Chapter 3 · Electricity

3.4 The Heating Effect of Current and Electric Power

A current always warms the conductor it flows through: the heat produced is $H=I^2Rt$. The rate at which a device converts electrical energy is its power, $P=VI=I^2R=\frac{V^2}{R}$, and the energy you pay for is counted in kilowatt-hours.

Switch on an electric kettle and within three minutes a litre of water is boiling. The kettle has no flame and no moving parts — just a coil of resistance wire carrying a current. Feel a phone charger, a laptop or even the cable of a heater after a while and you will find them warm too. Whenever charge flows through a resistance, some of the energy it carries from the source is turned into heat. In a kettle, an iron, a toaster or a geyser this heating effect is exactly what we want; in a motor or a computer it is wasted energy that must be removed by fans; and in house wiring it is a danger that a fuse guards against.

In this section we work out how much heat a current produces, see how heaters, filament bulbs and fuses use it, and define electric power — the rate at which an appliance uses energy, printed on every appliance as its wattage. Finally we meet the "unit" on your electricity bill, the kilowatt-hour, and learn to calculate what running each appliance costs.

The Heating Effect of Current

Think of a resistor connected to a battery, carrying a steady current $I$ with a potential difference $V$ across it. In a time $t$, a charge $Q=It$ passes through it. By the definition of potential difference (Section 3.1), the work done on this charge as it moves through the resistor is $W=VQ=VIt$. In a pure resistor this energy does not make anything move or light up usefully — the drifting electrons keep colliding with the vibrating atoms of the metal and make them vibrate harder, which means the resistor gets hotter. All the work done appears as heat $H$.

📐 Law 3.5 — Joule's law of heating

When a steady current $I$ flows for a time $t$ through a conductor of resistance $R$, with a potential difference $V$ across it, the heat produced is

$$H=VIt .$$

Using Ohm's law, $V=IR$, this becomes

$$H=I^2Rt ,$$

so for a given resistor the heat produced is (i) proportional to the square of the current, (ii) proportional to the resistance for a given current, and (iii) proportional to the time. Heat is measured in joules. When the potential difference is fixed instead of the current, it is often easier to use $H=\dfrac{V^2}{R}\,t$.

The square matters. Double the current through a heater coil and it produces four times as much heat every second; triple it and the heat goes up nine times. This is why a wire that is perfectly safe at its normal current can overheat dangerously when the current becomes a few times larger — and why thick cables are used for appliances that draw a large current.

✏️ Example 1 — Heat from a kettle element
The element of an electric kettle has a resistance of $44\ \Omega$ and carries a current of $5.0\ \text{A}$. How much heat does it produce in $2.0$ minutes?
Units $t=2.0\ \text{min}=120\ \text{s}$.
Joule's law $H=I^2Rt=(5.0)^2\times44\times120=1.32\times10^5\ \text{J}\approx1.3\times10^5\ \text{J}$.
Check $V=IR=5.0\times44=220\ \text{V}$, the mains voltage, and $H=VIt=220\times5.0\times120=1.32\times10^5\ \text{J}$. ✓
✏️ Example 2 — Which resistor gets hotter?
A $3.0\ \Omega$ and a $6.0\ \Omega$ resistor are connected to a $9.0\ \text{V}$ battery for $10\ \text{s}$. Find the heat produced in each when they are connected (a) in series, (b) in parallel.
(a) Series $I=\dfrac{9.0}{9.0}=1.0\ \text{A}$ through both. $H_3=1.0^2\times3.0\times10=30\ \text{J}$, $H_6=1.0^2\times6.0\times10=60\ \text{J}$: same current, so the larger resistance produces more heat.
(b) Parallel Each has $9.0\ \text{V}$ across it: $H_3=\dfrac{9.0^2}{3.0}\times10=270\ \text{J}$, $H_6=\dfrac{9.0^2}{6.0}\times10=135\ \text{J}$: same voltage, so the smaller resistance produces more heat.
Meaning Use $I^2Rt$ when the current is shared, $\dfrac{V^2}{R}t$ when the voltage is shared. The parallel pair produces far more heat in total ($405\ \text{J}$ against $90\ \text{J}$) because it draws more current.
Interactive 1 — The Joule Heating Lab

