Assertion (A): A wire of resistance 16 Ω is cut into four equal parts and these parts are connected in parallel. The equivalent resistance is 1 Ω. Reason (R): When the wire is cut into n equal parts, each part has resistance R/n, and for n equal resistors in parallel, equivalent resistance is R/n².

Both A and R are true and R is the correct explanation of A.

Both A and R are true but R is not the correct explanation of A.

Give two reasons why a series circuit is not suitable for domestic wiring.

1. In a series circuit, if one appliance fails, all appliances stop working because the circuit is broken. 2. The voltage divides, so each appliance gets only a fraction of the supply voltage, reducing efficiency.

State Ohm's law and write its mathematical expression.

Ohm's law states that the potential difference V across the ends of a conductor is directly proportional to the current I flowing through it, provided temperature and other physical conditions remain constant. Mathematically, V ∝ I or V = IR, where R is the resistance.

Two resistors 3 Ω and 6 Ω are connected in parallel. This combination is then connected in series with a 2 Ω resistor. Find the total resistance and the current drawn from a 12 V battery.

Total resistance = 4 Ω; current = 3 A.

Define resistivity. Name two factors on which the resistance of a conductor depends.

Resistivity is the resistance offered by a conductor of unit length and unit cross-sectional area. Two factors on which resistance depends are: length and cross-sectional area.

An electric heater is rated 220 V, 1500 W. Find the resistance of its heating element and the current drawn when operating at the rated voltage. If the heater is used on a 110 V supply, what power will it consume? Assume the resistance remains constant.

Rated voltage V₁ = 220 V, power P₁ = 1500 W. Resistance R = V₁²/P₁ = (220)²/1500 = 48400/1500 = 32.27 Ω (approx). Rated current I₁ = P₁/V₁ = 1500/220 ≈ 6.82 A. On 110 V, power P₂ = V₂²/R = (110)²/32.27 = 12100/32.27 ≈ 375 W.

Two resistors R1 and R2 are connected in series across a constant voltage source. The power consumed by R1 is 8 W and by R2 is 4 W. Find the ratio of their resistances. Now, if these two resistors are connected in parallel to the same voltage source, calculate the total power consumed by the combination.

In series, current I is same. Power P = I²R. Thus, P1/P2 = R1/R2 ⇒ 8/4 = R1/R2 ⇒ R1/R2 = 2:1. Hence R1 = 2R2. Let supply voltage = V. In series, total resistance R_s = R1+R2 = 2R2+R2 = 3R2. Current I = V/(3R2). Power in R1: P1 = I²R1 = (V²/(9R2²))·(2R2) = 2V²/(9R2) = 8 W ⇒ V²/R2 = (8×9)/2 = 36. So V²/R2 = 36. For parallel connection, R1 and R2 are in parallel: R_p = (R1·R2)/(R1+R2) = (2R2·R2)/(3R2) = (2/3)R2. Total power P_total = V²/R_p = V²/((2/3)R2) = (3/2)·(V²/R2) = (3/2)×36 = 54 W.

An electrician, while installing a heater, notes that it has a resistance of 22 Ω and is designed to operate on a 220 V supply. He recalls Ohm's law to predict the current. (1) State Ohm's law. (2) Calculate the current drawn by the heater when connected to the 220 V supply. (3) If the supply voltage drops to 110 V, what will be the new current?

1. V ∝ I at constant temperature. 2. 10 A. 3. 5 A.

Class 10 Science Chapter 11: Electricity — Important Questions & Sample Paper

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Yes — this page has 44+ original Class 10 Science Chapter 11 (“Electricity”) important questions with answers (Multiple Choice (MCQ), Assertion–Reason, Short Answer, Short Answer, Long Answer, Case Study). Practise them free, or generate a full CBSE board-pattern sample paper (80 marks) and export it to PDF or Word — in English & Hindi, for 2026-27.

