ScienceCBSE Class 10 · Chapter 11

Electricity Class 10: Notes, Formulas and Solved Numericals

Updated 14 October 2026 · Fact-checked

Electricity is the Class 10 Physics chapter on electric current, potential difference, Ohm's law, resistance and resistivity, series and parallel circuits, the heating effect of current and electric power. The board asks numericals on these formulas, reasons such as why alloys are used in heaters, V-I graph and circuit questions, assertion-reason items and case-based questions.

What Electricity is about

This chapter explains what electric current is, what pushes it round a circuit, and what controls how much of it flows. You meet three quantities that are tied together: current (I), potential difference (V) and resistance (R). Ohm's law links them, and the formula $R = ρ \frac{l}{A}$ tells you why a long, thin wire resists more than a short, thick one. You then learn how resistors behave when joined in series or in parallel, why heat is produced whenever current flows through a resistor, and how to calculate power and the energy you pay for in kilowatt hours.

The ideas here are used again in the next chapter on the magnetic effects of current, and the energy ideas link back to work, energy and power from Class 9. Skills you build here, such as rearranging a formula, converting units and checking an answer for sense, help across all of Physics.

Board questions on this chapter come in many shapes: one-mark MCQs on a formula or a unit, assertion-reason items on facts such as alloys in heaters, two- and three-mark numericals on Ohm's law, resistivity and combinations of resistors, V-I graph and practical-based questions, reasoning questions (tungsten filaments, parallel wiring at home, fuses), case-based passages set in a home, and five-mark long answers that join a definition, a derivation of series or parallel resistance and a numerical.

Weightage: This chapter is part of Unit IV, Effects of Current, which carries 13 of the 80 theory marks together with Magnetic Effects of Electric Current.Source: CBSE Science syllabus 2026-27, Unit IV Effects of Current

In real board papers: asked in 21 of the 21 board papers we hold (2023 to 2026), on average 8.1 marks per paper. See the questions

Key concepts

Electric current and electric circuit

NCERT section 11.1 Electric Current and Circuit

An electric current is a flow of electric charge through a conductor. In a metal wire, the moving charges are electrons. Current measures how much charge passes a cross-section of the wire each second, so $I = \frac{Q}{t}$ . Its SI unit is the ampere: 1 A means 1 coulomb of charge passing each second. Small currents are given in milliampere ( $1 m A = 1 0^{- 3} A$ ) and microampere ( $1 μ A = 1 0^{- 6} A$ ).

Current flows only through a circuit, which is a continuous, closed path. A switch or plug key completes or breaks that path; if the path is broken anywhere, current stops everywhere.

Why is current said to flow from the positive terminal to the negative terminal when electrons actually move the other way? Electricity was studied before electrons were known, so current was taken as the flow of positive charge. We keep that convention: current direction is opposite to electron flow.

Current is measured with an ammeter, always connected in series so that the same current passes through it. One electron carries $1.6 \times 1 0^{- 19} C$ , so 1 C is the charge of nearly $6 \times 1 0^{18}$ electrons.

Worked example

A current of 0.4 A flows through a torch bulb for 2 minutes. (a) How much charge flows through the bulb? (b) How many electrons pass through it in this time? (Charge on an electron = $1.6 \times 1 0^{- 19} C$ .) [2 marks]

Try it first, then show the solutionSolution

Convert time to seconds: $t = 2 \times 60 = 120 s$ . (½)
Charge: $Q = I t = 0.4 A \times 120 s = 48 C$ . (½)
Number of electrons: $n = \frac{Q}{e} = \frac{48}{1.6 \times 1 0 ^{- 19}}$ . (½)
$n = 30 \times 1 0^{19} = 3 \times 1 0^{20}$ electrons. (½)

Answer: (a) 48 C (b) $3 \times 1 0^{20}$ electrons

Electric potential difference

NCERT section 11.2 Electric Potential and Potential Difference

Charges do not move along a wire by themselves, just as water does not flow along a perfectly level pipe. Water flows when one end is joined to a raised tank, because there is a pressure difference. In a wire, the electrons move when there is a difference of 'electric pressure' between the ends. This is the potential difference.

A cell or battery creates this difference by chemical action inside it. The difference exists across its terminals even when no current is drawn. When a circuit is connected, the cell uses up its stored chemical energy to keep the current going.

Potential difference between two points tells you how much work is needed for each coulomb of charge carried from the first point to the second: $V = \frac{W}{Q}$ . Its SI unit is the volt. One volt means 1 joule of work is done to move 1 coulomb of charge between the two points, so $1 V = 1 J C^{- 1}$ . That is why a 6 V battery gives 6 J of energy to every coulomb passing through it.

Potential difference is measured with a voltmeter, always connected in parallel across the two points.

Worked example

A battery keeps a current of 0.5 A flowing through a resistor for 1 minute. In this time the battery does 360 J of work in moving the charge through the resistor. (a) Find the charge that flows. (b) Find the potential difference across the resistor. (c) How much energy does each coulomb receive from the battery? [3 marks]

Try it first, then show the solutionSolution

$t = 60 s$ , so $Q = I t = 0.5 A \times 60 s = 30 C$ . (1)
$V = \frac{W}{Q} = \frac{360 J}{30 C} = 12 V$ . (1)
By definition, a potential difference of 12 V means 12 J of work is done on each coulomb, so each coulomb receives 12 J. (1)

Answer: (a) 30 C (b) 12 V (c) 12 J per coulomb

Ohm's law and resistance

NCERT section 11.4 Ohm's Law

Ohm's law says: the potential difference across a metallic wire is directly proportional to the current through it, provided its temperature stays the same. So $V \propto I$ , or $\frac{V}{I}$ is a constant, called the resistance $R$ . This gives $V = I R$ .

In Activity 11.1, a nichrome wire is connected to one, two, three and then four cells of 1.5 V each. The ammeter and voltmeter readings rise together, and $\frac{V}{I}$ comes out nearly the same each time. The V-I graph is a straight line through the origin. The slope of this line ( $\frac{V}{I}$ ) is the resistance.

Resistance is the property of a conductor that opposes the flow of charge. Electrons are not fully free inside a conductor; the atoms around them hold them back. Its SI unit is the ohm ( $Ω$ ): $1 Ω = \frac{1 V}{1 A}$ .

From $I = \frac{V}{R}$ , current is inversely proportional to resistance for a fixed voltage: double $R$ and the current halves. A rheostat (variable resistance) changes the resistance in a circuit to adjust the current without changing the source.

Worked example

In the practical on Ohm's law, a student records these readings for a resistor:

V (volt): 1.0, 2.0, 3.0, 4.0, 5.0
I (ampere): 0.25, 0.50, 0.90, 1.00, 1.25

(a) Which reading is most likely faulty? Justify. (b) Find the resistance of the resistor. (c) Describe the V-I graph expected. [3 marks]

Try it first, then show the solutionSolution

Calculate $\frac{V}{I}$ for each pair: 4, 4, 3.33, 4, 4 ohm. (½)
The third reading (3.0 V, 0.90 A) breaks the constant ratio, so it is faulty; for 3.0 V the current should be 0.75 A. (1)
Resistance from the consistent readings: $R = \frac{V}{I} = \frac{4.0 V}{1.00 A} = 4 Ω$ . (½)
Graph: V on the y-axis, I on the x-axis; points lie on a straight line passing through the origin (the faulty point lies off the line). (½)
The straight line through the origin shows $V \propto I$ , which is Ohm's law; its slope gives $R = 4 Ω$ . (½)

Answer: (a) The 3.0 V, 0.90 A reading is faulty (V/I is 3.33 instead of 4). (b) $R = 4 Ω$ . (c) A straight line through the origin with slope 4 Ω.

Factors affecting resistance and resistivity

NCERT section 11.5 Factors on which the Resistance of a Conductor Depends

Activity 11.3 shows three things. When a nichrome wire of double length is used, the current halves, so resistance has doubled: $R \propto l$ . When a thicker nichrome wire of the same length is used, the current rises, so resistance has fallen: $R \propto \frac{1}{A}$ . When a copper wire of the same size replaces nichrome, the current changes, so the material matters too.

Combining these, $R = ρ \frac{l}{A}$ . The constant $ρ$ is the resistivity, a property of the material alone. Its SI unit is $Ω m$ . A longer wire gives electrons a longer obstructed path; a thicker wire gives them a wider path, so more can pass.

Key point: cutting, stretching or thickening a wire changes its resistance, but not its resistivity, unless the material or temperature changes. Both resistance and resistivity vary with temperature.

Metals and alloys have low resistivity ( $1 0^{- 8}$ to $1 0^{- 6} Ω m$ ); insulators such as rubber and glass have about $1 0^{12}$ to $1 0^{17} Ω m$ . Alloys have higher resistivity than their metals and do not oxidise easily when hot, so they are used in irons and toasters. Copper and aluminium, with low resistivity, are used for transmission lines.

Worked example

Wire P has length $l$ , area of cross-section $A$ and resistance $6 Ω$ . Wire Q is made of the same material, has length $2 l$ and twice the diameter of P. (a) Find the resistance of Q. (b) Compare the resistivities of P and Q, giving a reason. [3 marks]

Try it first, then show the solutionSolution

For P: $R_{P} = ρ \frac{l}{A} = 6 Ω$ . (½)
Area depends on the square of diameter, so doubling the diameter makes the area $4 A$ . (½)
For Q: $R_{Q} = ρ \frac{2 l}{4 A} = \frac{1}{2} ρ \frac{l}{A}$ . (½)
$R_{Q} = \frac{1}{2} \times 6 = 3 Ω$ . (½)
Resistivities are equal, $ρ_{P} : ρ_{Q} = 1 : 1$ . (½)
Reason: resistivity depends only on the material (and temperature), not on length or thickness. (½)

Answer: (a) $3 Ω$ (b) Equal resistivities, because both wires are of the same material.

