Class 10 Science · Chapter 11

Electricity

363 Class 10 Science questions on Electricity from the NCERT textbook, NCERT Exemplar and CBSE board papers 2018–2026, each with its answer and a step-by-step solution.

NCERT exercises

Exercises (18) Intext Questions (Page 172) (3) Intext Questions (Page 174) (3) Intext Questions (Page 181) (5) Intext Questions (Page 185) (2) Intext Questions (Page 188) (5) Intext Questions (Page 190) (3) Intext Questions (Page 192) (2)

NCERT Exemplar exercises

Multiple Choice Questions (18) Short Answer Questions (10) Long Answer Questions (7)

CBSE Class 10 Science previous year papers, with answers →

Questions · page 6 of 19

Q101 CBSE 2026
$V$ - $I$ graph for a nichrome wire is shown in the diagram.
1. (i)Prove that the graph follows Ohm's law and calculate the resistance of nichrome wire from the given $V$ - $I$ graph.
2. (ii)(I) Calculate the equivalent resistance of the combination of resistors connected in the given circuit. (II) Find the value of the total current drawn from the battery.
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$\frac{V}{I} = 4 Ω$ at every reading, so the graph obeys Ohm's law; $R_{eq} = 15 Ω$ and the battery current is $\frac{1}{3}$ A.
(i) The plotted points lie on a straight line through the origin and the ratio $V / I$ has the same value at the readings, $\frac{0.4}{0.1} = \frac{0.8}{0.2} = \frac{1.2}{0.3} = \frac{1.6}{0.4} = \frac{2.0}{0.5} = 4$ , so $V \propto I$ , which is Ohm's law; the resistance of the nichrome wire is the slope, $R = \frac{Δ V}{Δ I} = \frac{1.6}{0.4} = 4 Ω$ . (ii) (I) The $3 Ω$ and $7 Ω$ are in series through the apex and give $10 Ω$ , which is in parallel with the $10 Ω$ bridge and gives $5 Ω$ ; with the two $5 Ω$ arms in series the equivalent resistance is $R_{eq} = 5 + 5 + 5 = 15 Ω$ . (II) The total current drawn from the battery is $I = \frac{V}{R _{eq}} = \frac{5}{15} = \frac{1}{3} \approx 0.33$ A.

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Q102 CBSE 2026
1. (i)Derive an expression for the equivalent resistance of three resistors connected in parallel with a battery as shown in figure.
2. (ii)Find the values of the electric currents flowing through the resistors $R_{1}$ , $R_{2}$ and $R_{3}$ in the given electric circuit.
Show answer

The equivalent resistance obeys $\frac{1}{R _{p}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \frac{1}{R _{3}}$ , and the three resistors carry $6$ A, $3$ A and $2$ A.
(i) The three resistors bridge the same two points, so each has the same potential difference $V$ across it while the branch currents add: $I = I_{1} + I_{2} + I_{3}$ . With $I_{1} = V / R_{1}$ , $I_{2} = V / R_{2}$ , $I_{3} = V / R_{3}$ and $I = V / R_{p}$ this gives $\frac{V}{R _{p}} = \frac{V}{R _{1}} + \frac{V}{R _{2}} + \frac{V}{R _{3}}$ , and dividing right through by $V$ leaves $\frac{1}{R _{p}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \frac{1}{R _{3}}$ . (ii) The whole $12$ V of the battery appears between X and Y, so every resistor has $12$ V across it: $I_{1} = 12/2 = 6$ A through $R_{1}$ , $I_{2} = 12/4 = 3$ A through $R_{2}$ and $I_{3} = 12/6 = 2$ A through $R_{3}$ . Their sum, $11$ A, is what the ammeter reads, and it agrees with $V / R_{p} = 12 \div \frac{12}{11} = 11$ A, since here $\frac{1}{R _{p}} = \frac{1}{2} + \frac{1}{4} + \frac{1}{6} = \frac{11}{12}$ , that is $R_{p} = \frac{12}{11} \approx 1.09 Ω$ .

