April 10, 2026
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Chapter 12: Electricity
Complete NCERT Solutions (Intext + Exercise) | CBSE Class 10 Science | Verbatim Questions & Marks
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Intext Questions (Page 200)
11 mark
What does an electric circuit mean?
Answer: An electric circuit is a closed loop through which electric current flows, consisting of a battery (or cell), wires, switch, and a load (like a bulb).
21 mark
Define the unit of current.
Answer: SI unit of current is ampere (A). 1 ampere = 1 coulomb per second.
32 marks
Calculate the number of electrons constituting one coulomb of charge.
Answer: Charge on one electron = 1.6 × 10⁻¹⁹ C. Number of electrons = 1 / (1.6 × 10⁻¹⁹) = 6.25 × 10¹⁸ electrons.
Intext Questions (Page 202)
42 marks
Name a device that helps to maintain a potential difference across a conductor.
Answer: A cell or battery.
52 marks
What is meant by saying that the potential difference between two points is 1 V?
Answer: 1 volt means 1 joule of work is done to move 1 coulomb of charge from one point to another.
62 marks
How much energy is given to each coulomb of charge passing through a 6 V battery?
Answer: Energy = V × Q = 6 V × 1 C = 6 J.
Intext Questions (Page 209)
71 mark
On what factors does the resistance of a conductor depend?
Answer: Length (l), cross-sectional area (A), material (resistivity ρ), and temperature. R = ρl/A.
82 marks
Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?
Answer: Thick wire has larger cross-sectional area, lower resistance, so current flows more easily.
92 marks
Let the resistance of an electrical component remain constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?
Answer: Since V = IR, current also becomes half (I ∝ V when R constant).
102 marks
Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?
Answer: Alloys have higher resistivity, do not oxidise easily at high temperatures, and have high melting points.
112 marks
Use the data in Table 12.2 (NCERT) to answer the following: (a) Which among iron and mercury is a better conductor? (b) Which material is the best conductor?
Answer: (a) Iron (resistivity 10×10⁻⁸ Ωm) is better than mercury (94×10⁻⁸ Ωm). (b) Silver is the best conductor.
Intext Questions (Page 213)
122 marks
Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 Ω resistor, an 8 Ω resistor, and a 12 Ω resistor, and a plug key, all connected in series.
Answer: [Diagram description: Three cells in series (total 6V), then a plug key, then 5Ω, 8Ω, 12Ω resistors all in series.]
133 marks
Redraw the circuit of Question 12, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12 Ω resistor. What would be the readings in the ammeter and the voltmeter?
Answer: Total R = 5+8+12 = 25Ω, V=6V, I = V/R = 6/25 = 0.24 A. Ammeter reads 0.24 A. Voltmeter across 12Ω: V = IR = 0.24 × 12 = 2.88 V.
Intext Questions (Page 216)
142 marks
Judge the equivalent resistance when the following are connected in parallel: (a) 1 Ω and 10⁶ Ω, (b) 1 Ω and 10³ Ω and 10⁶ Ω.
Answer: (a) Equivalent resistance slightly less than 1 Ω (~1 Ω). (b) Slightly less than 1 Ω.
152 marks
An electric lamp of 100 Ω, a toaster of resistance 50 Ω, and a water filter of resistance 500 Ω are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?
Answer: 1/Rp = 1/100 + 1/50 + 1/500 = (5+10+1)/500 = 16/500 => Rp = 500/16 = 31.25 Ω. Total current I = V/Rp = 220/31.25 = 7.04 A. Iron resistance = 31.25 Ω (since same current).
162 marks
What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?
Answer: In parallel, each device gets full voltage, independent operation, and if one fails, others work. In series, voltage divides, and failure breaks the circuit.
172 marks
How can three resistors of resistances 2 Ω, 3 Ω, and 6 Ω be connected to give a total resistance of (a) 4 Ω, (b) 1 Ω?
Answer: (a) 3Ω and 6Ω in parallel gives 2Ω, then in series with 2Ω gives 4Ω. (b) All three in parallel gives 1Ω (1/2+1/3+1/6 = 1).
182 marks
What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 Ω, 8 Ω, 12 Ω, 24 Ω?
Answer: (a) Highest = series = 4+8+12+24 = 48 Ω. (b) Lowest = parallel: 1/R = 1/4+1/8+1/12+1/24 = (6+3+2+1)/24 = 12/24 => R = 2 Ω.
NCERT Exercise Questions (Page 221)
Ex 11 mark
A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R', then the ratio R/R' is (a) 1/25 (b) 1/5 (c) 5 (d) 25.
Answer: (d) 25. Each part resistance = R/5. Parallel of five = (R/5)/5 = R/25. So R/R' = R/(R/25) = 25.