A nichrome coil in $50\ \text{g}$ of water is connected to an adjustable battery. Set the voltage and the coil's resistance; the electrons in the circuit speed up with the current, the coil glows hotter, and the thermometer rises. Play runs the clock for $60\ \text{s}$ and draws the heat produced, $H$, against time; each finished run stays on the graph. Play sweeps $3\ \text{V}$, $6\ \text{V}$, $12\ \text{V}$ — each doubling the current — so you can see the heat go up four times.

6
current $I$—
power $I^2R$—
time $t$—
heat $H=I^2Rt$—
water—

Using the Heating Effect: Heaters, Bulbs and Fuses

Heating appliances. Irons, toasters, kettles, room heaters and geysers all contain an element of a high-resistivity alloy such as nichrome. The element and the copper cable feeding it are in series, so they carry the same current — but the element's resistance is hundreds of times larger, so by $H=I^2Rt$ nearly all the heat is produced in the element, and the cable stays cool. Nichrome is chosen because it has a high resistivity, a high melting point, and does not oxidise (burn) when red-hot.

Filament bulbs. In an old-style bulb, a thin coil of tungsten gets so hot (about $2500\,^\circ\text{C}$) that it glows white. Tungsten is used because it melts only at about $3400\,^\circ\text{C}$. The bulb is filled with an unreactive gas such as nitrogen or argon, which stops the hot filament from burning and makes it last longer. Even so, most of the energy comes out as heat rather than light — which is why filament bulbs are being replaced everywhere by LEDs that give the same light for about a sixth of the power.

🔌 The electric fuse

A fuse is a short piece of wire, made of a metal or alloy with a low melting point (for example an alloy of tin and lead), placed in a holder of porcelain or glass and connected in series with the appliance or circuit it protects. If the current becomes larger than the fuse's rating — because of a fault, or too many appliances on one circuit — the heat $I^2Rt$ in the fuse wire melts it, the circuit breaks, and the current stops before the wiring or the appliance can overheat or catch fire. Fuses for household use are rated at $1\ \text{A}$, $2\ \text{A}$, $3\ \text{A}$, $5\ \text{A}$, $10\ \text{A}$, $15\ \text{A}$ and so on; the right fuse has a rating a little above the normal current of the appliance. (Modern homes often use a resettable switch, the MCB, for the same job — see Chapter 4.)

✏️ Example 3 — Choosing a fuse
An electric iron rated $1.0\ \text{kW}$ and a room heater rated $2.0\ \text{kW}$ both work on $220\ \text{V}$. Fuses of $3\ \text{A}$, $5\ \text{A}$, $10\ \text{A}$ and $15\ \text{A}$ are available. (a) Which fuse should protect each appliance? (b) Both are plugged into one circuit protected by a $10\ \text{A}$ fuse and switched on together. What happens?
Idea The normal current is $I=\dfrac PV$ (from $P=VI$, explained in the next subsection); the fuse rating must be just above it.
(a) Iron: $I=\dfrac{1000}{220}=4.5\ \text{A}$, so a $5\ \text{A}$ fuse. Heater: $I=\dfrac{2000}{220}=9.1\ \text{A}$, so a $10\ \text{A}$ fuse. (A $3\ \text{A}$ fuse would blow every time the iron is switched on; a $15\ \text{A}$ fuse would let a dangerous fault current through.)
(b) In parallel the currents add: $4.5+9.1=13.6\ \text{A}\gt10\ \text{A}$, so the fuse wire overheats and melts, and both appliances go off. This is overloading.

Electric Power

Two appliances can use energy at very different rates: a phone charger uses a few joules every second, a geyser two thousand. In Class 9 you defined power as the rate of doing work. The same idea applies to electrical energy.