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Q103 CBSE 2026
1. (i)Due to change in length and area of cross-section of a conductor, resistance of conductor changes while resistivity does not change. Why ?
2. (ii)Conductors of electric toasters and electric iron are made of an alloy rather than a pure metal. Why ?
3. (iii)Define the S.I. unit of electric current.
Show answer

Resistance depends on dimensions, resistivity only on material; alloys have high resistivity and do not oxidise readily; 1 A is 1 C per second.
(i) Resistance depends on the dimensions of the conductor, $R = ρ \frac{L}{A}$ , so it changes when the length or the area of cross-section is changed. Resistivity $ρ = \frac{R A}{L}$ is a characteristic property of the material, depending only on the nature of the material and its temperature; when $L$ or $A$ changes, $R$ changes in the same ratio as $\frac{L}{A}$ , so $ρ$ stays the same. (ii) An alloy has a much higher resistivity than a pure metal, so even a short coil of it has a large resistance, and it is in this high resistance that the heat $H = I^{2} R t$ is produced; also, an alloy does not oxidise (burn) readily at the high temperature at which the element works. (iii) The S.I. unit of electric current is the ampere (A). One ampere is the current when one coulomb of charge flows through any cross-section of a conductor in one second: $1 A = \frac{1 C}{1 s}$ .

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Q104 CBSE 2026
1. (i)Consider the following electric circuit : Calculate the values of the following : (I) Total resistance of the circuit. (II) The total electric current drawn from the source. (III) Potential difference across $3 Ω$ resistor.
2. (ii)Two bulbs, rated as 100 W ; 220 V and 60 W ; 220 V are connected in parallel to an electric main supply of 220 V. Calculate the electric current drawn from the mains.
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The circuit has $R = 5 Ω$ , $I = 2$ A and $6$ V across the $3 Ω$ resistor; the two bulbs draw $0.73$ A.
(i) (I) Total resistance of the circuit $= 5 Ω$ ; (II) Current drawn from the source $= 2$ A; (III) Potential difference across the $3 Ω$ resistor $= 6$ V. (ii) Current drawn from the mains $= \frac{8}{11}$ A $\approx 0.73$ A.

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Q105 CBSE 2026
Study the given electric circuit in which 2 A electric current is flowing between points X and Y.
1. (i)Using the battery, key, voltmeter and ammeter in this given electric circuit, redraw a circuit diagram in which $2 Ω$ and $4 Ω$ resistors are connected between X and Y in parallel combination.
2. (ii)Circuit drawn by you, in which resistors are connected in parallel combination, calculate the electric current flowing through $4 Ω$ resistor.
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Connect the $2 Ω$ and $4 Ω$ resistors in parallel between X and Y; the $4 Ω$ resistor then carries $3$ A.
(i) Both the $2 Ω$ and the $4 Ω$ resistor are connected between X and Y, side by side; the battery, the key and the ammeter stay in series in the main line (ammeter + towards the battery's + terminal), and the voltmeter is connected across X and Y (+ at Y). (ii) From the given series circuit the battery's potential difference is $V = 2 A \times (2 + 4) Ω = 12$ V. In the parallel circuit the $4 Ω$ resistor has the whole $12$ V across it, so the current through it is $I = \frac{12 V}{4 Ω} = 3$ A. (The $2 Ω$ resistor carries $6$ A and the ammeter reads $9$ A.)

