Electrical Energy, Power & Joule Heating

Every device you own is, at bottom, a machine for converting electrical energy at some rate — and that rate has a name, a unit, and a three-way formula that decides half the marks in this chapter. A phone charger works at about 20 watts, an electric kettle at 2000, a railway locomotive at several million; the difference is not what electricity does but how fast it does it. Power is the tempo of the energy transfer, and $P = VI$ sets that tempo from two numbers you already own: the energy each coulomb carries, and how many coulombs arrive each second. Buried in that product is a genuinely consequential freedom. The same power can be delivered as high voltage with a trickle of current, or low voltage with a torrent — and the choice matters enormously, because the heat wasted in any wire grows as the square of the current through it. That single quadratic decides why power grids run at hundreds of kilovolts, why a fuse is rated in amperes, and why the thin filament inside a toaster glows while the thick cable feeding it stays cold. This module builds the master formula from first principles, derives its two resistive costumes $I^2R$ and $V^2/R$ along with the doctrine of which one when — the doctrine most students learn only after losing marks to it — quantifies Joule heating, defines the kilowatt-hour your electricity meter counts, and settles the most famous brightness question in exam history: what really happens when a 60 W and a 100 W bulb share a series circuit.

Key takeaways

Electric power is the rate of energy conversion: $P = W/t$ , measured in watts ( $1\ \mathrm{W} = 1\ \mathrm{J/s}$ ).
The master formula $P = VI$ holds for any component — resistor, motor, LED — because volts are joules per coulomb and amperes are coulombs per second.
For resistors, Ohm's law gives two more costumes: $P = I^2R$ (current known — series thinking) and $P = V^2/R$ (voltage known — parallel/mains thinking).
Joule heating: a resistor carrying current $I$ for time $t$ delivers heat $H = I^2Rt$ — the working principle of heaters, fuses and filament bulbs.
The kilowatt-hour is an energy unit, not a power unit: $1\ \mathrm{kWh} = 3.6 \times 10^6\ \mathrm{J}$ — the "unit" your electricity meter counts.
Key idea: joules per coulomb times coulombs per second is joules per second — every power formula is P = VI wearing an Ohm's-law disguise.

The core insight

Electric power $P = VI$ follows directly from the definitions of potential difference (energy per charge) and current (charge per time); substituting Ohm's law yields $P = I^2R$ and $P = V^2/R$ , and choosing the form whose variables are genuinely fixed for the component in question is the entire craft of power problems.

Where this lives today

Power is the binding constraint of the digital age. Artificial-intelligence data centres are planned in gigawatts — a single campus can demand the output of a large power station — and the limiting engineering problem is exactly this module's: delivering colossal $P$ at manageable $I$ , then removing the $I^2R$ heat. Inside every chip the same battle repeats in miniature: a modern processor sinks hundreds of watts at about one volt, which by $P = VI$ means hundreds of amperes squeezed through millimetres of copper. Chip designers publish "power maps" of their silicon the way meteorologists publish heat maps. At the other end of the scale, an electric-vehicle fast charger advertises its physics on the sign: 800 volts at 500 amperes is 400 kilowatts, the high-voltage choice made precisely so the cable stays light enough for one hand. And every one of those machines is billed, in the end, in Watt-and-Joule's units: kilowatts of tempo, kilowatt-hours on the meter.

The physical picture

Picture the circuit as a delivery fleet. Each coulomb of charge is a courier; the battery loads every courier with the same parcel of energy — $V$ joules apiece, that is what "potential difference" means; and the current is the traffic count — $I$ couriers passing per second. The rate at which energy arrives at a component is then inevitable: parcels per courier times couriers per second, $VI$ joules per second. The couriers themselves are never consumed — the same charges cycle endlessly, as Electric Current & Charge Flow established — but every parcel is dropped off: as heat in a resistor, light in an LED, motion in a motor. Where exactly does a resistor's parcel go? Drift Velocity & the Microscopic Picture drew the scene: electrons accelerate briefly in the field, slam into vibrating lattice ions, and hand over their gained energy collision by collision. The lattice shakes harder — the wire warms. That is Joule heating, and it is not a side effect; in a heater, a fuse, a filament, it is the entire job description. The fleet picture also explains the grid in one glance. You can ship 1000 joules per second as a thousand couriers with one-joule parcels, or one courier with a thousand-joule parcel. In the wire, what hurts is traffic: heating grows as the square of the current. Fewer, heavily-laden couriers — high voltage, low current — slip through nearly loss-free. That is the one-line reason transmission lines run at 400,000 volts; the full grid story, transformers and all, belongs to the AC chapter.