Resistors in series

NCERT section 11.6.1 Resistors in Series

Resistors are in series when they are joined end to end, so there is only one path for the current.

Activity 11.4 shows that an ammeter placed anywhere in a series circuit gives the same reading: the current is the same through every resistor. Activity 11.5 shows that the voltmeter reading across the whole combination equals the sum of the readings across each resistor: $V = V_{1} + V_{2} + V_{3}$ .

Now apply Ohm's law. Each resistor has $V_{1} = I R_{1}$ , $V_{2} = I R_{2}$ , $V_{3} = I R_{3}$ . For the single equivalent resistor, $V = I R_{s}$ . Substituting, $I R_{s} = I R_{1} + I R_{2} + I R_{3}$ ; cancelling $I$ gives $R_{s} = R_{1} + R_{2} + R_{3}$ .

So the equivalent resistance in series is larger than the largest single resistor. This makes sense: the charge has to pass through every obstacle one after another.

Notice too that the voltage is shared in proportion to resistance: the bigger resistor gets the bigger share, since $V = I R$ with the same $I$ . Drawbacks: all devices get the same current, and if one fails, the circuit breaks and all stop working.

Worked example

Resistors of $2 Ω$ , $4 Ω$ and $6 Ω$ are connected in series with a 6 V battery and a plug key. (a) Find the equivalent resistance. (b) Find the current. (c) Find the potential difference across each resistor and show that they add up to the battery voltage. (d) What will a voltmeter across the $6 Ω$ resistor read? [5 marks]

Try it first, then show the solutionSolution

$R_{s} = 2 + 4 + 6 = 12 Ω$ . (1)
$I = \frac{V}{R _{s}} = \frac{6 V}{12 Ω} = 0.5 A$ , the same through each resistor. (1)
$V_{1} = 0.5 \times 2 = 1 V$ ; $V_{2} = 0.5 \times 4 = 2 V$ ; $V_{3} = 0.5 \times 6 = 3 V$ . (1½)
Check: $1 + 2 + 3 = 6 V$ , equal to the battery voltage. (½)
Voltmeter across the $6 Ω$ resistor reads 3 V. (1)

Answer: (a) $12 Ω$ (b) 0.5 A (c) 1 V, 2 V, 3 V; total 6 V (d) 3 V

Resistors in parallel and their use in daily life

NCERT section 11.6.2 Resistors in Parallel

Resistors are in parallel when they are connected between the same two points, so each gives a separate path for current.

In Activity 11.6, a voltmeter across each resistor shows the same potential difference $V$ across every branch. An ammeter in each branch shows that the total current divides: $I = I_{1} + I_{2} + I_{3}$ .

Apply Ohm's law to each branch: $I_{1} = \frac{V}{R _{1}}$ and so on, and to the whole: $I = \frac{V}{R _{p}}$ . Substituting and cancelling $V$ : $\frac{1}{R _{p}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \frac{1}{R _{3}}$ .

The equivalent resistance is smaller than the smallest resistor, because every extra branch opens another path, so more current can flow for the same voltage.

This is why homes are wired in parallel. Each appliance gets the full supply voltage. Each draws the current it needs (a bulb needs little, a heater a lot, which would be impossible in series). Each can be switched on or off on its own, and if one fails, the others keep working. Decorative series lights show the opposite: one dead bulb puts out the whole string.

Worked example

A $4 Ω$ and a $12 Ω$ resistor are joined in parallel. This combination is connected in series with a $3 Ω$ resistor across a 12 V battery. Find (a) the total resistance, (b) the current from the battery, (c) the potential difference across the parallel part, and (d) the current in each parallel resistor. [5 marks]

Try it first, then show the solutionSolution

Parallel part: $\frac{1}{R _{p}} = \frac{1}{4} + \frac{1}{12} = \frac{3 + 1}{12} = \frac{4}{12}$ , so $R_{p} = 3 Ω$ . (1)
Total resistance: $R = 3 + 3 = 6 Ω$ . (½)
Battery current: $I = \frac{12 V}{6 Ω} = 2 A$ . (1)
p.d. across the parallel part: $V_{p} = I R_{p} = 2 \times 3 = 6 V$ . (1)
Current in $4 Ω$ : $\frac{6}{4} = 1.5 A$ ; in $12 Ω$ : $\frac{6}{12} = 0.5 A$ . (1)
Check: $1.5 + 0.5 = 2 A$ , equal to the battery current. (½)

Answer: (a) $6 Ω$ (b) 2 A (c) 6 V (d) 1.5 A in the $4 Ω$ resistor and 0.5 A in the $12 Ω$ resistor

Heating effect of electric current and its applications

NCERT section 11.7 Heating Effect of Electric Current

A source must keep spending energy to maintain a current. Part may do useful work (turning a fan) and the rest becomes heat; that is why a fan warms up. In a purely resistive circuit, all the energy turns into heat. This is the heating effect of electric current.

Work done moving charge $Q$ through p.d. $V$ is $V Q$ , and $Q = I t$ , so heat $H = V I t$ . Using $V = I R$ , $H = I^{2} R t$ . This is Joule's law of heating: heat is proportional to the square of the current, to the resistance, and to the time. Doubling the current gives four times the heat.

Useful applications: electric iron, toaster, oven, kettle and heater. In an electric bulb, the filament gets so hot it glows. It is made of tungsten, which has a very high melting point (3380°C); the bulb is filled with inactive nitrogen and argon to make the filament last longer.

The fuse is also based on heating. It is placed in series with the device. If the current exceeds its rating, the fuse wire heats, melts and breaks the circuit. Unwanted heating, though, wastes energy and can change the properties of components.

Worked example

A $10 Ω$ resistor is connected across a 20 V supply for 30 s. (a) Calculate the heat produced. (b) If the supply is changed to 40 V, how many times more heat is produced in the same time? Explain using Joule's law. [3 marks]

Try it first, then show the solutionSolution

Current: $I = \frac{V}{R} = \frac{20 V}{10 Ω} = 2 A$ . (½)
$H = I^{2} R t = 2^{2} \times 10 \times 30 = 1200 J$ . (1)
At 40 V, $I = \frac{40}{10} = 4 A$ , double the earlier current. (½)
$H = 4^{2} \times 10 \times 30 = 4800 J$ , which is 4 times 1200 J, because heat is proportional to the square of the current for the same R and t. (1)

Answer: (a) 1200 J (b) 4 times as much heat (4800 J)

Electric power and electrical energy

NCERT section 11.8 Electric Power

Power is the rate of doing work, which here is the rate at which electrical energy is used. Since energy supplied in time $t$ is $V I t$ , power is $P = V I$ . Using Ohm's law gives two more forms: $P = I^{2} R$ and $P = \frac{V ^{2}}{R}$ . Pick the form that uses the quantities you know.

The SI unit is the watt: 1 W is used when 1 A flows at 1 V. As the watt is small, the kilowatt (1000 W) is common.

Energy is power multiplied by time. The commercial unit is the kilowatt hour (kW h), called one 'unit' on a bill: $1 k W h = 1000 W \times 3600 s = 3.6 \times 1 0^{6} J$ .

A rating like '220 V, 100 W' means the device uses 100 W only when run at 220 V. Its resistance stays (nearly) fixed, so at a lower voltage use $P = \frac{V ^{2}}{R}$ : halving the voltage makes the power one quarter. Remember that we pay for energy, not for electrons; electrons are not used up in a circuit.

Worked example

An electric heater is rated 1100 W, 220 V. What power will it consume if it is run on a 110 V supply? (Assume its resistance stays the same.)
(a) 1100 W
(b) 550 W
(c) 275 W
(d) 137.5 W

Try it first, then show the solutionSolution

Resistance from the rating: $R = \frac{V ^{2}}{P} = \frac{220 \times 220}{1100} = 44 Ω$ .
At 110 V: $P = \frac{V ^{2}}{R} = \frac{110 \times 110}{44} = 275 W$ .
Check: voltage halved, so power becomes one quarter: $\frac{1100}{4} = 275 W$ .

Answer: (c) 275 W

Science in real life: Electricity

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Resistors in series

Imagine you buy a string of decorative lights for Diwali. It has 20 identical small bulbs joined in series and plugs into the 220 V supply at home. One evening, one bulb fuses and the whole string goes dark.

Suppose each bulb has a resistance of $22 Ω$ . (a) Find the current in the string. (b) Find the p.d. across each bulb. (c) Explain why the whole string goes off when only one bulb fuses.

Try it first, then show the solutionSolution

Total resistance: $R_{s} = 20 \times 22 = 440 Ω$ .
Current: $I = \frac{220 V}{440 Ω} = 0.5 A$ , the same through every bulb.
p.d. across each bulb: $V = I R = 0.5 \times 22 = 11 V$ (check: $20 \times 11 = 220 V$ ).
In series there is only one path for current. A fused bulb breaks that path, so current stops in the whole string and no bulb glows.

Answer: (a) 0.5 A (b) 11 V each (c) The single path is broken, so no current flows anywhere in the string.

Memory hook: One dead bulb, whole string dark: series has only one road.

Resistors in parallel and their use in daily life

Suppose in your room a 110 W lamp and an 1100 W room heater are both plugged into the 220 V home supply, which is wired in parallel. You switch both on at once on a winter night.

(a) Find the current drawn by each appliance. (b) Find the total current from the supply and the equivalent resistance of the two together. (c) Give two reasons why these appliances would not work properly if connected in series.

Try it first, then show the solutionSolution

Lamp: $I_{1} = \frac{P}{V} = \frac{110}{220} = 0.5 A$ .
Heater: $I_{2} = \frac{1100}{220} = 5 A$ .
Total current: $I = 0.5 + 5 = 5.5 A$ .
Equivalent resistance: $R_{p} = \frac{V}{I} = \frac{220}{5.5} = 40 Ω$ .
Reason 1: in series both would carry the same current, but they need very different currents (0.5 A and 5 A) to work properly.
Reason 2: in series the supply voltage would be shared, so neither gets 220 V; also switching off or failure of one would stop the other.