Chapter 11 'Electricity' in CBSE Class 10 Science introduces fundamental concepts that are critical for understanding electrical circuits. Students learn about electric current, potential difference, and the relationship defined by Ohm's law. The chapter explains resistance and its dependence on factors like length, cross-sectional area, material (resistivity), and temperature. It covers how to derive and apply formulas for combining resistors in series and parallel, and explores the heating effect of electric current through Joule’s law. These concepts lead to practical calculations of electric power and energy. Exam questions typically test numerical problem-solving skills, such as finding effective resistance in networks, calculating power consumed by appliances, determining current drawn, and comparing bulb brightness in different circuit arrangements. Conceptual questions often focus on how stretching a wire changes its resistance or why alloys are used in heating elements. A clear understanding of these principles, supported by regular practice of derivations and numericals, is essential for performing well in the CBSE board exams.

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Science — Electricity

Class 10Time: 3 hrsMax Marks: 80

SECTION A

1.
An electric bulb is rated 100 W, 220 V. The resistance of its filament is:
(a) 484 Ω(b) 2.2 Ω(c) 22000 Ω(d) 100 Ω
1
2.
The resistance of a conductor depends on several factors. Which one of the following does NOT affect the resistance of a metallic conductor at constant temperature?
(a) Length of the conductor(b) Area of cross-section(c) Color of the conductor(d) Material of the conductor
1
3.
Two resistors of resistances 6 Ω and 12 Ω are connected in parallel. The equivalent resistance is:
(a) 18 Ω(b) 4 Ω(c) 0.25 Ω(d) 72 Ω
1

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In a CBSE exam, this chapter typically contributes questions across the following types. The last column shows how many original questions of each type we have ready in our bank for this chapter:

Question type	Marks each	In our bank
Multiple Choice (MCQ)	1 mark	13
Assertion–Reason	1 mark	6
Short Answer	2 marks	8
Short Answer	3 marks	6
Long Answer	5 marks	5
Case Study	4 marks	6

44 original, exam-style questions in our bank for this chapter — with answers.

Important & sample questions (with answers)