Figure

Power is parcels times traffic — two ways to deliver 1000 watts

From joules per coulomb to watts

Two devices can convert the same total energy and still be utterly different machines: a phone charger and a lightning bolt both handle megajoules — one spread over a night, one crammed into a millisecond. What separates them is the tempo, and physics names it by dividing energy by time:

P = \frac{W}{t} \qquad (1)

where: $P$ — power, in watts ( $1\ \mathrm{W} = 1\ \mathrm{J/s}$ ); $W$ — energy converted, in joules; $t$ — the time taken, in seconds. Every wattage rating you have ever read — a 10 W LED bulb, a 2000 W kettle — is this fraction, a promise of so many joules per second for as long as the device runs. Larger units earn their keep fast: $1\ \mathrm{kW} = 10^3$ W, $1\ \mathrm{MW} = 10^6$ W.

Derivation AThe master formula from two definitions

We want $P = VI$ built from things already established, with no new physics — the derivation every board expects when it says "obtain the expression for electric power." The strategy: track the energy delivered to a component by following the charge through it, using only the definition of potential difference from Electric Potential & Potential Difference and the definition of current from Electric Current & Charge Flow. Watch how the coulombs cancel in the final step — that cancellation is the whole proof.

Step 1. Let a steady current $I$ flow through a component with potential difference $V$ across it, for a time $t$ . The charge that passes through is $Q = It$ — the definition of current, run backwards.

Step 2. Each coulomb passing through the component drops $V$ joules there, because potential difference is energy transferred per unit charge. So the energy delivered in time $t$ is $W = QV = VIt$ . Why the energy is "dropped" in the component: between its two terminals the charge falls through a potential difference V, and the electrical energy it loses must appear in that component — as heat, light or work — since it has nowhere else to go.

Step 3. Divide by the time. $P = W/t = VIt/t$ , giving the master formula:

P = VI \qquad (2)

Step 4. Check the units do the work: $\mathrm{(J/C)} \times \mathrm{(C/s)} = \mathrm{J/s}$ . The coulombs cancel, watts remain — the key idea of the module, verified dimensionally. Verified: with V = 230 V and I = 5.0 A, equation (2) gives 1150 W, and Step 2's route gives the same via W = 230 × 5.0 × 1 = 1150 J each second — the two readings of the formula agree.

Equation (2) is the general law — it holds for a motor, an LED, a battery being charged, anything with a voltage across it and a current through it, ohmic or not. For resistors, Ohm's law lets $V$ and $I$ trade places, and the exam-critical costumes appear.

Go deeper · Derivation B · 1 minThe two resistive costumes and Joule's heating law

Derivation BThe two resistive costumes and Joule's heating law

We want the resistor-only forms of power and the heat they predict — the formulas behind every heater, fuse and filament problem. The strategy is bare substitution of $V = IR$ (valid for the resistor in question, per Ohm's Law & Resistance) into the master formula, once each way, then multiplying by time to get the heat. The thing to watch is jurisdiction: each substituted form is true only when its $V$ , $I$ and $R$ all belong to the same resistor.

Step 1. Substitute $V = IR$ into $P = VI$ : the voltage disappears in favour of the current, $P = (IR) \times I$ :

P = I^2R \qquad (3)

Why this is the "series lens": elements in series share one current, so when I is the quantity you know — a wire in a chain, a cable carrying a load's current — equation (3) prices each element's share directly.

Step 2. Substitute $I = V/R$ instead: now the current disappears, $P = V \times (V/R)$ :

P = \frac{V^2}{R} \qquad (4)

Why this is the "parallel lens": elements in parallel share one voltage — every appliance plugged into the 230 V mains sees the same V — so when V is the fixed quantity, equation (4) reads off each branch's power at sight.

Step 3. Multiply the resistive power by the running time to get the energy delivered as heat. A resistor carrying current $I$ for time $t$ produces

H = I^2Rt \qquad (5)

where: $H$ — heat, in joules; this is Joule's law of heating, established experimentally by James Prescott Joule in 1841. Doubling the current quadruples the heating rate; doubling the time merely doubles the heat. The collisions of the drift picture are the microscopic machinery delivering it. Verified: (3) and (4) agree with each other and with (2) through V = IR — for V = 230 V, I = 5.0 A, R = 46 Ω: VI = 1150 W, I²R = 25 × 46 = 1150 W, V²/R = 52900/46 = 1150 W. Equation (5) over 40 minutes gives 1150 × 2400 = 2.76 × 10⁶ J, recomputed below.

The doctrine that decides marks: choose the form whose variables are genuinely fixed for the component in question. Series problem, shared current → think $I^2R$ . Mains appliance or parallel branch, shared voltage → think $V^2/R$ . Anything else, or any non-resistor → retreat to the always-true $P = VI$ . Nearly every "paradox" in this topic (does power rise or fall with $R$ ?) is two students using different lenses on different setups — with $I$ fixed, power grows with $R$ ; with $V$ fixed, power shrinks with $R$ . Both are right, in their own jurisdictions.

