CBSE Class 10 Science Electricity Assignment Set 05

Read and download the CBSE Class 10 Science Electricity Assignment Set 05 for the 2026-27 academic session. We have provided comprehensive Class 10 Science school assignments that have important solved questions and answers for Chapter 11 Electricity. These resources have been carefuly prepared by expert teachers as per the latest NCERT, CBSE, and KVS syllabus guidelines.

Solved Assignment for Class 10 Science Chapter 11 Electricity

Practicing these Class 10 Science problems daily is must to improve your conceptual understanding and score better marks in school examinations. These printable assignments are a perfect assessment tool for Chapter 11 Electricity, covering both basic and advanced level questions to help you get more marks in exams.

Chapter 11 Electricity Class 10 Solved Questions and Answers

Very Short Answer Type Questions :

Question. When is potential difference between two points said to be 1 volt?
Answer: Potential difference betweeen two point is said to be 1 Volt if the amount of work done in bringing 1 C charge between them is 1 J.

Question. An electric iron draws a current of 0.5 A when the voltage is 200 volt. Calculate the amount of electric charge flowing through it in one hour.
Answer: We have I = 0.5 A
V = 200 volt
t = 1 hr = 3600 s
Q = I x t = 0.5#3600 A – s = 1800 C

Question. Two resistors of 10 Ω and 15 Ω are connected in series to a battery of 6 V. How can the values of current passing through them be compared?
Answer: In series, same current flows through each resistor. So, ratio of current is 1 : 1.

Question. What happens to the resistance of a conductor when temperature is increased?
Answer: The resistance of a conductor increases with rise in temperature.

Question. Name a device that helps to maintain a potential difference across a conductor.
Answer: Cell or battery

Question. Write relation between heat energy produced in a conductor when a potential difference V is applied across its terminals and a current I flows through for ‘t’
Answer: Heat produced, H = VIt

Question. An electric appliance draws a current of 0.4 A when the voltage is 200 volt. Calculate the amount of charge flowing through it in one hour.
Answer: Q = It
Given, I = 0.4 A
V = 200 Volt
t = 1 hr = 3600 s
` Q = 0.4 x 3600 C
Q = 1440 C

Question. State Ohm’s law.
Answer: If the physical conditions of a conductor remain same then current through a conductor is directly proportional to the potential difference b/w the two ends of the conductor.
I ? V & V = IR

Question. V-I graphs for two wires A and B are shown in the figure. If both the wires are made of the same material and are of equal thickness, which of the two is of more length? Give justification for your answer.
Answer: We know for identical wire more length more resistance and vice versa slope of wire A is more than B. Hence resistance of A is more and its length also.

Short Answer Type Questions :

Question. Define 1 volt. Express it in terms of SI unit of work and charge calculate the amount of energy consumed in carrying a charge of 1 coulomb through a battery of 3 V.
Answer: When 1 joule of work is done in carrying 1 coulomb of charge, from infinity to a point in the electric field, then potential at that point is called 1 volt. Potential difference between two points is

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Question. Name and define SI unit of resistance. Calculate the resistance of a resistor if the current flowing through it is 200 mA, when the applied potential difference is 0.8 V.
Answer: a. SI unit of resistance is ohm Ω
1 Ω = 1V/1A
b. The resistance of a conductor is said to be 1 ohm if a current of one amp flows through it when a potential differences of 1 volt is applied across it.
c. R = V/I = 0.8/200 X 10-5 = 4 Ω

Question. How much current will an electric bulb draw from 220 V source if the resistance of the bulb is 1200Ω? If in place of bulb, a heater of resistance 100 Ω is connected to the sources, calculate the current drawn by it.
Answer: 

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Question. What is an electric circuit? Distinguish between an open and a closed circuit.
Answer: An arrangement for maintaining the continuous flow of electric current by the electrical energy source through the various electrical components connected with each other by conducting wires is termed as electric circuit. An open circuit does not carry any current, while a closed circuit carries current.