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Q106 CBSE 2026
A smart home uses energy monitoring system to track electricity usage of appliances. In a month of 30 days, the following devices are used : 1. An electric heater of $1500$ W used for $4$ hours daily. 2. A microwave oven rated at $1.5$ kW used for $3$ hours daily.
1. (i)Define electric power and write its SI unit.
2. (ii)Calculate the total energy consumed in joules and kWh in a whole month.
3. (iii)Find out the total cost of electricity for running both appliances for a month of 30 days if the cost of 1 unit is ₹ 7.
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Power is energy consumed per unit time (unit: watt); the month's energy is $315$ kWh $= 1.134 \times 1 0^{9}$ J, costing ₹2205.
(i) Electric power is the rate at which electric energy is consumed, that is, the electrical work done per unit time: $P = \frac{W}{t} = V I$ . Its SI unit is the watt (W), where $1$ W $= 1$ J/s $= 1$ V $\times$ $1$ A. (ii) Heater: $1.5$ kW for $4 \times 30 = 120$ h, so $E_{1} = 180$ kWh. Microwave oven: $1.5$ kW for $3 \times 30 = 90$ h, so $E_{2} = 135$ kWh. Total energy $E = 180 + 135 = 315$ kWh $= 315 \times 3.6 \times 1 0^{6} = 1.134 \times 1 0^{9}$ J. (iii) The month's energy is $315$ units, so the total cost $= 315 \times 7 = 2205$ , that is ₹ 2205.

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Q107 CBSE 2026
1. (i)Write the definition of the SI unit of power.
2. (ii)Two resistors $R_{1} = 1 Ω$ and $R_{2} = 2 Ω$ are connected in an electric circuit with a battery of $12$ V. Calculate the power consumed, when $R_{1}$ and $R_{2}$ are connected in (I) Parallel combination, and (II) Series combination.
3. (iii)Calculate the minimum number of identical electric bulbs of rating $4$ V, $6$ W; that can work safely with their optimum brightness, when connected in series with a $240$ V mains supply.
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One watt is $1 V \times 1 A$ ; the power is $216$ W in parallel and $48$ W in series; $60$ bulbs are needed.
(i) The SI unit of power is the watt (W). $1$ watt is the power consumed by a device that carries a current of $1$ A when operated at a potential difference of $1$ V, so $1 W = 1 V \times 1 A$ ; equivalently, it is the power when $1$ J of electrical energy is consumed in $1$ s. (ii) (I) In parallel, $\frac{1}{R _{p}} = \frac{1}{1} + \frac{1}{2} = \frac{3}{2}$ , so $R_{p} = \frac{2}{3} Ω$ and $P = \frac{V ^{2}}{R _{p}} = \frac{1 2 ^{2}}{2/3} = 216$ W. (II) In series, $R_{s} = 1 + 2 = 3 Ω$ and $P = \frac{V ^{2}}{R _{s}} = \frac{1 2 ^{2}}{3} = 48$ W. (iii) One bulb has resistance $R_{b} = \frac{V ^{2}}{P} = \frac{4 ^{2}}{6} = \frac{8}{3} Ω$ and rated current $I = \frac{P}{V} = \frac{6}{4} = \frac{3}{2}$ A. The same current flows through every bulb in series, so the total resistance of the chain on $240$ V must be $\frac{240}{3/2} = 160 Ω$ , and the number of bulbs is $n = \frac{160}{8/3} = 60$ . (Equivalently, each bulb must get its rated $4$ V, so $n = \frac{240}{4} = 60$ .)

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Q108 CBSE 2026
1. (i)What is meant by resistance in an electric circuit ?
2. (ii)In a series electric circuit comprising a resistor made of a metallic wire, the ammeter reads 500 mA. If the length of wire is doubled and area of cross-section is made half, what will be the new reading of ammeter ?
3. (iii)How will the resistivity of a wire be affected if both, the length of the wire and its area of cross-section, are doubled ? Justify your answer.
Show answer