Answer: (a) Lamp 0.5 A, heater 5 A (b) 5.5 A and $40 Ω$ (c) They need different currents and full voltage each; in series one failing stops both.

Memory hook: Every socket at home gets full 220 V because the house is wired in parallel.

Factors affecting resistance and resistivity

Imagine you use a room heater at home. The heating element glows red, but the cord connecting it to the socket stays cool even though the same current passes through both.

Suppose the nichrome element has a resistance of $44 Ω$ , the copper cord has a resistance of $0.1 Ω$ , and a current of 5 A flows. (a) Calculate the heat produced each second in the element and in the cord. (b) Explain, using resistivity, why the element glows but the cord does not.

Try it first, then show the solutionSolution

Element: $H = I^{2} R t = 25 \times 44 \times 1 = 1100 J$ each second.
Cord: $H = 25 \times 0.1 \times 1 = 2.5 J$ each second.
Same current flows in both because they are in series, so heat depends on resistance.
Nichrome, an alloy, has very high resistivity, so the element has a high resistance and produces a lot of heat; it also does not oxidise easily when hot.
Copper has very low resistivity, so the cord has a tiny resistance and produces very little heat.

Answer: (a) 1100 J per second in the element, 2.5 J per second in the cord (b) High-resistivity nichrome produces far more heat than low-resistivity copper for the same current.

Memory hook: Same current, but the high-resistivity alloy gets hot and the copper cord stays cool.

Heating effect of electric current and its applications

Suppose your family buys an electric iron rated 880 W, 220 V. The electrician asks which fuse to put in its circuit. Domestic fuses come in ratings of 1 A, 2 A, 3 A, 5 A and 10 A.

(a) Find the normal current drawn by the iron. (b) Which fuse should be used and why? (c) Why would a 3 A fuse and a 10 A fuse both be wrong choices?

Try it first, then show the solutionSolution

$I = \frac{P}{V} = \frac{880 W}{220 V} = 4 A$ .
Choose the fuse with rating just above the normal current: 5 A.
The fuse is in series with the iron; if current rises above 5 A, the fuse wire heats up, melts and breaks the circuit.
A 3 A fuse would melt during normal use, since 4 A is more than 3 A.
A 10 A fuse would not melt for a dangerously high current between 5 A and 10 A, so it would not protect the iron.

Answer: (a) 4 A (b) 5 A fuse (c) 3 A melts in normal use; 10 A fails to protect against unduly high current.

Memory hook: Fuse rating: just above the normal current, never far above.

Electric power and electrical energy

Imagine your home has a 2 kW water geyser that is used for 30 minutes every morning. At the end of a month your parents look at the electricity bill, which charges by the 'unit'.

Suppose the rate is Rs 6 per unit. (a) Find the energy used by the geyser in 30 days in kW h. (b) Find the cost. (c) Express the energy in joules.

Try it first, then show the solutionSolution

Energy per day: $E = P \times t = 2 k W \times 0.5 h = 1 k W h$ .
In 30 days: $1 \times 30 = 30 k W h = 30$ units.
Cost: $30 \times 6 =$ Rs 180.
In joules: $30 \times 3.6 \times 1 0^{6} = 1.08 \times 1 0^{8} J$ .

Answer: (a) 30 kW h (b) Rs 180 (c) $1.08 \times 1 0^{8} J$

Memory hook: Kilowatts × hours = units on the bill; 1 unit = 3.6 million joules.

Topics in this chapter

Electricity: formulas and key facts

Electric current: $I = \frac{Q}{t}$; Charge Q in coulomb, t in seconds, I in ampere. Use Q = It to find charge; divide by $1.6 \times 1 0^{- 19} C$ to get the number of electrons.
Unit of current: $1 A = \frac{1 C}{1 s}$; Also $1 m A = 1 0^{- 3} A$ and $1 μ A = 1 0^{- 6} A$ . Convert mA to A before using Ohm's law.
Potential difference: $V = \frac{W}{Q}$; Work W in joule. Use W = VQ to find work or energy given to charge. $1 V = 1 J C^{- 1}$ .
Ohm's law: $V = I R$; Holds for a metallic conductor at constant temperature. Rearrange as $I = \frac{V}{R}$ or $R = \frac{V}{I}$ . $1 Ω = \frac{1 V}{1 A}$ .
Resistance in terms of resistivity: $R = ρ \frac{l}{A}$; l in metre, A in square metre, $ρ$ in $Ω m$ . For a round wire $A = \frac{π d ^{2}}{4}$ . Use for ratio questions on changing length or thickness.
Resistivity: $ρ = \frac{R A}{l}$; Depends only on material and temperature. Low for metals ( $1 0^{- 8}$ to $1 0^{- 6} Ω m$ ), very high for insulators.
Series combination: $R_{s} = R_{1} + R_{2} + R_{3}$; Same current through each resistor; voltages add: $V = V_{1} + V_{2} + V_{3}$ . $R_{s}$ is greater than any single resistor.
Parallel combination: $\frac{1}{R _{p}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \frac{1}{R _{3}}$; Same p.d. across each branch; currents add: $I = I_{1} + I_{2} + I_{3}$ . $R_{p}$ is less than the smallest resistor. Remember to invert at the end.
Joule's law of heating: $H = I^{2} R t$; Also $H = V I t$ . Heat in joule. Heat is proportional to $I^{2}$ , to R and to t.
Electric power: $P = V I = I^{2} R = \frac{V ^{2}}{R}$; Use $I^{2} R$ when current is common (series), $\frac{V ^{2}}{R}$ when voltage is common (parallel, or a rated appliance at a new voltage).
Unit of power: $1 W = 1 V \times 1 A$; 1 kW = 1000 W. A rating such as 220 V, 100 W gives the power at that voltage only.
Electrical energy: $E = P \times t$; In joule if P is in W and t in s; in kW h if P is in kW and t in hours. Bill cost = units (kW h) × rate per unit.
Commercial unit of energy: $1 k W h = 3.6 \times 1 0^{6} J$; 1 kW h is called 1 'unit'. Multiply kW h by $3.6 \times 1 0^{6}$ to get joules.
Measuring instruments: Ammeter in $ser i es$ ; voltmeter in $p a r a l l e l$; Ammeter measures current through a part; voltmeter measures p.d. across two points. Apply this every time you draw a circuit with meters.
Material choices: Alloys (e.g. nichrome) for heating elements; tungsten for filaments; copper and aluminium for transmission lines; Alloys: high resistivity and do not oxidise easily when hot. Tungsten: very high melting point (3380°C). Copper and aluminium: low resistivity.

Electricity: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

Circuit to verify Ohm's law (study of V and I for a resistor)

"Draw a circuit diagram to study the relation between potential difference and current" or practical-based questions on the V-I experiment.

Draw

Draw a battery of cells (long line positive, short line negative) on the left.
From the positive terminal, draw a wire to a plug key, then to an ammeter (circle with A), then to a rheostat (resistor symbol with an arrow across it).
Draw the resistor or nichrome wire XY next in the same loop and close the loop back to the negative terminal: everything so far is in series.
Draw a voltmeter (circle with V) connected directly across the two ends X and Y of the resistor, in parallel with it.
Mark the direction of current with an arrow from the positive terminal through the external circuit to the negative terminal.

Label

Battery (+ and − terminals)
Plug key (K)
Ammeter (A)
Rheostat (Rh)
Resistor / nichrome wire XY
Voltmeter (V)
Direction of current (arrow)

Watch out: Do not connect the voltmeter in series or the ammeter in parallel. Ammeter in series, voltmeter across the resistor only. Mark + and − on the meters: the + terminal goes towards the positive side of the battery.

V-I graph for a metallic conductor (Ohm's law)

"Plot a graph between V and I" questions with a data table, or "what is the nature of the V-I graph".

Draw

Draw two perpendicular axes meeting at the origin O.
Put current I (in A) on the x-axis and potential difference V (in V) on the y-axis, with an even scale on each.
Plot the given points and draw one straight line through the origin that passes through or close to them.
To find R, take two points on the line far apart and work out slope = change in V ÷ change in I.

Label

x-axis: Current I (A)
y-axis: Potential difference V (V)
Origin O
Straight line through origin
Scale used on each axis

Watch out: Do not join the points zig-zag. Draw one best straight line through the origin. If V is on the y-axis, slope = R; if you swap the axes, slope = 1/R, so say clearly which you used.

Resistors in series with ammeter and voltmeter

"Draw a schematic diagram of resistors in series" or questions that ask you to redraw a series circuit with meters added.

Draw

Draw a battery and a plug key.
Draw the resistors R1, R2, R3 joined end to end in one line, so there is only one path for current.
Put an ammeter in series anywhere in the loop.
Draw a voltmeter across the whole combination (points X and Y), or across one resistor if the question asks for the potential difference across that resistor.
Show the current direction from + terminal through the resistors to − terminal.

Label

Battery
Plug key (K)
R1, R2, R3 with their values
Ammeter (A)
Voltmeter (V)
Points X and Y (ends of combination)
Current direction arrow

Watch out: Write the resistance value beside each resistor. Do not place the voltmeter across the battery and the resistor together; connect it only across the points named in the question.

Resistors in parallel with ammeter and voltmeter

"Draw a circuit showing resistors in parallel" or practical questions on finding equivalent resistance in parallel.

Draw

Draw a battery, a plug key and an ammeter in series in the main line.
At point X, split the wire into three branches; place R1, R2 and R3 one in each branch, and join the branches again at point Y.
Connect a voltmeter across X and Y, in parallel with the combination.
To measure a branch current, show an ammeter in series with that resistor inside its own branch.
Mark the main current dividing into I1, I2, I3 at X and joining again at Y.

