Important Questions Ch11 Class 10 Science Electricity

The chapter Electricity is one of the most scoring and conceptually important topics in Class 10 Physics. It deals with the study of electric current, potential difference, resistance, and the laws governing electric circuits.

This chapter forms the foundation for higher-level topics like current electricity, magnetism, and electrical measurements studied in Classes 11 and 12. It focuses on basic concepts and their applications through formulas, numerical problems, and circuit analysis.

The Electricity Important Questions for Class 10 Physics are designed to help you:

• Strengthen your understanding of electric current and circuits.
• Practice both theoretical and numerical questions.
• Prepare thoroughly for CBSE board exams with a clear grasp of formulas and laws.

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Chapter 11 Electricity: Important Questions

1. Read the information given below and answer four out of five following questions.

Suresh bought a packet of 100 LEDs to make his own lights for decoration in his house. The packet on the LEDs had the following printed on a label:

LED 2835, 0.2 W, 30 Lumens, 3 V

To understand how he should connect the LEDs, he referred to the following circuit diagram on a website.

1.1 Which of the following describes how the LEDs are connected in the circuit diagram? 

a. all in series 

b. all in parallel

c. 8 each in a series combination, and the two combinations in parallel

d. 8 each in a parallel combination, and the two combinations in series

Answer: (C) 8 each in a series combination, and the two combinations in parallel.

Explanation: In the circuit diagram, LEDs are grouped into two rows. Each row contains 8 LEDs connected in series, and the two rows are connected in parallel to the mains supply.

1.2  If the LED marked 'LED 2' in the diagram stops working, which other LEDs will also stop working?

(Note: When an LED stops working, current cannot flow across it.)

a. only LED 3 to LED 8

b. only LED 3 to LED 8 and LED 1

c. all the other LEDs in the circuit

d. none of the other LEDs in the circuit

Answer: (B) Only LED 3 to LED 8 and LED 1.

Explanation: In a series connection, if one LED stops working, the entire current path for that series segment is broken. Therefore, all LEDs in that specific series combination (LEDs 1 to 8) will stop working, while LEDs in the other parallel branch (LEDs 9 to 16) will continue to work.

1.3  Suresh decided to connect all the LEDs in his lights in a series combination.

How many LEDs will he need to connect if he is going to connect the lights to a 240 V mains supply so that the LEDs work at their power rating?

a. 16

b. 80

c. 240

d. 1200

Answer: (B) 80 LEDs.

Explanation: Each LED operates at 3 V. In a series connection, the total voltage is divided among all the LEDs.

Number of LEDs= Total Voltage/Voltage per LED

= 240/3

=80

1.4  What will happen if he connects 100 LEDs, all in a parallel combination, to the 240 V mains supply?

a. Each LED will work as expected since the available voltage is more than 3 V.

b. Each LED will have a potential difference of 220 V and therefore they will get damaged.

c. Each LED will glow but the ones closer in the circuit to the main supply will glow brighter.

d. Each LED will have a potential difference of 2.4 V across it and therefore will glow dimmer than normal.

Answer:  (B) Each LED will have a potential difference of 220 V and therefore they will get damaged.

Explanation: In a parallel connection, each branch gets the full voltage of the mains supply. If the LEDs are directly connected to a 240 V supply instead of their rated 3 V, the LEDs will burn out due to excessive voltage.

1.5  How much current is each LED expected to draw when used according to the ratings given in the label?

a. 0.067 A

b. 0.600 A

c. 10 A

d. 15 A

Answer: (A) 0.067 A

Explanation: The power rating of each LED is 0.2 W, and the voltage rating is 3 V. Using the formula:

I = P/V

= 0.2 / 3

=0.067A

2. The diagram below shows how Amita had connected a circuit to verify Ohm's law.

(a) Identify which of the devices in the circuit is an ammeter. Justify your answer.

(b) Draw a circuit diagram with appropriate symbols for the circuit shown in the diagram above.