Resistance opposes current, $R = \frac{V}{I}$ ; the ammeter reads 125 mA; resistivity is unchanged as it depends only on the material and temperature.
(i) Resistance is the property of a conductor by which it opposes the flow of electric current through it. It is measured by the ratio of the potential difference across the conductor to the current through it, $R = V / I$ , and its SI unit is the ohm ( $Ω$ ); a conductor has a resistance of $1 Ω$ when a potential difference of $1$ V across it drives a current of $1$ A. (ii) For a wire, $R = ρl / A$ . With the length doubled and the area of cross-section halved, $R^{'} = ρ (2 l) / (A /2) = 4 ρl / A = 4 R$ . The battery is unchanged, so the same potential difference acts and $I^{'} = V / R^{'} = I /4 = 500/4 = 125$ mA. The ammeter will read $125$ mA, that is $0.125$ A. (iii) The resistivity is not affected at all - it stays exactly the same. Resistivity is a property of the **material** of the wire (and of its temperature) only, not of the size or shape of the piece. (Here the resistance is unchanged too: doubling the length doubles it and doubling the area halves it, so $ρ = R A / l = R (2 A) / (2 l)$ keeps its value.)

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Q109 CBSE 2026
1. (i)In a house, 5 fans, rated 100 W; 220 V each are used for 4 hours daily and an electric heater rated 1000 W; 220 V is used for 5 hours daily. Calculate the cost of electrical energy consumed by the fans and the electric heater in a month of 30 days if the cost of 1 unit of electrical energy is ₹ 5.
2. (ii)You have two fuse wires X and Y with electric current rating 5 A and 10 A respectively. Which of the two would you use with a heater rated 2000 W; 220 V ? Justify your answer.
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The fans and heater use 210 kWh a month costing ₹ 1050; fuse Y (10 A) suits the heater, which draws about 9.09 A.
(i) Power of the five fans $= 5 \times 100 = 500$ W $= 0.5$ kW; power of the heater $= 1000$ W $= 1$ kW. Energy per day: fans $0.5 \times 4 = 2$ kWh, heater $1 \times 5 = 5$ kWh, total $7$ kWh. Energy in 30 days $= 7 \times 30 = 210$ kWh $= 210$ units (fans $60$ units, heater $150$ units). Cost $= 210$ units at ₹ 5 per unit $=$ ₹ 1050 (fans ₹ 300, heater ₹ 750). (ii) Fuse wire Y ( $10$ A). The heater draws $I = \frac{P}{V} = \frac{2000}{220} \approx 9.09$ A. This is more than the $5$ A rating of X, so X would melt as soon as the heater is switched on; it is less than the $10$ A rating of Y, so Y carries the normal current and melts only if the current exceeds $10$ A, which protects the circuit from overloading and short-circuiting.

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Q110 CBSE 2026
1. (i)What is meant by the resistance of a conductor ? Derive an expression for the equivalent resistance of an electric circuit in which two resistors, $R_{1}$ and $R_{2}$ are connected in parallel.
2. (ii)Calculate the power consumed per second if two resistors, $R_{1} = 2 Ω$ and $R_{2} = 4 Ω$ are connected in (I) a series combination, and (II) a parallel combination, with a battery of $8$ V potential difference.
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Parallel resistors obey $\frac{1}{R _{p}} = \frac{1}{R _{1}} + \frac{1}{R _{2}}$ ; power consumed per second is about 10.67 W in series and 48 W in parallel.
(i) Resistance is the property of a conductor by which it opposes the flow of electric current through it; it is the ratio of the potential difference across the conductor to the current through it, $R = \frac{V}{I}$ , and its SI unit is the ohm ( $Ω$ ). Circuit diagram: $R_{1}$ and $R_{2}$ are both joined between the same two points A and B, and a battery of potential difference $V$ is connected across A and B, the current $I$ dividing into $I_{1}$ through $R_{1}$ and $I_{2}$ through $R_{2}$ (as in the figure). Each resistor has the same $V$ across it and the total current is $I = I_{1} + I_{2}$ , so $\frac{V}{R _{p}} = \frac{V}{R _{1}} + \frac{V}{R _{2}}$ , which gives $\frac{1}{R _{p}} = \frac{1}{R _{1}} + \frac{1}{R _{2}}$ , that is, $R_{p} = \frac{R _{1} R _{2}}{R _{1} + R _{2}}$ . (ii) (I) Series: $R_{s} = 2 + 4 = 6 Ω$ and $I = \frac{8}{6} = \frac{4}{3}$ A, so $P = V I = \frac{32}{3} \approx 10.67$ W - about $10.67$ J of energy consumed every second. (II) Parallel: $R_{p} = \frac{2 \times 4}{2 + 4} = \frac{4}{3} Ω$ and $I = 6$ A, so $P = V I = 48$ W - $48$ J of energy consumed every second.