Label

Battery
Plug key (K)
Ammeter (A) in main line
R1, R2, R3
Junction points X and Y (wire joints shown as dots)
Voltmeter (V) across XY
Currents I, I1, I2, I3

Watch out: Show junctions with a dot (wire joint symbol). Without dots, crossing wires mean they do not join. Do not put the main ammeter inside one branch.

Symbols of common circuit components

"Draw the symbols of" questions (1-2 marks) and in every circuit diagram.

Draw

Cell: one long thin line (positive) and one short thick line (negative), parallel to each other.
Battery: two or more cells joined in a row.
Plug key open: a small key gap shown open; closed: shown plugged in or joined.
Wire joint: two wires meeting with a dot. Wires crossing without joining: one crosses the other with no dot (or a small hump).
Bulb: a circle with a cross or a loop inside; resistor: a zig-zag line or a small rectangle; rheostat: resistor with an arrow through it.
Ammeter: circle with A; voltmeter: circle with V, each with + and − marked.

Label

Cell (+, −)
Battery
Plug key (open / closed)
Wire joint
Wires crossing without joining
Electric bulb
Resistor R
Rheostat / variable resistance
Ammeter
Voltmeter

Watch out: Keep the long line of a cell as positive. Mixing up a wire joint and a crossing without joining changes the whole circuit.

How the board asks Electricity

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Ohm's law and basic numericals	One- to three-mark questions: find current, voltage, resistance or charge; often with mA or minutes to convert, or 'what happens to the current if V is halved'.	List the given values with units, convert to SI (mA to A, minutes to s), write the formula, substitute with units and box the answer with its unit.
Resistivity and ratio questions	A wire is cut, joined, made longer or thicker, or a new wire of the same material is compared; MCQs on how R or $ρ$ changes; finding length of wire for a given resistance.	Write $R = ρ \frac{l}{A}$ for both wires and divide one by the other so $ρ$ cancels. Remember area goes as diameter squared, and resistivity does not change unless the material changes.
Series, parallel and mixed circuits	Find total resistance, current from the battery, current in a branch or p.d. across a part; arrange given resistors to get a stated value; the highest and lowest possible resistance.	Reduce the circuit step by step: combine parallel groups first, then add series parts. Find the battery current, then work back to each part using $V = I R$ . Check that branch currents add up to the total.
Heating effect and power numericals	Heat produced in a resistor, power of an appliance from its rating, power at a lower voltage, comparing heat in series and parallel, choosing a fuse.	Pick the right form: $H = I^{2} R t$ or $V I t$ ; $P = V I$ , $I^{2} R$ or $\frac{V ^{2}}{R}$ . For a rated device, first find R from the rating, then use it at the new voltage.
Energy and kW h cost	Energy used by appliances over days and the cost at a rate per unit; converting kW h into joules; which appliance uses more energy.	Convert power to kW and time to hours, multiply to get kW h, then multiply by the rate. Use $1 k W h = 3.6 \times 1 0^{6} J$ for conversions.
Reasoning and 'give reasons' questions	Why alloys in heaters, why tungsten in bulbs, why homes use parallel wiring, why a heater cord does not glow, why copper is used for wires, how a fuse works.	Give the property and then link it to the use in one more sentence. For example: high resistivity and no oxidation at high temperature, so the element gets hot without burning.
V-I graph and practical-based questions	Plot V against I from a table and find R from the slope; spot a faulty reading; state how the ammeter and voltmeter are connected; precautions; equivalent resistance in the series and parallel practical.	Put V on the y-axis and I on the x-axis, draw a straight line through the origin, take slope as $\frac{V}{I}$ . Mention precautions: tight connections, take the key out after each reading so the wire does not heat up.
Assertion-reason and case-based questions	A statement about a fact (alloys, parallel wiring, resistivity) with a reason to judge; a passage about appliances at home with sub-questions on power, current and heating.	For A-R, judge A and R separately as true or false, then ask whether R actually explains A. For case-based, pull the numbers out of the passage, then use the same formulas as in numericals.

Tricks to remember Electricity

Short memory aids. Say each one aloud once, then use it on a question the same day.

The VIR triangle
Write V on top and I and R side by side below. Cover the one you want: covering V leaves I × R; covering I leaves V over R; covering R leaves V over I.
Use it when: Rearrange Ohm's law quickly without algebra slips.
Series shares Voltage, Parallel shares current
Series: Same current, voltages add. Parallel: same Potential, currents add. Think 'Series-Same-I, Parallel-Same-V'.
Use it when: Decide straight away which quantity is common before solving any circuit.
Answer sense-check for combinations
Series total must be bigger than the biggest resistor. Parallel total must be smaller than the smallest resistor. For two in parallel use product over sum; for n equal resistors R in parallel, the answer is R/n.
Use it when: Catch the common error of forgetting to flip $\frac{1}{R _{p}}$ and get two-resistor answers in seconds.
Long and thin resists
'Long, thin, alloy = high resistance; short, thick, copper = low.' Length on top, area at the bottom of $ρ \frac{l}{A}$ . Double diameter means four times area.
Use it when: Ratio questions on wires and reasons for choosing materials.
Which power formula?
Same current (series)? Use $I^{2} R$ . Same voltage (parallel or home supply)? Use $\frac{V ^{2}}{R}$ . Both known? Use $V I$ .
Use it when: Comparing heat or brightness in series and parallel, and rated appliances at a new voltage.
1 unit = 36 lakh joules
$1 k W h = 1000 \times 3600 = 3.6 \times 1 0^{6} J$ , which is 36 lakh joules.
Use it when: Converting between kW h and joules in bill and energy questions.

Electricity: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Equivalent resistance in parallel · 1 mark · MCQBoard level

A 6 Ω resistor and a 3 Ω resistor are connected in parallel. Their equivalent resistance is:
(a) 9 Ω
(b) 4.5 Ω
(c) 2 Ω
(d) 0.5 Ω

Try it first, then show the solutionSolution

$\frac{1}{R _{p}} = \frac{1}{6} + \frac{1}{3} = \frac{1 + 2}{6} = \frac{3}{6}$ , so $R_{p} = 2 Ω$ (also, the answer must be less than 3 Ω, which rules out (a) and (b))

Answer: (c) 2 Ω

Example 2: Current and resistance in series · 1 mark · Assertion-reasonBoard level

Assertion (A): In a series combination of resistors, the same current flows through each resistor.
Reason (R): The equivalent resistance of resistors in series is less than the smallest individual resistance.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Try it first, then show the solutionSolution

A is true: a series circuit has only one path, so the current is the same in every part. R is false: in series $R_{s} = R_{1} + R_{2} + R_{3}$ , which is greater than any individual resistance; it is the parallel combination whose resistance is less than the smallest.

Answer: (c) A is true but R is false.

Example 3: Give reasons: wiring and materials · 3 marks · Short answerHarder board level

Give reasons:
(a) Series arrangement is not used for domestic circuits.
(b) Copper and aluminium wires are used for electricity transmission.
(c) The heating elements of electric irons and toasters are made of an alloy rather than a pure metal.

Try it first, then show the solutionSolution

(a) In series the same current flows through every appliance, but a bulb and a heater need very different currents. Also, if one appliance fails, the circuit breaks and all others stop working.
(b) Copper and aluminium have very low resistivity (about $1.62 \times 1 0^{- 8}$ and $2.63 \times 1 0^{- 8} Ω m$ ), so the wires have low resistance and little energy is wasted as heat ( $H = I^{2} R t$ ).
(c) An alloy has a higher resistivity than its constituent metals, so it produces more heat, and it does not oxidise (burn) readily at high temperatures.

Answer: (a) Different appliances need different currents, and one failure switches everything off. (b) Low resistivity means less heating loss. (c) Alloys have high resistivity and do not oxidise easily when hot.

Example 4: Ohm's law practical: meters and readings · 3 marks · Short answerHarder board level

In the experiment to study how the potential difference V across a resistor depends on the current I through it:
(a) How are the ammeter and voltmeter connected?
(b) Why should the plug key be taken out after each reading?
(c) A student gets V = 1.2 V when I = 0.3 A, and the V-I graph is a straight line through the origin. Find the resistance and the current when V = 2.0 V.

Try it first, then show the solutionSolution

(a) Ammeter in series with the resistor; voltmeter in parallel across the ends of the resistor.
(b) If current keeps flowing, the heating effect raises the temperature of the wire and its resistance changes. Ohm's law holds only when temperature stays the same, so the key is removed between readings.
(c) $R = \frac{V}{I} = \frac{1.2}{0.3} = 4 Ω$ ; at 2.0 V, $I = \frac{V}{R} = \frac{2.0}{4} = 0.5 A$

Answer: (a) Ammeter in series, voltmeter in parallel. (b) To prevent heating, which would change the resistance. (c) R = 4 Ω; I = 0.5 A

Example 5: Mixed circuit: current, voltage and power · 5 marks · Long answerHarder board level

A 3 Ω resistor is connected in series with a parallel combination of a 6 Ω and a 3 Ω resistor. The whole arrangement is connected to a 15 V battery through a plug key. Find:
(a) the equivalent resistance of the circuit,
(b) the current drawn from the battery,
(c) the potential difference across the parallel part and the current through each of its resistors,
(d) the power supplied by the battery and the power used in the series 3 Ω resistor.

