NCERT Solutions for Class 10 Science Chapter 11: Electricity

These Class 10 Science Chapter 11 solutions cover Electricity from the NCERT textbook (session 2026–27). You will find every in-text “Questions” set and the complete end-of-chapter “Exercises” reproduced word-for-word and solved step by step — including all Ohm’s-law, resistance, heating-effect and electric-power numericals worked out with correct units and verified answers.

Class: 10 Subject: Science (Physics) Chapter: 11 Chapter Name: Electricity Exercises: 18 questions + in-text sets Session: 2026–27
On this page

Class 10 Science Chapter 11 Solutions – Overview

Chapter 11, Electricity, explains what constitutes electric current, how charges flow in a circuit, and the factors that control that flow. You learn that current is the rate of flow of charge, that a potential difference (set up by a cell or battery) drives the charges, and that every conductor offers some resistance. Ohm’s law links these three quantities (V = IR). The chapter then shows how resistance depends on a wire’s length, area of cross-section and material (resistivity), how resistors combine in series and parallel, and finally the heating effect of current (Joule’s law, H = I2Rt) and electric power (P = VI = I2R = V2/R), with the commercial unit kilowatt-hour. It is one of the most numerical-heavy chapters of Class 10 physics.

Key Concepts & Definitions

Electric current (I): the rate of flow of electric charge through a cross-section of a conductor, I = Q/t. SI unit: ampere (A); 1 A = 1 C s−1. Conventional current flows opposite to electron flow.

Electric charge (Q): measured in coulomb (C); 1 C ≈ charge of 6 × 1018 electrons; charge on one electron = 1.6 × 10−19 C.

Potential difference (V): work done to move a unit charge between two points, V = W/Q. SI unit: volt (V); 1 V = 1 J C−1. Measured by a voltmeter, connected in parallel.

Resistance (R): the property of a conductor that opposes the flow of charge, R = V/I. SI unit: ohm (Ω). Measured indirectly; current is measured by an ammeter connected in series.

Ohm’s law: at constant temperature, V ∝ I, so V = IR. The V–I graph is a straight line through the origin.

Resistivity (ρ): R = ρl/A; ρ is a property of the material (SI unit Ω m). Metals/alloys have low ρ; insulators very high ρ.

Electric power (P): rate of consumption of electrical energy, P = VI = I2R = V2/R. SI unit: watt (W). Commercial unit of energy: kilowatt-hour (kW h), 1 kW h = 3.6 × 106 J.

Important Formulas (Chapter 11)

Current: I = Q/t  •  Potential difference: V = W/Q

Ohm’s law: V = IR  •  R = V/I  •  I = V/R

Resistivity: R = ρl/A  (A = πr2 = πd2/4 for a round wire)

Series: Rs = R1 + R2 + R3 + … (same current through each)

Parallel: 1/Rp = 1/R1 + 1/R2 + 1/R3 + … (same voltage across each)

Heating (Joule’s law): H = VIt = I2Rt  •  Power: P = VI = I2R = V2/R

Energy: 1 kW h = 1000 W × 3600 s = 3.6 × 106 J

In-text “Questions” — Answers

The questions below are reproduced verbatim from the NCERT textbook, grouped exactly as they appear after each section.

Page 172

1. What does an electric circuit mean?

ANSWERAn electric circuit is a continuous and closed conducting path along which an electric current flows. It typically contains a source (cell or battery), connecting wires, a switch (plug key) and one or more electrical components such as a bulb or resistor. If the path is broken anywhere, the current stops.

2. Define the unit of current.

ANSWERThe SI unit of current is the ampere (A). One ampere is the current produced when one coulomb of charge flows through a cross-section of a conductor in one second, i.e. 1 A = 1 C s−1.

3. Calculate the number of electrons constituting one coulomb of charge.

ANSWER Charge on one electron, e = 1.6 × 10−19 C; total charge Q = 1 C. Number of electrons n = Q/e = 1 / (1.6 × 10−19) = 6.25 × 1018 electrons.