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Q111 CBSE 2026
1. (i)How much electric current will an electric iron draw from 220 V source if the resistance of its heating element when hot, is 55 $Ω$ ? Calculate the power consumed by the electric iron when it is operated at 220 V.
2. (ii)In a house, 3 bulbs of 100 watt each, are lit for 5 hours daily and an electric heater of 1.0 kW is used for half an hour daily. Calculate the total energy consumed in a month of 30 days and its cost at the rate of ₹ 3.60 per kWh.
3. (iii)With reason explain, why are alloys commonly used to make elements of electrical heating devices.
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The iron draws $4$ A and consumes $880$ W; the house uses $60$ kWh costing ₹216; alloys have high resistivity and melting point.
(i) The iron draws $I = V / R = 220/55 = 4$ A, and the power it consumes is $P = V I = 220 \times 4 = 880$ W, that is $0.88$ kW. (ii) Using $E = P \times t$ (with $P$ in kW and $t$ in h), the three bulbs use $3 \times 0.1 \times 5 = 1.5$ kWh a day and the heater $1.0 \times 0.5 = 0.5$ kWh a day, so the house uses $2.0$ kWh a day and $2.0 \times 30 = 60$ kWh, that is 60 units, in the month; at ₹3.60 per kWh the cost is $60 \times 3.60 =$ ₹216. (iii) Alloys such as nichrome are used because their resistivity is much higher than that of the pure metals they are made from, so a short length of wire has a large resistance and gives out a great deal of heat ( $H = I^{2} R t$ ); and because an alloy does not oxidise or burn readily even at high temperature and has a high melting point, so the element is not destroyed by being made red hot again and again.

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Q112 CBSE 2026
1. (i)Define the SI unit of power. Express the electric power of an electric appliance in terms of potential difference and current flowing through it.
2. (ii)In a house, 3 bulbs of 100 W each, are lit for 5 hours daily and an electric heater of 1.0 kW is used for half an hour daily. Calculate the total energy consumed in a month of 30 days and its cost at the rate of ₹ 3.60 per kWh.
3. (iii)Convert the commercial unit of electric energy into Joule (J).
Show answer

Power is $P = V I$ (unit watt); the house uses 60 kWh a month costing ₹ 216; $1 kWh = 3.6 \times 1 0^{6}$ J.
(i) The SI unit of power is the watt (W): an appliance has a power of $1$ watt when it consumes energy at the rate of $1$ joule per second, that is when it carries a current of $1$ ampere at a potential difference of $1$ volt. Since the work done in carrying a charge $Q$ through a potential difference $V$ is $W = V Q$ and $Q = I t$ , the electric power is $P = W / t = V I t / t = V I$ . (ii) The three bulbs use $3 \times 0.1 \times 5 = 1.5$ kWh a day and the heater $1.0 \times 0.5 = 0.5$ kWh a day, so the house uses $2.0$ kWh a day and $2.0 \times 30 = 60$ kWh in the month; at ₹ 3.60 per kWh the cost is ₹216. (iii) The commercial unit of electrical energy is the kilowatt-hour, and $1$ kWh $= 1000$ W $\times 3600$ s $= 3.6 \times 1 0^{6}$ J.