Try it first, then show the solutionSolution

(a) Parallel part: $\frac{1}{R _{p}} = \frac{1}{6} + \frac{1}{3} = \frac{1}{2}$ , so $R_{p} = 2 Ω$ ; total $R = 3 + 2 = 5 Ω$
(b) $I = \frac{V}{R} = \frac{15 V}{5 Ω} = 3 A$
(c) Voltage across series 3 Ω: $V_{1} = I R = 3 \times 3 = 9 V$ ; so across the parallel part $V_{p} = 15 - 9 = 6 V$ (or $3 A \times 2 Ω$ )
Currents: 6 Ω branch $\frac{6}{6} = 1 A$ ; 3 Ω branch $\frac{6}{3} = 2 A$ ; they add to 3 A
(d) Battery power $P = V I = 15 \times 3 = 45 W$ ; power in series 3 Ω: $P = I^{2} R = 3^{2} \times 3 = 27 W$

Answer: (a) 5 Ω; (b) 3 A; (c) 6 V; 1 A in 6 Ω, 2 A in 3 Ω; (d) 45 W from the battery, 27 W in the series 3 Ω resistor

Example 6: Joule's law, fuse and energy bill · 5 marks · Long answerHarder board level

(a) State Joule's law of heating and write the three ways in which the heat produced depends on the quantities in it.
(b) An electric iron is rated 440 W, 220 V. Find the current it draws and its resistance.
(c) Find the heat it produces in 1 minute. Which fuse should be used in its circuit from the ratings 1 A, 2 A, 3 A, 5 A, 10 A?
(d) If it is used for 1 hour every day for 30 days, find the energy used in kW h and its cost at Rs 5 per kW h.

Try it first, then show the solutionSolution

(a) Heat produced in a resistor is $H = I^{2} R t$ (Joule's law of heating).
H is directly proportional to the square of the current (for given R), to the resistance (for given I), and to the time for which current flows.
(b) $I = \frac{P}{V} = \frac{440}{220} = 2 A$ ; $R = \frac{V}{I} = \frac{220}{2} = 110 Ω$
(c) $H = P t = 440 W \times 60 s = 26400 J$ ; a 3 A fuse: the nearest rating above the working current of 2 A, so normal current does not melt it but a larger current does
(d) Energy $= 0.44 k W \times 1 h \times 30 = 13.2 k W h$ ; cost $= 13.2 \times 5 = R s 66$

Answer: (b) 2 A, 110 Ω; (c) 26400 J, 3 A fuse; (d) 13.2 kW h, Rs 66

Example 7: Finding two resistances from series and parallel readings · 3 marks · Short answerHOTS / competency

In the practical on equivalent resistance, a student finds that two resistors give an equivalent resistance of 9 Ω in series and 2 Ω in parallel. Find the value of each resistor and check your answer.

Try it first, then show the solutionSolution

Series: $R_{1} + R_{2} = 9$
Parallel: $R_{p} = \frac{R _{1} R _{2}}{R _{1} + R _{2}} = 2$ , so $R_{1} R_{2} = 2 \times 9 = 18$
Two numbers with sum 9 and product 18 are 3 and 6, so $R_{1} = 3 Ω$ , $R_{2} = 6 Ω$ ; check: $3 + 6 = 9$ and $\frac{1}{3} + \frac{1}{6} = \frac{1}{2}$ gives 2 Ω

Answer: 3 Ω and 6 Ω

Example 8: Which bulb glows brighter in series? · 3 marks · Short answerHOTS / competency

Two bulbs rated 100 W, 220 V and 25 W, 220 V are connected in series across a 220 V supply. Assume their resistances stay the same. (a) Find the resistance of each bulb. (b) Find the current in the circuit. (c) Find the power used by each bulb and say which glows brighter. Is this the same bulb that glows brighter when they are connected in parallel?

Try it first, then show the solutionSolution

(a) $R = \frac{V ^{2}}{P}$ : 100 W bulb $= \frac{48400}{100} = 484 Ω$ ; 25 W bulb $= \frac{48400}{25} = 1936 Ω$
(b) Series: $R_{s} = 484 + 1936 = 2420 Ω$ ; $I = \frac{220}{2420} = \frac{1}{11} A$
(c) $P = I^{2} R$ : 25 W bulb $= \frac{1936}{121} = 16 W$ ; 100 W bulb $= \frac{484}{121} = 4 W$ . In series the 25 W bulb glows brighter (same current, larger R). In parallel each gets 220 V, so the 100 W bulb glows brighter.

Answer: (a) 484 Ω and 1936 Ω; (b) 1/11 A (about 0.09 A); (c) 25 W bulb uses 16 W, 100 W bulb uses 4 W, so the 25 W bulb is brighter in series; in parallel the 100 W bulb is brighter.

Case-study practice (Section E style)

Read the situation, then answer every part. In the board paper this is a 4-mark question split into 1 + 1 + 2 marks.

The fuse that melted

Suppose Riya's family plugs three appliances into the same 220 V household circuit, which is wired in parallel and protected by a 10 A fuse. The ratings marked on them are:

electric kettle: 1100 W, 220 V
lamp: 44 W, 220 V
room heater: 1100 W, 220 V

One cold morning the kettle and the lamp are on. Riya's brother then switches on the heater as well, and a moment later every appliance on the circuit goes off. When the electrician opens the fuse holder, he finds that the fuse wire has melted.

(i) · 1 mark
What current does the kettle draw when it works normally?
Answer
Answer: 5 A
$I = \frac{P}{V} = \frac{1100 W}{220 V} = 5 A$
(ii) · 1 mark
Give one reason why household appliances are connected in parallel and not in series.
Answer
Answer: In parallel each appliance gets the full 220 V and draws the current it needs, and each can be switched on or off without affecting the others.
In series the same current would pass through all, though a lamp and a heater need very different currents, and one failure would stop all of them.
(iii) · 2 marks
Using calculations, explain why the fuse melted when the heater was switched on.
Answer
Answer: Total current became 10.2 A, more than the 10 A rating, so the fuse wire heated, melted and broke the circuit.
Lamp current $= \frac{44}{220} = 0.2 A$ ; heater current $= 5 A$ . Total $= 5 + 0.2 + 5 = 10.2 A$ , which is more than 10 A. The larger current raises the temperature of the fuse wire (Joule heating), it melts and breaks the circuit, protecting the wiring and appliances.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Two resistors of 10 Ω and 15 Ω are connected in parallel. Find their equivalent resistance. How does it compare with the smaller of the two resistances? [2 marks] (2 marks)

Step	Marks
Writes $\frac{1}{R _{p}} = \frac{1}{R _{1}} + \frac{1}{R _{2}}$ and substitutes: $\frac{1}{R _{p}} = \frac{1}{10} + \frac{1}{15} = \frac{3 + 2}{30} = \frac{5}{30}$	1
$R_{p} = 6 Ω$ , which is less than the smaller resistance (10 Ω), as expected for a parallel combination	1

(a) What is meant by the heating effect of electric current? (b) Name two devices that work on this effect. (c) Why is tungsten used for the filaments of electric bulbs? [3 marks] (3 marks)

Step	Marks
(a) When current flows through a purely resistive circuit, the energy from the source is entirely given out as heat. This is the heating effect of electric current.	1
(b) Any two of: electric iron, electric toaster, electric oven, electric kettle, electric heater (½ mark each)	1
(c) Tungsten is a strong metal with a very high melting point (3380°C), so the filament can get hot enough to glow without melting	1

(a) Three resistors $R_{1}$ , $R_{2}$ and $R_{3}$ are connected in parallel to a battery of potential difference V. Using Ohm's law, obtain an expression for their equivalent resistance. (b) Resistors of 10 Ω, 20 Ω and 20 Ω are connected in parallel across a 10 V battery. Find the equivalent resistance, the total current and the current in each resistor. [5 marks] (5 marks)

Step	Marks
(a) The potential difference across each resistor is the same, V. The total current divides: $I = I_{1} + I_{2} + I_{3}$	1
Applies Ohm's law to each resistor: $I_{1} = \frac{V}{R _{1}}$ , $I_{2} = \frac{V}{R _{2}}$ , $I_{3} = \frac{V}{R _{3}}$ , and to the combination: $I = \frac{V}{R _{p}}$	1
Substitutes and divides by V: $\frac{V}{R _{p}} = \frac{V}{R _{1}} + \frac{V}{R _{2}} + \frac{V}{R _{3}}$ , so $\frac{1}{R _{p}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \frac{1}{R _{3}}$	1
(b) $\frac{1}{R _{p}} = \frac{1}{10} + \frac{1}{20} + \frac{1}{20} = \frac{4}{20}$ , so $R_{p} = 5 Ω$ ; total current $I = \frac{10 V}{5 Ω} = 2 A$	1
Currents: 10 Ω: $\frac{10}{10} = 1 A$ ; each 20 Ω: $\frac{10}{20} = 0.5 A$ ; check: $1 + 0.5 + 0.5 = 2 A$	1

Where students lose marks in Electricity

The slips examiners see most often, and what to write instead.

