp consumes energy at the rate of 36 J/s.

11. What current will be taken by a 920 W appliance if the supply voltage is 230 V.

Answer

→ Here P = 920W, V = 230V, I = ?
We know that
P = V x I
920 = 230 x I
I = 920/230 = 4amp

Short Answer Type Questions-Pg-58

12. Define watt. Write down an equation linking watts, volts and amperes.

Answer

→ One watt is defined as the amount of electrical energy consumed by electrical appliances at the rate of 1 joule per second.
1 watt = 1 volt x 1 ampere

13. Define watt-hour. How many joules are equal to 1 watt-hour?

Answer

→ The amount of electrical energy consumed by an electrical appliance of 1 watt power for 1 hour is called watt-hour.
1 watt hour is equal to 3600 joules

14. How much energy is consumed when a current of 5 amperes flows through the filament (or element) of a heater having resistance of 100 ohms for two hours? Express it in joules.

Answer

→ Here, current (I) = 5amp, Resistance(R) = 100 ohms and time = 2h
As we know that
Electric energy consumed = P x t = I²Rt
= 5² x 100 x 2
= 5000 Wh
= 5 kwh
We know that
1kwh = 3.6 x 10⁶ J
Therefore, 5kwh = 5 x 3.6 x 10⁶ J = 18 x 10⁶ J

15. An electric bulb is connected to a 220 V power supply line. If the bulb draws a current of 0.5 A, calculate the power of the bulb.

Answer

→ Here, Potential difference (V) = 220V, Current (I) = 0.5amp, Power = ?
We know that
P = VI = 220 X 0.5
P = 110 watt.

16. In which of the following cases ore electrical energy is consumed per hour?
(i) A current of 1 ampere passed through a resistance of 300 ohms.
(ii) A current of 2 amperes passed through a resistance of 100 ohms.

Answer

→ (i) Here current (I) = 1 A, Resistance (R) = 300 ohm, time(t) = 1h
We know that
P = I²R = 1² x 300 = 300 W
E = P x t = 300 x 1 = 300 Wh
(ii) Here Current (I) = 2 A Resistance(R) = 100 ohm, time (t) = 1h
We know that
P = I²R = 2² x 100 = 400 W
E = P x t = 400 x 1 = 400 Wh
Hence, in case (ii), the electrical energy consumed per hour is more.

17. An electric kettle rated at 220 V, 2.2 kW, works for 3 hours. Find the energy consumed and the current drawn.

Answer

→ Here potential difference(V) = 220V, Power (P) = 2.2kW = 2200W, time(t) = 3h
We know that
Electrical energy consumed = P × t = 2.2 ×3 = 6.6 kWh
We have, P = V x I
I = P/V
I = 2200/220
= 10amp

18. In a house two 60 W electric bulbs are lighted for 4 hours, and three 100 W bulbs for 5 hours every day. Calculate the electric energy consumed in 30 days.

Answer

→ When two electric bulb of 60W are lighted for 4 hours then
Here, p₁ = 60w
Number, n₁ = 2
Time for daily use, t₁ = 4h
Total electrical consumed everyday, E₁ = n₁×p₁×t₁
= 2×60×4 = 480
Energy consumed in 30 days = 30×0.48 = 14.4Kwh (As 1kW=1000W)
When three electric bulb of 100W are lighted for 5 hours then
Here, p₂ = 100w
Number, n₂ = 3
Time for daily use, t₂ = 5h
Total electrical consumed every day, E₁ = n₂×p₂×t₂
= 3×100×5 = 1500
Energy consumed in 30 days = 30× 1.5 = 45Kwh (As 1kW= 1000 W)
Total electrical consumed in 30 days = 14.4Kwh + 45Kwh = 59.4 kWh

19. A bulb is rated as 250 V; 0.4 A Find its:
(i) power, and (ii) resistance.

Answer

→ (i) We know that
V = 250V, I = 0.4amp
Power = VI = 250X0.4 = 100watt
(ii) We have,
V = 250V, I = 0.4amp
We know that
P = I²R
100 = (0.4)²XR
R = 625 ohm

20. For a heater rated at 4 kW and 220 V, calculate:
(a) the current,
(b) the resistance of the heater,
(c) the energy consumed in 2 hours, and
(d) the cost if 1 kWh is priced at ₹ 4.60

Answer

→ (a) Here, P = 4kw, V = 220v
I = ?
We know that
Power = VI = 220 × I
4000 = 220I
I = 4000/220= 18.18A
(b) Here, P = 4kw, V = 220v
R = ?
We know that,
P = I²R
P = (18.18)²×R
R = 4000/(18.18)²
R = 12.10 Ω
(c) Here, P = 4kw, time = 2hr
Energy consumed in two hour = P × t
= 4 × 2
= 8kw-hr
(d) If 1kwh = Rs 4.6
Here, Energy consumed in two hour = 8kwh
So, total cost = 8 x 4.6 = Rs 36.8

21. An electric motor take 5 amperes current from a 220 volts supply line. Calculate the power of the motor and electrical energy consumed by it in 2 hours.

Answer

→ Here current (I) = 5amp, Potential difference(V) = 220 volt, time = 2h,
We have to find
Power (P) = ?
Energy (E) = ?
We know that
P = V×I
= 220×5
= 1100 watt
= 1.1 kW
Energy consumed, E = P × t
= 1.1 × 2
= 2.2 kWh

22. Which users more energy: a 250 W TV set in 1 hour or a 1200 W toaster in 10 minutes?

Answer

→ TV set uses 0.25 kWh energy whereas toaster uses 1.20 kWh energy. So, toaster uses more energy.

23. Calculate the power used in the 2 Ω resistor in each of the following circuits:
(i) A 6V battery in series with 1 Ω and 2 Ω resistors.
(ii) A 4V battery in parallel with 12 Ω and 2Ω resistors.

Answer

→ (a) Here V = 6 volt, R₁ = 1 ohm, R₂ = 2 ohm
The connection is in series. So Equivalent resistance = R₁+R₂ = 1 + 2 = 3 ohm
We know that
Total current I = V/R = 6/3 = 2A
Current through R₂ = I₂ = I = 2A
Voltage across R₂ = V₂ = I₂R₂ = 2×2 = 4 ohm
Power used in R₂ = I₂V₂ = 2×4 = 8W
(b) Here, V = 4 volt,R₁ = 12 ohm, R₂ = 2 ohm
We know that, as the connection is in parallel so voltage across R₂ = V₂ = V = 4V
Current across R₂ = I₂ = V₂/R₂ = 4/2 = 2A
As power = IV
Power used in R₂ = I₂V₂ = 2×4 = 8W

24. Two lamps, one rated 40 W at 220 V and the other 60 W and 220 V, are connected in parallel to the electric supply at 220 V.
(a) Draw a circuit diagram to show the connections.
(b)Calculate the current drawn from the electric supply.
(b) Calculate the total energy consumed by the two lamps together when they operate for one hour.