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Q113 CBSE 2026
1. (i)Define electrical resistivity of a material.
2. (ii)How will the resistivity and resistance of a conducting wire change if the length and area of cross-section of wire both are doubled ? Justify your answer.
3. (iii)Calculate the electric current flowing through a $4 Ω$ resistor that produces $100$ J of heat every second.
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Resistivity stays the same and resistance is unchanged when length and area both double; the current in the resistor is $5$ A.
(i) The electrical resistivity $ρ$ of a material is the resistance of a conductor of that material of unit length and unit area of cross-section; from $R = ρ l / A$ it is $ρ = R A / l$ , its SI unit is the ohm metre ( $Ω$ m), and it depends only on the material and its temperature. (ii) The resistivity does **not** change, because it is a property of the material and the material is the same. The resistance does not change either: $R^{'} = ρ \frac{2 l}{2 A} = ρ \frac{l}{A} = R$ . The justification is that doubling the length on its own would double $R$ while doubling the area on its own would halve it, and $2 \times \frac{1}{2} = 1$ - equivalently, $R$ depends on the shape only through the ratio $l / A$ , which doubling both leaves unchanged. (iii) From $H = I^{2} R t$ , $I^{2} = \frac{H}{R t} = \frac{100}{4 \times 1} = 25$ , so $I = 5$ A.

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Q114 CBSE 2026
1. (i)How many bulbs of resistance $8 Ω$ each should be connected in parallel combination to draw a current of 2 A from a battery of 4 V ?
2. (ii)Name the device used for measuring electric current. How is it connected in a circuit ?
3. (iii)State Joule's law of heating.
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Four bulbs are needed; current is measured by an ammeter connected in series; Joule's law gives the heat as $H = I^{2} R t$ .
(i) Four bulbs ( $n = 4$ ). The combination must have resistance $R_{p} = \frac{V}{I} = \frac{4}{2} = 2 Ω$ , and $n$ identical $8 Ω$ bulbs in parallel have $R_{p} = \frac{8}{n} Ω$ , so $n = 4$ ; each bulb draws $\frac{4}{8} = 0.5$ A, and $4 \times 0.5 = 2$ A. (ii) The device is an ammeter. It is connected in series with the part of the circuit whose current is to be measured, with its positive terminal towards the positive terminal of the battery. (iii) Joule's law of heating: the heat produced in a resistor is directly proportional to the square of the current for a given resistance, directly proportional to the resistance for a given current, and directly proportional to the time for which the current flows through the resistor; that is, $H = I^{2} R t$ .

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Q115 CBSE 2026
1. (i)State Ohm’s law and draw V-I graph for a conductor which follows Ohm’s law. Show that the slope of V-I graph gives resistance of conductor.
2. (ii)Derive an expression for the equivalent resistance of a series combination of three resistors having resistances $R_{1}$ , $R_{2}$ and $R_{3}$ .
Show answer

The V-I graph of an ohmic conductor is a straight line of slope $R$ ; resistors in series give $R_{s} = R_{1} + R_{2} + R_{3}$ .
(i) Ohm’s law: the electric current flowing through a metallic conductor is directly proportional to the potential difference across its ends, provided its temperature remains the same, i.e. $V \propto I$ , or $V = I R$ . The V-I graph (V on the vertical axis, I on the horizontal axis) is a straight line through the origin, and its slope $= \frac{Δ V}{Δ I} = \frac{I _{2} R - I _{1} R}{I _{2} - I _{1}} = R$ , the resistance of the conductor. (ii) In series the same current $I$ flows through each resistor and $V = V_{1} + V_{2} + V_{3}$ . With $V_{1} = I R_{1}$ , $V_{2} = I R_{2}$ , $V_{3} = I R_{3}$ and $V = I R_{s}$ : $I R_{s} = I (R_{1} + R_{2} + R_{3})$ , so $R_{s} = R_{1} + R_{2} + R_{3}$ .

