ScienceCBSE Class 10 · Electricity

Heating Effect of Electric Current and Joule's Law

Updated 13 October 2026 · Fact-checked

The heating effect of electric current is the production of heat when current flows through a resistor. Joule's law gives the heat as H = I squared R t. The board asks you to state the law, solve heat and fuse numericals, and explain uses such as heaters, irons, tungsten bulb filaments and fuses.

Energy from the battery turns into heat in the resistor: H = I²Rt, so doubling the current gives four times the heat.

Illustration drawn for QuestionBuddy.

Understand Heating Effect of Electric Current

A cell or battery is a store of energy. Its chemical action creates a potential difference that keeps charges moving. To keep a current going, the source must keep spending energy. Where does that energy go? Part of it may do useful work, such as turning a fan's blades. The rest turns into heat, which is why a fan feels warm after running for a long time.

If a circuit contains only resistors, all the energy from the source becomes heat. This is the heating effect of electric current. Electrons moving through a conductor are held back by the atoms around them, and this resistance is where the energy is converted.

You can work out the heat. Moving charge $Q$ through a potential difference $V$ needs work $V Q$ . Since $Q = I t$ , the energy supplied in time $t$ is $V I t$ , so $H = V I t$ . Using Ohm's law, $V = I R$ , this becomes $H = I^{2} R t$ . This is Joule's law of heating. Heat is directly proportional to the square of the current (for fixed R), to the resistance (for fixed I), and to the time. Doubling the current gives four times the heat, not twice.

Which form should you use? $H = I^{2} R t$ compares resistors carrying the same current, as in series. When an appliance is joined to a known voltage, first find its current with $I = \frac{V}{R}$ , then use Joule's law. This links the topic to Ohm's law, resistivity, and electric power, since heat produced per second is the power.

Heating is sometimes a nuisance: it wastes energy and can change the properties of circuit parts. But it has many uses. Irons, toasters, ovens, kettles and heaters use elements made of alloys, which have high resistivity and do not oxidise easily when hot. A bulb filament must get hot enough to glow without melting, so tungsten (melting point 3380°C) is used, and the bulb is filled with inactive nitrogen and argon to make it last longer. Most of its power still appears as heat; only a small part is light.

A fuse also works by heating. It is a wire of suitable melting point placed in series with the device. If the current goes above its rating, the wire heats up, melts and breaks the circuit, protecting the appliance.

Heating Effect of Electric Current: formulas and key facts

Heat in terms of V, I and t: $H = V I t$; V in volt, I in ampere, t in second, H in joule. Comes from work done $W = V Q$ with $Q = I t$ .
Joule's law of heating: $H = I^{2} R t$; Heat is proportional to $I^{2}$ (fixed R), to R (fixed I) and to t. Best when the current is known or common.
Heat with voltage known: $H = \frac{V ^{2}}{R} t$; Follows from $I = \frac{V}{R}$ . Useful when the same voltage acts across resistors, as in parallel.
Heat per second (power): $P = \frac{H}{t} = V I = I^{2} R$; Rate of heat production in a purely resistive circuit equals the electric power.
Fuse choice: Current drawn $I = \frac{P}{V}$ ; choose the standard rating equal to or just above the working current; Domestic fuses are rated 1 A, 2 A, 3 A, 5 A, 10 A. A 1 kW iron at 220 V draws about 4.54 A, so a 5 A fuse.
Materials for heating uses: Heating elements: alloys such as nichrome; filaments: tungsten; fuse wire: metal or alloy of suitable melting point; Alloys: high resistivity, do not oxidise readily when hot. Tungsten: strong, melting point 3380°C. Fuse is always in series.

How to solve Heating Effect of Electric Current questions

Use this method for any heating-effect numerical or reasoning question.

1Write down the given quantities with units and mark what is asked (H, I, R, V or t).
2Convert to SI units: minutes and hours to seconds, kW to W, mA to A.
3Decide which form fits: $H = I^{2} R t$ if I and R are known, $H = V I t$ if V and I are known, $H = \frac{V ^{2}}{R} t$ if V and R are known.
4If only voltage and resistance (or a rating) are given, first find the current using $I = \frac{V}{R}$ or $I = \frac{P}{V}$ .
5Substitute with units and give the answer in joule.
6For ratio questions, write the formula for both cases and divide so the unchanged quantities cancel; remember current is squared.
7For 'give reasons' questions, name the property (high resistivity, high melting point, low melting point) and then link it to the use in one more sentence.