Page 173

1. Name a device that helps to maintain a potential difference across a conductor.

ANSWERA cell, a battery (combination of cells), or any source of e.m.f. maintains a potential difference across a conductor.

2. What is meant by saying that the potential difference between two points is 1 V?

ANSWERIt means that 1 joule of work is done in moving a charge of 1 coulomb from one point to the other, since V = W/Q and 1 V = 1 J C−1.

3. How much energy is given to each coulomb of charge passing through a 6 V battery?

ANSWER Energy (work done) W = VQ = 6 V × 1 C = 6 J. So each coulomb of charge gains 6 joules of energy.

Page 181

1. On what factors does the resistance of a conductor depend?

ANSWER The resistance of a conductor depends on: (i) its length l (R ∝ l), (ii) its area of cross-section A (R ∝ 1/A), (iii) the nature of its material (resistivity ρ), and (iv) its temperature.

2. Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?

ANSWERCurrent flows more easily through the thick wire. Since R = ρl/A, a larger area of cross-section gives lower resistance; for the same voltage, lower resistance means greater current (I = V/R).

3. Let the resistance of an electrical component remains 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 By Ohm’s law I = V/R. With R constant, I ∝ V. If V becomes V/2, the new current I′ = (V/2)/R = I/2. The current is halved.

4. Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?

ANSWERBecause alloys (like nichrome) have a much higher resistivity than pure metals, so they produce more heat for a given current, and they do not oxidise (burn) readily at high temperature. This makes them ideal for heating elements.

5. Use the data in Table 11.2 to answer the following –(a) Which among iron and mercury is a better conductor?(b) Which material is the best conductor?

ANSWER (a) Resistivity of iron = 10.0 × 10−8 Ω m; of mercury = 94.0 × 10−8 Ω m. Iron has lower resistivity, so iron is the better conductor. (b) The lowest resistivity in the table is that of silver (1.60 × 10−8 Ω m), so silver is the best conductor.

Page 185

1. 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 Draw a single closed loop containing, in series: a battery of three 2 V cells (drawn as three cell symbols giving 6 V), a plug key, and the three resistors 5 Ω, 8 Ω and 12 Ω one after another, all joined end to end by connecting wires. Total voltage = 3 × 2 V = 6 V; total resistance = 5 + 8 + 12 = 25 Ω (useful for the next question).

2. Redraw the circuit of Question 1, 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 Connect the ammeter in series in the loop and the voltmeter in parallel across the 12 Ω resistor. Total resistance R = 5 + 8 + 12 = 25 Ω; supply V = 6 V. Ammeter reading I = V/R = 6 V / 25 Ω = 0.24 A. Voltmeter reading (across 12 Ω) = IR = 0.24 A × 12 Ω = 2.88 V.

Page 187

1. 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 In parallel the equivalent resistance is always less than the smallest resistor, and a very large resistor in parallel hardly changes the result. (a) 1/Rp = 1/1 + 1/106 ≈ 1 ⇒ Rp ≈ 1 Ω (just under 1 Ω). (b) 1/Rp = 1/1 + 1/103 + 1/106 ≈ 1 ⇒ Rp ≈ 1 Ω (just under 1 Ω).

2. 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 For the parallel combination: 1/Rp = 1/100 + 1/50 + 1/500 = 5/500 + 10/500 + 1/500 = 16/500. Rp = 500/16 = 31.25 Ω. Total current I = V/Rp = 220 V / 31.25 Ω = 7.04 A. The electric iron that draws the same current must have resistance R = V/I = 220 V / 7.04 A = 31.25 Ω, and the current through it is 7.04 A.

3. What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?

ANSWER (i) Each device gets the full supply voltage and works at its rated value. (ii) Each device can be switched on/off independently; if one fails, the others keep working. (iii) The total resistance is reduced, so a larger total current is available, letting each device draw the current it needs.

4. 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) Connect 3 Ω and 6 Ω in parallel, then add 2 Ω in series: parallel part = (3 × 6)/(3 + 6) = 18/9 = 2 Ω; total = 2 + 2 = 4 Ω. (b) Connect all three in parallel: 1/R = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1 ⇒ R = 1 Ω.