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Q116 CBSE 2026
1. (i)Two lamps ‘A’ and ‘B’ of rating 50 W; 220 V and 100 W, 220 V are connected in series combination. Find out the ratio of the resistances ( $R_{A} : R_{B}$ ) of these lamps.
2. (ii)Derive the expression for the equivalent resistance of three resistors $R_{1}$ , $R_{2}$ and $R_{3}$ connected in parallel combination.
Show answer

$R_{A} : R_{B} = 2 : 1$ , and the equivalent resistance of three parallel resistors follows $\frac{1}{R _{p}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \frac{1}{R _{3}}$ .
(i) $R = \frac{V ^{2}}{P}$ gives $R_{A} = \frac{( 220 ) ^{2}}{50} = 968 Ω$ and $R_{B} = \frac{( 220 ) ^{2}}{100} = 484 Ω$ , so $R_{A} : R_{B} = 2 : 1$ (connecting the lamps in series does not change their resistances). (ii) Let $R_{1}$ , $R_{2}$ and $R_{3}$ be connected in parallel across a potential difference $V$ . Each resistor has the same $V$ across it, and the main current divides as $I = I_{1} + I_{2} + I_{3}$ . By Ohm's law $I_{1} = \frac{V}{R _{1}}$ , $I_{2} = \frac{V}{R _{2}}$ , $I_{3} = \frac{V}{R _{3}}$ and $I = \frac{V}{R _{p}}$ , so $\frac{V}{R _{p}} = \frac{V}{R _{1}} + \frac{V}{R _{2}} + \frac{V}{R _{3}}$ ; dividing by $V$ gives $\frac{1}{R _{p}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \frac{1}{R _{3}}$ .

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Q117 CBSE 2026
1. (i)What is meant by electrical resistivity of a material ?
2. (ii)A series electric circuit comprising a resistor made of a metallic wire has an ammeter which reads $300$ mA. If the length of the wire is made three times, what will be the new reading in the ammeter ?
3. (iii)Calculate the resistance of a $500$ m long aluminium wire of cross-sectional area $1 \times 1 0^{- 3} m^{2}$ and resistivity $2.65 \times 1 0^{- 8} Ω m$ .
Show answer

Resistivity is the resistance of unit length and unit area of a material; the ammeter then reads $100$ mA and $R = 0.0133 Ω$ .
(i) The electrical resistivity of a material is the resistance offered by a conductor of that material having unit length and unit area of cross-section, $ρ = \frac{R A}{L}$ . It depends on the material and its temperature but not on the length or thickness of the particular piece, and its SI unit is the ohm metre ( $Ω$ m). (ii) The new reading is $100$ mA ( $0.1$ A). Making the length three times makes the resistance $3 R$ , and since the same battery keeps $V$ unchanged, $I^{'} = \frac{V}{3 R} = \frac{I}{3} = \frac{300}{3} = 100$ mA. (iii) $R = \frac{ρ L}{A} = \frac{2.65 \times 1 0 ^{- 8} \times 500}{1 \times 1 0 ^{- 3}} = 1.325 \times 1 0^{- 2} Ω \approx 0.0133 Ω$ .

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Q118 CBSE 2026
1. (i)State Ohm's law. $V$ - $I$ graph for Ohm's law is always a straight line. Justify.
2. (ii) $V$ - $I$ graphs for two wires of identical dimensions but made of different materials A and B are shown in the figure. Out of A and B, which one is a better choice for making a heating element ? Justify your answer.
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Ohm's law gives a straight V-I line because $R$ is constant; B is the better heating element as its steeper line means higher resistance.
(i) Ohm's law: the potential difference across the ends of a metallic conductor is directly proportional to the current flowing through it, provided its temperature and other physical conditions remain unchanged - that is, $V \propto I$ , or $V = I R$ , where the constant $R$ is the resistance. The graph is a straight line because $R$ is a constant, so $V$ is directly proportional to $I$ ; the ratio $V / I$ is the same at every point, which plots as a straight line through the origin whose slope is $R$ . (ii) B is the better choice. On this graph the slope is $V / I$ , which is the resistance, and B's line is steeper, so $R_{B} > R_{A}$ . The wires have identical length and cross-section, so by $R = ρl / A$ the resistivity of B is greater than that of A. A heating element must have a high resistance and high resistivity, because the heat it gives out at a given current is $H = I^{2} R t$ - hence B.

