Electricity — Class 10 Physics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 10 Physics chapter "Electricity" — 10 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 10 Physics chapter "Electricity" — 10 important questions with detailed answers for CBSE board exam prepar…
Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated
NCERT Solutions for Class 10 Physics — Electricity. Step-by-step answers to all 10 textbook questions from this chapter, written for CBSE board preparation and free to use.
Electricity — All 10 Questions Solved
Q1. A current of 0.5 A is drawn by a filament of an electric bulb for 10 minutes. Find the amount of electric charge that flows through the circuit.
To find the amount of electric charge, we use the relationship between current, charge, and time.
Given:
Current (I) = 0.5 A
Time (t) = 10 minutes
First, convert the time from minutes to seconds, as the SI unit of time is seconds:
t = 10 minutes × 60 seconds/minute = 600 seconds
The formula relating current, charge, and time is:
I = Q / t
Where Q is the electric charge.
To find Q, we rearrange the formula:
Q = I × t
Now, substitute the given values:
Q = 0.5 A × 600 s
Q = 300 C
Therefore, the amount of electric charge that flows through the circuit is 300 Coulombs.
Q2. How much work is done in moving a charge of 2 Coulombs across two points having a potential difference of 12 V?
To find the work done, we use the definition of potential difference.
Given:
Charge (Q) = 2 C
Potential difference (V) = 12 V
The formula for potential difference is:
V = W / Q
Where W is the work done.
To find W, we rearrange the formula:
W = V × Q
Now, substitute the given values:
W = 12 V × 2 C
W = 24 J
Therefore, 24 Joules of work is done in moving the charge.
Q3. An electric heater draws a current of 5 A when connected to a 220 V supply. What is the resistance of the heater element?
We can find the resistance using Ohm's Law, which relates voltage, current, and resistance.
Given:
Voltage (V) = 220 V
Current (I) = 5 A
According to Ohm's Law:
V = I × R
Where R is the resistance.
To find R, we rearrange the formula:
R = V / I
Now, substitute the given values:
R = 220 V / 5 A
R = 44 Ω
Therefore, the resistance of the heater element is 44 Ohms.
Q4. A copper wire has a diameter of 0.5 mm and resistivity of 1.6 × 10⁻⁸ Ω m. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?
Part 1: Calculate the length of the wire.
Given:
Diameter (d) = 0.5 mm = 0.5 × 10⁻³ m
Resistivity (ρ) = 1.6 × 10⁻⁸ Ω m
Desired Resistance (R) = 10 Ω
First, calculate the cross-sectional area (A) of the wire.
The radius (r) = d/2 = (0.5 × 10⁻³ m) / 2 = 0.25 × 10⁻³ m
Area (A) = πr² = π (0.25 × 10⁻³ m)²
A = 3.14159 × (0.0625 × 10⁻⁶ m²)
A ≈ 0.19635 × 10⁻⁶ m²
Now, use the formula for resistance:
R = ρL / A
Where L is the length of the wire.
To find L, rearrange the formula:
L = R × A / ρ
Substitute the values:
L = (10 Ω) × (0.19635 × 10⁻⁶ m²) / (1.6 × 10⁻⁸ Ω m)
L = (1.9635 × 10⁻⁶ Ω m²) / (1.6 × 10⁻⁸ Ω m)
L = 122.7 m (approx)
Therefore, the length of the wire required is approximately 122.7 meters.
Part 2: How much does the resistance change if the diameter is doubled?
Original diameter (d₁) = 0.5 mm
New diameter (d₂) = 2 × d₁ = 2 × 0.5 mm = 1.0 mm
The resistance is inversely proportional to the cross-sectional area (R ∝ 1/A).
The area is proportional to the square of the diameter (A ∝ d²).
So, Resistance R ∝ 1/d².
Let the original resistance be R₁ and the new resistance be R₂.
R₂ / R₁ = (d₁)² / (d₂)²
R₂ / R₁ = (d₁)² / (2d₁)²
R₂ / R₁ = (d₁)² / (4d₁)²
R₂ / R₁ = 1/4
So, R₂ = R₁ / 4
If the original resistance (R₁) was 10 Ω, then the new resistance (R₂) will be:
R₂ = 10 Ω / 4 = 2.5 Ω
Therefore, if the diameter is doubled, the resistance becomes one-fourth of its original value (i.e., it decreases by a factor of 4).