Ex 21 mark
Which of the following terms does not represent electrical power in a circuit? (a) I²R (b) IR² (c) VI (d) V²/R.
Answer: (b) IR² is not power (I²R is, IR² is dimensionally incorrect).
Ex 31 mark
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) 25 W. Resistance R = V²/P = (220)²/100 = 484 Ω. Power at 110 V = (110)²/484 = 12100/484 = 25 W.
Ex 41 mark
Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be (a) 1:2 (b) 2:1 (c) 1:4 (d) 4:1.
Answer: (c) 1:4. H = V²t/R. Series R = 2R, parallel R = R/2. H_series/H_parallel = (V²t/2R) / (V²t/(R/2)) = (1/2R) × (R/2) = 1/4.
Ex 52 marks
How is a voltmeter connected in the circuit to measure the potential difference between two points?
Answer: Voltmeter is connected in parallel across the two points.
Ex 62 marks
A copper wire has diameter 0.5 mm and resistivity 1.6 × 10⁻⁸ Ωm. 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?
Answer: R = ρl/A => l = RA/ρ. A = πr² = π(0.25×10⁻³)² = π×6.25×10⁻⁸ = 1.9625×10⁻⁷ m². l = (10 × 1.9625×10⁻⁷)/(1.6×10⁻⁸) = 1.9625×10⁻⁶/1.6×10⁻⁸ ≈ 122.66 m. If diameter doubled, area ×4, resistance becomes 1/4 = 2.5 Ω.
Ex 72 marks
The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below. Plot a graph between V and I and calculate the resistance of that resistor. I (A): 0.5, 1.0, 2.0, 3.0, 4.0; V (V): 1.6, 3.4, 6.7, 10.2, 13.2.
Answer: Graph V vs I is a straight line. Slope = V/I gives resistance ≈ 3.3 Ω (average).
Ex 83 marks
When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.
Answer: R = V/I = 12 / (2.5 × 10⁻³) = 4800 Ω = 4.8 kΩ.
Ex 93 marks
A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω. How much current would flow through the 12 Ω resistor?
Answer: Total R = 0.2+0.3+0.4+0.5+12 = 13.4 Ω. I = V/R = 9/13.4 ≈ 0.6716 A. Same current through all in series.
Ex 103 marks
How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?
Answer: Equivalent resistance R = V/I = 220/5 = 44 Ω. For n resistors of 176Ω in parallel: 176/n = 44 => n = 4.
Ex 113 marks
Show how you would connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of (i) 9 Ω, (ii) 4 Ω.
Answer: (i) Two in parallel (3Ω) then series with third (6Ω) gives 9Ω. (ii) Two in series (12Ω) then parallel with third (6Ω) gives 4Ω (1/12+1/6=1/4).
Ex 123 marks
Several electric bulbs designed to be used on a 220 V electric supply line are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A?
Answer: Each bulb current I = P/V = 10/220 = 1/22 A. Number = total current / per bulb current = 5 / (1/22) = 110 bulbs.
Ex 132 marks
A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 Ω resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?
Answer: (i) Separate: I = 220/24 ≈ 9.17 A. (ii) Series: R = 48Ω, I = 220/48 ≈ 4.58 A. (iii) Parallel: R = 12Ω, I = 220/12 ≈ 18.33 A.
Ex 142 marks
Compare the power used in the 2 Ω resistor in each of the following circuits: (i) a 6 V battery in series with 1 Ω and 2 Ω resistors, (ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.
Answer: (i) I = 6/(1+2)=2A, P = I²R = 4×2=8W. (ii) For parallel, voltage across 2Ω = 4V, P = V²/R = 16/2=8W. Same power (8W).
Ex 152 marks
Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to the electric mains supply. What current is drawn from the line if the supply voltage is 220 V?
Answer: Total power = 100+60=160W. I = P/V = 160/220 ≈ 0.727 A.
Ex 163 marks
A 100 W electric bulb is used for 10 hours per day. How many 'units' of energy are consumed in 30 days? (1 unit = 1 kWh)
Answer: Energy per day = 100W × 10h = 1000 Wh = 1 kWh. For 30 days = 30 units.
Additional Important Questions
A12 marks
State Ohm's law. Draw a circuit diagram to verify it.
Answer: Ohm's law: At constant temperature, current through a conductor is directly proportional to potential difference across its ends. V = IR.
A22 marks
What is resistivity? Write its SI unit.
Answer: Resistivity is the resistance of a conductor of unit length and unit cross-sectional area. SI unit: ohm-metre (Ωm).
NCERT Class 10 Science | Chapter 12: Electricity | Font: Lato 14px | Complete solutions with page numbers (200, 202, 209, 213, 216, 221).
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