Mistake: Not converting minutes, hours or milliampere to SI units
Students substitute numbers straight from the question in a hurry.
Wrong working
Current 0.5 A for 10 min: $Q = 0.5 \times 10 = 5 C$
Right working
$t = 10 \times 60 = 600 s$ , so $Q = 0.5 \times 600 = 300 C$
Do this: Before any formula, write the given values and convert each to SI: s, A, m. Then substitute.
Usually costs: Usually ½ to 1 mark, and the final answer mark.
Mistake: Forgetting to invert in a parallel combination
After adding the reciprocals, students stop and write that value as the resistance.
Wrong working
$\frac{1}{R _{p}} = \frac{1}{3} + \frac{1}{6} = \frac{1}{2}$ , so $R_{p} = 0.5 Ω$
Right working
$\frac{1}{R _{p}} = \frac{1}{2}$ , so $R_{p} = 2 Ω$ (smaller than 3 Ω, as it should be)
Do this: Always write the last line ' $R_{p} =$ ...' after flipping, and check it is less than the smallest resistor.
Usually costs: 1 mark, and every later step that uses $R_p$.
Mistake: Using diameter in place of radius, or not squaring it, in area
Students write $A = π d$ or $A = π d^{2}$ without dividing by 4.
Wrong working
Wire of diameter 0.4 mm: $A = π \times (0.4 \times 1 0^{- 3})^{2}$
Right working
$A = \frac{π d ^{2}}{4} = \frac{π \times ( 0.4 \times 1 0 ^{- 3} ) ^{2}}{4}$ , or use radius $r = 0.2 \times 1 0^{- 3} m$ in $π r^{2}$
Do this: Write $A = π r^{2}$ first, find r from d, convert mm to m, then substitute.
Usually costs: 1 to 2 marks in a resistivity numerical.
Mistake: Thinking heat doubles when current doubles
Students treat Joule's law as a straight proportion in I and forget the square.
Wrong working
Current doubled, so heat produced becomes 2 times.
Right working
$H = I^{2} R t$ ; with I doubled, $H \propto (2 I)^{2} = 4 I^{2}$ , so heat becomes 4 times.
Do this: Underline the square in $I^{2} R t$ and write the ratio step clearly in every 'how does it change' question.
Usually costs: The full 1 mark in an MCQ or A-R question.
Mistake: Using the rated power directly at a different voltage
Students assume a 100 W bulb always uses 100 W.
Wrong working
Bulb rated 220 V, 100 W used on 110 V: power = 100 W, or half of it, 50 W.
Right working
$R = \frac{22 0 ^{2}}{100} = 484 Ω$ ; at 110 V, $P = \frac{11 0 ^{2}}{484} = 25 W$
Do this: From a rating, find R first (it stays fixed), then use $P = \frac{V ^{2}}{R}$ with the new voltage.
Usually costs: 1 mark in an MCQ; 1 to 2 marks in a short answer.
Mistake: Connecting the ammeter in parallel or the voltmeter in series in a circuit diagram
Students mix up the two meters while drawing in a hurry.
Wrong working
Circuit drawn with the voltmeter in the main line and the ammeter across the resistor.
Right working
Ammeter in series in the main line (current passes through it); voltmeter connected across the two ends of the resistor.
Do this: Say it aloud: 'Ammeter: through it, in series. Voltmeter: across it, in parallel.' Also mark the + terminal of each meter towards the + of the battery.
Usually costs: 1 mark or more in a diagram question.
Mistake: Leaving power in watts and time in seconds when asked for kW h
Students mix the SI unit of energy with the commercial unit.
Wrong working
400 W for 8 h: $E = 400 \times 8 = 3200$ units
Right working
$E = 0.4 k W \times 8 h = 3.2 k W h$ = 3.2 units
Do this: For bills, convert W to kW (divide by 1000) and keep time in hours.
Usually costs: 1 mark and the final answer in a cost question.

NCERT exercise guide: Electricity

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

In-text questions after sections 11.1 and 11.2

What it tests: Meaning of an electric circuit, the unit of current (ampere), the link between charge and number of electrons, the meaning of 1 volt, the device that maintains potential difference, and energy given per coulomb.

How to approach it: Learn the two basic relations, $I = \frac{Q}{t}$ and $V = \frac{W}{Q}$ , and the unit definitions that come from them. For electron counting, divide the total charge by the charge on one electron. A 'V volt' battery gives V joules to each coulomb, so read voltage as 'joules per coulomb'.

Mirrors the charge-electron and energy-per-coulomb questions

(a) How many electrons make up a charge of 2 C? (b) A 9 V battery moves this 2 C of charge through a circuit. How much energy does the battery give to the charge? (Charge on an electron = $1.6 \times 1 0^{- 19} C$ .)

Show the solutionSolution

(a) Number of electrons $n = \frac{Q}{e} = \frac{2}{1.6 \times 1 0 ^{- 19}}$
$n = 1.25 \times 1 0^{19}$ electrons
(b) A 9 V battery gives 9 J to each coulomb, so $W = V Q = 9 V \times 2 C$
$W = 18 J$

Answer: (a) $1.25 \times 1 0^{19}$ electrons; (b) 18 J

In-text questions after section 11.5

What it tests: Factors on which resistance depends (length, area of cross-section, material), thick versus thin wires, the effect of halving voltage on current, why heating coils use alloys, and reading the resistivity table.

How to approach it: Always start from $R = ρ \frac{l}{A}$ . For ratio questions, write the formula for both wires and divide, so $ρ$ cancels when the material is the same. For 'which is a better conductor', the material with the lower resistivity wins. Give a reason in every 'why' answer, not just the fact.

Mirrors the resistance-ratio questions using R = ρl/A

A wire of length l and area of cross-section A has a resistance of 8 Ω. Find the resistance of a wire of the same material with length 3l and area of cross-section 2A.

Show the solutionSolution

For the first wire: $R_{1} = ρ \frac{l}{A} = 8 Ω$
For the second wire: $R_{2} = ρ \frac{3 l}{2 A} = \frac{3}{2} \times ρ \frac{l}{A}$
$R_{2} = \frac{3}{2} \times 8 = 12 Ω$

Answer: 12 Ω

In-text questions after sections 11.6.1 and 11.6.2

What it tests: Drawing series circuits with ammeter and voltmeter, readings of meters, judging parallel resistance, replacing parallel appliances by one equivalent, advantages of parallel connection, and combining given resistors to get a required value.

How to approach it: In series add resistances; current is the same everywhere. In parallel add reciprocals; voltage is the same across each branch, and the answer is always smaller than the smallest resistor. For 'how to connect to get X ohm', try series and parallel mixes and check by calculation. Always state the arrangement in words as well as the value.

Mirrors the 'connect given resistors to get a stated total' question

You have three resistors of 4 Ω each. Show how to connect them to get (a) 6 Ω, (b) $\frac{4}{3}$ Ω.

Show the solutionSolution

(a) Join two 4 Ω resistors in parallel: $\frac{1}{R} = \frac{1}{4} + \frac{1}{4} = \frac{2}{4}$ , so $R = 2 Ω$
Connect this pair in series with the third 4 Ω: $R = 2 + 4 = 6 Ω$
(b) Join all three in parallel: $\frac{1}{R _{p}} = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \frac{3}{4}$
$R_{p} = \frac{4}{3} Ω$

Answer: (a) Two in parallel, then in series with the third: 6 Ω. (b) All three in parallel: 4/3 Ω

In-text questions after sections 11.7 and 11.8

What it tests: Why a heater's element glows but its cord does not, heat produced from charge and voltage, heat from $H = I^{2} R t$ , what decides the rate of energy delivery, and power and energy of a device.

How to approach it: Pick the heat formula that matches the data: $H = V Q$ if charge is given, $H = V I t$ or $H = I^{2} R t$ otherwise. Convert time to seconds for joules. For energy in 'units', use power in kW and time in hours. Power is the rate of energy use, $P = V I$ .

Mirrors the 'heat from charge transferred through a potential difference' question

A charge of 7200 C flows through a resistor in 1 hour under a potential difference of 20 V. Find (a) the heat produced and (b) the power.

Show the solutionSolution

(a) Heat $H = V Q = 20 V \times 7200 C$
$H = 144000 J = 1.44 \times 1 0^{5} J$
(b) Time $t = 1 h = 3600 s$ ; $P = \frac{H}{t} = \frac{144000}{3600}$
$P = 40 W$

Answer: (a) $1.44 \times 1 0^{5}$ J; (b) 40 W

End-of-chapter exercises

What it tests: MCQs on cut-and-joined wires, power expressions, power at lower voltage and heat in series versus parallel; voltmeter connection; resistivity numericals; V-I graph from data; series and parallel numericals; number of lamps or resistors on a line; power comparison; energy comparison; and 'explain why' questions on tungsten, alloys, domestic wiring and transmission wires.

How to approach it: Write the formula first, then substitute with units, then box the answer with its unit: this is how marks are given. For 'how many appliances' questions, find the allowed total current or power and divide. For rated appliances, first find resistance using $R = \frac{V ^{2}}{P}$ , which stays fixed. For 'explain' parts, link each fact to a property (melting point, resistivity, oxidation).

Mirrors the 'how many lamps on a line with a maximum allowed current' question

Bulbs rated 40 W, 220 V are to be connected in parallel across a 220 V line. The maximum allowed current is 2 A. How many bulbs can be connected?

Show the solutionSolution

Maximum power the line can supply: $P = V I = 220 V \times 2 A = 440 W$
Each bulb takes 40 W at 220 V (parallel, so each gets 220 V)
Number of bulbs $= \frac{440}{40} = 11$

Answer: 11 bulbs

Activities 11.1 to 11.6

What it tests: Activity 11.1: V and I for a nichrome wire and the V-I graph. Activity 11.2: different currents through different components. Activity 11.3: effect of length, thickness and material on current. Activity 11.4: same current everywhere in series. Activity 11.5: potential differences add in series. Activity 11.6: currents add in parallel, same voltage across each branch.

How to approach it: For each activity, know what is set up, what is measured, what is seen and what it proves. Remember: ammeter in series, voltmeter in parallel, and take the key out after each reading so the wire does not heat up and change its resistance. Practical questions often give readings and ask for the conclusion or a missing value.

Mirrors Activity 11.6 (resistors in parallel)

In an activity like Activity 11.6, a 6 Ω and a 3 Ω resistor are joined in parallel across a 6 V battery. What will an ammeter read (a) in series with the 6 Ω resistor, (b) in series with the 3 Ω resistor, (c) in the main line? (d) Use the readings to find the equivalent resistance.

Show the solutionSolution

Voltage across each branch is 6 V
(a) $I_{1} = \frac{6}{6} = 1 A$
(b) $I_{2} = \frac{6}{3} = 2 A$
(c) $I = I_{1} + I_{2} = 3 A$
(d) $R_{p} = \frac{V}{I} = \frac{6}{3} = 2 Ω$ ; check: $\frac{1}{6} + \frac{1}{3} = \frac{1}{2}$

Answer: (a) 1 A; (b) 2 A; (c) 3 A; (d) 2 Ω

Electricity: previous year questions

Asked in 21 of the 21 board papers we hold (2023 to 2026), on average 8.1 marks per paper; 78 questions in all, counting OR alternatives. In 4 of 4 sample papers.