Q5. Three resistors of 5 Ω, 8 Ω, and 12 Ω are connected in series to a 6 V battery. Calculate: (a) The total resistance of the circuit. (b) The current flowing through the circuit. (c) The potential difference across the 8 Ω resistor.
Given:
R₁ = 5 Ω
R₂ = 8 Ω
R₃ = 12 Ω
Battery Voltage (V) = 6 V
(a) Total resistance of the circuit (Rs) for series combination:
In a series circuit, the total resistance is the sum of individual resistances.
Rs = R₁ + R₂ + R₃
Rs = 5 Ω + 8 Ω + 12 Ω
Rs = 25 Ω
(b) The current flowing through the circuit (I):
Using Ohm's Law (V = I × Rs) for the entire circuit:
I = V / Rs
I = 6 V / 25 Ω
I = 0.24 A
(c) The potential difference across the 8 Ω resistor (V₂):
In a series circuit, the current is the same through each resistor. So, the current through the 8 Ω resistor is 0.24 A.
Using Ohm's Law for the 8 Ω resistor (V₂ = I × R₂):
V₂ = 0.24 A × 8 Ω
V₂ = 1.92 V
Therefore:
(a) The total resistance of the circuit is 25 Ω.
(b) The current flowing through the circuit is 0.24 A.
(c) The potential difference across the 8 Ω resistor is 1.92 V.
Q6. Two resistors of 6 Ω and 12 Ω are connected in parallel to a 6 V battery. Calculate: (a) The equivalent resistance of the parallel combination. (b) The total current drawn from the battery. (c) The current flowing through the 6 Ω resistor.
Given:
R₁ = 6 Ω
R₂ = 12 Ω
Battery Voltage (V) = 6 V
(a) The equivalent resistance of the parallel combination (Rp):
For resistors in parallel, the reciprocal of the equivalent resistance is the sum of the reciprocals of individual resistances.
1/Rp = 1/R₁ + 1/R₂
1/Rp = 1/6 Ω + 1/12 Ω
1/Rp = 2/12 Ω + 1/12 Ω
1/Rp = 3/12 Ω
1/Rp = 1/4 Ω
Rp = 4 Ω
(b) The total current drawn from the battery (I_total):
Using Ohm's Law (V = I_total × Rp) for the entire circuit:
I_total = V / Rp
I_total = 6 V / 4 Ω
I_total = 1.5 A
(c) The current flowing through the 6 Ω resistor (I₁):
In a parallel circuit, the potential difference across each resistor is the same as the battery voltage.
So, V₁ = V = 6 V.
Using Ohm's Law for the 6 Ω resistor (V₁ = I₁ × R₁):
I₁ = V₁ / R₁
I₁ = 6 V / 6 Ω
I₁ = 1 A
Therefore:
(a) The equivalent resistance of the parallel combination is 4 Ω.
(b) The total current drawn from the battery is 1.5 A.
(c) The current flowing through the 6 Ω resistor is 1 A.
Q7. In the circuit diagram given below, R1 = 5 Ω, R2 = 10 Ω, R3 = 15 Ω, and a 12 V battery is connected. Resistors R2 and R3 are in parallel, and this combination is in series with R1. (a) Calculate the equivalent resistance of the circuit. (b) Calculate the total current flowing from the battery. (c) Calculate the current through R2. (d) Calculate the voltage drop across R1.
Given:
R₁ = 5 Ω
R₂ = 10 Ω
R₃ = 15 Ω
Battery Voltage (V) = 12 V
(a) Calculate the equivalent resistance of the circuit (R_eq):
First, find the equivalent resistance of R₂ and R₃ which are in parallel (R_p):
1/R_p = 1/R₂ + 1/R₃
1/R_p = 1/10 Ω + 1/15 Ω
1/R_p = (3 + 2) / 30 Ω
1/R_p = 5 / 30 Ω
1/R_p = 1 / 6 Ω
R_p = 6 Ω
Now, R_p is in series with R₁.
The total equivalent resistance (R_eq) of the circuit is:
R_eq = R₁ + R_p
R_eq = 5 Ω + 6 Ω
R_eq = 11 Ω
(b) Calculate the total current flowing from the battery (I_total):
Using Ohm's Law (V = I_total × R_eq):
I_total = V / R_eq
I_total = 12 V / 11 Ω
I_total ≈ 1.09 A
(c) Calculate the current through R2 (I₂):
First, find the voltage across the parallel combination (V_p), which is the voltage across R₂ and R₃.
