Electricity is one of the most important chapters in CBSE Board Class 10 Science because it explains how electric current flows, how circuits work, and how electrical energy is used in everyday life. This chapter covers key concepts like electric current, potential difference, Ohm’s Law, resistance, heating effect of current, and electric power.
Practicing Electricity Class 10 Science MCQs helps students strengthen their concepts, improve formula application, and prepare effectively for exams. Students can also practice more chapter-wise questions through our Class 10 Science MCQs collection, explore subject-wise Class 10 MCQs, and solve more MCQs to improve their overall exam preparation.
These questions are created according to important NCERT concepts and help students revise theory-based, numerical, and application-based topics from the Electricity chapter with better clarity.
Electricity Class 10 Science MCQs with Answers
Practice the following MCQs from Class 10 Science Chapter 12 Electricity to check your understanding of key concepts, formulas, and applications. These questions are based on important topics from the NCERT syllabus and help you prepare for school tests and board examinations.
Q. A current of 0.75 A is drawn by a filament of an electric bulb for 12 minutes. Find the amount of electric charge that flows through the circuit.
A) 9 C
B) 90 C
C) 540 C
D) 720 C
Answer: C
Explanation:
Given: Current (I) = 0.75 A, Time (t) = 12 minutes = 12 × 60 seconds = 720 s.
Using Formula: Q = I × t = 0.75 × 720 = 540 C.
Q. Which of the following correctly defines the direction of conventional electric current?
A) In the direction of flow of electrons.
B) From the negative terminal to the positive terminal of a cell.
C) In the direction of flow of positive charges.
D) It is randomly oriented depending on the conductor type.
Answer: C
Explanation:
By historical convention, the direction of electric current is considered to be the direction of flow of positive charges, which moves from the positive terminal to the negative terminal. This is exactly opposite to the physical drift direction of free electrons.
Q. How much work is done in moving a charge of 3 C across two points having a potential difference of 15 V?
A) 5 J
B) 45 J
C) 0.2 J
D) 30 J
Answer: B
Explanation:
Given: Charge (Q) = 3 C, Potential Difference (V) = 15 V.
Using Formula: Work Done (W) = V × Q = 15 × 3 = 45 Joules.
Q. The physical quantity that remains constant when resistors are connected in a series circuit is:
A) Potential difference across each resistor
B) Electric current through each resistor
C) Both current and potential difference
D) Total electric resistance
Answer: B
Explanation:
In a series circuit configuration, there is only one continuous path for charge carriers to flow. Therefore, the volume of electric current passing through each resistor remains identical, whereas the voltage drops divide across individual loads.
Q. If the length of a metallic wire is doubled while its cross-sectional area is halved, its new resistance will become:
A) Two times
B) Four times
C) Half
D) Unchanged
Answer: B
Explanation:
Resistance is given by R = ρ(L/A). If the new length L' = 2L and new cross-sectional area A' = A/2, then the new resistance R' = ρ(2L / (A/2)) = 4 × ρ(L/A) = 4R. Hence, total loop resistance scales by four times.
Q. An electrical appliance has a resistance of 44 ohms and operates on a 220 V supply line. What is the current drawn by it?
A) 5 A
B) 0.2 A
C) 10 A
D) 2.5 A
Answer: A
Explanation:
Given: Resistance (R) = 44 Ω, Voltage (V) = 220 V.
By Ohm's Law: I = V / R = 220 / 44 = 5 A.
Q. The resistivity of a certain material wire does not depend on which of the following parameters?
A) Nature of the material
B) Temperature of the conductor
C) Length of the wire
D) All of the above affect resistivity
Answer: C
Explanation:
Resistivity (ρ) is an intrinsic micro-property of a material. It changes only with the material's structural nature and operating temperature. It is entirely independent of the macroscopic dimensions like the length or cross-sectional area of the wire sample.
Q. Three resistors of resistances 2 Ω, 4 Ω, and 6 Ω are connected in parallel. The equivalent resistance of the combination will be:
A) Greater than 6 Ω
B) Exactly 12 Ω
C) Less than 2 Ω
D) Between 4 Ω and 6 Ω
Answer: C
Explanation:
In a parallel network, the reciprocal equivalent resistance formula (1/Rp = 1/R1 + 1/R2 + ...) guarantees that the net total resistance will always drop below the value of the smallest single individual branch resistor in the loop. Since the smallest value here is 2 Ω, the total network resistance must be less than 2 Ω.
Q. Which device is used to measure electric potential difference in a circuit and how is it connected?
A) Ammeter, connected in series
B) Voltmeter, connected in parallel
C) Ammeter, connected in parallel
D) Voltmeter, connected in series
Answer: B
Explanation:
A voltmeter measures the electric potential difference across two unique circuit nodes. It is structurally built with an exceptionally high internal impedance and must always be placed in parallel with the targeted device to avoid drawing significant bypass line current.