(c) Amita forgot to put a switch in the circuit. During the experiment, the wire labelled 'Unknown resistor' became hot. The resistivity of the material of the wire increases with temperature. Draw two potential difference vs current graphs (in the same diagram):

(i) as expected by Amita,

(ii) as based on actual observation she would make.

Answer:  In the circuit:

• Meter 1 is connected in series with the circuit, which is a requirement for an ammeter (since it measures the current flowing through the circuit).
• Meter 2, on the other hand, is connected in parallel across the unknown resistor, making it a voltmeter (since it measures the potential difference).

Thus, Meter 1 is the ammeter because it is positioned to measure the total current flowing through the circuit.

(b) 

(c)

Graph expected by Amita (Ohm's law):

Ohm's law states V=IR, where RRR is constant.

The graph of potential difference (V) vs current (I) would be a straight line passing through the origin, indicating a constant resistance.

Graph based on actual observation:

As the wire heats up, its resistivity increases, causing the resistance R to increase with temperature.

The graph would curve downward as the current decreases for a given voltage, deviating from the straight line.

3. An incandescent bulb works on the heating effect of electric current. When a current passes through the filament of a bulb it heats the filament to a high temperature which causes the filament to glow.

The graph below shows the variation in the current through a bulb immediately after it is switched on. The current decreases from 1 A at time t=0 to 0.5 A at t=t,. The voltage of the power supply is 200 V and remains constant throughout.

(a) Based on the graph, state how the resistance of the bulb filament changes as the temperature increases from time t=0 to t=t₁.

(b) What is the power consumed by the bulb when it is glowing at its full brightness?

Answer: (a)

At t=0, I=1A,

R=V/I

200/1

200Ω.

At t=t₁, I=0.5A,

R=V/I

200/0.5

400Ω.

(b)

P=V×I

=200×0.5

=100W.

4. Observe the circuit shown below. All the three switches are open.

Identify the switch/switches that on being closed will cause the fuse to blow.

Answer:  Switch 1 and Switch 3, when closed together, will cause the fuse to blow due to a short circuit.

Closing only Switch 1 and Switch 2 may also blow the fuse if the bulb's resistance allows excessive current, but this depends on the fuse's rating and the bulb's specifications.

5. You are given three identical 10 ohm resistors and a 12 V cell.

Draw the circuit diagram to show how the resistors can be connected with the 12 V cell so that the total heat produced in the circuit is the MINIMUM.

Answer:  To minimize the heat produced in the circuit, the total current flowing through the circuit should be minimized. This is achieved by maximizing the total resistance of the circuit. To do this, the three resistors should be connected in series because the total resistance of resistors connected in series is the sum of their individual resistances, which is higher than in any other configuration.

6. Study the circuit diagram given below. You are given one extra resistor. By drawing a new circuit diagram, show how you can connect the extra resistor to increase the reading on the ammeter in the circuit below.

Answer: To increase the current in the circuit, you need to reduce the overall resistance. This can be done by connecting the extra resistor in parallel with the existing resistors in the circuit. When resistors are connected in parallel, the equivalent resistance decreases.

This reduces the total resistance and, according to Ohm's Law (I=V/R), increases the current flowing through the circuit.

7. Priya has a copper wire and an aluminium wire of the same length.

Can the electrical resistance of the two wires be the same? Justify your answer.

Answer: Yes, the electrical resistance of the copper wire and the aluminum wire can be the same, but certain conditions must be satisfied.

The resistance of a wire is given by the formula:

R=ρ (L/A)

Since the two wires have the same length (L), the resistance depends on their resistivity (ρ) and cross-sectional area (A).

Copper has a lower resistivity (ρ_copper=1.68×10⁻⁸ Ω m} than aluminum (ρ_aluminum=2.82×10⁻⁸ Ω m}. To achieve the same resistance, the aluminum wire must have a larger cross-sectional area than the copper wire to compensate for its higher resistivity.