The total current (I_total) flows through R₁ and the parallel combination (R_p) since they are in series.
V_p = I_total × R_p
V_p = (12/11 A) × 6 Ω
V_p = 72/11 V ≈ 6.55 V
Now, the current through R₂ (I₂) can be found using Ohm's Law (V_p = I₂ × R₂):
I₂ = V_p / R₂
I₂ = (72/11 V) / 10 Ω
I₂ = 72 / 110 A
I₂ = 36 / 55 A ≈ 0.65 A
(d) Calculate the voltage drop across R1 (V₁):
Using Ohm's Law for R₁ (V₁ = I_total × R₁):
V₁ = (12/11 A) × 5 Ω
V₁ = 60 / 11 V ≈ 5.45 V
Therefore:
(a) The equivalent resistance of the circuit is 11 Ω.
(b) The total current flowing from the battery is approximately 1.09 A.
(c) The current through R2 is approximately 0.65 A.
(d) The voltage drop across R1 is approximately 5.45 V.
Q8. An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 seconds.
To calculate the heat developed, we use Joule's Law of Heating.
Given:
Resistance (R) = 20 Ω
Current (I) = 5 A
Time (t) = 30 s
The formula for heat developed (H) is:
H = I²Rt
Now, substitute the given values:
H = (5 A)² × 20 Ω × 30 s
H = 25 A² × 20 Ω × 30 s
H = 500 × 30 J
H = 15000 J
We can also express this in kilojoules:
H = 15 kJ
Therefore, the heat developed in 30 seconds is 15000 Joules (or 15 kJ).
Q9. An electric refrigerator rated 400 W operates 8 hours/day. Calculate the cost of the energy to operate it for 30 days at ₹ 3.00 per kWh.
To calculate the cost of energy, we first need to find the total energy consumed in kilowatt-hours (kWh).
Given:
Power (P) = 400 W
Operating time per day (t_daily) = 8 hours
Number of days = 30 days
Cost per kWh = ₹ 3.00
Step 1: Convert power from Watts to kilowatts (kW).
P = 400 W / 1000 = 0.4 kW
Step 2: Calculate the energy consumed per day.
Energy_daily = Power × t_daily
Energy_daily = 0.4 kW × 8 hours
Energy_daily = 3.2 kWh
Step 3: Calculate the total energy consumed for 30 days.
Total Energy = Energy_daily × Number of days
Total Energy = 3.2 kWh/day × 30 days
Total Energy = 96 kWh
Step 4: Calculate the total cost.
Total Cost = Total Energy × Cost per kWh
Total Cost = 96 kWh × ₹ 3.00/kWh
Total Cost = ₹ 288.00
Therefore, the cost of the energy to operate the refrigerator for 30 days is ₹ 288.00.
Q10. (a) Why are household appliances connected in parallel? (b) An electric fuse is rated 5 A. What does it mean? Can a 5 A fuse be used with an appliance rated 2 kW, 220 V?
(a) Household appliances are connected in parallel for the following reasons:
1. Independent Operation: In a parallel circuit, each appliance has its own separate connection to the main power supply. This means they can be switched on or off independently without affecting other appliances. If they were in series, switching one off would break the entire circuit, turning all others off.
2. Same Voltage: All appliances in a parallel circuit receive the full supply voltage (e.g., 220 V in India). This is essential for their proper functioning, as they are designed to operate at a specific voltage. In a series circuit, the voltage would divide among the appliances, making them operate below their rated voltage.
3. Lower Total Resistance: When appliances are connected in parallel, the overall resistance of the circuit decreases. This allows a higher total current to be drawn from the supply, which is necessary to power multiple appliances simultaneously.
4. No Overload if one fails: If one appliance stops working (e.g., due to a burnt filament), it doesn't interrupt the circuit for other appliances, allowing them to continue functioning.
(b) An electric fuse rated 5 A means that it is designed to safely allow a maximum current of 5 Amperes to flow through the circuit. If the current in the circuit exceeds 5 A (due to a short circuit or overload), the fuse wire will melt and break the circuit, thus protecting the appliance and the electrical wiring from damage.