📘 Definition 3.7 — Electric power

Electric power $P$ is the rate at which electrical energy is converted into other forms (heat, light, motion) in a circuit or appliance. For a device with a potential difference $V$ across it and a current $I$ through it, the energy converted in time $t$ is $W=VIt$, so

$$P=\frac Wt=VI .$$

For a resistor, Ohm's law gives two more forms:

$$P=I^2R=\frac{V^2}{R} .$$

The SI unit of power is the watt (W): $1\ \text{W}=1\ \text{J/s}=1\ \text{V}\times1\ \text{A}$, the power of a device that carries $1\ \text{A}$ with $1\ \text{V}$ across it. For larger powers, $1\ \text{kW}=1000\ \text{W}$.

Reading a rating. A bulb marked "$100\ \text{W}$, $220\ \text{V}$" uses energy at $100\ \text{J}$ every second when connected to $220\ \text{V}$. From the rating you can find its current, $I=\dfrac PV$, and its resistance, $R=\dfrac{V^2}{P}$. The resistance stays (roughly) fixed, so on a different voltage the power changes as $V^2$. Note what follows from $R=\dfrac{V^2}{P}$: for appliances made for the same voltage, the higher the power, the lower the resistance.

✏️ Example 4 — Reading a bulb's rating
A bulb is rated $100\ \text{W}$, $220\ \text{V}$. Find (a) the current it draws, (b) the resistance of its hot filament, (c) the energy it uses in $10$ hours.
(a) $I=\dfrac PV=\dfrac{100}{220}=0.455\ \text{A}\approx0.45\ \text{A}$.
(b) $R=\dfrac{V^2}{P}=\dfrac{220^2}{100}=484\ \Omega$. (Check: $\dfrac VI=\dfrac{220}{0.4545}=484\ \Omega$ ✓.)
(c) $W=Pt=100\times(10\times3600)=3.6\times10^6\ \text{J}$.
Meaning That is exactly $1$ "unit" of electricity — the kilowatt-hour of the next subsection: $100\ \text{W}$ for $10\ \text{h}$ is $1\ \text{kW}$ for $1\ \text{h}$.
✏️ Example 5 — The same bulb on a lower voltage
(a) The $100\ \text{W}$, $220\ \text{V}$ bulb of Example 4 is connected to a $110\ \text{V}$ supply. What power does it use (assume its resistance does not change)? (b) Two such bulbs are connected in series across $220\ \text{V}$. Find the power of each, and the total. (c) What if they are connected in parallel across $220\ \text{V}$?
(a) $P=\dfrac{V^2}{R}=\dfrac{110^2}{484}=25\ \text{W}$ — half the voltage gives a quarter of the power.
(b) In series the two equal bulbs share the $220\ \text{V}$: $110\ \text{V}$ each, so $25\ \text{W}$ each and $50\ \text{W}$ in total. (Or: $I=\dfrac{220}{968}=0.227\ \text{A}$, $P=I^2R=0.227^2\times484=25\ \text{W}$.)
(c) In parallel each has the full $220\ \text{V}$: $100\ \text{W}$ each, $200\ \text{W}$ in total — eight times the series total.
Check Series: $R_s=968\ \Omega$, $\dfrac{220^2}{968}=50\ \text{W}$ ✓. Parallel: $R_p=242\ \Omega$, $\dfrac{220^2}{242}=200\ \text{W}$ ✓.

Electrical Energy and the Kilowatt-hour

The energy an appliance uses is its power multiplied by the time it runs, $W=Pt$. In joules the numbers quickly become huge — a $2\ \text{kW}$ geyser running for an hour uses $7\,200\,000\ \text{J}$ — so electricity companies use a much larger unit.

📘 Definition 3.8 — The kilowatt-hour

One kilowatt-hour (kWh) is the electrical energy used in one hour by an appliance of power one kilowatt:

$$1\ \text{kWh}=1000\ \text{W}\times3600\ \text{s}=3.6\times10^6\ \text{J}.$$

It is the commercial unit of electrical energy, called one "unit" on the electricity bill. Energy in kWh $=$ power in kW $\times$ time in hours. Note that the kilowatt-hour is a unit of energy, not of power.