Heating Effect of Electric Current in real life

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.

Joule's law of heating

Suppose you have a small electric kettle in your hostel room with a heating coil of resistance 50 Ω. It runs on a 220 V supply. You switch it on for 2 minutes to warm water for tea.

Find the current through the coil and the heat produced in 2 minutes. If the coil were replaced by one of 25 Ω on the same supply, how would the heat in the same time change?

Try it first, then show the solutionSolution

Current $I = \frac{V}{R} = \frac{220 V}{50 Ω} = 4.4 A$ .
Time $t = 2 \times 60 = 120 s$ .
Heat $H = I^{2} R t = (4.4)^{2} \times 50 \times 120 = 19.36 \times 6000 = 116160 J$ .
Check with $H = V I t = 220 \times 4.4 \times 120 = 116160 J$ .
At fixed voltage $H = \frac{V ^{2}}{R} t$ , so halving R doubles the heat.

Answer: Current 4.4 A; heat about 1.16 × 10^5 J (116160 J). With a 25 Ω coil the heat doubles to 232320 J.

Memory hook: Same plug, lower resistance coil, more current, faster tea.

Tricks to remember Heating Effect of Electric Current

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

I-squared is the boss
In $H = I^{2} R t$ , only I is squared: 2 × current = 4 × heat, 3 × current = 9 × heat.
Use it when: Ratio MCQs where the current is changed.
Three things heat depends on
'I Really Toast': I squared, Resistance, Time.
Use it when: Stating Joule's law and its three proportionalities.
Material match
Heater = Alloy (does not burn), Bulb = Tungsten (does not melt), Fuse = melts on purpose.
Use it when: Give-reason questions on choice of materials.

Heating Effect of Electric Current: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Heat in a room heater coil · 3 marks · Short answerBoard level

A room heater coil of resistance 40 Ω carries a steady current of 5 A for 20 minutes. Find the heat produced in the coil and express it in kilojoule.

Try it first, then show the solutionSolution

Convert time: $t = 20 \times 60 = 1200 s$ (½ mark).
Formula: $H = I^{2} R t$ (½ mark).
Substitution: $H = (5)^{2} \times 40 \times 1200$ (½ mark).
$H = 25 \times 48000 = 1200000 J$ (1 mark).
In kilojoule: $H = \frac{1200000}{1000} = 1200 k J$ (½ mark).

Answer: Heat produced = 1.2 × 10^6 J = 1200 kJ.

Example 2: Joule's law and current change · 3 marks · Short answerHarder board level

(a) State Joule's law of heating.
(b) The current through a resistor is doubled and the time is halved, keeping the resistance the same. By what factor does the heat produced change?
(c) A thick copper supply wire and a nichrome coil carry the same current. Explain why only the coil becomes red hot.

Try it first, then show the solutionSolution

(a) Heat produced in a resistor is directly proportional to the square of the current, to the resistance and to the time for which current flows: $H = I^{2} R t$ (1 mark).
(b) New heat $H^{'} = (2 I)^{2} R \frac{t}{2} = 2 I^{2} R t$ , so heat doubles (1 mark).
(c) The current is the same, so heat depends on resistance ( $H = I^{2} R t$ ). Copper has very low resistivity and the thick wire has a large area, so by $R = ρ \frac{l}{A}$ its resistance and heat are small; nichrome has high resistivity, so the coil has high resistance, produces much more heat and becomes red hot (1 mark).

Answer: (a) H = I²Rt with the three proportionalities. (b) Heat becomes 2 times. (c) The supply wire has much lower resistance, so much less heat is produced in it; the high-resistance coil gets red hot.

Example 3: Applications: bulb and fuse · 5 marks · Long answerHarder board level

(a) Give two reasons why tungsten is used for bulb filaments and state why bulbs are filled with nitrogen and argon.
(b) Explain how a fuse protects an appliance. Where is it connected?
(c) An electric heater is rated 2000 W, 220 V. Which fuse from 1 A, 2 A, 3 A, 5 A, 10 A, 15 A should be used? Show working.