Answer


(b) As the connection is in parallel.
So voltage across both the bulbs is same and is equal to 220V.
Using the Formula P=VI,We can get I as I=P/V Where:-
P= Power, I=Current, V=Voltage/Potential Difference
Current through AB i.e 40W lamp = I₁ = P₁/V = 40/220 A
Current through CD i.e. 60W lamp = I₂ = P₂/V = 60/220 A
Total current drawn from the electric supply = 40/220 + 60/220 = 0.45 A
(c)We know that Energy= Power x TimeTotal Power to two bulbs together= 40W+60W = 100W According to the question both of the bulbs are being operated together at 1 hour∴ Energy= 100W x 1hr Energy= 100Whr
We know 1Whr = 3600 Joules ∴100Whr = 3600 X 100 =360000Joules  =360kJoules.

25. An electric kettle connected to the 230 V mains supply draws a current of 10 A. Calculate:
(a) the power of the kettle.
(b) the energy transferred in 1 minute.

Answer

→ (a) Here, V = 230V, I = 10amp
We know that, P = VI
P = 230X10
P = 2300watt = 2300 J/s
(b) The energy transferred in 1 minute. = P x t = 2300 J/s x 60s = 138000 J

26. A 2 kW heater, a 200 W TV and three 100 W lamps are all switched on from 6 p.m. to 10 p.m. What is the total cost at Rs. 5.50 per kWh?

Answer

→ Here for heater we have, P = 2kW, t = 4h
So energy consumed by heater, E = P x t = 2x4 = 8kWh
Here for TV, we have, P = 200W = 0.2kW, time 10pm – 6pm = 4h
So energy consumed by TV, E = P x t = 0.2x4 = 0.8kWh
Here for TV, we have, P = 100W = 0.1kW, t = 4h, number of lamp(n) = 3
So energy consumed by three lamps, E = n x P x t = 3x0.1x4 = 1.2kWh
Total energy consumed by heater, TV and three lamps = 8+0.8+1.2 = 10kWh
Cost of 1kWh = Rs. 5.50
Cost of 10kWh = Rs. 5.50 x 10 = Rs. 55

27. What is the maximum power in kilowatts of the appliance that can be connected safely to a 13 A; 230 V mains socket?

Answer

→ Here I = 13amp, V = 230V
As we know that
Power = VI
= 230 x 13
= 2990W
P = 2.99kW

28. An electric fan runs from the 230 V mains. The current flowing through it is 0.4 A. At what rate is electrical energy transferred by the fan?

Answer

→ Here, V = 230V, I = 0.4amp
We know that power is the rate at which electric energy is transferred
P = V x I
= 230 x 0.4
= 92 W = 92 J/s

Long Answer Type Questions-Pg-59

29A. What is meant by “electric power”? Write the formula for electric power in terms of potential difference and current.

Answer

→ The rate at which electrical energy is consumed, or the rate of doing electric work is known as electric power.
We know that
P = VI

29B. The diagram below shows a circuit containing a lamp L, a voltmeter and an ammeter. The voltmeter reading is 3 V and the ammeter reading is 0.5 A.

(i) What is the resistance of the lamp?
(ii) What is the power of the lamp?

Answer

→ (i) resistance = V/I = 3/0.5 = 6 Ω
(ii) P = VI = 3 × 0.5 = 1.5 Watt

29C. Define kilowatt-hour. How many joules are the in one kilowatt-hour?

Answer

→ The amount of electrical energy consumed when an electrical appliance having a power rating of 1 kilowatt is used for 1 hour is called One kilowatt hour.
1kWh = 3.6 x 10⁶J

29D. Calculate the cost of operating a heater of 500 W for 20 hours at the rate of ₹ 3.90 per unit.

Answer

→ Here, Power of heater (P) = 500W = 0.5kW, Time(t) = 20hr
We know that
Energy consumed = P x t = 0.5 X 20
= 10kwh
Total cost = 10Xcost per unit
Cost per unit = Rs. 3.9 per unit
Therefore, total cost = 10X 3.9 = Rs 39

Multiple Choice Questions (MCQs)-Pg-59

30. When an electric lamp is connected to 12 V battery, it draws a current of 0.5 A. The power of the lamp is :
A. 0.5 W
B. 6W
C. 12 W
D. 24W

Answer

→ We know that
P = VI
P = 12× 0.5 = 6W

31. The unit for expressing electric power is:
A. volt
B. joule
C. coulomb
D. watt

Answer

→ The unit for expressing electric power is watt

32. Which of the following is likely to be the correct wattage for an electric iron used in our homes?
A. 60 W
B. 250 W
C. 850 W
D. 2000 W

Answer

→ The correct wattage for an electric iron used in our homes is 850 W.

33. An electric heater is rated at 2kW. Electrical energy costs ₹ 4 per kWh. What is the cost of using the heater for 3 hours?
A. ₹ 12
B. ₹ 24
C. ₹ 36
D. ₹ 48

Answer

→ Energy consumed = P x t = 2 X 3 = 6kWh
Total cost = 6 x cost per unit
Cost per unit = Rs. 4 per kWh
Therefore, total cost = 6 x 4 = Rs 24

34. The SI unit of energy is:
A. joule
B. coulomb
C. watt
D. ohm-metre

Answer

→ The SI unit of energy is joule.

35. The commercial unit of energy is:
A. watt
B. watt-hour
C. kilowatt-hour
D. kilo-joule

Answer

→ The commercial unit of energy is kilowatt-hour.

36. How much energy does a 100 W electric bulb transfer in 1 minute?
A. 100 J
B. 600 J
C. 3600 J
D. 6000 J

Answer

→ Energy consumed = P x t = 100 ×60 = 6000 joule
As in 1 minute = 60 second.

37. An electric kettle for use on a 230 V supply is rated at 3000 W. For safe working, the cable connected to it should be able to carry at least:
A. 2 A
B. 5 A
C. 10 A
D. 15 A

Answer

→ As P = VI
I = P/V

38. How many joules of electrical energy are transferred per second by a 6V; 0.5 A lamp?
A. 30 J/s
B. 12 J/s
C. 0.83 J/s
D. 3 J/s

Answer

→ As P = VI

39. At a given time, a house is supplied with 100 A at 220 V. How many 75 W, 220 V light bulbs could be switched on in the house at the same time (if they are all connected in parallel)?
A. 93
B. 193
C. 293
D. 393

Answer

→ When they are connected in parallel 293 bulb will glow in the house at the same time.

40. If the potential difference between the ends of a fixed resistor is halved, the electric power will become:
A. double
B. half
C. four times
D. one-fourth

Answer

→ If the potential difference between the ends of a fixed resistor is halved, the electric power will become one-fourth.

Questions Based on High Order Thinking Skills (HOTS)-Pg-60

41. State whether an electric heater will consumer more electrical energy or less electrical energy per second when the length of its heating element is reduced. Given reasons for our answer.

Answer

→ An electric heater will consume more electrical energy as power is inversely proportional to the resistance.

42. The table below shows the current in three different electrical appliances when connected to the 240 V mains supply:

(a) Which appliance has the greatest electrical resistance? How does the data show this?
(b) The lamp is connected to the mains supply by using a thin, twin-cored cable consisting of live and neutral wires. State two reasons why this cable should not be used for connecting the kettle to the mains supply.
(c) Calculate the power rating of the kettle when it is operated from the 240 V mains supply.
(d) A man takes the kettle abroad where the mains supply is 120 V. What is the current in the kettle when it is operated from the 120 V supply?