5. 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 resistance — connect all four in series: R = 4 + 8 + 12 + 24 = 48 Ω. (b) Lowest resistance — connect all four in parallel: 1/R = 1/4 + 1/8 + 1/12 + 1/24 = 6/24 + 3/24 + 2/24 + 1/24 = 12/24 = 1/2 ⇒ R = 2 Ω.

Page 190

1. Why does the cord of an electric heater not glow while the heating element does?

ANSWERThe heating element is made of a high-resistance alloy (nichrome), so for the same current it produces a large amount of heat (H = I2Rt) and becomes white-hot and glows. The cord is made of thick copper of very low resistance, so it produces very little heat and does not glow.

2. Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.

ANSWER Given Q = 96000 C, V = 50 V, t = 1 h = 3600 s. Heat H = VQ = 50 V × 96000 C = 4 800 000 J = 4.8 × 106 J.

3. An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 s.

ANSWER Given R = 20 Ω, I = 5 A, t = 30 s. H = I2Rt = (5)2 × 20 × 30 = 25 × 20 × 30 = 15 000 J = 1.5 × 104 J.

Page 192

1. What determines the rate at which energy is delivered by a current?

ANSWERThe rate at which energy is delivered is the electric power, P = VI. It is determined by the potential difference across the device and the current through it (equivalently P = I2R = V2/R).

2. An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.

ANSWER Power P = VI = 220 V × 5 A = 1100 W = 1.1 kW. Energy = P × t = 1100 W × (2 × 3600 s) = 1100 × 7200 = 7.92 × 106 J. In commercial units: 1.1 kW × 2 h = 2.2 kW h (= 7.92 × 106 J).

End-of-chapter “Exercises” — Solutions

1. 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 of the five equal parts has resistance R/5. Five such parts in parallel: R′ = (R/5) / 5 = R/25. Therefore R/R′ = R / (R/25) = 25.

2. 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 Power can be written as P = VI = I2R = V2/R, so options (a), (c) and (d) all give power. The expression IR2 is not a valid power formula, so it does not represent power.

3. 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 of the bulb (constant) R = V2/P = (220)2/100 = 48400/100 = 484 Ω. At 110 V: P′ = V′2/R = (110)2/484 = 12100/484 = 25 W. (Shortcut: halving the voltage makes power one-fourth, 100/4 = 25 W.)

4. 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 Let each wire have resistance R. Series: Rs = 2R; Parallel: Rp = R/2. For the same voltage V and time t, H = V2t/R, so H ∝ 1/R. Hs : Hp = (1/2R) : (1/(R/2)) = (1/2R) : (2/R) = 1 : 4 = 1:4.

5. How is a voltmeter connected in the circuit to measure the potential difference between two points?

ANSWERA voltmeter is always connected in parallel across the two points between which the potential difference is to be measured. (It has a very high resistance so it draws negligible current.)

6. A copper wire has diameter 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?

ANSWER Given d = 0.5 mm = 5 × 10−4 m, so radius r = 2.5 × 10−4 m; ρ = 1.6 × 10−8 Ω m; R = 10 Ω. Area A = πr2 = 3.14 × (2.5 × 10−4)2 = 3.14 × 6.25 × 10−8 = 1.963 × 10−7 m2. From R = ρl/A ⇒ l = RA/ρ = (10 × 1.963 × 10−7) / (1.6 × 10−8) = (1.963 × 10−6) / (1.6 × 10−8) = 122.7 m (≈ 122.7 m). If the diameter is doubled, the area becomes 4 times (A ∝ d2). Since R ∝ 1/A, the new resistance = 10/4 = 2.5 Ω, i.e. the resistance becomes one-fourth of its original value.

7. 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.