Q. According to Joule's heating law, the heat produced in a resistor is:
A) Inversely proportional to the square of the current
B) Directly proportional to the square of the current
C) Inversely proportional to the resistance
D) Directly proportional to the square root of time
Answer: B
Explanation:
Joule's Law of heating states that H = I²Rt. This clearly asserts that heat generated in a standard conductor varies directly as the square of the electric current (I²) flowing through that pathway.
Q. An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be:
A) 100 W
B) 75 W
C) 50 W
D) 25 W
Answer: D
Explanation:
Resistance of the bulb filament is R = V² / P = (220)² / 100 = 484 Ω. When operated at the newly reduced voltage V' = 110 V, the actual power consumed drops to: P' = (V')² / R = (110)² / 484 = 12,100 / 484 = 25 W.
Q. The commercial unit of electric energy is kilowatt-hour (kWh). 1 kWh is equal to:
A) 3.6 × 10⁵ J
B) 3.6 × 10⁶ J
C) 0.36 × 10⁶ J
D) 36 × 10⁶ J
Answer: B
Explanation:
1 kWh = 1 kilowatt × 1 hour = 1000 Watts × 3600 seconds = 3,600,000 Joules = 3.6 × 10⁶ J.
Q. Why are the filaments of incandescent electric lamps generally made of tungsten?
A) It has a very low melting point.
B) It reacts readily with atmospheric oxygen at high temperatures.
C) It has a very high melting point and high resistivity.
D) It is a poor conductor of electricity at room temperature.
Answer: C
Explanation:
Tungsten is selectively preferred for standard incandescent lightbulbs because it possesses an extraordinarily high melting point (approx 3422°C) combined with high operational thermal stability, preventing filament collapse or rapid oxidation at white-hot luminous temperatures.
Q. A cylindrical conductor of length 'L' and uniform area of cross-section 'A' has a resistance 'R'. Another conductor of length '2L' and resistance 'R' of the same material has an area of cross-section equal to:
A) A/2
B) 2A
C) 4A
D) A
Answer: B
Explanation:
Resistance R = ρL/A. For a secondary wire of the same material (so ρ matches) with double length (2L) to hold the identical absolute resistance R, we balance the relation: R = ρ(2L)/A'. Equating both formulas gives ρL/A = ρ(2L)/A', which maps directly to A' = 2A.
Q. What happens to the total circuit current when more resistors are added in a parallel combination across a fixed voltage source?
A) The current increases.
B) The current decreases.
C) The current remains exactly the same.
D) The current fluctuates unpredictably.
Answer: A
Explanation:
Adding extra alternative pathways in a parallel setup decreases the overall net equivalent resistance of the active circuit network. Since net resistance drops while the main driving source voltage remains fixed, the primary current drawn directly from the battery must increase via Ohm's law (I = V/R).
Q. If 10²⁰ electrons pass through a given cross-section of a wire in 4 seconds, what is the current flowing through it? (Charge of an electron = 1.6 × 10^-19 C)
A) 4 A
B) 16 A
C) 2.5 A
D) 40 A
Answer: A
Explanation:
Total net charge Q = n × e = 10²⁰ × 1.6 × 10−19 C = 16 C.
Current (I) = Q / t = 16 C / 4 s = 4 A.
Q. An electrical fuse works on the principle of:
A) Chemical effect of electric current
B) Magnetic effect of electric current
C) Heating effect of electric current
D) Optical effect of electric current
Answer: C
Explanation:
A safety circuit fuse incorporates a specialized low-melting-point metallic ribbon. When internal current loads spike past safe engineering thresholds, rapid internal thermal energy builds via the heating effect of electric current (Joule's heating), melting the element to cleanly split the loop open.
Q. Two copper wires A and B have lengths in the ratio 1:2 and areas of cross-section in the ratio 2:1. The ratio of their resistivities is:
A) 1:1
B) 1:4
C) 4:1
D) 1:2
Answer: A
Explanation:
Resistivity is an intrinsic macro-independent constant dependent exclusively on the structural chemical material matrix and thermal environment. Since both wires A and B are explicitly stated to be copper, their resistivity coefficients are identical, preserving a 1:1 ratio.
Q. Ohm's law is valid only when the ________ of the conductor remains constant.
A) Current
B) Potential difference
C) Temperature
D) Resistance
Answer: C
Explanation:
Ohm's Law (V = IR) holds structurally accurate only under static experimental parameters where environmental and physical variables—specifically the physical temperature of the metallic conductor matrix—are strictly kept uniform during logging.