Answer

→ (a) Lamp has the greatest electrical resistance as resistance oppose the flow of current, and the lamp has least amount of current.
(b) According to the given data large amount of current is needed by kettle, so it must be connected to the earthing.
(c) As we know that
P = VI
V = 240V, I = 8.5A
P = 240X8.5 = 2040 W = 2.04 kW
(d) When kettle is connected to 240 V supply, then the P = 2040 W

43. A boy noted the readings on his home’s electricity meter on Sunday at 8 AM and again on Monday at 8 AM (see Figures below).

(a) What was the meter reading on Sunday ?
(b) What was the meter reading on Monday?
(c) How many units of electricity have been used ?
(d) If the rate is Rs. 5 per unit, what is the cost of electricity used during this time?

Answer

→ (a) The meter reading on Sunday is 42919.
(b) The meter reading on Monday is 42935.
(c) Units of electricity used = 42935- 42919 = 16 units.
(d) The cost of 1 unit is Rs 5. So the cost of 16 units = 16×5 = 80 rupees

44. An electric bulb is rated as 10 W, 220V. How many of these bulbs can be connected in parallel across the two wires of 220 V supply line if the maximum current which can be drawn in 5A ?

Answer

→ Here it is given P = 10W, V = 220V, I = 5A
As we know that
P = VI
= 220X5
P = 1100W
Power of one bulb = 10W
So the total no. of bulbs that can be connected = 1100/10 = 110

45. Two exactly similar electric lamps are arranged (i) in parallel, and (ii) in series. If the parallel and series combination of lamps are connected to 220 V supply line one by one, what will be the ratio of electric power consumed by them?

Answer

→ As we know that
Electric power consumed = V²/R
Here V is given we have to find the value of R
So when the connection is in parallel
The equivalent resistance is

Electric power consumed in parallel connection

When the connection is in Series
The equivalent resistance = R+R = 2R
Electric power consumed in series connection

The ratio of electric power consumed by them


Very Short Answer Type Questions-Pg-66

1. How does the heat H produced by a current passing through a fixed resistance wire depend on the magnitude of current I?

Answer

→ Heat produced, H = I²Rt
As heat produced is directly proportional to the square of current.

2. If the current passing through a conductor is doubled, what will be the change in heat produced?

Answer

→ Heat produced, H = I²Rt
So, If the current passing through a conductor is doubled then heat produced becomes four times.

3. Name two effects produced by electric current.

Answer

→ The two effects produced by electric current are:
(a) Heating effect of current
(b) Magnetic effect of current

4. Which effect of current is utilized in an electric light bulb?

Answer

→ Heating effect of current is utilized in an electric light bulb.

5. Which effect of current is utilized in the working of an electric fuse?

Answer

→ Heating effect of current is utilized in the working of an electric fuse.

6. Name two devices which work on the heating effect of electric current.

Answer

→ The two devices which work on the heating effect of electric current are Electric heater and electric fuse.

7. Name two gases which are filled in filament type electric light bulbs.

Answer

→ Argon and nitrogen are filled in filament type electric light bulbs.

8. Explain why, filament type electric bulbs are not power efficient.

Answer

→ Due to the heating effect of current lots of electric power is consumed by the filament of a bulb.

9. Why does the connecting cord of an electric heater not glow hot while the heating element does ?

Answer

→ Due to the low resistance of copper the connecting cord of the heater does not glow because negligible heat is produced in it by passing current but the heating element made of nichrome glows because it becomes red-hot due to the large amount of heat produced on passing current (because of its high resistance).

Short Answer Type Questions-Pg-66

10A. Write down the formula for the heat produced when a current I is passed through a resistor R for time t.

Answer

→ The heat produced when a current I is passed through a resistor R for time t will be
Heat produced, H = I²Rt

10B. An electric iron of resistance 20 ohms draws a current of 5 amperes. Calculate the heat produced in 30 seconds.

Answer

→ Here,
R = 20ohm, I = 5amp, t = 30s
We know that H = I²Rt
H = 5²X20X30
H = 15000 J

11. State three factors on which the heat produced by an electric current depends. How does it depend on these factors?

Answer

→ Heat produced by an electric current is given as H = I²Rt, that means
(i) Heat produced is directly proportional to square of current.
(ii) Heat produced is directly proportional to resistance.
(iii) Heat produced is directly proportional to the time for which current flows.

12A. State and explain Joule's law of heating.

Answer

→ Joule's law of heating states that when a current of I amperes flows in a wire of resistance R ohms for time t seconds then the heat produced in joules and is given by H = I²Rt
The law implies that heat produced in a resistor is (i) directly proportional to the square of current for a given resistance, (ii) directly proportional to resistance to resistance for a given current and (iii) directly proportional to the time for which current flows through the resistors.

12B. A resistance of 40 ohms and one of 60 ohms are arranged in series across 220 volt supply. Find the heat in joules produced by this combination of resistances in half a minute.

Answer

→ Here it is given that R₁ = 40 ohms, R₂ = 60 ohms, V = 220V, t = 30 sec
Here R₁ and R₂ are arranged in series connection.
So the equivalent resistance = R = 40+60 = 100 ohms
According to Ohm's law, we know that
V = IR
I = V/R
I = 220/100 = 2.2amp
As we know that heat produced in joules is given by H = I²Rt
Substituting the values of I, R and t in eq. H = I²Rt
H = (2.2)² X 100 X 30
H = 14520 J

13. Why is an electric light bulb not filled with air? Explain why argon or nitrogen is filled in an electric bulb.

Answer

→ As the filament of the bulb is made from tungsten, so when we fill air inside the electric bulb this tungsten filament will burn quickly. So we fill non-reactive gas like argon or nitrogen in the electric bulb to prolong the life of filament.

14. Explain why, tungsten is used for making the filaments of electric bulbs.

Answer

→ Due to its very high melting point of tungsten it is used for making the filaments of electric bulbs.

15. Explain why, the current that makes the heater element very hot, only slightly warms the connecting wires leading to the heater.

Answer

→ Due to the low resistance of the connecting wires of the heater negligible heat is produced in them by passing current so it is slightly warm.

16. When a current of 4.0 A passes through a certain resistor for 10 minutes, 2.88x10⁴ J of heat are produced. Calculate:
(a) the power of the resistor.
(b) the voltage across the resistor.

Answer

→ (a) Here it is given,
I = 4 amp,
t = 10 min = 10 x 60 = 600sec, H = 2.88x10⁴J
We Know that according to Joule's law of heating
H = I²Rt
28800 = 42xRx600
R = 3ohms
We know that
P = I²xR
= 4²x3
P = 48W
(b) Here we have to find the potential difference, V = ?
We know that
V = IR
V = 4x3
V = 12V

17. A heating coil has a resistance of 200 Ω. At what rate will heat be produced in it when a current of 2.5 A flows through it ?

Answer

→ Here Resistance of the coil(R) = 200 ohms, Current (I) = 2.5 amp, time(t) = 1 sec
We know that according to Joule's law of heating
H = I²Rt
H = (2.5)²×200×1
H = 1250 J/s

18. An electric heater of resistance 8 Ω takes a current of 15 A from the mains supply line. Calculate the rate at which heat is developed in the heater.