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Q1. An electric bulb is rated 100 W, 220 V. The resistance of its filament is:
1 mark
Multiple Choice (MCQ)
(A) 484 Ω(B) 2.2 Ω(C) 22000 Ω(D) 100 Ω
▸ Answer▾ Answer
484 Ω
Q2. The resistance of a conductor depends on several factors. Which one of the following does NOT affect the resistance of a metallic conductor at constant temperature?
1 mark
Multiple Choice (MCQ)
(A) Length of the conductor(B) Area of cross-section(C) Color of the conductor(D) Material of the conductor
▸ Answer▾ Answer
Color of the conductor
Q3. Two resistors of resistances 6 Ω and 12 Ω are connected in parallel. The equivalent resistance is:
1 mark
Multiple Choice (MCQ)
(A) 18 Ω(B) 4 Ω(C) 0.25 Ω(D) 72 Ω
▸ Answer▾ Answer
4 Ω
Q4. Two bulbs rated (100 W, 200 V) and (200 W, 200 V) are connected in series across a 200 V supply. Which bulb will glow brighter?
1 mark
Multiple Choice (MCQ)
(A) 100 W bulb(B) 200 W bulb(C) Both glow equally bright(D) Neither glows
▸ Answer▾ Answer
100 W bulb
Q5. Assertion (A): A wire of resistance 16 Ω is cut into four equal parts and these parts are connected in parallel. The equivalent resistance is 1 Ω. Reason (R): When the wire is cut into n equal parts, each part has resistance R/n, and for n equal resistors in parallel, equivalent resistance is R/n².
1 mark
Assertion–Reason
(A) Both A and R are true and R is the correct explanation of A.(B) Both A and R are true but R is not the correct explanation of A.(C) A is true but R is false.(D) A is false but R is true.
▸ Answer▾ Answer
Both A and R are true and R is the correct explanation of A.
Q6. Give two reasons why a series circuit is not suitable for domestic wiring.
2 marks
Short Answer
▸ Answer▾ Answer
1. In a series circuit, if one appliance fails, all appliances stop working because the circuit is broken. 2. The voltage divides, so each appliance gets only a fraction of the supply voltage, reducing efficiency.
Q7. State Ohm's law and write its mathematical expression.
2 marks
Short Answer
▸ Answer▾ Answer
Ohm's law states that the potential difference V across the ends of a conductor is directly proportional to the current I flowing through it, provided temperature and other physical conditions remain constant. Mathematically, V ∝ I or V = IR, where R is the resistance.
Q8. Two resistors 3 Ω and 6 Ω are connected in parallel. This combination is then connected in series with a 2 Ω resistor. Find the total resistance and the current drawn from a 12 V battery.
3 marks
Short Answer
▸ Answer▾ Answer
Total resistance = 4 Ω; current = 3 A.
Q9. Define resistivity. Name two factors on which the resistance of a conductor depends.
3 marks
Short Answer
▸ Answer▾ Answer
Resistivity is the resistance offered by a conductor of unit length and unit cross-sectional area. Two factors on which resistance depends are: length and cross-sectional area.
Q10. An electric heater is rated 220 V, 1500 W. Find the resistance of its heating element and the current drawn when operating at the rated voltage. If the heater is used on a 110 V supply, what power will it consume? Assume the resistance remains constant.
5 marks
Long Answer
▸ Answer▾ Answer
Rated voltage V₁ = 220 V, power P₁ = 1500 W. Resistance R = V₁²/P₁ = (220)²/1500 = 48400/1500 = 32.27 Ω (approx). Rated current I₁ = P₁/V₁ = 1500/220 ≈ 6.82 A. On 110 V, power P₂ = V₂²/R = (110)²/32.27 = 12100/32.27 ≈ 375 W.
Q11. Two resistors R1 and R2 are connected in series across a constant voltage source. The power consumed by R1 is 8 W and by R2 is 4 W. Find the ratio of their resistances. Now, if these two resistors are connected in parallel to the same voltage source, calculate the total power consumed by the combination.
5 marks
Long Answer
▸ Answer▾ Answer
In series, current I is same. Power P = I²R. Thus, P1/P2 = R1/R2 ⇒ 8/4 = R1/R2 ⇒ R1/R2 = 2:1. Hence R1 = 2R2. Let supply voltage = V. In series, total resistance R_s = R1+R2 = 2R2+R2 = 3R2. Current I = V/(3R2). Power in R1: P1 = I²R1 = (V²/(9R2²))·(2R2) = 2V²/(9R2) = 8 W ⇒ V²/R2 = (8×9)/2 = 36. So V²/R2 = 36. For parallel connection, R1 and R2 are in parallel: R_p = (R1·R2)/(R1+R2) = (2R2·R2)/(3R2) = (2/3)R2. Total power P_total = V²/R_p = V²/((2/3)R2) = (3/2)·(V²/R2) = (3/2)×36 = 54 W.
Q12. An electrician, while installing a heater, notes that it has a resistance of 22 Ω and is designed to operate on a 220 V supply. He recalls Ohm's law to predict the current.
4 marks
Case Study
1. (i) State Ohm's law.1 mark
2. (ii) Calculate the current drawn by the heater when connected to the 220 V supply.2 marks
3. (iii) If the supply voltage drops to 110 V, what will be the new current?1 mark
▸ Answer▾ Answer
1. V ∝ I at constant temperature. 2. 10 A. 3. 5 A.

Frequently asked questions

What are the key formulas and concepts I must revise for the Electricity chapter?

Focus on Ohm's law (V=IR), resistance dependence (R=ρL/A), series and parallel resistance formulas, Joule's law of heating (H=I²Rt), and electric power (P=VI=I²R=V²/R). Also understand the difference between resistance and resistivity, and how to derive equivalent resistance for combinations.

How do I solve problems involving combinations of resistors in series and parallel?

First, identify the configuration. For series, directly add resistances: R_total = R1 + R2 + ... For parallel, use the reciprocal formula: 1/R_total = 1/R1 + 1/R2 + ... Simplify step by step. Often, complex circuits can be broken down into simpler series and parallel parts. Remember that in series, current is the same through each resistor; in parallel, voltage is the same.

Why does the resistance of a wire increase when it is stretched?

When a wire is stretched, its length increases and its cross-sectional area decreases, keeping the volume constant. Since resistance R is proportional to length L and inversely proportional to area A (R = ρL/A), the combined effect leads to a higher resistance. For example, doubling the length while halving the area quadruples the resistance.

How do I compare the brightness of bulbs connected in series versus parallel?

The brightness depends on the power dissipated (P = I²R or V²/R). In series, the bulb with higher resistance glows brighter because current is the same. In parallel, the bulb with lower resistance glows brighter because voltage is the same. For two bulbs of different wattage ratings, calculate their resistances using R = V_rated² / P_rated, then analyze the circuit.

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