Q. A wire of resistance R is cut into five equal pieces. These pieces are then connected in parallel. If the equivalent resistance of this combination is R', then the ratio R/R' is:
A) 1/25
B) 5
C) 1/5
D) 25
Answer: D
Explanation:
When a standard wire of resistance R is cleanly sectioned into 5 identical pieces, each piece holds an individual resistance value of r = R/5. When these 5 individual segments are linked across a parallel mesh: 1/R' = 5 × (1 / r) = 5 × (5 / R) = 25/R. Rearranging this relationship yields the exact ratio: R / R' = 25.
Q. The V-I graph for a certain metallic wire at two different temperatures T1 and T2 is given. If the straight line for T1 is steeper (closer to the V-axis) than T2, which temperature is higher? (Assume V is on the y-axis and I is on the x-axis)
A) T1 > T2
B) T2 > T1
C) T1 = T2
D) Cannot be determined from the graph
Answer: A
Explanation:
On a V-I graph where Potential Difference (V) tracks on the y-axis, the instantaneous line slope directly represents the Resistance value (R = V/I). A steeper slope marks a higher resistance. Because the electrical resistance of pure metallic conductors scales upward with temperature, the higher resistance curve (T1) signals a higher corresponding thermal state (T1 > T2).
Q. An electric iron draws a current of 5 A when connected to a 220 V line. Calculate the heat developed in 30 seconds.
A) 33,000 J
B) 1,100 J
C) 6,600 J
D) 2,200 J
Answer: A
Explanation:
Given: Current (I) = 5 A, Voltage (V) = 220 V, Time (t) = 30 seconds.
Using formula: Heat (H) = V × I × t = 220 × 5 × 30 = 33,000 Joules.
Q. What is the maximum resistance that can be made using five resistors each of value 1/5 Ω?
A) 1/5 Ω
B) 1 Ω
C) 5 Ω
D) 25 Ω
Answer: B
Explanation:
Maximum prospective resistance is achieved when elements are linked in a strict series combination. Therefore, R_max = 1/5 + 1/5 + 1/5 + 1/5 + 1/5 = 5/5 = 1 Ω.
Q. What is the minimum resistance that can be made using five resistors each of value 1/5 Ω?
A) 1/25 Ω
B) 1/5 Ω
C) 1 Ω
D) 1/125 Ω
Answer: A
Explanation:
Minimum prospective network resistance is created when all component pieces are wired in a fully parallel configuration: 1/Rmin = 5 × (1 / (1/5)) = 5 × 5 = 25. Inverting this reciprocal structural term gives Rmin = 1/25 Ω.
Q. Which of the following variables is plotted on a graph to verify Ohm's law, and what is the shape of the resulting curve for an ohmic conductor?
A) V vs I, Parabolic curve
B) V vs I, Straight line passing through the origin
C) P vs I, Straight line passing through the origin
D) V vs R, Exponential curve
Answer: B
Explanation:
Ohm's fundamental law establishes that V is directly proportional to I under standard limits. Tracking Potential Difference (V) against line Current (I) generates a clean linear straight-line relationship tracking precisely through the grid origin coordinates (0,0).
Q. A current of 2 A flows through a 10 Ω resistor for 1 minute. The electrical energy consumed is:
A) 240 J
B) 2400 J
C) 40 J
D) 400 J
Answer: B
Explanation:
Given: I = 2 A, R = 10 Ω, t = 1 minute = 60 seconds.
Energy = I² × R × t = (2)² × 10 × 60 = 4 × 10 × 60 = 2400 Joules.
Q. Which of the following substances has the lowest electrical resistivity at room temperature?
A) Constantan
B) Glass
C) Silver
D) Silicon
Answer: C
Explanation:
Silver is universally recognized as holding the premier electrical conductivity index among common crystalline metals at standard room temperatures, boasting the absolute lowest resistivity values (~1.59 × 10−8 Ω·m) to limit path losses.
Q. Two electric bulbs have specifications 220 V, 60 W and 220 V, 100 W. The filament of which bulb will have a higher resistance value?
A) The 100 W bulb
B) The 60 W bulb
C) Both will have equal resistance since voltage rating is identical
D) It depends on the current passing through the circuit
Answer: B
Explanation:
Using the physical relationship R = V² / P, resistance values map inversely to wattage ratings when active operational voltage thresholds match. Therefore, the lower-wattage lighting appliance (60 W) will possess a higher absolute warm filament resistance than the 100 W element.
Q. When a variable resistor (rheostat) is connected in a circuit, its main purpose is to:
A) Change the voltage without altering the current
B) Change the current without altering the source voltage
C) Measure the exact resistance value of the load
D) Safeguard the circuit against ground faults
Answer: B
Explanation:
A variable inline rheostat regulates dynamic circuit parameters by shifting its active physical internal path length to alter current loads smoothly without changing the underlying fixed main source feed voltage.