Answer

→ Here Resistance (R) = 8 ohms, Current(I) = 15 amp, time (t) = 1 sec
We know that according to Joule's law of heating
H = I²Rt
H = 15²X8X1
H = 1800J/s

19. A resistance of 25 Ω is connected to a 12 V battery. Calculate the heat energy in joules generated per minute.

Answer

→ Here R = 25ohms, V = 12V, H = ?, t = 60sec
According to Ohm’s law
V = IR
12 = 25XI
I = 0.48amp
We know that according to Joule's law of heating
H = I²Rt
H = (0.48)²X25X60
H = 345.6J

20. 100 joules of heat is produced per second in a 4 ohm resistor. What is the potential difference across the resistor?

Answer

→ Here, it is given H = 100J, t = 1sec, R = 4ohms,
We know that according to Joule's law of heating
H = I²Rt
100 = (I)²×4×1
100/4 = I2
I = 5amp
Again we know that
V = IR
V = 5X4
= 20V

21A. Derive the expression for the heat produced due to a current 'I' flowing for a time interval 't' through a resistor 'R' having a potential difference 'V' across its ends. With which name is this relation known?

Answer

→ All materials offer resistance to the flow of current through them. So some external energy is required to make the current flow. This energy is provided by the battery. Some of this energy gets dissipated as heat energy, so the resistor becomes hot.
Work done in carrying a charge Q through a potential difference V is given as
Also, Q = I t
Using Ohm's law, V = I R
W = I²Rt
This work done in carrying the charge through the wire appears as the heat produced.
i.e. H = V I t =  I²Rt
this energy is dissipated as heat energy.
This law is called Joule law of heating effect.

21B. How much heat will an instrument of 12 W produce in one minute if it is connected to a battery of 12 V?

Answer

→ Here, it is given that P = 12W, V = 12V, t = 60sec
We know that
P = VI
I = P/V = 12/12 = 1A
Again, we know that
V = IR
R = V/I = 12/1 = 12ohm
So according to the Joule’s heating law, we know that
H = I²Rt
H = 1²x12x60
H = 720J

21C. The current passing through a room heater has been halved. What will happen to the heat produced by it?

Answer

→ Joule heating is given by
H = i²Rt
H is the heat produced, i is the current through the wire of resistance R for time t. If i is halved i.e i' = i/2
H become 1/4 th

21D. What is meant by the heating effect of current? Give two applications of the heating effect of current.

Answer

→ When electric current is supplied to a purely resistive conductor, the energy of electric current is dissipated entirely in the form of heat and as a result, resistor gets heated. The heating of resistor because of dissipation of electrical energy is commonly known as Heating Effect of Electric Current. Some examples are as follows:
When electric energy is supplied to an electric bulb, the filament gets heated because of which it gives light. The heating of electric bulb happens because of heating effect of electric current.
When an electric iron is connected to an electric circuit, the element of electric iron gets heated because of dissipation of electric energy, which heats the electric iron. The element of electric iron is a purely resistive conductor. This happens because of heating effect of electric current.

21E. Name the material which is used for making the filaments of an electric bulb.

Answer

→ Pure tungsten has some amazing properties including the highest melting point (3695 K), lowest vapor pressure, and greatest tensile strength out of all the metals. Because of these properties it is the most commonly used material for light bulb filaments.

Multiple Choice Questions (MCQs)-Pg-66

22. The heat produced by passing an electric current through a fixed resistor is proportional to the square of:
A. magnitude of resistance of the resistor
B. temperature of the resistor
C. magnitude of current
D. time for which current is passed

Answer

→ The heat produced by passing an electric current through a fixed resistor is proportional to the square of magnitude of current.
As we know that, H = V I t = I²Rt

23. The current passing through an electric kettle has been doubled. The heat produced will become:
A. half
B. double
C. four times
D. one-fourth

Answer

→ The current passing through an electric kettle has been doubled. The heat produced will become four times.
As H = I²Rt

24. An electric fuse works on the:
A. chemical effect of current
B. magnetic effect of current
C. lighting effect of current
D. heating effect of current

Answer

→ When the circuit current exceeds a specified value due to voltage fluctuations or short-circuiting, the fuse wire gets heated and melts.

25. The elements of electrical heating devices are usually made of:
A. tungsten
B. bronze
C. nichrome
D. argon

Answer

→ An electric heating element is generally made from nichrome and can come in the shape of either a coil, ribbon or wire strip. When electricity is introduced into the heating element, its internal temperature increases and grows red hot as it begins to radiate heat outward.

26. The heat produced in a wire of resistance ‘x’ when a current ‘y’ flows through it in time ‘z’ is given by :
A. x² x y x z
B. x × z x y²
C. y x z² x x
D. y x z x x

Answer

→ As H = I²Rt
So, the heat produced in a wire of resistance ‘x’ when a current ‘y’ flows through it in time ‘z’ is given by x × z x y²

27. Which of the following characteristic is not suitable for a fuse wire?
A. thin and short
B. thick and short
C. low melting point
D. higher resistance than res of wiring

Answer

→ In electronics and electrical engineering, a fuse is a type of low resistance resistor that acts as a sacrificial device to provide over current protection, of either the load or source circuit.

28. In a filament type light bulb, most of the electric power consumed appears as :
A. visible light
B. infra-red-rays
C. ultraviolet rays
D. fluorescent light

Answer

→ In a filament type light bulb, most of the electric power consumed appears as infra-red-rays.

29. Which of the following is the most likely temperature of the filament of an electric light bulb when it is working on the normal 220 V supply line?
A. 500° C
B. 1500°C
C. 2500°C
D. 4500°C

Answer

→ The automated mass manufacturing of incandescent filaments intended for operation at over 2500 K is a significant technical achievement.

30. If the current flowing through a fixed resistor is halved, the heat produced in it will become:
A. double
B. one-half
C. one-fourth
D. four times

Answer

→ If the current flowing through a fixed resistor is halved, the heat produced in it will become one-fourth
As H = I²Rt

Questions Based on High Order Thinking Skills (HOTS)-Pg-67

31. The electrical resistivities of four materials P, Q, R and S are given below :
P - 6.84 × 10^-8 Ω m
Q - 1.70 × 10^-8 Ω m
R - 1.0 × 10¹⁵ Ω m
S - 11.0 × 10^-7 Ω m
Which material will you use for making:
(a) heating element of electric iron
(b) connecting wires of electric iron
(c) covering of connecting wires? Give reason for your choice in each case.

Answer

→ (a) The metals and alloy have very low resistivity in the range of 10^-8 ohm meter to 10^-6 ohm meter, so they are good conductor of electricity. While insulators have resistivity of the order of 10¹² to 10¹⁷ ohm meter. Alloys have generally higher resistivity than its constituent metals and consider good for the heating elements. So S is used for heating element of electric iron.
(b) Q has very low resistivity of 1.7 X10^-8ohm-m (it is actually copper). So it is used in connecting wires of electric iron.
(c) R has very high resistivity of 1.0X10¹⁵ohm-m (it is actually rubber). So it is used for covering of connecting wires.

32B. What property of the filament wire accounts for this difference?

Answer

→ This is because of the high resistance offered by filament wire.

33. Two exactly similar heating resistances are connected (i) in series, and (ii) in parallel, in two different circuits, one by one. If the same current is passed through both the combinations, is more heat obtained per minute when they are connected in series or when they are connected in parallel? Give reason for your answer.