The energy meter fixed where the mains cable enters a house counts the kilowatt-hours used. Older meters have an aluminium disc that turns faster the more power is being drawn; digital meters flash a light for each small amount of energy. The bill is the number of units used in the billing period multiplied by the price per unit (often with a fixed charge added).

✏️ Example 6 — A month's bill
A household runs a $1.5\ \text{kW}$ air conditioner for $6.0$ hours a day, four $10\ \text{W}$ LED lamps for $6.0$ hours a day and a $150\ \text{W}$ refrigerator compressor for $12$ hours a day. Find the energy used in a $30$-day month and its cost at $\text{₹}\,7.50$ per unit.
Per day AC: $1.5\ \text{kW}\times6.0\ \text{h}=9.0\ \text{kWh}$. Lamps: $4\times0.010\ \text{kW}\times6.0\ \text{h}=0.24\ \text{kWh}$. Fridge: $0.150\ \text{kW}\times12\ \text{h}=1.8\ \text{kWh}$. Total $11.04\ \text{kWh}$ a day.
Per month $11.04\times30=331.2\ \text{kWh}\approx331$ units.
Cost $331.2\times7.50=\text{₹}\,2484$.
Meaning The air conditioner alone is $\dfrac{9.0}{11.04}=82\%$ of the bill; the four LED lamps cost only about $\text{₹}\,54$ a month. To save money, look at the high-power appliances first.
Interactive 2 — Fuse, Power and the Electricity Bill

Appliances are connected in parallel across the $220\ \text{V}$ mains, through a fuse and an energy meter. Switch them on and off: the currents add, the fuse wire warms up as the total current approaches its rating — and melts if you overload it. The meter disc turns at a speed proportional to the power; one second here is one hour of use, so the meter counts kilowatt-hours and the bill grows. Play switches appliances on one by one.

7.0
total power—
mains current—
fuse—
energy (units)—
bill—
⚠️ Common mistakes

1. Forgetting to square. $H=I^2Rt$: doubling the current gives four times the heat, not twice.

2. Choosing the wrong form. "Larger resistance, more heat" is true only when the current is the same (series). When the voltage is the same (parallel, or appliances on the mains), $P=\dfrac{V^2}{R}$ and the smaller resistance gives more power.

3. Mixing units. In $W=Pt$ use watts and seconds for joules, or kilowatts and hours for kWh — never watts with hours. $1\ \text{kWh}=3.6\times10^6\ \text{J}$.

4. kW versus kWh. The kilowatt measures power (a rate); the kilowatt-hour measures energy. "The heater uses $2\ \text{kWh}$" means energy; "the heater is $2\ \text{kW}$" means power.

5. Fuse in the wrong place or of the wrong rating. A fuse must be in series (in the live wire) and rated just above the normal current. Replacing a blown fuse with a thick copper wire removes the protection completely.

🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — optional: why power lines use very high voltages

In JEE Main, NEET and JEE Advanced 2026 (electrical power and the heating effect of current); not in the CBSE Class 10 syllabus.

A power station must deliver a power $P=VI$ along long cables of resistance $R_{\text{line}}$. The power wasted as heat in the cables is $I^2R_{\text{line}}$. For the same $P$, raising the voltage lowers the current in proportion, so the loss falls as the square of the voltage. Example: to deliver $22\ \text{kW}$ through cables of total resistance $2.0\ \Omega$, at $220\ \text{V}$ the current would be $100\ \text{A}$ and the loss $100^2\times2.0=20\ \text{kW}$ — almost everything. At $22\ \text{kV}$ the current is only $1.0\ \text{A}$ and the loss is $1.0^2\times2.0=2.0\ \text{W}$. That is why electricity travels across the country at hundreds of kilovolts and is stepped down near your home. (Here we ignore the voltage drop along the line; the devices that change the voltage, transformers, are studied in Class 12 Physics.)