Counted from the real papers on our previous year papers page, not estimated. Average marks are over every board paper we hold, with an OR pair counted once. Chapter tags are ours.

Solved board questions from Electricity

Real questions from recent papers. Try each one first, then open our step-by-step solution with marks per step.

2026, set 31/1/1, Q34(a) · 2 marks · Very short answer

The resistance of a wire of 0.01 cm radius and 1.0 cm length is 7 $Ω$ . Calculate its resistivity.

Try it first, then show our solutionSolution

Given: $r = 0.01$ cm $= 1 \times 1 0^{- 4}$ m, $l = 1$ cm $= 0.01$ m, $R = 7 Ω$ . Formula: $R = ρ \frac{l}{A}$ , so $ρ = \frac{R A}{l} = \frac{R π r ^{2}}{l}$
½ mark
Substituting: $ρ = \frac{7 \times \frac{22}{7} \times ( 1 0 ^{- 4} ) ^{2}}{0.01} = \frac{22 \times 1 0 ^{- 8}}{1 0 ^{- 2}}$
1 mark
$ρ = 22 \times 1 0^{- 6} Ω m = 2.2 \times 1 0^{- 5} Ω m$
½ mark

Answer: ρ = 2.2 × 10⁻⁵ Ω m

Where the marks are

Writes R = ρl/A with values in SI units (½ mark)
Correct substitution with A = πr² (1 mark)
Result 2.2 × 10⁻⁵ Ω m with unit (½ mark)

See it in the full 2026 board paper, set 31/1/1

2026, set 31/1/1, Q39(a) · 5 marks · Long answer

Answer the following.

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

Try it first, then show our solutionSolution

(i) Resistance is $R = ρ \frac{l}{A}$ . It is directly proportional to the length $l$ and inversely proportional to the area of cross-section $A$ . So changing $l$ or $A$ changes $R$ .
1 mark
(i) Resistivity $ρ$ is a characteristic property of the material (at a given temperature). It does not depend on the dimensions of the conductor, so it does not change.
1 mark
(ii) An alloy has a higher resistivity than its constituent pure metals, so it produces more heat for the same current.
1 mark
(ii) Alloys also do not oxidise (burn) readily at high temperatures. This makes them ideal for heating elements such as those in toasters and irons.
1 mark
(iii) The SI unit of electric current is the ampere (A). One ampere is the current when 1 coulomb of charge flows through a cross-section of a conductor in 1 second: $1 A = \frac{1 C}{1 s}$ .
1 mark

Answer: (i) R = ρl/A depends on l and A, while ρ is a property of the material only; (ii) alloys have high resistivity and do not oxidise at high temperature; (iii) 1 A = flow of 1 C of charge per second

Where the marks are

(i) R depends on length and area via $R = ρl / A$ (1 mark)
(i) Resistivity is a characteristic of the material, so unchanged (1 mark)
(ii) Higher resistivity of alloys / do not oxidise readily at high temperature (2 marks)
(iii) 1 ampere = 1 coulomb per second (1 mark)

See it in the full 2026 board paper, set 31/1/1

2026, set 31/2/1, Q37 · 3 marks · Short answer

The values of current $I$ flowing in a given resistor for the corresponding values of potential difference $V$ applied across the ends of resistor are given below in the table :

$I$ (Amperes)	0·4	0·8	1·2	1·6	2·0
$V$ (Volts)	1·2	2·4	3·6	4·8	6·0

Plot a graph between $V$ and $I$ and calculate the resistance of that resistor.

Try it first, then show our solutionSolution

Graph: Take current $I$ (A) on the x-axis and potential difference $V$ (V) on the y-axis with a suitable scale (e.g. 1 cm = 0.2 A, 1 cm = 0.6 V). Plot the points (0.4, 1.2), (0.8, 2.4), (1.2, 3.6), (1.6, 4.8), (2.0, 6.0). Joining them gives a straight line passing through the origin, which shows $V \propto I$ (Ohm's law).
2 marks
Resistance = slope of the $V$ – $I$ graph. Take points A(0.4 A, 1.2 V) and B(2.0 A, 6.0 V) on the line, with C at (2.0 A, 1.2 V):
$R = \frac{B C}{A C} = \frac{6.0 - 1.2}{2.0 - 0.4}$
½ mark
$R = \frac{4.8}{1.6} = 3 Ω$
½ mark

Answer: Straight-line $V$ – $I$ graph through the origin; $R = 3 Ω$

Where the marks are

Correctly plotted straight-line $V$ – $I$ graph with labelled axes (2 marks)
Uses slope = $Δ V /Δ I$ with substitution (½ mark)
$R = 3 Ω$ (½ mark)

See it in the full 2026 board paper, set 31/2/1

Every Electricity question, by year

For 2027

Sample paper 2026-27: Q31 (1 mark, solved), Q34(A) (2 marks, solved), Q34(B) (2 marks, solved), Q37 (3 marks, solved)

2026

Set 31/1/1: Q34(a) (2 marks, solved), Q34(b) (2 marks, solved), Q39(a) (5 marks, solved), Q39(b) (5 marks, solved)

Set 31/2/1: Q37 (3 marks, solved), Q38 (4 marks, solved)

Set 31/3/1: Q22 (1 mark, solved), Q30 (1 mark, solved), Q36 (3 marks, solved), Q37 (3 marks, solved)

Set 31/4/1: Q33(A) (2 marks, solved), Q34 (2 marks, solved), Q39(A) (5 marks, solved), Q39(B) (5 marks, solved)

Set 31/5/1: Q37 (3 marks, solved), Q38 (4 marks, solved)

Sample paper 2025-26: Q34(A) (2 marks, solved), Q34(B) (2 marks, solved), Q36 (3 marks, solved), Q39(A) (5 marks, solved), Q39(B) (5 marks, solved)

2025

Set 31/1/1: Q25(a) (2 marks, solved), Q25(b) (2 marks, solved), Q32 (3 marks, solved), Q33 (3 marks, solved), Q39 (4 marks, solved)

Set 31/2/1: Q14 (1 mark, solved), Q15 (1 mark, solved), Q19 (1 mark, solved), Q26 (2 marks, solved), Q32 (3 marks, solved)

Set 31/3/1: Q13 (1 mark, solved), Q26 (2 marks, solved), Q39 (4 marks, solved)

Set 31/4/1: Q12 (1 mark, solved), Q13 (1 mark, solved), Q14 (1 mark, solved), Q32 (3 marks, solved)

Set 31/5/1: Q25 (2 marks, solved), Q32 (3 marks, solved), Q33 (3 marks, solved)

Set 31/6/1: Q19 (1 mark, solved), Q33 (3 marks, solved), Q39 (4 marks, solved)

Sample paper 2024-25: Q25(A) (2 marks, solved), Q25(B) (2 marks, solved), Q32 (3 marks, solved), Q36(A) (5 marks, solved), Q36(B) (5 marks, solved)

2024

Set 31/1/1: Q25(A) (2 marks, solved), Q25(B) (2 marks, solved), Q36(A) (5 marks, solved), Q36(B) (5 marks, solved)

Set 31/2/1: Q14 (1 mark, solved), Q15 (1 mark, solved), Q25 (2 marks, solved), Q39(a) (4 marks, solved)

Set 31/3/1: Q14 (1 mark, solved), Q32 (3 marks, solved), Q33 (3 marks, solved), Q39 (4 marks, solved)

Set 31/4/1: Q3 (1 mark, solved), Q13 (1 mark, solved), Q14 (1 mark, solved), Q26 (2 marks, solved), Q39 (4 marks, solved)

Set 31/5/1: Q19 (1 mark, solved), Q26 (2 marks, solved), Q36(a) (5 marks, solved), Q36(b) (5 marks, solved)

Sample paper 2023-24: Q25(a) (2 marks, solved), Q32 (3 marks, solved), Q33 (3 marks, solved), Q39 (4 marks, solved)

2023

Set 31/1/1: Q13 (1 mark, solved), Q14 (1 mark, solved), Q15 (1 mark, solved), Q36 (5 marks, solved)

Set 31/2/1: Q14 (1 mark, solved), Q15 (1 mark, solved), Q25 (2 marks, solved), Q32 (3 marks, solved), Q39 (4 marks, solved)

Set 31/4/1: Q13 (1 mark, solved), Q15 (1 mark, solved), Q36 (5 marks, solved)

Set 31/5/1: Q13 (1 mark, solved), Q15 (1 mark, solved), Q36 (5 marks, solved)

Set 31/6/1: Q24 (2 marks, solved), Q36(a) (5 marks, solved), Q36(b) (5 marks, solved)