Answer

→ Resultant resistance in a series combination is greater than any individual resistance, and resultant resistance in a parallel combination is smaller than any individual resistance.
Since His proportional to R; so more the resistance, more the heat. Therefore, the resistances must be connected in series to obtain more heat per minute.

34. An electric iron is connected to the mains power supply of 220 V. When the electric iron is adjusted at 'minimum heating' it consumes a power of 360 W but at 'maximum heating' it takes a power of 840 W. Calculate the current and resistance in each case.

Answer

→ To find the heat produced by iron in the problem, we need to use an expression for produced power, which depends on the resistance of the iron and the current flowing through the iron.
P_min = VI
360 = 220XI
I = 1.63amp
R = V/I
R = 220/1.63
R = 134.96ohms
In case of maximum heating we get
We know that
Pmax = VI
840 = 220XI
I = 3.81amp
R = V/I
R = 220/3.81
R = 57.74ohms

35. Which electric heating devices in your home do you think have resistors which control the flow of electricity?

Answer

→ Electric fuse, Room heater, Electric iron are some of the device which have resistor which control the flow of electricity.

1. State any two properties of magnetic field lines.

Answer

→ The two properties of magnetic field
Lines are:
(a) the field lines start with the north
pole of the magnet and ends at the south pole of the very same.
(b) the closer the field lines will be
the stronger will be the magnetic

field.

2. What are the two ways in which you can trace the magnetic field pattern of a bar magnet?

Answer

→ The two ways are :
(a) with the help of a compass, as it always indicate the north pole.
(b) it can be traced by sprinkling iron dust.

3. You are given the magnetic field pattern of a magnet. How will you find out from it where the magnetic field is the strongest ?

Answer

→ As we know, the magnetic field is Stronger where the field line is Closest. So the area with the Closest field line in the given Pattern have the strongest Magnetic field.

4. State whether the following statement is true or false :
The axis of earth's imaginary magnet and the geographical axis coincide with each other.

Answer

→ False
The two axis do not coincide with each other as they are separated by an angle of about 17 degree.

5. Why does a compass needle get deflected when brought near a bar magnet?

Answer

→ A compass needle is a small bar magnet. when it is brought near to another bar magnet their field lines interact with each other.
Hence, the compass needle shows deflection.

6. Where do the manufacturers use a magnetic strip in the refrigerator? Why is this magnetic strip used?

Answer

→ To keep the door closed properly manufacturers use a strip of magnets at the doors of refrigerators.

7. Fill in the following blanks with suitable words :
(a) Magnetic field lines leave the..........pole of a bar magnet and enter at its........pole.
(b) The earth's magnetic field is rather like that of a ......... magnet with its........pole in the northern hemisphere.

Answer

→ (a) North, south (magnetic field lines begins with north pole and end at the south pole.)
(b) bar, south.

Short Answer Type Questions-Pg-73

8. Draw a diagram to show the magnetic field lines around a bar magnet

Answer


The field lines begins with north pole and ends at south pole.

9. What is a magnetic field? How can the direction of magnetic field lines at a place be determined?

Answer

→ The area in which the force of magnet is exerted is called magnetic field line to determine the magnetic field line direction you to place a compass on a plane the direction of the needle of a compass shows the north pole, and so will be the direction of the magnetic field line which begins with the north pole and end at south pole.

10. Explain why, two magnetic field lines do not intersect each other.

Answer

→ As a matter of fact the resultant of forces at north pole can only be in one direction. But if magnetic field lines intersect each other then the north pole will show resultant forces in two direction, which is not possible.

11. When an electric current is passed through any wire, a magnetic field is produced around it. Then why an electric iron connecting cable does not attract nearby iron objects when electric current is switched on through it?

Answer

→ This is because, the field produced is very weak.

Long Answer Type Questions-Pg-73

12A. Define magnetic field lines. Describe an activity to draw a magnetic field line outside a bar magnet from one pole to another pole.

Answer

→ A magnetic field lines are lines which are drawn around the magnet and which always starts with the north pole and ending at the south pole.
Activity:
> to draw a magnetic field line around the bar magnet.
> take a bar magnet and a compass and place them on a plane over a hard sheet.
> the boundary of the marker should be marked.
> Place the compass near the north pole of the magnet. The south pole of the needle points towards the north pole of the magnet. The north pole of the compass is directed away from the north pole of the magnet.

> mark the both end of the needle.
> now move the needle to a new position such that the previously occupied north position should now occupy the south position and vice versa.
> continue this process till you the south pole of the bar magnet.
> now by joining all the points by a curve will represent the magnetic field line.

12B. Explain why, a freely suspended magnet always points in the north-south direction.

Answer

→ The earth itself act as a magnet .and its geographical north is in south pole and the geographical south in north pole .so a freely suspended magnet always points in north-south direction.

Multiple Choice Questions (MCQs)-Pg-73

13. A strong bar magnet is placed vertically above a horizontal wooden board. The magnetic lines of force will be:
A. only in horizontal plane around the magnet
B. only in vertical plane around the magnet
C. in horizontal as well as in vertical planes around the magnet
D. in all the planes around the magnet

Answer


→ This is because magnet exert force in every direction.

14. The magnetic field lines produced by a bar magnet :
A. originate from the south pole and end at its north pole
B. originate from the north pole and end at its east pole
C. originate from the north pole and end at its south pole
D. originate from the south pole and end at its west pole

Answer

→ The magnetic field line always originates from the north pole and ends at south pole.

15. Which of the following is not attracted by a magnet ?
A. steel
B. cobalt
C. brass
D. nickel

Answer

→ Because brass is an alloy.

16. The magnetic field lines :
A. intersect at right angles to one another
B. intersect at an angle of 45° to each other
C. do not cross one another
D. cross at an angle of 60° to one another

Answer

→ Magnetic field line never intersect each other.

17. The north pole of earth's magnet is in the:
A. geographical south
B. geographical east
C. geographical west
D. geographical north

Answer

→ Geographical south is in the north pole.

18. The axis of earth's magnetic field is inclined with the geographical axis at an angle of about:
A. 5°
B. 15°
C. 25°
D. 35°

Answer

→ 15°

19. The shape of the earth's magnetic field resembles that of an imaginary:
A. U-shaped magnet
B. Straight conductor carrying current
C. Current-carrying circular coil
D. Bar magnet

Answer

→ Bar magnet.

20. A magnet attracts:
A. plastics
B. any metal
C. aluminium
D. iron and steel

Answer

→ Iron and steel shows magnetic behaviour.

21. A plotting compass is placed near the south pole of a bar magnet. The pointer of plotting compass will :
A. point away from the south pole
B. point parallel to the south pole
C. point towards the south pole
D. point at right angles to the south pole

Answer

→ The plotting compass is placed near the south pole of a bar magnet. The pointer of plotting compass will point at right angles to the south pole.

22. The metallic pointer of a plotting compass gets deflected only when it is placed near a bar magnet because the pointer has :
A. electromagnetism
B. permanent magnetism
C. induced magnetism
D. ferromagnetism

Answer

→ The metallic pointer of a plotting compass get deflected only when it is placed near a bar magnet because the pointer has permanent magnetism.