Q. The charge flowing through a circuit per unit time is called:
A) Conductance
B) Potential energy
C) Electric current
D) Resistivity
Answer: C
Explanation:
By strict definition, the volumetric rate at which an active electrical charge stream drifts past an observed structural cross-sectional line reference per unit time interval is classified as electric current (I = Q/t).
Quick Overview of Electricity Class 10 Science Chapter 12
Electricity deals with the movement of electric charges and the different effects produced by electric current. The chapter explains the relationship between current, voltage, and resistance along with their practical applications.
| Important Points | Details |
|---|---|
| Chapter Name | Electricity |
| Subject | Class 10 Science (Physics) |
| Main Concepts | Current, Resistance, Circuits, Electric Power |
| Important Formula | V = IR |
| Applications | Electrical appliances, circuits, heating devices |
This chapter builds the foundation for understanding electrical systems and helps students connect Physics concepts with real-life examples.
Why Practice Electricity Class 10 Science MCQs?
MCQ practice is not only about remembering answers. It helps you understand how different concepts are connected and improves your ability to apply formulas correctly.
Regular practice of Electricity MCQs helps students:
- Revise important NCERT concepts quickly
- Improve understanding of circuit-based questions
- Learn correct application of formulas
- Build speed in numerical calculations
- Identify weak areas before exams
- Prepare for competency-based questions
Since CBSE exams focus more on concept clarity, practicing different types of MCQs can improve accuracy and confidence.
Topics Covered in Electricity Class 10 MCQs
The MCQs from this chapter cover all major concepts mentioned in the NCERT Class 10 Science syllabus.
| Topic | What You Will Practice |
|---|---|
| Electric Current | Charge flow, current calculation, SI units, direction of current |
| Electric Potential Difference | Voltage, potential difference, and related formulas |
| Ohm’s Law | Relation between voltage, current, and resistance |
| Resistance | Factors affecting resistance and resistivity concepts |
| Combination of Resistors | Series and parallel circuits |
| Heating Effect of Current | Joule’s law of heating and applications |
| Electric Power | Power, energy consumption, and electrical units |
These topics help students prepare both theory-based and numerical MCQs from the Electricity chapter.
Quick Revision Notes for Electricity Class 10
Use these short revision points before attempting the MCQs:
- Electric current is the rate of flow of electric charge.
- SI unit of electric current is Ampere (A).
- Potential difference is measured in Volt (V).
- Ohm’s Law shows the relationship between voltage, current, and resistance.
Ohm’s Law Formula:
- V = IR
- Where:
- V = Potential difference
- I = Electric current
- R = Resistance
Resistance depends on length, area, material, and temperature of the conductor.
In a series combination, the same current flows through all resistors.
In a parallel combination, the potential difference remains the same.
Joule’s law explains the heating effect produced by electric current.
Electric power represents the rate of electrical energy consumption.
Smart Tips to Solve Electricity MCQs Faster
Electricity includes many formula-based and application-based questions. A clear strategy can help you avoid confusion during exams.
Follow these tips while solving MCQs:
- Read the question carefully and identify the given values first.
- Write the correct formula before calculation.
- Always check units like Ampere, Volt, Ohm, and Watt.
- Understand circuit diagrams instead of memorizing answers.
- Practice resistance combination questions separately.
- Revise important formulas regularly.
- Analyze incorrect answers to understand the mistake.
A strong command over basic formulas makes Electricity one of the highest-scoring chapters in Class 10 Science.
Common Mistakes Students Make in Electricity Chapter
Many students understand the concepts but lose marks because of small mistakes. Avoid these common errors while solving MCQs:
| Mistake | Correct Approach |
|---|---|
| Confusing current and voltage | Remember their definitions and units clearly |
| Using wrong formula | Identify the required quantity before solving |
| Ignoring unit conversion | Convert values properly before calculation |
| Mixing series and parallel rules | Learn the difference with examples |
| Memorizing circuits without understanding | Focus on current flow and connections |
Correcting these mistakes improves both speed and accuracy.
Important Exam Focus Areas from Electricity
Students should pay extra attention to these topics while preparing Chapter 12 Electricity:
- Ohm’s Law and its applications
- Relationship between current, voltage, and resistance
- Factors affecting resistance
- Resistance in series and parallel combinations
- Circuit diagrams
- Heating effect of electric current
- Joule’s law calculations
- Electric power and commercial unit of energy
Questions from these areas are commonly asked in different formats, including MCQs, numerical problems, and competency-based questions.
Conclusion
Electricity Class 10 Science MCQs are a useful way to revise important concepts and improve problem-solving skills. This chapter connects theoretical Physics concepts with practical applications, making it important for both exams and everyday understanding.
Practice different types of questions, revise formulas regularly, and focus on understanding concepts instead of memorizing answers. This approach will help you solve Electricity MCQs with better confidence and accuracy.