Now, let's check if a 5 A fuse can be used with an appliance rated 2 kW, 220 V.
Given:
Power (P) of appliance = 2 kW = 2000 W
Voltage (V) = 220 V
First, calculate the current (I) drawn by the appliance using the power formula:
P = V × I
I = P / V
I = 2000 W / 220 V
I ≈ 9.09 A
The current drawn by the appliance is approximately 9.09 A.
Since the current drawn by the appliance (9.09 A) is greater than the fuse rating (5 A), a 5 A fuse cannot be used with this appliance. If used, the fuse would blow immediately upon the appliance being switched on, as the appliance draws more current than the fuse can safely handle. An appropriate fuse with a rating higher than 9.09 A (e.g., a 10 A fuse) would be required.
Electricity — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. Which of the following statements correctly defines electric current?
- A. The rate of flow of electric potential.
- B. The rate of flow of electric charge.
- C. The rate of flow of electric energy.
- D. The rate of flow of electric power.
Answer: B. The rate of flow of electric charge. — Electric current is defined as the rate of flow of electric charge through a conductor. Its SI unit is Ampere.
MCQ 2. What is the SI unit of electric potential difference?
- A. Ampere (A)
- B. Ohm (Ω)
- C. Volt (V)
- D. Coulomb (C)
Answer: C. Volt (V) — The SI unit of electric potential difference is Volt (V). Ampere is for current, Ohm for resistance, and Coulomb for charge.
MCQ 3. An ammeter is always connected in a circuit in:
- A. Parallel
- B. Series
- C. Either series or parallel
- D. Neither series nor parallel
Answer: B. Series — An ammeter is used to measure current and is always connected in series in a circuit to ensure that the entire current flows through it.
MCQ 4. What happens to the resistance of a conductor if its length is doubled while keeping its cross-sectional area constant?
- A. It halves.
- B. It doubles.
- C. It remains the same.
- D. It quadruples.
Answer: B. It doubles. — The resistance of a conductor is directly proportional to its length (R ∝ L). Therefore, if the length is doubled, the resistance also doubles.
MCQ 5. According to Ohm's Law, the relationship between potential difference (V), current (I), and resistance (R) is:
- A. V = I/R
- B. I = V*R
- C. R = V/I
- D. V = R/I
Answer: C. R = V/I — Ohm's Law states that V = IR, where V is potential difference, I is current, and R is resistance. Rearranging this gives R = V/I.
MCQ 6. Which of the following materials is typically used for making the filament of an incandescent light bulb?
- A. Copper
- B. Aluminium
- C. Tungsten
- D. Nichrome
Answer: C. Tungsten — Tungsten is used for the filament of incandescent light bulbs due to its very high melting point and high resistivity, allowing it to glow brightly without melting.
MCQ 7. The commercial unit of electrical energy is:
- A. Joule (J)
- B. Watt (W)
- C. Kilowatt-hour (kWh)
- D. Volt (V)
Answer: C. Kilowatt-hour (kWh) — The commercial unit of electrical energy is kilowatt-hour (kWh), often simply called 'unit'. Joule is the SI unit, and Watt is the unit of power.
MCQ 8. Which instrument is used to measure potential difference across two points in a circuit?
- A. Ammeter
- B. Voltmeter
- C. Galvanometer
- D. Rheostat
Answer: B. Voltmeter — A voltmeter is used to measure the potential difference between two points in an electrical circuit. It is always connected in parallel across the component.
MCQ 9. What is the SI unit of electric resistance?
- A. Volt
- B. Ampere
- C. Ohm
- D. Coulomb
Answer: C. Ohm — The SI unit of electric resistance is Ohm (Ω). Volt is for potential difference, Ampere for current, and Coulomb for charge.
MCQ 10. Joule's Law of Heating states that the heat produced (H) in a resistor is directly proportional to:
- A. I * R * t
- B. I^2 * R * t
- C. I * R^2 * t
- D. I^2 * R^2 * t
Answer: B. I^2 * R * t — Joule's Law of Heating states that H = I²Rt, where H is the heat produced, I is the current, R is the resistance, and t is the time.
MCQ 11. A resistor of 5 Ω is connected across a battery of 10 V. What is the current flowing through the resistor?