23. Which of the following statements is incorrect regarding magnetic field lines ?
A. The direction of magnetic field at a point is taken to be the direction in which the north pole of a magnetic compass needle points.
B. Magnetic field lines are closed curves
C. If magnetic field lines are parallel and equidistant, they represent zero field strength
D. Relative strength of magnetic field is shown by the degree of closeness of the field lines.

Answer

→ The strength of field is zero, when the magnetic field lines are equidistance and parallel.

Questions Based on High Order Thinking Skills (HOTS)-Pg-74

24. Copy the figure given below which shows a plotting compass and a magnet. Label the N pole of the magnet and draw the field line on which the compass lies.


Answer
→


As the needle of the compass is tending to move toward the other end of the bar magnet, so the nearer end will show the north pole and the other end will show the south pole.

25A. The diagram shows a bar magnet surrounded by four plotting compasses. Copy the diagram and mark in it direction of the compass needle for each of the cases B, C and D.


Answer

→ Arrows shows the direction due to the deflection.
As the


25B. Which is the north pole, X or Y ?


Answer

→ As we see in the figure, the pointer is pointing away from the magnet. It means X is the NORTH POLE.

26. The three diagrams in the following figure shows the lines of force (field lines) between the poles of two magnets. Identify the poles A, B, C, D, E and F.


Answer

→ A=N, B=N, C=S, D=S, E=N, and F=S.
The field lines originate from the north pole and end at the south pole.

27. The figure given below shows the magnetic field between two magnets:

(i) Copy the diagram and label the other poles of the magnets.
(ii) Which is the weaker magnet?

Answer

→ (i) S-N, N-S

The field line always emits form the north pole and ends at the south pole.
(ii) Magnet 2 is the weaker magnet.
It has small magnetic field.

Very Short Answer Type Questions-Pg-81

1. Which effect of current can be utilised in detecting a current-carrying wire cancelled in a wall?

Answer


→ Magnetic effect of a current can be utilised in detecting a current carrying wire cancelled in a wall.

2. What conclusion do you get from the observation that a current-carrying wire deflects a compass needle placed near it?

Answer

→ By the deflection of a compass needle, when placed near a current carrying wire gives a conclusion that there is a magnetic field produced around the wire.

3. Name the scientist who discovered the magnetic effect of current.

Answer

→ Oersted, was the scientist who first discovered the magnetic effect of current.

4. State qualitatively the effect of inserting an iron core into a current-carrying solenoid.

Answer

→ The magnetic field become very strong when the iron core is inserted into a current carrying conductor.

5. Name the rule for finding the direction of magnetic field produced by a straight current-carrying conductor.

Answer

→ The rule for finding the direction of magnetic field produced by a straight current carrying conductor was given by Maxwell’s right hand thumb rule.

6. State the form of magnetic field lines around a straight current--++carrying conductor.

Answer

→ A concentric circle is formed around a straight current carrying conductor and its centre lies in the conductor.

7. What is the other name of Maxwell’s right-hand thumb rule?

Answer

→ Maxwell’s corkscrew rule is the other name for the Maxwell’s right-hand thumb rule.

8. State whether the following statement is true or false:
The magnetic field inside a long circular coil carrying current will be parallel straight lines.

Answer

→ True, Inside a long circular coil carrying current the magnetic field lines are always parallel with each other.

9. What is the shape of current-carrying conductor whose magnetic field pattern resembles that of a bar magnet?

Answer

→ A current carrying conductor is of solenoid shape whose magnetic field pattern seems to that of the bar magnet.

10. State three ways in which the strength of an electromagnet can be increased.

Answer

→ The three ways are :-
(1) By increasing the number of turns in the coil.
(2) Increase the current flow in the coil.
(3) The air gaps between the poles should must be reduced.

11. Fill in the following blanks with suitable words:
(a) The lines of _____ round a straight-carrying conductor are in the shape of _____
(b) For a current-carrying solenoid, the magnetic field is like that of a _____
(d) The magnetic effect of a coil can be increased by increasing the number of _____, increasing the _____ or inserting an _____ core.
(e) If a coil is viewed from one end and the current flows in an anticlockwise direction, then this end is a _____ pole.
(f) If a coil is viewed from one end, and the current flows in a clockwise direction, then this end is a _____ pole.

Answer

→ (a) Magnetic field, concentric circles
(b) Bar magnet
(d) Turns, current, coil
(e) North pole
(f) South pole
Explanation: (a) a concentric circle is formed by the lines of a straight current carrying conductor.
(b) when current is passes through a solenoid it starts to act as a bar magnet.
(d) by increasing the no. of turns or increasing the flow of current or inserting the coil in a core the magnetic effect can be increased.
(e) north pole is the direction, if a coil is viewed from one end and the flow of current is in anticlockwise direction.
(f) when the current flow direction is clockwise when view from one end of the coil then that end is south pole.

12. Describe how you will locate a current-carrying wire concealed in a wall.

Answer

→ To locate a current carrying wire concealed in a wall one should use a plotting compass. by moving plotting compass on the wall the deflection in its needle shows the place where wire is concealed in the wall.

13. Describe some experiment to show that the magnetic field is associated with an electric current.

Answer


Let us take a copper wire AB on a plane such that it faces North-South. Now place a plotting compass under it. connect both end to a battery with a switch to put it on and off.
When the switch is on then the plotting compass needle shows deflection. When its turned off then the compass stop showing any deflection. This shows that the magnetic field is associated with electric field.

14A. Draw a sketch to show the magnetic lines of force due to a current-carrying straight conductor.

Answer



14B. Name and state the rule to determine the direction of magnetic field around a straight current-carrying conductor.

Answer

→ The Fleming's left rule is used to determine the direction of magnetic field around a straight current-carrying conductor. According to the rule,
The F(Thumb) represents the direction of Force of the conductorThe B(Forefinger) represents the direction of the Magnetic fieldThe I(Centre finger) represents the direction of the Current.


15. State and explain Maxwell's right-hand thumb rule.

Answer

→ Maxwell’s right hand thumb rule. According to this law if we hold our thumb of right hand in the direction of magnetic field and close our finger then our enclosed finger shows the magnetic field line around the conductor.

Let AB be a straight wire. Placed vertically A to B. To find the direction of magnetic field we assume that we are holding the current carrying wire in such a way that our thumb of right hand shows the direction of flow of current. Then the finger enclosing the wire. The direction in which our finger is enclosed shows the direction of magnetic field which is anticlockwise.

16. What is Maxwell's corkscrew rule? For what purpose is it used?

Answer

→ Maxwell’s right hand thumb rule is also known as Maxwell’s corkscrew rule. It is used to find out the direction of magnetic field along a straight current carrying conductor.

17A. Draw the magnetic lines of force due to a circular wire carrying current.

Answer



17B. What are the various ways in which the strength of magnetic field produced by a current-carrying circular coil can be increased?

Answer

→ There are two ways of doing this :-
(1) must increase the number of turns of the coil.
(2) Must increase the current flow in the coil.