Exercises

I
Joule's law of heating.
  • 1 Find the heat produced in a $10\ \Omega$ resistor carrying $2.0\ \text{A}$ for $30\ \text{s}$.
  • 2 A $110\ \Omega$ heating element is connected to $220\ \text{V}$ for $5.0$ minutes. How much heat does it produce?
  • 3 The current through a resistor is doubled while its resistance and the time stay the same. By what factor does the heat produced change?
  • 4 A heater coil carrying $5.0\ \text{A}$ produces $9000\ \text{J}$ of heat in $30\ \text{s}$. Find its resistance.
  • 5 A $2.0\ \Omega$ and a $4.0\ \Omega$ resistor in series are connected to $6.0\ \text{V}$ for $1.0$ minute. Find the heat produced in each.
  • 6 The same two resistors are connected in parallel across $6.0\ \text{V}$ for $1.0$ minute. Find the heat produced in each.
  • 7 The element of a heater glows red-hot but the cable connecting it to the socket stays cool. Explain.
  • 8 How long does a $50\ \Omega$ resistor carrying $2.0\ \text{A}$ take to produce $6000\ \text{J}$ of heat?
1 $H=2.0^2\times10\times30=1200\ \text{J}=1.2\times10^3\ \text{J}$.
2 $H=\tfrac{V^2}{R}t=\tfrac{220^2}{110}\times300=1.32\times10^5\ \text{J}$.
3 $H\propto I^2$: four times as much.
4 $R=\tfrac{H}{I^2t}=\tfrac{9000}{25\times30}=12\ \Omega$.
5 $I=\tfrac{6.0}{6.0}=1.0\ \text{A}$; $H_2=1.0^2\times2.0\times60=120\ \text{J}$, $H_4=240\ \text{J}$.
6 $H_2=\tfrac{6.0^2}{2.0}\times60=1080\ \text{J}$, $H_4=\tfrac{36}{4.0}\times60=540\ \text{J}$.
7 The element and the cable are in series and carry the same current. The element's resistance is far larger, so by $H=I^2Rt$ it produces nearly all the heat; the low-resistance copper cable produces very little.
8 $t=\tfrac{H}{I^2R}=\tfrac{6000}{4.0\times50}=30\ \text{s}$.
II
Electric power.
  • 1 An appliance draws $0.50\ \text{A}$ from the $220\ \text{V}$ mains. Find its power.
  • 2 Find the resistance of a bulb rated $60\ \text{W}$, $220\ \text{V}$.
  • 3 What current does a $1.1\ \text{kW}$ electric iron draw from $220\ \text{V}$?
  • 4 A car headlamp is rated $48\ \text{W}$, $12\ \text{V}$. Find its current and resistance.
  • 5 Bulbs rated $40\ \text{W}$ and $100\ \text{W}$ are both made for $220\ \text{V}$. Which has the larger resistance? Calculate both.
  • 6 A heater rated $1000\ \text{W}$, $220\ \text{V}$ is used on $110\ \text{V}$. What power does it give (resistance unchanged)?
  • 7 Starting from $P=VI$ and Ohm's law, show that $P=I^2R=\dfrac{V^2}{R}$.
  • 8 Two bulbs rated $100\ \text{W}$, $220\ \text{V}$ are connected in series across $220\ \text{V}$. Find the total power.
Check: enter any two of $V$, $I$, $R$, $P$ (leave the others blank)
$V$—
$I$—
$R$—
$P$—
1 $P=VI=220\times0.50=110\ \text{W}$.
2 $R=\tfrac{V^2}{P}=\tfrac{48400}{60}=807\ \Omega\approx8.1\times10^2\ \Omega$.
3 $I=\tfrac PV=\tfrac{1100}{220}=5.0\ \text{A}$.
4 $I=\tfrac{48}{12}=4.0\ \text{A}$; $R=\tfrac VI=\tfrac{12}{4.0}=3.0\ \Omega$.
5 $R=\tfrac{V^2}{P}$: $40\ \text{W}$: $\tfrac{48400}{40}=1210\ \Omega$; $100\ \text{W}$: $484\ \Omega$. The lower-power bulb has the larger resistance.
6 $R=\tfrac{220^2}{1000}=48.4\ \Omega$; $P=\tfrac{110^2}{48.4}=250\ \text{W}$ — a quarter.
7 $V=IR$ gives $P=(IR)I=I^2R$; $I=\tfrac VR$ gives $P=V\cdot\tfrac VR=\tfrac{V^2}{R}$.
8 Each is $484\ \Omega$; in series $968\ \Omega$; $P=\tfrac{220^2}{968}=50\ \text{W}$.
III
Energy and cost.
  • 1 Express $1\ \text{kWh}$ in joules, showing your working.
  • 2 How many units does a $2.0\ \text{kW}$ geyser use in $1.5$ hours?
  • 3 Five $20\ \text{W}$ LED lamps are used for $8.0$ hours a day for $30$ days. Find the energy used in kWh.