ANSWER The data are:
I (amperes)0.51.02.03.04.0
V (volts)1.63.46.710.213.2
SOLUTION Plotting V (y-axis) against I (x-axis) gives a straight line passing through the origin — this verifies Ohm’s law. The resistance equals the slope V/I. Taking a representative point, e.g. I = 3.0 A, V = 10.2 V: R = V/I = 10.2 / 3.0 = 3.4 Ω. The ratios for the other points are also close to 3.4 (1.6/0.5 = 3.2, 3.4/1.0 = 3.4, 6.7/2.0 = 3.35, 13.2/4.0 = 3.3). Hence the resistance R ≈ 3.4 Ω.

8. 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 Given V = 12 V, I = 2.5 mA = 2.5 × 10−3 A. R = V/I = 12 / (2.5 × 10−3) = 4800 Ω = 4.8 × 103 Ω (4.8 kΩ).

9. 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, respectively. How much current would flow through the 12 Ω resistor?

ANSWER In series, total resistance R = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 Ω. Current I = V/R = 9 / 13.4 = 0.67 A. Since it is a series circuit, the same current (0.67 A) flows through every resistor, including the 12 Ω resistor.

10. How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?

ANSWER Total resistance needed: R = V/I = 220 / 5 = 44 Ω. For n equal resistors of 176 Ω in parallel, Rp = 176/n. Set 176/n = 44. n = 176/44 = 4 resistors.

11. 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) For 9 Ω: connect two 6 Ω resistors in parallel (giving 6/2 = 3 Ω), then add the third 6 Ω in series: 3 + 6 = 9 Ω. (ii) For 4 Ω: connect two 6 Ω resistors in series (giving 12 Ω), then connect this in parallel with the third 6 Ω: (12 × 6)/(12 + 6) = 72/18 = 4 Ω.

12. 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 Current drawn by one lamp: I1 = P/V = 10/220 = 0.04545 A. Number of lamps n = total allowable current / current per lamp = 5 / 0.04545 = 110 lamps.

13. 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 Used separately (one 24 Ω coil): I = V/R = 220/24 = 9.17 A. In series (24 + 24 = 48 Ω): I = 220/48 = 4.58 A. In parallel (24/2 = 12 Ω): I = 220/12 = 18.33 A.

14. 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, and (ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.

ANSWER Circuit (i) — series: total R = 1 + 2 = 3 Ω; current I = V/R = 6/3 = 2 A (same through both). Power in 2 Ω: P = I2R = (2)2 × 2 = 8 W. Circuit (ii) — parallel: each resistor gets the full 4 V. Power in 2 Ω: P = V2/R = (4)2/2 = 16/2 = 8 W. Comparison: the power used in the 2 Ω resistor is the same (8 W) in both circuits, ratio 1:1.

15. Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V?

ANSWER Current through 100 W lamp: I1 = P/V = 100/220 = 0.4545 A. Current through 60 W lamp: I2 = 60/220 = 0.2727 A. Total current (parallel) I = I1 + I2 = 0.4545 + 0.2727 = 0.727 A (≈ 0.73 A).

16. Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?

ANSWER Energy = power × time. TV: E = 250 W × 1 h = 250 W h = 9.0 × 105 J. Toaster: E = 1200 W × (10/60) h = 1200 × (1/6) = 200 W h = 7.2 × 105 J. Since 250 W h > 200 W h, the TV set uses more energy.

17. An electric heater of resistance 44 Ω draws 5 A from the service mains for 2 hours. Calculate the rate at which heat is developed in the heater.

ANSWER The rate at which heat is developed is the power, P = I2R. P = (5)2 × 44 = 25 × 44 = 1100 W = 1100 J s−1. (The “2 hours” affects total energy, not the rate; the rate of heating is constant at 1100 W.)

18. Explain the following.(a) Why is the tungsten used almost exclusively for filament of electric lamps?(b) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal?(c) Why is the series arrangement not used for domestic circuits?(d) How does the resistance of a wire vary with its area of cross-section?(e) Why are copper and aluminium wires usually employed for electricity transmission?