Try it first, then show the solutionSolution

(a) Tungsten is a strong metal with a very high melting point (3380°C), so it can get hot enough to glow without melting (1 mark).
(a) Nitrogen and argon are chemically inactive and prolong the life of the filament (1 mark).
(b) The fuse is a wire of suitable melting point connected in series with the device; if the current exceeds its rated value, the wire heats, melts and breaks the circuit, stopping the excess current (1½ marks).
(c) $I = \frac{P}{V} = \frac{2000 W}{220 V} \approx 9.09 A$ (1 mark).
(c) The fuse rating must be equal to or just above the working current; 5 A is below 9.09 A and would melt, so use the 10 A fuse (½ mark).

Answer: (a) High melting point, strength; inert gases prolong filament life. (b) Series fuse melts on excess current and breaks the circuit. (c) Current is about 9.09 A, so a 10 A fuse.

Example 4: Heat in series and parallel · 3 marks · Short answerBoard level

Two identical resistors of 10 Ω each are connected first in series and then in parallel across a 20 V supply. Find the total heat produced in 10 s in each case and the ratio of heat in series to heat in parallel.

Try it first, then show the solutionSolution

Series: $R_{s} = 20 Ω$ , $H_{s} = \frac{V ^{2}}{R _{s}} t = \frac{400}{20} \times 10 = 200 J$ (1 mark).
Parallel: $R_{p} = 5 Ω$ , $H_{p} = \frac{400}{5} \times 10 = 800 J$ (1 mark).
Ratio $H_{s} : H_{p} = 200 : 800 = 1 : 4$ (1 mark).

Answer: Series 200 J, parallel 800 J; ratio 1 : 4.

Example 5: Case: two heating coils in a hot plate · 4 marks · Case-basedHOTS / competency

Read the passage and answer.

Riya's school lab has a hot plate with two coils, P of 40 Ω and Q of 40 Ω, which can be used alone, in series or in parallel on a 200 V supply. Her teacher asks which setting heats water fastest and why.

(a) Find the current through the supply when P alone is used. (1 mark)
(b) Find the heat produced per second with P alone. (1 mark)
(c) Find the heat produced per second when P and Q are in series and when in parallel, and state which setting is fastest. (2 marks)

Try it first, then show the solutionSolution

(a) $I = \frac{V}{R} = \frac{200}{40} = 5 A$ (1 mark).
(b) Heat per second $= I^{2} R = 25 \times 40 = 1000 J$ each second (1 mark).
(c) Series: $R = 80 Ω$ , heat per second $= \frac{V ^{2}}{R} = \frac{40000}{80} = 500 J$ ; parallel: $R = 20 Ω$ , heat per second $= \frac{40000}{20} = 2000 J$ (1 mark).
(c) Parallel setting has lowest resistance, so at fixed voltage it draws the most current and produces most heat; it heats fastest (1 mark).

Answer: (a) 5 A. (b) 1000 J per second. (c) Series 500 J/s, parallel 2000 J/s; parallel is fastest.

Example 6: Mixed network: current and heat · 3 marks · Short answerHarder board level

A 12 V battery is connected to a network of three resistors. A 3 Ω resistor is joined in series with a parallel pair made of a 4 Ω resistor and a 12 Ω resistor. (a) Find the total current supplied by the battery. (b) Find the heat produced in the 4 Ω resistor in 10 s.

Try it first, then show the solutionSolution

Parallel pair: $\frac{1}{R _{p}} = \frac{1}{4} + \frac{1}{12} = \frac{4}{12}$ , so $R_{p} = 3 Ω$ (½ mark).
Total resistance: $R = 3 Ω + 3 Ω = 6 Ω$ (½ mark).
Total current: $I = \frac{V}{R} = \frac{12 V}{6 Ω} = 2 A$ (½ mark).
Potential difference across the parallel pair: $V_{p} = I R_{p} = 2 \times 3 = 6 V$ (½ mark).
Current in the 4 Ω resistor: $I_{1} = \frac{6 V}{4 Ω} = 1.5 A$ (½ mark).
Heat: $H = I_{1}^{2} R t = (1.5)^{2} \times 4 \times 10 = 90 J$ (½ mark).

Answer: (a) Total current = 2 A. (b) Heat in the 4 Ω resistor in 10 s = 90 J.