Question. Suppose a 6-volt battery is connected across a lamp whose resistance is 20 ohm the current in the circuit is 0.25 A, calculate the value of the resistance from the resistor which must be used.
Answer: Lamp resistance, R = 20 ohm
Extra resistance from resistor, R = ?
(to be calculated)
For R and R’ in series,
Total circuit resistance, Rs = R + R’

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= 24 ohm
But      Rs = R + R’
Hence  R’ = Rs – R
= 24 – 20 = 4 ohm
Extra resistance from resistor,
R’ = 4 ohm.

Question. A wire of length l and resistance R is stretched so that the length is doubled and area of cross-section is halved. How will its:
a. Resistance change?
b. Resistivity change?
Answer: a. Let initial length, area of cross-section and resistance of wire are l , A and R respectively.
When length is stretched two times, let its area of cross-section becomes Al. Its initial volume of wire = final volume of the wire.
Al = Al2l
or A’/A = 1/2 or A’ = A/2
R’ = P,(21)/A’ = P,21 X 2/A
= 4P,l/A
= 4 R
b. Resistivity does not change because it is property of the material of a conductor.

Question. Explain the role of fuse in series with any electrical appliance in an electric circuit. Why should a fuse with defined rating for an electric circuit not be replaced by one with a larger rating?
Answer: Fuse wire is a safety device connected in series with the live wire of circuit. It has high resistivity and low melting point. It melts when a sudden urge of large current passes through it and disconnects the entire circuit from the electrical supply. But, in case if we use a larger rating instead of a defined rating, then it will not protect the circuit as high current will easily pass through it and it will not melt.

Question. Draw a schematic diagram of an electric circuit consisting of a battery of five 2 V cells, a 20 Ω resistor, a 30 Ω resistor, a plug key, all connected in series. Calculate the value of current flowing through the 20 Ω resistor and the power consumed by the 30 Ωresistor.
Answer: Req = 20 + 30 = 50 Ω img
Current through both 20 Ωand 30 Ω= I =10/50
= 0.2 A
Power consumed by 30 Ω = I2R = (0.2)2 x 30=1.2W

Question. (a) List the factors on which the resistance of a conductor in the shape of a wire depends.
(b) Why are metals good conductors of electricity whereas glass is a bad conductor of electricity? Give reason.
(c) Why are alloys commonly used in electrical heating devices? Give reason.
Answer: a. Factors on which resistance of a wire depends:
i. Resistance is directly proportional to length.
ii. Resistance is inversely proportional to area of cross-section.
R ∝ l , R ∝ 1/A
R ∝ 1/A or R = P,1/A
b. Metal are good conductor due to having large number of free electrons and their low resistivity.
Glass is a bad conductor because it has no free electrons and its resistivity is higher.
c. Alloys are commonly used in electrical heating devices due to their high resistivity and high melting point which produces more heat.

Question. A wire of resistance 20 Ω is bent to form a closed square. What is the resistance across a diagonal of the square?
Answer: 

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Question. Define resistance of a conductor. State the factors on which resistance of a conductor depends. Name the device which is often used to change the resistance without changing the voltage source in an electric circuit.
Calculate the resistance of 50 m length of wire of cross-sectional area 0.01 square mm and of resistivity 5X10-8 Wm .
Answer: Resistance is the opposition offered in the path of flow of current by the atoms or molecules of a conductor.
Factors affecting resistances:
a. length R ∝ l
b. area of cross-section R ∝ 1/A
c. nature of material.
Rheostat is used to change the current in the circuit without changing the voltage source.
Given: l = 50 m
A = 0.01mm2 = 0.01 X 10-6 m2
r = 5X10-8 Wm
As R = P,lA = 5 X 10-5 X 50/0.01 X 10-5
R = 250 Ω