ANSWER (a) Tungsten has a very high melting point (about 3380°C) and high resistivity, so its filament becomes white-hot and emits light without melting, and it retains heat well. (b) Alloys have high resistivity (so they produce more heat) and they do not oxidise (burn) readily at high temperatures, unlike pure metals; this makes them durable heating elements. (c) In a series circuit all devices share the same current and the same line, so (i) each device would not get its full rated voltage, and (ii) if one device fails the whole circuit breaks and every device stops. Also devices needing different currents cannot work together. Hence series is unsuitable for homes; parallel wiring is used. (d) Resistance is inversely proportional to the area of cross-section (R ∝ 1/A): a thicker wire (larger A) has lower resistance. (e) Copper and aluminium have very low resistivity (good conductors) and are relatively cheap and ductile, so they carry current with little energy loss in transmission lines.

Extra Practice Questions

Short Answer Type Questions

Q1. State Ohm’s law and write its mathematical form.

ANSWEROhm’s law: at constant temperature, the potential difference across a conductor is directly proportional to the current through it, V ∝ I, i.e. V = IR, where R is the resistance.

Q2. A current of 4 A flows through a 12 Ω resistor. Find the potential difference across it.

ANSWERV = IR = 4 A × 12 Ω = 48 V.

Q3. Why is an ammeter connected in series and a voltmeter in parallel?

ANSWERAn ammeter measures the current that must pass through it, so it is in series and has very low resistance. A voltmeter measures the p.d. across two points, so it is in parallel and has very high resistance, drawing negligible current.

Q4. How much charge flows when a current of 0.5 A passes for 4 minutes?

ANSWERQ = It = 0.5 A × (4 × 60 s) = 0.5 × 240 = 120 C.

Q5. Define electric power and give its SI unit.

ANSWERElectric power is the rate of consumption of electrical energy, P = VI = I2R = V2/R. Its SI unit is the watt (W); 1 W = 1 V × 1 A.

Long Answer Type Questions

Q1. Derive the expression for the equivalent resistance of three resistors connected in series.

ANSWERIn series the same current I flows through R1, R2, R3, and the total p.d. is the sum of the individual p.d.s: V = V1 + V2 + V3. By Ohm’s law V1 = IR1, V2 = IR2, V3 = IR3, and for the equivalent resistor V = IRs. Substituting: IRs = IR1 + IR2 + IR3. Dividing by I gives Rs = R1 + R2 + R3. The equivalent resistance in series is greater than any individual resistance.

Q2. Derive the expression for the equivalent resistance of three resistors connected in parallel.

ANSWERIn parallel each resistor has the same p.d. V, and the main current splits: I = I1 + I2 + I3. By Ohm’s law I1 = V/R1, I2 = V/R2, I3 = V/R3, and for the equivalent resistor I = V/Rp. Substituting: V/Rp = V/R1 + V/R2 + V/R3. Dividing by V gives 1/Rp = 1/R1 + 1/R2 + 1/R3. The equivalent resistance in parallel is less than the smallest individual resistance.

Q3. State Joule’s law of heating and describe two practical applications of the heating effect of current.

ANSWERJoule’s law: the heat produced in a resistor is H = I2Rt — directly proportional to the square of current, to the resistance, and to the time of flow. Applications: (i) Heating appliances such as electric irons, toasters, heaters and geysers use high-resistance alloy elements that get hot. (ii) The electric fuse is a short piece of low-melting-point wire placed in series; if the current exceeds a safe value, the I2Rt heating melts the fuse and breaks the circuit, protecting appliances. The electric bulb (tungsten filament glowing white-hot) is another example.

MCQs & Assertion–Reason

1. The SI unit of electric charge is the:

(a) ampere    (b) volt    (c) coulomb    (d) ohm

2. Ohm’s law gives the relation:

(a) V = I/R    (b) V = IR    (c) I = VR    (d) R = IV

3. The resistance of a wire is doubled when its:

(a) length is halved    (b) length is doubled    (c) area is doubled    (d) diameter is doubled