  • 4 Find the cost of the energy in part 3 at $\text{₹}\,6.00$ per unit.
  • 5 A $60\ \text{W}$ fan runs for $2.0$ hours. Find the energy used in joules and in kWh.
  • 6 A $120\ \text{W}$ television is used for $5.0$ hours a day. Find the cost for a $30$-day month at $\text{₹}\,7.00$ per unit.
  • 7 An energy meter reads $4823$ units at the start of a month and $5054$ units at the end. Find the bill at $\text{₹}\,7.50$ per unit.
  • 8 Applied. A $100\ \text{W}$ filament bulb is replaced by a $15\ \text{W}$ LED lamp giving the same light. Both would be used $4.0$ hours a day. How much money is saved in a year ($365$ days) at $\text{₹}\,8.00$ per unit?
1 $1\ \text{kWh}=1000\ \text{J/s}\times3600\ \text{s}=3.6\times10^6\ \text{J}$.
2 $2.0\times1.5=3.0\ \text{kWh}=3.0$ units.
3 $5\times0.020\ \text{kW}\times8.0\ \text{h}\times30=24\ \text{kWh}$.
4 $24\times6.00=\text{₹}\,144$.
5 $W=60\times7200=4.32\times10^5\ \text{J}$; in kWh $0.060\times2.0=0.12\ \text{kWh}$ (and $\tfrac{4.32\times10^5}{3.6\times10^6}=0.12$ ✓).
6 $0.120\times5.0\times30=18\ \text{kWh}$; cost $18\times7.00=\text{₹}\,126$.
7 $5054-4823=231$ units; $231\times7.50=\text{₹}\,1732.50$.
8 Power saved $85\ \text{W}$; energy saved $0.085\times4.0\times365=124.1\ \text{kWh}$; money saved $124.1\times8.00=\text{₹}\,993$ a year.
IV
Fuses, heaters and bulbs.
  • 1 Why is a fuse wire made of a material with a low melting point, and why is it connected in series with the appliance?
  • 2 Which fuse — $3\ \text{A}$, $5\ \text{A}$, $10\ \text{A}$ or $15\ \text{A}$ — should be used for a $1.5\ \text{kW}$ kettle on $220\ \text{V}$?
  • 3 A $2.0\ \text{kW}$ electric oven is plugged into a circuit protected by a $5\ \text{A}$ fuse on $220\ \text{V}$. What happens? Why?
  • 4 Why is the filament of a bulb made of tungsten, and why is the bulb filled with nitrogen or argon rather than air?
  • 5 Challenge. A heater coil is cut to half its length and the half is used on the same mains supply. How does its power change? Is this safe?
  • 6 Challenge. Heaters rated $1000\ \text{W}$ and $500\ \text{W}$ (both $220\ \text{V}$) are connected in series across $220\ \text{V}$. Find the total power and the power of each. Which gives more heat?
1 It must melt quickly when the heat $I^2Rt$ rises above normal, breaking the circuit before the wiring overheats; it is in series so that the whole current of the appliance passes through it and is cut off when it melts.
2 $I=\tfrac{1500}{220}=6.8\ \text{A}$, so the $10\ \text{A}$ fuse (just above the normal current).
3 $I=\tfrac{2000}{220}=9.1\ \text{A}\gt5\ \text{A}$: the fuse wire melts and the oven switches off. The circuit's fuse is too small for this appliance (the circuit is meant for light loads).
4 Tungsten has a very high melting point (about $3400\,^\circ\text{C}$), so it can glow white-hot without melting. Oxygen in air would make the hot filament burn; an unreactive gas prevents this and the filament lasts longer.
5 Half the length means half the resistance; at the same $V$, $P=\tfrac{V^2}{R}$ doubles. It is not safe: the coil carries twice the current, may overheat and burn out, and may blow the fuse.
6 $R_1=\tfrac{220^2}{1000}=48.4\ \Omega$, $R_2=\tfrac{220^2}{500}=96.8\ \Omega$; $R_s=145.2\ \Omega$; $I=\tfrac{220}{145.2}=1.52\ \text{A}$; total $P=\tfrac{220^2}{145.2}=333\ \text{W}$. $P_1=I^2R_1=111\ \text{W}$, $P_2=I^2R_2=222\ \text{W}$: the $500\ \text{W}$ heater gives more heat in series (same current, larger resistance).
🎯 Exam corner — board and entrance-exam practice BoardNEETJEE Main