Where students lose marks in Heating Effect of Electric Current

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

Mistake: Forgetting to square the current
Students treat heat as proportional to I, like R and t.
Wrong working
Current doubled, so heat doubles.
Right working
$H \propto I^{2}$ , so $H^{'} = (2)^{2} H = 4 H$ .
Do this: Circle the square on I every time you write Joule's law.
Usually costs: Full mark in 1-mark MCQs and ratio parts.
Mistake: Leaving time in minutes or hours
Given data is in minutes and is substituted directly.
Wrong working
$H = 5^{2} \times 20 \times 2 = 1000 J$ for 2 minutes.
Right working
$t = 120 s$ , $H = 25 \times 20 \times 120 = 60000 J$ .
Do this: Convert all time to seconds before substituting.
Usually costs: The final answer mark and often the substitution mark.
Mistake: Using the wrong form at fixed voltage
Students think more resistance always means more heat.
Wrong working
A 100 Ω heater makes more heat than a 50 Ω one on 220 V because H ∝ R.
Right working
At fixed V, $H = \frac{V ^{2}}{R} t$ , so the 50 Ω heater makes twice the heat.
Do this: H ∝ R only when current is fixed; at fixed voltage H ∝ 1/R.
Usually costs: 1-2 marks in reasoning and comparison questions.
Mistake: Choosing a fuse rated below the working current
Students pick the closest number, even when it is lower than the current drawn.
Wrong working
Iron draws 4.54 A, so use a 3 A fuse.
Right working
Iron draws 4.54 A, so use a 5 A fuse.
Do this: Choose the standard rating equal to or just above the working current; a lower rating melts in normal use.
Usually costs: The final mark of fuse questions.

Exam tips

When stating Joule's law, write the formula and all three proportionalities, each with its condition (for given R, for given I).
In numericals, always show formula, substitution with units and the final answer in joule; marking schemes give separate marks.
For 'give reasons' questions on bulbs, heaters and fuses, name the property and link it to the use; one word answers lose marks.
In case-based questions set at home or in a lab, find the current first from the rating or Ohm's law, then apply Joule's law.
Read whether the current or the voltage is fixed before deciding if more resistance means more or less heat.