4. The commercial unit of electrical energy is the:

(a) watt    (b) joule    (c) kilowatt-hour    (d) volt

5. In a series combination of resistors, the quantity that stays the same through each resistor is the:

(a) potential difference    (b) current    (c) power    (d) resistance

6. Three resistors of 2 Ω each in parallel give an equivalent resistance of:

(a) 6 Ω    (b) 2 Ω    (c) 1 Ω    (d) 2/3 Ω

7. The heat produced in a resistor is given by:

(a) IRt    (b) I2Rt    (c) IR2t    (d) I2R/t

8. A device that protects a circuit from excessively high current is the:

(a) ammeter    (b) voltmeter    (c) rheostat    (d) fuse

9. 1 kilowatt-hour is equal to:

(a) 3.6 × 103 J    (b) 3.6 × 106 J    (c) 1000 J    (d) 3600 J

10. A bulb rated 60 W, 220 V has a resistance (approx.) of:

(a) 807 Ω    (b) 220 Ω    (c) 484 Ω    (d) 60 Ω

Answer key: 1-(c), 2-(b), 3-(b), 4-(c), 5-(b), 6-(d), 7-(b), 8-(d), 9-(b), 10-(a).

For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.

A-R 1. Assertion: An ammeter is always connected in series in a circuit.

Reason: An ammeter measures the current that flows through it and has very low resistance.

A-R 2. Assertion: The equivalent resistance in a parallel combination is less than the smallest resistor.

Reason: Adding more parallel paths provides more routes for current and lowers the total resistance.

A-R 3. Assertion: Heating elements of irons and toasters are made of nichrome.

Reason: Nichrome has a low melting point and oxidises quickly at high temperature.

A-R 4. Assertion: A thick wire has lower resistance than a thin wire of the same material and length.

Reason: Resistance is inversely proportional to the area of cross-section of the wire.

A-R 5. Assertion: Domestic appliances are connected in parallel.

Reason: Parallel connection lets each appliance get the full supply voltage and work independently.

Answer key: 1-(A), 2-(A), 3-(C), 4-(A), 5-(A).

Common Mistakes & Exam Tips

Common mistakes to avoid

  • Forgetting to convert units — minutes to seconds, mA to A, mm to m — before substituting in formulas.
  • Adding resistances directly in parallel; remember you add the reciprocals, then invert (1/Rp = Σ 1/R).
  • Connecting the voltmeter in series or the ammeter in parallel — it is the other way round.
  • Using the wrong power formula: when voltage is constant use P = V2/R; when current is constant use P = I2R.
  • Assuming bulb power stays 100 W when the voltage changes — the resistance is fixed, so recalculate power from R.
  • Confusing area with diameter: A ∝ d2, so doubling the diameter makes the area 4 times and resistance one-fourth.

How to score full marks in this chapter

Always write the formula, substitute values with units, and box the final answer with the correct unit. For combination problems, redraw the circuit and reduce step by step (parallel groups first, then series). State the units for each quantity (A, V, Ω, W, J). For graph questions, label both axes, draw a straight line through the origin and find resistance as the slope V/I. Learn the three forms of power (VI, I2R, V2/R) and choose the one that fits the given data.

Frequently Asked Questions

What is Class 10 Science Chapter 11 Electricity about?

Chapter 11 explains electric current, potential difference, resistance and Ohm’s law (V = IR), how resistance depends on length, area and material (resistivity), how resistors combine in series and parallel, and the heating effect of current (H = I2Rt) and electric power (P = VI = I2R = V2/R).

How many numericals are there in Chapter 11 Electricity?

The end-of-chapter Exercises have 18 questions (most of them numericals on Ohm’s law, resistance, series/parallel combinations, heating and power), plus several in-text “Questions” sets. All of them are solved step by step with units on this page.

What is the unit of resistance and how is it defined?

The SI unit of resistance is the ohm (Ω). A conductor has a resistance of 1 Ω if a potential difference of 1 V across its ends drives a current of 1 A through it (R = V/I).

Are these Class 10 Science Chapter 11 solutions free?

Yes. All solutions are free and follow the official NCERT Science textbook for session 2026–27, with every exercise and numerical solved and verified.

← Chapter 10 All Class 10 Science Chapters Chapter 12 →
Scroll to Top