Practice questions on Electricity

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
One kilowatt hour is equal to:
(a) $3.6 \times 1 0^{3}$ J
(b) $3.6 \times 1 0^{5}$ J
(c) $3.6 \times 1 0^{6}$ J
(d) 1000 J
Hint
Multiply 1000 W by the number of seconds in one hour.
Answer
(c) $3.6 \times 1 0^{6}$ J, because $1 k W h = 1000 W \times 3600 s = 3.6 \times 1 0^{6} J$ .
Q2 · 2 marks · Very short answerBoard level
When a 6 V battery is connected across a resistor, a current of 3 mA flows. Find the resistance. What will the current be if the voltage is reduced to 3 V?
Hint
Change mA to A before using Ohm's law.
Answer
$I = 3 m A = 3 \times 1 0^{- 3} A$ ; $R = \frac{V}{I} = \frac{6}{3 \times 1 0 ^{- 3}} = 2000 Ω$ .
At 3 V, current halves: $I = \frac{3}{2000} = 1.5 \times 1 0^{- 3} A = 1.5 m A$ .
Q3 · 1 mark · Assertion-reasonBoard level
Assertion (A): Nichrome is used for the heating elements of electric toasters.
Reason (R): Alloys have higher resistivity than their constituent metals and do not oxidise readily at high temperatures.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Ask whether R tells you why the toaster uses nichrome.
Answer
(a) Both A and R are true and R is the correct explanation of A. High resistivity gives more heat, and not burning when hot makes the element last.
Q4 · 3 marks · Short answerHarder board level
A copper wire is 50 m long and has an area of cross-section of $1 \times 1 0^{- 6} m^{2}$ . The resistivity of copper is $1.6 \times 1 0^{- 8} Ω m$ . (a) Find its resistance. (b) What will be the resistance of a copper wire of the same length but half the area of cross-section? (c) Does the resistivity change? Why?
Hint
Use $R = ρ \frac{l}{A}$ ; resistance goes inversely with area.
Answer
(a) $R = \frac{1.6 \times 1 0 ^{- 8} \times 50}{1 \times 1 0 ^{- 6}} = 0.8 Ω$ .
(b) Half the area doubles the resistance: 1.6 Ω.
(c) No. Resistivity depends only on the material (at a given temperature), not on its length or thickness.
Q5 · 3 marks · Short answerHarder board level
You have four resistors of 4 Ω each. Find (a) the highest and (b) the lowest resistance you can get using all four. Show how to connect them to get (c) 10 Ω and (d) 4 Ω.
Hint
All in series gives the highest; all in parallel gives the lowest.
Answer
(a) All in series: $4 \times 4 = 16 Ω$ . (b) All in parallel: $\frac{1}{R} = \frac{4}{4}$ , so 1 Ω.
(c) Two in series (8 Ω) with a parallel pair (2 Ω) in series: $8 + 2 = 10 Ω$ .
(d) Two series pairs (8 Ω each) joined in parallel: $\frac{8 \times 8}{16} = 4 Ω$ .
Q6 · 5 marks · Long answerHarder board level
Resistors of 1 Ω, 2 Ω and 6 Ω are connected in series with a 9 V battery and a plug key. (a) Describe in words a circuit diagram with an ammeter to measure the current and a voltmeter to measure the potential difference across the 6 Ω resistor. (b) Find the ammeter and voltmeter readings. (c) Find the power used in each resistor and show that the total equals the power supplied by the battery.
Hint
Find the current first; it is the same in every resistor.
Answer
(a) Battery, key, ammeter and the three resistors all in one loop; voltmeter joined across the two ends of the 6 Ω resistor only.
(b) $R_{s} = 1 + 2 + 6 = 9 Ω$ ; $I = \frac{9}{9} = 1 A$ (ammeter). Voltmeter: $V = 1 \times 6 = 6 V$ .
(c) $P = I^{2} R$ : 1 W, 2 W, 6 W; total 9 W. Battery: $P = V I = 9 \times 1 = 9 W$ . They match.
Q7 · 3 marks · Short answerHarder board level
Wires A and B are made of the same material and have the same length. The diameter of B is twice that of A. (a) Find the ratio of their resistances $R_{A} : R_{B}$ . (b) If they are connected in parallel across the same battery, find the ratio of heat produced in them in the same time.
Hint
Doubling the diameter makes the area four times. In parallel, use $H = \frac{V ^{2}}{R} t$ .
Answer
(a) Area of B $= 4 \times$ area of A, so $R_{A} : R_{B} = 4 : 1$ .
(b) Same V and t, so $H \propto \frac{1}{R}$ : $H_{A} : H_{B} = 1 : 4$ . The thicker wire B produces four times as much heat.
Q8 · 5 marks · Long answerHarder board level
Suppose a home uses an electric oven of 1000 W for 1 hour a day and five bulbs of 40 W each for 5 hours a day. (a) Find the energy used per day in kW h. (b) Find the cost for 30 days at Rs 6 per kW h. (c) Express the monthly energy in joules. (d) Which fuse (1 A, 2 A, 3 A, 5 A, 10 A) suits the oven on a 220 V supply? Justify.
Hint
Energy in kW h = power in kW × time in hours.
Answer
(a) Oven: $1 k W \times 1 h = 1 k W h$ ; bulbs: $5 \times 0.04 \times 5 = 1 k W h$ ; total 2 kW h per day.
(b) $2 \times 30 = 60 k W h$ ; cost $= 60 \times 6 = R s 360$ .
(c) $60 \times 3.6 \times 1 0^{6} = 2.16 \times 1 0^{8} J$ .
(d) $I = \frac{1000}{220} \approx 4.55 A$ , just below 5 A, so a 5 A fuse.
Q9 · 2 marks · Very short answerHOTS / competency
A uniform wire of resistance 16 Ω is bent into a complete circle. What is the resistance between two points at the opposite ends of a diameter?
Hint
The two halves of the circle form two paths between the points.
Answer
Each half of the wire has half the length, so 8 Ω. The two halves are in parallel between the points: $\frac{8 \times 8}{8 + 8} = 4 Ω$ .
Q10 · 3 marks · Short answerHOTS / competency
In an activity like Activity 11.3, the same cell and ammeter are used each time. A student records: nichrome wire of length l: 0.6 A; nichrome wire of length 2l, same thickness: 0.3 A; thicker nichrome wire of length l: 1.2 A. (a) What does the first pair of readings show? (b) How does the area of cross-section of the thicker wire compare with the first wire? (c) A copper wire of the same length and thickness as the first wire gives a much larger current. Explain.
Hint
Current halves means resistance doubles, by Ohm's law.
Answer
(a) Doubling length halves the current, so resistance doubles: $R \propto l$ .
(b) Current doubles, so resistance halves; since $R \propto \frac{1}{A}$ , the thicker wire has twice the area.
(c) Copper has much lower resistivity than nichrome, so its resistance is much smaller and more current flows.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.State Ohm's law, including its condition.Show answer
The potential difference across a metallic conductor is directly proportional to the current through it, provided its temperature stays the same.
2.What is 1 volt?Show answer
The p.d. between two points when 1 J of work is done to move 1 C of charge between them.
3.On what three factors does the resistance of a conductor depend?Show answer
Length (directly), area of cross-section (inversely) and the nature of the material.
4.SI unit of resistivity?Show answer
Ohm metre, $Ω m$ .
5.In which combination is the current the same through each resistor?Show answer
Series.
6.State Joule's law of heating.Show answer
Heat produced is proportional to the square of current, to the resistance and to the time: $H = I^{2} R t$ .
7.How many joules make 1 kW h?Show answer
$3.6 \times 1 0^{6} J$ .
8.Why is tungsten used for bulb filaments?Show answer
It is a strong metal with a very high melting point (3380°C), so it can glow white-hot without melting.

Best order to study it

Start with current, charge and potential difference (I = Q/t, V = W/Q), the units and how ammeters and voltmeters are connected.
Learn Ohm's law with Activity 11.1 and the V-I graph, then practise rearranging V = IR with unit conversions.
Move to resistance and resistivity: the three factors, R = ρl/A, ratio questions and the choice of materials.
Study series and then parallel combinations: the derivations, the rules for current and voltage, mixed circuits and why homes use parallel wiring.
Learn the heating effect and Joule's law with its applications (heaters, bulb filament, fuse).
Finish with electric power, its three forms, rated appliances and kW h bill calculations, then mix numericals from all sections.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you define 1 ampere, 1 volt and 1 ohm and write each in terms of other units?
Can you find the number of electrons in a given charge using $1.6 \times 1 0^{- 19} C$ per electron?
Can you state Ohm's law with its condition and describe the V-I graph and how to get R from it?
Can you write $R = ρ \frac{l}{A}$ and say what happens to R when length is doubled or diameter is doubled?
Can you derive $R_{s} = R_{1} + R_{2} + R_{3}$ and $\frac{1}{R _{p}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \frac{1}{R _{3}}$ from Ohm's law?
Can you solve a mixed circuit and find the current in each branch?
Can you give three reasons why domestic wiring is in parallel?
Can you state Joule's law and explain how a fuse works and how to choose its rating?
Can you write all three forms of power and find the power of a rated appliance at a different voltage?
Can you calculate a monthly bill in kW h and convert kW h into joules?

Electricity: frequently asked questions

Is the box on the flow of charges inside a wire part of the 2026-27 syllabus?

No. The box item on the flow of charges inside a wire has been dropped and is not in the 2026-27 syllabus. Also remember that information in NCERT text boxes is not assessed in the board exam.

Do I need to learn the derivations of series and parallel resistance?

Yes, they come straight from Ohm's law and are a natural part of a long answer. For series, use the same current and add the voltages; for parallel, use the same voltage and add the currents, then cancel the common quantity.

Do I have to memorise the resistivity values in the NCERT table?

No. NCERT says you need not memorise them; they are given for solving problems. You should, however, know the pattern: metals and alloys have low resistivity, alloys more than their metals, and insulators very high.

Which power formula should I use in a numerical?

Use P equals VI when both voltage and current are known. Use I squared R when the current is common, as in series, and V squared over R when the voltage is common, as in parallel or for a rated appliance on a new supply.

What is the difference between resistance and resistivity?

Resistance belongs to a particular wire and changes with its length, thickness, material and temperature. Resistivity belongs to the material only, so cutting or stretching a wire changes its resistance but not its resistivity.

Why does the heating element of a heater glow but not its cord?

The same current flows through both, so heat depends on resistance. The element is made of a high-resistivity alloy and has high resistance, so it gets very hot, while the copper cord has very low resistance and produces little heat.

Based on: CBSE Science (086) syllabus 2026-27; NCERT Science, Textbook for Class X (Reprint 2026-27)