Practice questions on Heating Effect of Electric Current

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

Q1 · 3 marks · Short answerHarder board level
A heating coil is connected across a 30 V supply. In 10 minutes, 1200 C of charge passes through it. Find (a) the heat produced, (b) the current through the coil and (c) its resistance. Check your heat value using Joule's law.
Hint
Heat equals work done, $H = V Q$ . Current is $I = \frac{Q}{t}$ with t in seconds.
Answer
(a) H = VQ = 30 × 1200 = 36000 J. (b) t = 600 s, I = Q/t = 1200/600 = 2 A. (c) R = V/I = 30/2 = 15 Ω. Check: H = I²Rt = 4 × 15 × 600 = 36000 J.
Q2 · 1 mark · Assertion-reasonBoard level
Assertion (A): The heating elements of electric toasters are made of an alloy and not a pure metal.
Reason (R): Alloys have lower resistivity than their constituent metals.
(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
Recall how alloy resistivity compares with that of pure metals.
Answer
(c) A is true but R is false. Alloys have higher resistivity and do not oxidise readily at high temperature.
Q3 · 3 marks · Short answerBoard level
100 J of heat is produced each second in a resistor when the potential difference across it is 10 V. Find the current, the resistance, and the heat produced each second if the potential difference is raised to 20 V (resistance unchanged).
Hint
Heat per second is VI. Then use R = V/I and H = V²t/R.
Answer
VI = 100, so I = 10 A; R = 10/10 = 1 Ω; at 20 V, heat per second = 400/1 = 400 J (four times).
Q4 · 5 marks · Long answerHarder board level
(a) Derive H = I²Rt starting from the work done in moving charge through a potential difference.
(b) A 4 Ω resistor and a 6 Ω resistor are in series with a 20 V battery. Find the heat produced in each in 1 minute and say which gets hotter.
(c) Give one disadvantage of the heating effect of current.
Hint
W = VQ, Q = It, V = IR. In series, the current is the same, so heat is proportional to R.
Answer
(a) W = VQ, Q = It, so H = VIt; V = IR gives H = I²Rt (2 marks). (b) R = 10 Ω, I = 2 A; H(4 Ω) = 4 × 4 × 60 = 960 J; H(6 Ω) = 4 × 6 × 60 = 1440 J; the 6 Ω gets hotter (2 marks). (c) It wastes useful electrical energy as heat and can raise the temperature of components and alter their properties (1 mark).
Q5 · 3 marks · Short answerHOTS / competency
A student says: 'A thin fuse wire and a thick fuse wire of the same material and length carry the same current, so they heat up equally. Either can be used.' Find the error in the reasoning and explain which wire melts at a lower current.
Hint
Use R = ρl/A and H = I²Rt with the same current.
Answer
Same material and length, but the thin wire has smaller area, so higher resistance (R = ρl/A). With the same current, H = I²Rt is larger in the thin wire, so it heats more and melts at a lower current. The two are not interchangeable; the wire must suit the fuse rating.
Q6 · 4 marks · Case-basedHOTS / competency
Read the passage and answer.
Suppose a family uses a room heater rated 880 W, 220 V in winter. The heater is protected by a fuse, and the electrician has fuses of 3 A, 5 A and 15 A.
(a) Find the current drawn by the heater. (1 mark)
(b) Find the resistance of its heating element. (1 mark)
(c) Find the heat produced by the heater in 10 minutes. (2 marks)
OR
(d) Which fuse should be used? What would happen if the 3 A fuse were fitted? (2 marks)
Hint
Use I = P/V first, then R = V/I. A fuse wire melts when the current is larger than its rating.
Answer
(a) I = P/V = 880/220 = 4 A. (b) R = V/I = 220/4 = 55 Ω. (c) t = 600 s; H = Pt = 880 × 600 = 528000 J (check: I²Rt = 16 × 55 × 600 = 528000 J). (d) Use the 5 A fuse, the standard rating just above 4 A. With a 3 A fuse, the working current of 4 A is larger than its rating, so the fuse wire would heat, melt and break the circuit as soon as the heater runs.
Q7 · 5 marks · Long answerHarder board level
(a) A copper wire and a nichrome wire of the same length and the same area of cross-section are joined in series with a battery. Which wire gets hotter? Explain using $R = ρ \frac{l}{A}$ and $H = I^{2} R t$ . (2 marks)
(b) Give two reasons why the heating elements of electric irons and toasters are made of an alloy. (1 mark)
(c) A nichrome heating element of resistance 50 Ω carries a current of 2 A. Find the heat produced in 5 minutes. (1 mark)
(d) Why is tungsten used for the filaments of electric bulbs? (1 mark)
Hint
In series the current is the same, so heat depends on resistance, which depends on resistivity when l and A are equal.
Answer
(a) Same l and A, so R depends only on resistivity; nichrome has a far higher resistivity than copper, so its resistance is much higher. In series both carry the same current, so by H = I²Rt the nichrome wire produces much more heat and gets hotter (2 marks). (b) Alloys have higher resistivity than their constituent metals, and they do not oxidise (burn) readily at high temperatures (1 mark). (c) t = 300 s; H = I²Rt = 4 × 50 × 300 = 60000 J (1 mark). (d) 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 mark).

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 Joule's law of heating.Show answer
Heat produced in a resistor, $H = I^{2} R t$ : proportional to the square of current, to resistance and to time.
2.What happens to source energy in a purely resistive circuit?Show answer
All of it is dissipated as heat.
3.Melting point of tungsten given in NCERT?Show answer
3380°C.
4.How is a fuse connected?Show answer
In series with the device.
5.Why are bulbs filled with nitrogen and argon?Show answer
These inactive gases prolong the life of the filament.
6.Name four appliances based on Joule's heating.Show answer
Electric iron, toaster, oven, kettle (also electric heater).

Heating Effect of Electric Current: frequently asked questions

What is the heating effect of electric current in simple words?

When current passes through a resistor, the energy given by the source turns into heat. In a circuit with only resistors, all of it becomes heat. This is used in irons, heaters and kettles.

Is H = VIt the same as H = I squared R t?

Yes. H = VIt comes from the work done in moving charge. Putting V = IR into it gives H = I squared R t, which is Joule's law.

Why does a fuse wire melt but the house wiring does not?

The fuse wire is chosen with a suitable melting point and rating, so it heats and melts when the current goes above its rated value. This breaks the circuit before the excess current can harm the appliance.

Do I need to learn the 'flow of charges inside a wire' box?

No. That box item has been dropped and is not in the 2026-27 syllabus. Information in NCERT boxes is not assessed in the board exam.

Does an electric bulb give out mostly light?

No. Most of the power used by the filament appears as heat, and only a small part is radiated as light. The filament must get very hot to glow, which is why tungsten is used.