A_aluminum​=A_copper​⋅ρ_copper​/ρ_aluminum​​

Since ρ_aluminum>ρ_copper​, the aluminum wire's cross-sectional area (A_aluminum​) must be larger than that of the copper wire (A_copper).

The electrical resistance of the copper wire and aluminum wire of the same length can be made equal if the aluminum wire has a larger cross-sectional area than the copper wire, proportional to the ratio of their resistivities.

Some More Important Question Answers of Class 10 Electricity

Q1. What is electric current?

Ans.  Electric current is the rate at which electric charges flow through a conductor. In metals, current is due to the movement of free electrons.
Formula: I = Q / t
where Q = total charge (in coulombs), t = time (in seconds).
SI Unit: Ampere (A).
1 Ampere means that 1 coulomb of charge flows through the conductor in 1 second.
Example: If 10 C charge flows through a wire in 2 seconds, current = 10/2 = 5 A.

Q2. What is an electric circuit? Draw a simple diagram.

Ans.  An electric circuit is a closed path in which current flows when a voltage source like a cell or battery is connected. It consists of a source (cell), a switch, a device (resistor or bulb), and measuring instruments.

Simple circuit:

Here:

• Ammeter (A) measures current.
• Voltmeter (V) measures voltage across the resistor.
• The switch controls the flow.

Q3. State Ohm’s Law.
Ans.  Ohm’s Law states that the potential difference (V) across a conductor is directly proportional to the current (I) flowing through it, provided the temperature remains constant.
Mathematical form: V = IR
Where R = resistance of conductor.

Graph: Straight line shows V ∝ I.

Q4. A potential difference of 12 V is applied across a resistor of 6 Ω. Find current.

Ans.  Using Ohm’s law,
I = V / R = 12 / 6 = 2 A.
So a current of 2 amperes flows through the resistor.

Q5. What is resistance? On what factors does it depend?

Ans.  Resistance is the property of a conductor that opposes the flow of electric current. Greater resistance means less current for the same voltage.
Resistance depends on:

1. Length (l): Longer wire → more resistance.
2. Cross-sectional area (A): Thicker wire → less resistance.
3. Nature of material: Different materials have different resistivity.
4. Temperature: In metals, resistance increases with rise in temperature.

Formula: R = ρ × (l / A), Where ρ = resistivity of material.

Q6. Define resistivity.

Ans.  Resistivity is a property of a material that shows how strongly it resists current. It is defined as the resistance of a conductor of unit length and unit area.
Formula: ρ = R × (A / l)
SI Unit: ohm metre (Ω m).
Example: Resistivity of copper is 1.7 × 10⁻⁸ Ωm (very low), so it is a good conductor.

Q7. A wire of resistance 1 Ω is stretched to double its length. Find new resistance.

Ans.  When wire is stretched to double its length, volume remains constant, so area becomes half.
New resistance R’ = ρ × (2l) / (A/2) = 4 × (ρl/A) = 4R.
Since original resistance = 1 Ω, new resistance = 4 Ω.
Thus, stretching increases resistance by square of stretching factor.

Q8. Define conductors and insulators with examples.

Ans. 

• Conductors are substances that allow electricity to pass easily because they have free electrons. Examples: copper, aluminium.
• Insulators do not allow electricity to pass through because they lack free electrons. Examples: rubber, glass, wood.

Q9. What is the difference between potential difference and emf?

Ans. 

• Potential Difference (PD): The work done to move a unit charge between two points in a circuit.
• Electromotive Force (emf): The total energy supplied per unit charge by a source (battery) to drive charge around the complete circuit.

PD is measured across components, while emf is property of the source.

Q10. Derive equivalent resistance for resistors in series.

Ans.  When resistors R1, R2, R3 are connected in series:

• The current is the same through all.
• Voltage adds: V = V1 + V2 + V3.
  Using Ohm’s law:
  V = IR1 + IR2 + IR3 = I(R1 + R2 + R3).
  So equivalent resistance Rs = R1 + R2 + R3.