E1. Board · 3 marks A current $I$ flows through a resistor $R$ for a time $t$. Derive an expression for the heat produced. State how the heat depends on $I$, $R$ and $t$.

Charge In time $t$ a charge $Q=It$ flows through the resistor, which has a potential difference $V$ across it.
Work Work done on the charge $W=VQ=VIt$; in a resistor all of it becomes heat, so $H=VIt$. With $V=IR$, $H=I^2Rt$.
Dependence $H\propto I^2$ (for given $R$, $t$), $H\propto R$ (for given $I$, $t$), $H\propto t$ (for given $I$, $R$). Exam tip: state the condition "for a given…" with each proportionality — it carries the mark.

E2. Board · 3 marks A toaster is rated $880\ \text{W}$, $220\ \text{V}$. Find (a) the current it draws, (b) its resistance, (c) the energy it uses in $15$ minutes, in kWh and in joules.

(a) $I=\dfrac PV=\dfrac{880}{220}=4.0\ \text{A}$.
(b) $R=\dfrac VI=\dfrac{220}{4.0}=55\ \Omega$.
(c) $0.88\ \text{kW}\times0.25\ \text{h}=0.22\ \text{kWh}$; $0.22\times3.6\times10^6=7.92\times10^5\ \text{J}$ (or $880\times900$ ✓). Exam tip: convert minutes to hours for kWh and to seconds for joules.

E3. Board · Assertion–Reason Assertion (A): A fuse is connected in series with the appliance it protects. Reason (R): A fuse wire is made of a material with a high melting point. (A) Both A and R are true and R explains A. (B) Both are true but R does not explain A. (C) A is true but R is false. (D) A is false but R is true.

Idea A is true. R is false: a fuse wire must have a low melting point so that it melts when the current is too large.
Answer (C). Exam tip: high melting point is what a filament needs; low is what a fuse needs — examiners swap them.

E4. Board · Case study A family uses a $2.0\ \text{kW}$ geyser for $1.0$ hour a day and three $75\ \text{W}$ fans for $8.0$ hours a day, all on $220\ \text{V}$. (i) Find the energy used per day. [2 marks] (ii) Find the cost for a $30$-day month at $\text{₹}\,6.50$ per unit. [1 mark] (iii) Should the geyser circuit have a $5\ \text{A}$ or a $10\ \text{A}$ fuse? Give a reason. [1 mark]

(i) Geyser $2.0\times1.0=2.0\ \text{kWh}$; fans $3\times0.075\times8.0=1.8\ \text{kWh}$; total $3.8\ \text{kWh}$ a day.
(ii) $3.8\times30=114$ units; $114\times6.50=\text{₹}\,741$.
(iii) $I=\dfrac{2000}{220}=9.1\ \text{A}$, more than $5\ \text{A}$, so a $10\ \text{A}$ fuse. Exam tip: the rating must be just above the normal current.