- A. 0.5 A
- B. 2 A
- C. 5 A
- D. 10 A
Answer: B. 2 A — Using Ohm's Law, I = V/R. Here, V = 10 V and R = 5 Ω, so I = 10/5 = 2 A.
MCQ 12. Three resistors, each of 2 Ω, are connected in series. What is the equivalent resistance?
- A. 2 Ω
- B. 0.67 Ω
- C. 6 Ω
- D. 4 Ω
Answer: C. 6 Ω — For resistors in series, the equivalent resistance R_eq = R1 + R2 + R3. So, R_eq = 2 Ω + 2 Ω + 2 Ω = 6 Ω.
MCQ 13. Three resistors, each of 6 Ω, are connected in parallel. What is the equivalent resistance?
- A. 2 Ω
- B. 18 Ω
- C. 3 Ω
- D. 6 Ω
Answer: A. 2 Ω — For resistors in parallel, 1/R_eq = 1/R1 + 1/R2 + 1/R3. So, 1/R_eq = 1/6 + 1/6 + 1/6 = 3/6 = 1/2. Therefore, R_eq = 2 Ω.
MCQ 14. Why are alloys preferred over pure metals for making heating elements of electric toasters and electric irons?
- A. Alloys have lower resistivity.
- B. Alloys have higher melting points and don't oxidize easily at high temperatures.
- C. Alloys are cheaper than pure metals.
- D. Alloys are better conductors of electricity.
Answer: B. Alloys have higher melting points and don't oxidize easily at high temperatures. — Alloys have higher resistivity than their constituent metals, produce more heat for a given current, and do not oxidize (burn easily) at high temperatures, making them ideal for heating elements.
MCQ 15. An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, what will be the power consumed?
- A. 100 W
- B. 50 W
- C. 25 W
- D. 200 W
Answer: C. 25 W — First, find the resistance of the bulb: R = V²/P = (220 V)² / 100 W = 48400 / 100 = 484 Ω. When operated at 110 V, the power consumed is P' = V'²/R = (110 V)² / 484 Ω = 12100 / 484 = 25 W.
MCQ 16. If a wire of resistance R is stretched to double its length, then its resistance will become:
- A. 2R
- B. R/2
- C. 4R
- D. R/4
Answer: C. 4R — When a wire is stretched, its volume remains constant. If the length doubles (L' = 2L), its cross-sectional area halves (A' = A/2). Since R = ρL/A, the new resistance R' = ρ(2L)/(A/2) = 4ρL/A = 4R.
MCQ 17. Two bulbs are rated 60 W, 220 V and 100 W, 220 V. If they are connected in series to a 220 V supply, which bulb will glow brighter?
- A. The 60 W bulb
- B. The 100 W bulb
- C. Both will glow with the same brightness
- D. Neither will glow
Answer: A. The 60 W bulb — In a series circuit, the current is the same through both bulbs. The power (brightness) is given by P = I²R. The 60 W bulb has higher resistance (R = V²/P, so R_60 = 220²/60 and R_100 = 220²/100, meaning R_60 > R_100). Thus, the 60 W bulb, having higher resistance, will dissipate more power and glow brighter.
MCQ 18. An electric heater consumes 1500 W of power. If it is used for 2 hours daily for 30 days, what is the total energy consumed in kilowatt-hours?
- A. 90 kWh
- B. 900 kWh
- C. 30 kWh
- D. 150 kWh
Answer: A. 90 kWh — Total energy consumed = Power × Total time. Power = 1500 W = 1.5 kW. Total time = 2 hours/day × 30 days = 60 hours. Energy = 1.5 kW × 60 h = 90 kWh.
MCQ 19. Which of the following does not represent electrical power in a circuit?
- A. VI
- B. I²R
- C. V²/R
- D. I²R²
Answer: D. I²R² — The formulas for electrical power are P = VI, P = I²R, and P = V²/R. I²R² is not a standard formula for power; it would represent (IR)² or V².
MCQ 20. If the current flowing through a fixed resistor is doubled, how does the heat produced in it change?
- A. It doubles.
- B. It quadruples.
- C. It halves.
- D. It remains the same.
Answer: B. It quadruples. — According to Joule's Law of Heating (H = I²Rt), heat produced is directly proportional to the square of the current. If I is doubled, H becomes (2I)²Rt = 4I²Rt, so it quadruples.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.