18. State and explain the Clock face rule for determining the polarities of a circular wire carrying current.

Answer

→ According to this rule, we first look at one face of the coil and if the current through the coil seems to be flowing in the clockwise direction, then that face is considered as the South pole. And if the current seems to be flowing in the anti-clockwise direction, then the face of the coil is considered as the North pole.
a) For this face of the coil, the current is flowing in the clockwise direction, so it acts as South-Pole.
b) For this face of the coil, he current if flowing in he anti-clockwise direction, so it acts as North-Pole.


19. Name any two factors on which the strength of magnetic field produced by a current-carrying solenoid depends. How does it depend on these factors?

Answer

→ It depends on following factors:-
i) Current: Larger he current, stronger will be the magnetic field.
ii) Number of coils in the solenoid: Greater the number of coils in the solenoid, stronger will be the magnetic field produced.

20A. Draw a circuit diagram to show how a soft iron piece can be transformed into an electromagnet.

Answer

→ A coil C of insulated wire is wound around the soft iron core and the two ends of the coil are connected to a battery. This is how an electromagnet is made.


20B. Describe how an electromagnet could be used to separate copper from iron in a scrap yard.

Answer

→ Electromagnetic cranes are used to separate copper from iron in a scrap yard. Switch in on the curen energises the electromagnet, thus producing magnetic field. This magnetic field attracts the iron pieces, thus separating them from the copper.

21A. How does an electromagnet differ from a permanent magnet?

Answer

→ An electromagnet produces a temporary magnetic field which sustains till the current flows through the coils of the electromagnet, while a permanent magnet produces a permanent magnetic field.

21B. Name two devices in which electromagnets are used and two devices where permanent magnets are used.

Answer

→ Electromagnets are used in Electric bell and Electric motors.
Permanent magnets are used in Refrigerators doors and Toys.

Long Answer Type Questions-Pg-82

22A. What is a solenoid? Draw a sketch to show the magnetic field pattern produced by a current-carrying solenoid.

Answer


→ A solenoid is a long coil containing a large number of close turns of insulated copper wire.


22B. Name the type of magnet with which the magnetic field pattern of a current-carrying solenoid resembles.

Answer

→ The magnetic field produced by a current carrying solenoid is similar to that produced by a bar magnet.

22C. What is the shape of field lines inside a current-carrying solenoid? What does the pattern of field lines inside a current-carrying solenoid indicate?

Answer

→ Magnetic field lines inside a current carrying conductors are in the form of parallel straight lines.

22D. List three ways in which the magnetic field strength of a current-carrying solenoid can be increased?

Answer

→ Following are the three ways of increasing the strength of the magnetic field of current carrying solenoid:
i) Increasing the current flowing through the coil of solenoid.
ii) Increasing the number of turns in he solenoid.

22E. What type of core should be put inside a current-carrying solenoid to make an electromagnet ?

Answer

→ Soft iron core should be used.

23A. What is an electromagnet? Describe the construction and working of an electromagnet with the help of a labelled diagram.

Answer

→ An electromagnet is called a temporary magnet because it produces magnetic field lines only when the current flows through the coil of the electromagnet. When we stop the current through the coil, the magnetic field lines diminishes.

23B. Explain why, an electromagnet is called a temporary magnet.

Answer

→ Soft iron core is used in the electromagnets because it loses all its magnetism when the current through the coil stops flowing. While in case of steel, there is still some magnetism even after the current stops flowing through the coil of electromagnet.

23C. Explain why, the core of an electromagnet should be of soft iron and not of steel.

Answer

→ The strength of electromagnet depends on the following factors:
i) Number of turns in the coil: Increasing the number of turns in the coil increase the strength of magnetic field produced by the electromagnet.
ii) Current flowing through the coil: Increasing the current flowing through the coil increases the strength of magnetic field produced.
iii) Length of air gap between its poles: Decreasing he length of air gap between the poles of electromagnet increases the strength of electromagnet.

23D. State the factors on which the strength of an electromagnet depends. How does it depend on these factors ?

Answer

→ An electromagnet is a temporary magnet which works on the magnetic effect of the current. It consists of long coil of insulated copper wire wound around a soft iron core and he two ends of he coil are connected to a battery. As the current flows through the coil, the magnetic field lines ae produced. These field lines sustains till the current continues to flow through the coil of electromagnet.

23E. Write some of the important uses of electromagnets

Answer

→ Electromagnets are used in several devices such as Electric bell, Electric motors, Loudspeakers, etc. They are also used by doctors to remove iron or steel particles from patient’s eyes and to remove iron pieces from wounds.

Multiple Choice Questions (MCQs)-Pg-83

24. The strength of the magnetic field between the poles of an electromagnet would be unchanged if :
A. current in the electromagnet winding were doubled
B. direction of current in electromagnet winding were reversed
C. Distance between the poles of electromagnet were doubled
D. material of the core of electromagnet were changed

Answer

→ The strength of the magnetic field between the poles of an electromagnet would be unchanged if the direction of current in electromagnet winding were reversed.

25. The diagram given below represents magnetic field caused by a current-carrying conductor which is:

A. a long straight wire
B. a circular coil
C. a solenoid
D. a short straight wire

Answer

→ The diagram represents the magnetic field caused by a circular current carrying conductor.

26. The magnetic inside a long straight solenoid carrying current :
A. is zero
B. decreases as we move towards its end.
C. increases as we move towards its end.
D. is the same at all points.

Answer

→ The magnetism inside a long straight solenoid remains same at every point.

27. Which of the following correctly describes the magnetic field near a long straight wire?
A. The field consists of straight lines perpendicular to the wire.
B. The field consists of straight lines parallel to the wire.
C. The field consists of radial lines originating from the wire.
D. The field consists of concentric circles centred on the wire.

Answer

→ The magnetic field near a long straight wire consists of concentric circles centred on the wire.

28. The north-south polarities of an electromagnet can be found easily by using:
A. Fleming’s right-hand rule
B. Fleming’s left-hand rule
C. Clock face rule
D. Left-hand thumb rule

Answer

→ The north-south polarities of an electromagnet can be found easily by using clock face rule.

29. The direction of current in the coil at one end of an electromagnet is clockwise. This end of the electromagnet will be:
A. north pole
B. east pole
C. south pole
D. west pole

Answer

→ The direction of current in the coil at one end of an electromagnet is clockwise then that end is south pole.

30. If the direction of electric current in a solenoid when viewed from a particular end is anticlockwise, then this end of solenoid will be:
A. west pole
B. south pole
C. north pole
D. east pole

Answer

→ The direction of electric current is anticlockwise when viewed form the north pole.

31. The most suitable material for making the core of an electromagnet is:
A. soft iron
B. brass
C. aluminium
D. steel

Answer

→ Soft iron is mostly used for making the core of the electromagnet.

32. The magnetic effect of current was discovered by:
A. Maxwell
B. Fleming
C. Oersted
D. Faraday

Answer

→ The magnetic effect of the current was discovered by Oersted.

33. A soft iron bar is inserted inside a current-carrying solenoid. The magnetic field inside the solenoid:
A. will decrease
B. will increase
C. will become zero
D. will remain the same

Answer

→ When soft iron bar is inserted inside a current carrying solenoid then it increases the magnetic field inside the solenoid.

34. The magnetic field lines in the middle of the current-carrying solenoid are :
A. circles
B. spirals
C. parallel to the axis of the tube
D. perpendicular to the axis of the tube

Answer

→ The magnetic field lines in the middle of the current carrying solenoid are parallel to the axis of the tube.