Q11. Derive equivalent resistance for resistors in parallel.

Ans.  For resistors R1, R2, R3 in parallel:

• The potential difference is the same across all.
• Current divides: I = I1 + I2 + I3.
  Using Ohm’s law:
  I = V/R1 + V/R2 + V/R3 = V(1/R1 + 1/R2 + 1/R3).
  So 1/Rp = 1/R1 + 1/R2 + 1/R3.

Q12. Three resistors 2 Ω, 3 Ω, 6 Ω are connected in series. Find total resistance.

Ans. Rs = R1 + R2 + R3 = 2 + 3 + 6 = 11 Ω.

Q13. Same resistors (2 Ω, 3 Ω, 6 Ω) in parallel. Find total resistance.

Ans.  1/Rp = 1/2 + 1/3 + 1/6 = (3 + 2 + 1)/6 = 6/6 = 1, So Rp = 1 Ω.

Q14. State advantages of parallel connection in household wiring.
Ans. This is why homes are wired in parallel.

1. All appliances get the same voltage (220 V).
2. If one appliance fails, others continue working.
3. Parallel reduces overall resistance, so power distribution is efficient.

Q15. What is electric power?

Ans.  Electric power is the rate at which electrical energy is consumed or converted into other forms of energy.
Formula: P = VI = I²R = V²/R.
SI unit: Watt (W).
1 watt = power when 1 A current flows at 1 V potential difference.

Q16. A bulb is rated 100 W, 220 V. Find current and resistance.

Ans. Current I = P/V = 100/220 ≈ 0.45 A.
Resistance R = V/I = 220 / 0.45 ≈ 489 Ω.
Thus, the bulb draws about 0.45 A and has resistance of nearly 489 Ω.

Q17. Define electric energy.

Ans.  Electric energy is the total work done by current in a circuit in a given time.
E = P × t = VIt.
SI unit: Joule (J).
Commercial unit: kilowatt-hour (kWh).
1 kWh = 1 unit = energy consumed when 1 kW power is used for 1 hour = 3.6 × 10⁶ J.

Q18. Calculate cost of electricity if 5 bulbs of 100 W each are used 4 hours daily for 30 days at Rs 6 per unit.

Ans. Total power = 5 × 100 W = 500 W = 0.5 kW.
Energy used in 1 day = 0.5 × 4 = 2 kWh.
Energy used in 30 days = 2 × 30 = 60 kWh (60 units).
Cost = 60 × 6 = Rs 360.

Q19. Why is fuse wire made of an alloy of tin and lead?

Ans. Fuse wire must have low melting point and high resistivity. Alloy of tin and lead melts easily when current exceeds safe value. This breaks circuit and prevents appliances from damage due to overheating.

Q20. State Joule’s law of heating.

Ans.  According to Joule’s law, heat produced in a resistor is directly proportional to:

1. Square of current (I²),
2. Resistance (R),
3. Time (t).
4. Formula: H = I²Rt.

This explains the working of electric heater, iron, kettle, and fuse.

Video Lecture: Must-watch for Quick Revision

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Extra Questions for practice!

Q1. Define drift velocity of electrons.

Q2.  Draw V-I graph for a non-ohmic conductor.

Q3. Why is the current the same in series connection?

Q4. Why is copper used in household wiring?

Q5. Explain the relation between electric energy and electric power.

Q6. What happens to the resistance of a metal when temperature increases?

Q7. A heater of 1 kW is used for 3 hours daily. Find energy consumed in 20 days.

Q8. Why is parallel wiring used in homes instead of series?

Q9. Calculate current through a 4 Ω resistor when 12 V is applied.

Q10. Why is the fuse always connected in the live wire of household wiring?