Question. A hot plate of an electric oven connected to a 220 V line has two resistors A and B each of 22 W resistance.
These resistors may be used separately, in series or in parallel. Find the current flowing in all the three cases.
Answer: Given, V = 220 Volt
RA = 22 W
RB = 22 W
a. Current in resistances separately?
= 220/22 A = 10 A
b. Current in series combination
RS = 22 + 22 = 44 Ω
IS = 220/44 = 5 A
c. Current in parallel combination 
1/RP = 1/22+ 1/22 = 2/22 Ω
RP = 11Ω
IP = 220/11 = 20 A

Question. A given length of a wire is doubled on itself and this process is repeated once again. By what factor does the resistance of the wire change?
Answer: Am. Length becomes one-fourth of the original length and area of cross-section becomes four times that of original.

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Question. (a) Explain why a conductor offers resistance to the flow of current.
(b) Differentiate between conductor, resistor and resistance.
Answer: a. When a current is passed through a conductor, the atoms or molecule of the conductor produce an hindrance in the path of flow of electron. This hindrance in the path of flow of charge is called resistance of the conductor.
b. A substance which allow to pass the charges through them easily is called a conductor.
Resistor : A conductor having some value of resistance is called a resistor.
Resistance : It is the property of any conductor by virtue of which it opposes the flow of charge through it.

Question. (a) State the commercial unit of electric energy and find its relation with its SI unit.
(b) The current through a resistor is made three times its initial value. Calculate how it will affect the heat produced in the resistor.
(c) Find the increase in the amount of heat generated in a conductor if another conductor of double resistance is connected in the circuit keeping all other factors unchanged.
Answer: a. Commercial unit of electric energy = kWh
1 kWh = 3.6X106 J
b. Initial heat generated in the resistor = I2Rt
H1 = I2Rt …(1)
when current is made three times i.e. 3I now heat generated
H2 = ^3I h2Rt …(2)
H2 = 9I2Rt
In later case, heat generated is 9 times the initial heat
generated.
c. If another conductor of 2R is connected in series
then total resistance = R + 2R = 3R.
Now heat generated H = I2 ^3Rht
H = 3I2Rt
In this case, heat generated is three times.

Question. V-I graph for two wires A and B are shown in the figure. If both wires are of same length and same thickness, which of the two is made of a material of high resistivity? Give justification for your answer.

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Answer: Greater than slope of V-I graph, greater will be the resistance of given metallic wire. In the given graph, wire A has greater slope then B. Hence, wire A has greater resistance.
For the wires of same length and same thickness, resistance depends on the nature of material of the wire,

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Question. What is meant by “electrical resistance” of a conductor? State how resistance of a conductor is affected when
a. a low current passes through it for a short duration;
b. a heavy current passes through it for about 30 seconds.
Answer: Electrical resistance is the property of a conductor by virtue of which it opposes the flow of current through it. It is equal to the ratio of the potential difference applied across its ends to the current flowing through it.
R = V/I
a. When a low current is passed for a short duration, through a conductor, heat produced is almost negligible and hence no appreciable change in its resistance.
b. When heavy current is passed through the conductor for 30 s. Conductor may be get heated and its resistance and resistivity change.

Question. Name and define the SI unit of current. Calculate the number of electrons that flow through a conductor in 1 second to constitute a current of 1 ampere. (Charge on an electron = 1.6 X 10-19 coulomb)
Answer: SI unit of current is Ampere (A)
I = q/t
If q = 1 C, t = 1 s
then I = 1A
If 1 C charge flows in 1 s in a conductor then magnitude of current is said to be 1A.
q = ne
n = q/e = I x t/e
= 1A x 1e/1.6 x 10-19 = 100/16 x 1018
= 6.25 x 1018

Question. The figure below shows three cylindrical copper conductors along with their face areas and lengths. Discuss in which geometrical shape the resistance will be highest.