E5. NEET Bulbs rated $40\ \text{W}$, $220\ \text{V}$ and $100\ \text{W}$, $220\ \text{V}$ are connected in series across $220\ \text{V}$. Then (A) the $100\ \text{W}$ bulb glows brighter (B) the $40\ \text{W}$ bulb glows brighter (C) both glow equally (D) neither glows at all

Working $R_{40}=\dfrac{220^2}{40}=1210\ \Omega$, $R_{100}=484\ \Omega$. In series the current is the same, so $P=I^2R$ is larger for the larger resistance: $I=\dfrac{220}{1694}=0.130\ \text{A}$, $P_{40}=20\ \text{W}$, $P_{100}=8.2\ \text{W}$.
Answer (B). Exam tip: in series the lower-rated bulb glows brighter; in parallel (each at its rated voltage) the higher-rated one does.

E6. JEE Main A heater coil is cut into two equal halves and only one half is connected to the same supply. The heat produced in the same time becomes (A) half (B) double (C) four times (D) unchanged

Working $R'=\dfrac R2$ and $V$ is fixed, so $H=\dfrac{V^2}{R'}t=2\times\dfrac{V^2}{R}t$.
Answer (B). Exam tip: fixed supply voltage → use $\dfrac{V^2}{R}$, not $I^2R$ (the current changes too).
🔬 Research–Teaching NexusResearch-tutored · optional, about 60 minutes

Research question. How much of the electrical energy taken by an electric kettle actually ends up heating the water — and is it cheaper to boil water in a kettle or on an induction or gas hob?

Why it matters. Boiling water is one of the most common uses of energy in homes around the world. Energy-efficiency labels (the star ratings of India's Bureau of Energy Efficiency, the EU energy label, Energy Star in North America) are built on exactly this kind of measurement: energy in, useful energy out.

What to do.

  1. Read the power rating $P$ on the kettle's label. (With an adult's help, a plug-in energy monitor can measure the actual power and energy.)
  2. Put $1.0\ \text{kg}$ (one litre) of tap water in the kettle, measure its temperature, switch on and time how long it takes to boil.
  3. Calculate the electrical energy used, $E=Pt$, in joules and in kWh, and its cost at your local price per unit.
  4. The energy needed to warm $1.0\ \text{kg}$ of water by $1\,^\circ\text{C}$ is about $4200\ \text{J}$. Calculate the useful energy, $4200\times(100-T_{\text{start}})$ joules, and the efficiency $=\dfrac{\text{useful energy}}{E}\times100\%$.
  5. Repeat with $0.5\ \text{kg}$ of water. Is the efficiency the same? Where does the rest of the energy go?

Tools: electric kettle, measuring jug, kitchen thermometer, stopwatch, optionally a plug-in energy monitor.

Hand in: Your measurements, the energy, cost and efficiency calculations, and a one-paragraph recommendation for saving energy when boiling water.

Going further. Compare with an induction hob or a gas stove (for gas you need the price and energy content of the fuel). Which is cheapest per litre where you live?

Instructor note: a $2.0\ \text{kW}$ kettle typically boils $1.0\ \text{kg}$ from $25\,^\circ\text{C}$ in about $3$ minutes: $E\approx2000\times180=3.6\times10^5\ \text{J}=0.10\ \text{kWh}$; useful energy $4200\times75=3.15\times10^5\ \text{J}$; efficiency about $85$–$90\%$. Smaller volumes give lower efficiency because the kettle body and element must be heated each time. The specific heat capacity is used here only as a given number; it is studied formally in Class 11.

🧪 Self-check quiz

0/5 correct

Q1. The current through a resistor is doubled, with the resistance and the time unchanged. The heat produced becomes

Q2. One kilowatt-hour is equal to

Q3. The resistance of a bulb rated $100\ \text{W}$, $220\ \text{V}$ is

Q4. A fuse wire should have

Q5. Which expression for electric power is not correct?

← 3.3 Resistors in Series and in ParallelChapter 4 — Magnetic Effects of Electric Current →