How to Ace These Class 10 Electricity Important Questions

To score full marks in this chapter, it is essential to understand both the concepts and their mathematical applications. Follow this systematic approach for preparation:

Step 1: Understand the Concept of Electric Current

Start with the basics of electric current, potential difference, and resistance. Learn their definitions, symbols, SI units, and relationships between them.

Step 2: Study Ohm’s Law

Understand Ohm’s Law, which states that the current flowing through a conductor is directly proportional to the potential difference across it when the temperature remains constant. Practise solving problems based on this law using the formula V = IR.

Step 3: Revise the Factors Affecting Resistance

Know how resistance depends on the material, length, area of cross-section, and temperature of the conductor. Understand the concept of resistivity and its units.

Step 4: Learn Series and Parallel Circuits

Practise how to calculate equivalent resistance in series and parallel combinations. Remember that total resistance increases in series and decreases in parallel. Solve numerical problems for both types of connections.

Step 5: Study Electric Power and Energy

Learn formulas for electric power and energy:
P = VI, P = I²R, and P = V²/R
Also understand how electrical energy is measured in kilowatt-hours (kWh) and how to calculate electricity consumption from power ratings of appliances.

Step 6: Practise Circuit Diagrams

Draw neat and labeled diagrams showing series and parallel connections, ammeters, voltmeters, and resistors. Accuracy in circuit representation is important for full marks.

Step 7: Solve Numerical Problems Regularly

Numericals are a key scoring area. Practise different types of problems daily to strengthen your formula application and improve speed and accuracy.

Tips for Preparing Chapter 11 Class 10 CBSE: Electricity

To prepare for this chapter, focus on understanding you need to revise key concepts and relate them to real-life applications. By following the tips given below, it can be really helpful and easy for students to cover up the chapter Electricity:

Understand Basic Concepts: Focus on the fundamentals of current, voltage, resistance, and Ohm’s law. Understand the relationship between current, voltage, and resistance using the formula V=IR (Ohm’s Law).

Learn the Units and Their Conversion: Memorise the units for electrical quantities—amperes (A) for current, volts (V) for voltage, ohms (Ω) for resistance, and watts (W) for power. Practice converting between units, like milliamperes (mA) to amperes (A) and millivolts (mV) to volts (V).

Practice Circuit Diagrams: Draw and label different circuit diagrams, including series and parallel circuits. Understand how to calculate total resistance, current, and voltage in series and parallel circuits.

Work on Numericals: Solve problems involving Ohm’s Law, power calculation (P=VIP=VIP=VI), and resistances in series and parallel. Practice questions related to heating effects of current and their applications.

Understand Electrical Safety: Learn the importance of fuses, circuit breakers, and earthing in electrical circuits for safety. Study the principles behind electric power transmission and how resistivity plays a role in efficiency.

Revise Key Formulas: Ensure you are comfortable with important formulas such as P=I2R, V=IR, and RTotal for series and parallel circuits.

Solve Sample Papers and Past Year Questions: Practice with sample papers and past CBSE exam questions to get familiar with the types of problems that may appear in the exam.

We hope that you practice the above Electricity Class 10 Imp Questions and achieve your dream marks. All the best! 

FAQs

Q1. How many marks are generally allotted to this chapter in Class 10 board exams?

Ans. This chapter usually carries 8 to 10 marks, with a mix of theory, formula-based, and numerical questions.

Q2. Which topics are most important for exams?

Ans. Ohm’s Law, resistors in series and parallel, power and energy formulas, and circuit-based numerical problems are the most important.

Q3. Are derivations compulsory in this chapter?

Ans. Yes, short derivations such as those for the equivalent resistance in series and parallel are commonly asked and carry marks for steps.

Q4. How can I improve my speed in solving numericals?

Ans. Practise daily, write all steps clearly, and ensure units are consistent. Avoid skipping intermediate steps during practice.

Q5. Are diagrams necessary in every question?

Ans. Not always, but circuit diagrams are essential for questions involving connections or circuit setup. Neatly labeled diagrams help you score full marks.

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