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Answer: 

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Question. What is meant by resistance of a conductor? Name and define its SI unit. List the factors on which the resistance of a conductor depends. How is the resistance of a wire affected if:
a. its length is doubled,
b. its radius is doubled?
Answer: Property of any conductor by virtue of which it opposes the flow of current in the conductor is called its resistance.
SI unit of resistance is Ohm. If by applying a potential difference of 1 volt the current in the conductor is 1 A.
Then the resistance of the conductor is said to be 1 ohm.
Factor affecting resistances:
a. If length is double then resistance also becomes doubled.
b. If radius is doubled then area A = p π2rh2 becomes
4 times, then the resistance becomes 1/4.

Question. What do you mean by heating effect of electric current?
Explain the production of heat in a resistor by flow of electric current through it. Name two devices based on heating effect of current.
Answer: When current is passed through a conductor, heat is produced. Production of heat in a conductor on passing the current is called heating effect of current.
When a current is passed in a conductor by applying a potential difference, electrons get accelerated and collide with the atoms of the conductor. During the collision there is a loss of kinetic energy. This loss in K.E. appears in the form of heat energy in the conductor 1. Electric heater, 2. Bulb.

Question. a. What is meant by saying that the potential difference between two points is 1 volt?
b. Why does the connecting cord of an electrical heater not glow while the heating element does?
c. Electrical resistivity of some substance at 20°C are given below:

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Answer the following questions in relation to them:
i. Among silver and copper which one is better conductor? Why?
ii. Which material would you advise to use in electrical heating device? Why?
Answer: a. If the amount of work done in bringing a unit positive charge from one point to another is 1 J in an electric field then potential difference b/w the two points is said to be 1 volt. 1V = 1J/1C
b. Connected cord of an electrical heater does not glow because this is of copper metal, which has low resistivity and good conductivity. Heating element glows due to its high resistivity or poor conductivity.
c. (i) The resistivity of silver is lesser than copper so it is a good conductor.
(ii) The resistivity of nichrome is maximum in the given table so nichrome will be advised to use in electrical heating device.

Question. What is the lowest resistance that can be obtained by combining four coils of resistances 4 W , 8 W , 12 W and 24 W ?
Answer: If the resistances are combined in parallel then we shall get the lowest resistance. In parallel combination:

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Long Answer Type Questions :

Question. Consider the following circuit diagram. If R= R2 =R3 = R4= R5 = 3 Ω, find the equivalent resistance(Rs) of the circuit.

CBSE-Class-10-Science-Electricity-Assignment-Set-E-9

Answer: 

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Question. Show how would you join three resistors, each of resistance 9 W so that the equivalent resistance of the combination is (i) 13.5 W (ii) 6 W ?
                          or
(a) Write Joule’s law of heating.
(b) Two bulbs, one rated 100 W; 220 V, and the other 60 W; 220 V are connected in parallel to electric mains supply. Find the current drawn by two bulbs from the line if the supply voltage is 220 V.
Answer: 

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a. Joules Law of Heating : When a current I is passed through a resistor R for time t then heat produced
H ∝ I2
∝ R
∝ t
H ∝ I2Rt
or H = I2Rt
b. Given two bulbs of 100 W–200 V and 60W–220 V.
Current in 1st bulb:
I1 = P1/V = 100/220 A = 5/11 = 0.45 A
Current in second bulb.
I2 = P2/V = 60/220 A = 3/11 A = 0.27 A

Question. When a high resistance voltmeter is connected directly across an electric bulb, its reading is 2 V. An electric cell is sending the current of 0.4 ampere (measured by an ammeter) in the electric circuit.
a. Draw the circuit.
b. Find the resistance of the electric bulb,
c. State the law that is applied for making these calculation. If a graph is plotted between V and I, show the nature of the graph obtained.
Answer: 

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c. Ohm’s law: If the physical conditions of a conductor is kept constant then current in the circuit is directly proportional to the potential difference applied across the ends of the conductor.