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Q119 CBSE 2026
1. (i)State Ohm’s law.
2. (ii)Derive an expression for equivalent resistance of parallel combination of three resistors having resistances $R_{1}$ , $R_{2}$ and $R_{3}$ .
3. (iii)V-I graph is given for two wires having identical shapes and sizes, but made of two different materials A and B. Out of A and B, which one is a better choice for making a heating element ? Justify your answer.
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Ohm's law is $V = I R$ ; for parallel resistors $\frac{1}{R _{p}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \frac{1}{R _{3}}$ ; material B suits a heating element.
(i) Ohm's law: at constant temperature, the potential difference across the ends of a metallic conductor is directly proportional to the current flowing through it; $V \propto I$ , so $V = I R$ , where the constant $R$ is the resistance of the conductor. (ii) Draw a circuit diagram with $R_{1}$ , $R_{2}$ and $R_{3}$ connected between the same two points across a battery and a plug key, as in the figure, marking the main current $I$ and the branch currents $I_{1}$ , $I_{2}$ , $I_{3}$ . In this parallel combination the same potential difference $V$ acts across all three resistors and the main current divides, so $I = I_{1} + I_{2} + I_{3}$ . With $I_{1} = V / R_{1}$ , $I_{2} = V / R_{2}$ , $I_{3} = V / R_{3}$ and $I = V / R_{p}$ , this gives $\frac{V}{R _{p}} = \frac{V}{R _{1}} + \frac{V}{R _{2}} + \frac{V}{R _{3}}$ ; cancelling the common $V$ leaves $\frac{1}{R _{p}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \frac{1}{R _{3}}$ . (iii) Material B. On the given graph the slope of each line is $V / I$ , which is the resistance of that wire, and B's line is the steeper, so $R_{B} > R_{A}$ . The two wires have the same length and the same area of cross-section, so from $R = ρl / A$ B has the greater resistivity - and a high resistivity is the property needed in the material of a heating element. For the same current, $H = I^{2} R t$ is also larger for B.

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Q120 CBSE 2026
1. (i)State the relationship of the resistance ( $R$ ) of a wire with its area of cross-section ( $A$ ) and its length ( $l$ ). With the help of this relation, define resistivity of the material of the wire.
2. (ii)In an electric circuit, 2 A electric current flows through a $40 Ω$ resistor for 50 seconds. (I) Calculate the heat produced in the wire. (II) If another $40 Ω$ resistor is connected in parallel to the given $40 Ω$ resistor, calculate the heat produced in the circuit in 50 seconds.
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Resistivity is the resistance of a wire of unit length and unit area; heat is 8000 J, rising to 16000 J in parallel.
(i) For a wire of a given material at a given temperature, the resistance is directly proportional to its length and inversely proportional to its area of cross-section: $R \propto l$ and $R \propto \frac{1}{A}$ , so $R = ρ \frac{l}{A}$ , where $ρ$ is the resistivity of the material. Putting $l = 1$ m and $A = 1$ m $^{2}$ gives $ρ = R$ : the resistivity of a material is the resistance of a wire of that material of unit length and unit area of cross-section. Its SI unit is the ohm metre ( $Ω$ m). (ii) (I) $H = I^{2} R t = 2^{2} \times 40 \times 50 = 8000$ J. (II) The same supply keeps the potential difference $V = I R = 2 \times 40 = 80$ V across the resistors. In parallel, $\frac{1}{R _{p}} = \frac{1}{40} + \frac{1}{40}$ , so $R_{p} = 20 Ω$ ; the supply now delivers $\frac{80}{20} = 4$ A ( $2$ A through each resistor), and $H = \frac{V ^{2} t}{R _{p}} = \frac{8 0 ^{2} \times 50}{20} = 16000$ J, twice the heat in (I).

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