35. The front face of a circular wire carrying current behaves like a north pole. The direction of current in this face of the circular wire is:
A. clockwise
B. downwards
C. anticlockwise
D. upwards

Answer

→ When the front face of a circular wire carrying current behaves like north pole ,then the direction of current in this field is anticlockwise.

36. The back face of a circular loop of wire found to be south magnetic pole. The direction of current in this face of the circular loop of wire will be:
A. towards south
B. clockwise
C. anticlockwise
D. towards north

Answer

→ The back face of a circular loop of wire found to be south magnetic pole .the direction of the current in this face is clockwise.

Questions Based on High Order Thinking Skills (HOTS)-Pg-84

37. In the straight wire A, current is flowing in the vertically downward direction whereas in wire B the current is flowing in the vertically upward direction. What is the direction of magnetic field:
(a) in wire A?
(b) in wire B?
Name the rule which you have used to get the answer.

Answer


→ (a) Clockwise would be the direction of magnetic field in wire A.

(b) the direction of the magnetic field in the wire B is anticlockwise.

Maxwell’s right hand thumb rule.

38. The figure shows a solenoid wound on a core of soft iron. Will the end A be a N pole or S pole when the current flows in the direction shown?


Answer

→ According to the fig. end a is the south pole as the direction of the current flow is clockwise from this end.

39. A current-carrying straight wire is held in exactly vertical position. H the current passes through this wire in the vertically upward direction, what is the direction of magnetic field produced by it? Name the rule used to find out the direction of magnetic field.

Answer

→ The direction of the current flow is anticlockwise. This can be determined by using Maxwell’s right hand thumb rule.


40. For the coil in the diagram below, when the switch is pressed:
(a) What is the polarity of end A?
(b) Which way will the compass point then?


Answer

→ (a) End A becomes south pole. This is because from the end at the current flows in clockwise direction.
(b) The movement of the compass needle is away form the end B. As end B is the north pole.

41. A current flows downwards in a wire that passes vertically through a table top. Will the magnetic field lines around it go clockwise or anticlockwise when viewed from above the table?

Answer

→ According to the Maxwell’s right hand thumb rule the direction of the magnetic field will be in clockwise direction.

42. The directions of current flowing in the coil of an electromagnet at its two ends X and Y are as shown below:

(a) What is the polarity of end X?
(b) What is the polarity of end Y?
(c) Name and state the rule which you have used to determine the polarities.

Answer

→ The easy way to find the polarity is right-hand rule: " If the current flows in the direction of your fingers, the thumb points towards the north pole".

If we assume an electromagnet which has two ends i.e. X and Y, then,
(a) South is the polarity of end X as direction of current flow from that end is clockwise.
(b) The end Y will be the north pole as the current flow from this end shows the anticlockwise direction.
(c) The rule used here to find the polarity of the ends is clock face rule. According to this rule if the face from where it is being observed is showing the flow of current in anticlockwise direction then that end is N-pole but if the same face shows the flow of current in a clockwise direction then that end becomes S-pole.

43. The magnetic field associated with a current-carrying straight conductor is in anticlockwise direction. H the conductor was held along the east-west direction, what will be the direction of current through it? Name and state the rule applied to determine the direction of current?

Answer

→ East to west will be the direction of the current.
According to the Maxwell’s right hand thumb rule if the enclosed fingers shows the direction of the magnetic field then the thumb will always point in the direction of current flow.

44. A current-carrying conductor is held in exactly vertical direction. In order to produce a clockwise magnetic field around the conductor, the current should be passed in the conductor:
(a) from top towards bottom
(b) from left towards right
(c) from bottom towards top
(d) from right towards left

Answer

→ (a) the current should be passed from top to bottom to produce a clockwise magnetic field around the conductor.

45. A thick wire is hanging from a wooden table. An anticlockwise magnetic field is to be produced around the wire by passing current through this wire by using a battery. Which terminal of the battery should be connected to the:
(a) top end of wire?
(b) bottom end of wire?

Answer

→ The direction desired to pass the current is upward.

Very Short Answer Type Questions-Pg-91

1. What happens when a current-carrying conductor is placed in a magnetic field?

Answer


→ When a conductor is placed in a magnetic field then a mechanical force is exerted on the conductor, which eventually make the conductor move.

2. When is the force experienced by a current-carrying conductor placed in a magnetic field largest ?

Answer

→ The force experienced by a current carrying conductor is largest when it is placed at right angle in a magnetic field.

3. In a statement of Fleming's left-hand rule, what do the following represent?
(a) direction of centre finger.
(b) direction of forefinger.
(c) direction of thumb.

Answer

→ (a) By Fleming’s left hand rule direction of centre finger represent the current.
(b) Direction of the forefinger in the Fleming’s left hand rule represents the magnetic field.
(c) Force acting on the conductor is represented by the direction of thumb in Fleming’s left hand rule.

4. Name one device which works on the magnetic effect of current.

Answer

→ The device which work on the magnetic effect of current is electric bell. Electric bell works when electromagnetism is produced.

5. Name the device which converts electrical energy into mechanical energy.

Answer

→ Electrical motor is used to convert electrical energy in to mechanical energy.

6. A motor converts one form of energy into another. Name the two forms.

Answer

→ A motor converts electrical energy into mechanical energy. So the two forms are electrical and mechanical.

7. State whether the following statement is true or false:
An electric motor converts mechanical energy into electrical energy.

Answer

→ False, because electric motor is used to convert electrical energy in to mechanical energy.

8. For Fleming's left-hand rule, write down the three things that are 90° to each other, and next to each one write down the finger or thumb that represents it.

Answer

→ The three things that are 90° to each other, and next to each, and next to each one are:-
Current- it is denoted by the direction of the centre finger.
Magnetic field- ii is denoted by the direction of the fore finger.
Force or motion- it is denoted by the thumb.

9. Name the device which is used to reverse the direction of current in the coil of a motor.

Answer

→ Commutator is used to reverse the direction of current in the coil of a motor.

10. What is the other name of the split ring used in an electric motor?

Answer

→ Commutator is the other name for the split ring which is used in the coil of a motor.

11. What is the function of a commutator in an electric motor?

Answer

→ Its function is to reverse the current every time it reaches the vertical position in the coil.

12. Of what substance are the brushes of an electric motor made ?

Answer

The brushes of an electrical motor are made of carbon.

13. Of what substance is the core of the coil of an electric motor made?

Answer

→ The core of the coil of an electric motor is made of soft iron.

14. In an electric motor, which of the following remains fixed and which rotates with the coil?

Answer

→ The part of the electric motor which remain fixed is brush and the part which rotate with the coil is commutator.

15. What is the role of the split ring in an electric motor?

Answer

→ Every time in a revolution when the coil passes the vertical position the split ring is there to reverse the current.

16. Fill in the following blanks with suitable words:
(a) Fleming's Rule for the motor effect uses the ............... hand.
(b) A motor contains a kind of switch called a ............ which reverses the current every half ............ .

Answer

→ (a) Left
Fleming’s rule for the motor effect uses the left hand.
(b) Commutator; rotation
A motor contains a kind of switch called a commutator which reverse the current every half rotation.