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Since V ? I
Hence graph between V and I is a straight line for a conductor. Which passes through the origin O of the graph.

Question. (a) Derive an expression to find the equivalent resistance of three resistors connected in series.
Also, draw the schematic diagram of the circuit.
(b) Find the equivalent resistance of the following circuit:

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Answer: 

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Question. (a) Explain how does a cell maintain current in a circuit.
(b) In the circuit given below the resistance of the path xTy = 2 W and that of xZy = 6 W
(i) Find the equivalent resistance between x and y.
(ii) Find the current in the main circuit.
(iii) Calculate the current that flows through the path xTy and xZy.

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Answer: a. Potential difference b/w the two terminals of a cell is produced due to chemical reaction in the cell. This potential difference maintains a current in the circuit.
b. (i) The resistance b/w XZY and XTY are in parallel combination. Their equivalent resistance.

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CBSE Class 10 Science Chapter 11 Electricity Assignment

Access the latest Chapter 11 Electricity assignments designed as per the current CBSE syllabus for Class 10. We have included all question types, including MCQs, short answer questions, and long-form problems relating to Chapter 11 Electricity. You can easily download these assignments in PDF format for free. Our expert teachers have carefully looked at previous year exam patterns and have made sure that these questions help you prepare properly for your upcoming school tests.

Benefits of solving Assignments for Chapter 11 Electricity

Practicing these Class 10 Science assignments has many advantages for you:

  • Better Exam Scores: Regular practice will help you to understand Chapter 11 Electricity properly and  you will be able to answer exam questions correctly.
  • Latest Exam Pattern: All questions are aligned as per the latest CBSE sample papers and marking schemes.
  • Huge Variety of Questions: These Chapter 11 Electricity sets include Case Studies, objective questions, and various descriptive problems with answers.
  • Time Management: Solving these Chapter 11 Electricity test papers daily will improve your speed and accuracy.

How to solve Science Chapter 11 Electricity Assignments effectively?

  1. Read the Chapter First: Start with the NCERT book for Class 10 Science before attempting the assignment.
  2. Self-Assessment: Try solving the Chapter 11 Electricity questions by yourself and then check the solutions provided by us.
  3. Use Supporting Material: Refer to our Revision Notes and Class 10 worksheets if you get stuck on any topic.
  4. Track Mistakes: Maintain a notebook for tricky concepts and revise them using our online MCQ tests.

Best Practices for Class 10 Science Preparation

For the best results, solve one assignment for Chapter 11 Electricity on daily basis. Using a timer while practicing will further improve your problem-solving skills and prepare you for the actual CBSE exam.

Where can I download the latest CBSE Class 10 Science Chapter Chapter 11 Electricity assignments?

You can download free PDF assignments for Class 10 Science Chapter Chapter 11 Electricity from StudiesToday.com. These practice sheets have been updated for the 2026-27 session covering all concepts from latest NCERT textbook.

Do these Science Chapter Chapter 11 Electricity assignments include solved questions?

Yes, our teachers have given solutions for all questions in the Class 10 Science Chapter Chapter 11 Electricity assignments. This will help you to understand step-by-step methodology to get full marks in school tests and exams.

Are the assignments for Class 10 Science Chapter Chapter 11 Electricity based on the 2026 exam pattern?

Yes. These assignments are designed as per the latest CBSE syllabus for 2026. We have included huge variety of question formats such as MCQs, Case-study based questions and important diagram-based problems found in Chapter Chapter 11 Electricity.

How can practicing Chapter Chapter 11 Electricity assignments help in Science preparation?

Practicing topicw wise assignments will help Class 10 students understand every sub-topic of Chapter Chapter 11 Electricity. Daily practice will improve speed, accuracy and answering competency-based questions.

Can I download Science Chapter Chapter 11 Electricity assignments for free on mobile?

Yes, all printable assignments for Class 10 Science Chapter Chapter 11 Electricity are available for free download in mobile-friendly PDF format.