The kilowatt-hour: the unit your meter counts

Power ratings run the show, but your electricity provider bills you for energy — rate times duration. Running equation (1) backwards, a device of power $P$ operating for time $t$ consumes $W = Pt$ . In SI units that arrives in joules, and joules are inconveniently tiny for household bookkeeping: one kettle boil is already half a million of them. So the meter on your wall counts a larger unit — the energy of one kilowatt flowing for one hour:

1\ \mathrm{kWh} = 1000\ \mathrm{W} \times 3600\ \mathrm{s} = 3.6 \times 10^6\ \mathrm{J} \qquad (6)

where: the kWh — called simply a "unit" on Indian and UK bills — is an energy unit, despite the watt in its name. The conversion habit worth automating: $P$ in kilowatts times $t$ in hours gives kWh directly, no factors needed. A 2 kW heater running 90 minutes consumes $2 \times 1.5 = 3.0$ kWh. What a month of such habits costs, and how the meter is read, is the business of Household Circuits, kWh & Electrical Safety, next.

Fuses: the heating law hired as a bodyguard

Joule heating has a dark side — an overloaded wire is a fire waiting to happen — and its own cure. A fuse is a short length of wire with a deliberately low melting point, chosen so that Joule heating melts it when the current exceeds its rating. The rating is a current, in amperes: a "5 A fuse" carries 5 A indefinitely but melts soon above it, breaking the circuit before the cables cook. Choosing one is a power calculation: an appliance of power $P$ on mains voltage $V$ draws $I = P/V$ , and the right fuse is rated just above that — for a 1150 W iron on 230 V mains drawing 5.0 A, a 6 A fuse is correct; a 13 A fuse would legally carry a lethal fault. Where the fuse sits in the household circuit, and why it must be in the live wire, is Household Circuits' story.

Limiting cases

case	result	meaning
fixed $V$ , $R \to \infty$	$P = V^2/R \to 0$	an open circuit converts nothing
fixed $V$ , $R \to 0$	$P = V^2/R \to \infty$	why a short across the mains is violent
fixed $I$ , $R \to 0$	$P = I^2R \to 0$	an ideal connecting wire wastes nothing
current doubled, same $R$	$P$ quadruples	heating rides on the square — the fuse's working principle
same $P$ shipped at 100× the voltage	current ÷100, wire heating ÷10,000	the one-line case for high-voltage transmission (full story in the AC chapter)
non-ohmic device (motor, LED)	only $P = VI$ applies	the resistive costumes inherit Ohm's-law jurisdiction

Common trapA 100 W bulb always outshines a 60 W bulb

The rating on a bulb is a conditional promise: 100 W of conversion when the bulb sees its full rated voltage. Plug both bulbs into separate mains sockets (parallel) and the promise holds — the 100 W bulb wins. But wire them in series across the mains and everything reverses. The rated resistances tell the story: $R = V^2/P$ , so the 60 W bulb has the larger resistance ( $806.7\ \Omega$ against $484\ \Omega$ for the 220 V ratings the worked example below runs). In series both carry the same current, the $I^2R$ lens applies, and the larger resistance — the 60 W bulb — dissipates more and glows brighter. Ratings rank bulbs only at rated voltage; circuits rank them by which variable they actually share.

Figure

The series showdown — same current through both bulbs, the higher resistance wins the glow

Worked examples

Board level — the electric iron, end to end (CBSE/ISC pattern, 3 marks)

An electric iron is rated $1150\ \mathrm{W}$ at $230\ \mathrm{V}$ . Find (a) the current it draws and the resistance of its element, (b) the heat produced in 40 minutes of use, in joules and in kWh, and (c) a suitable fuse rating from the available set 2 A, 6 A, 13 A.

Step 1. Part (a), current from the master formula: $I = P/V = 1150/230 = \boxed{5.0\ \mathrm{A}}$ . Resistance from Ohm's law: $R = V/I = 230/5.0 = \boxed{46\ \Omega}$ . (Cross-check with the parallel lens: $V^2/R = 230^2/46 = 1150$ W ✓.)

Step 2. Part (b), heat by Joule's law, equation (5): $H = I^2Rt = (5.0)^2 \times 46 \times (40 \times 60) = 25 \times 46 \times 2400 = \boxed{2.76 \times 10^6\ \mathrm{J}}$ .

Step 3. The same energy in billing units, by equation (6): $H = 2.76 \times 10^6 / 3.6 \times 10^6 = 0.77\ \mathrm{kWh}$ — or directly, $1.15\ \mathrm{kW} \times \tfrac{2}{3}\ \mathrm{h} = \boxed{0.77\ \mathrm{kWh}}$ . The two routes agreeing is your built-in arithmetic check.

Step 4. Part (c): the iron draws 5.0 A steadily, so the fuse must tolerate that and no more than necessary — $\boxed{6\ \mathrm{A}}$ . A 2 A fuse melts at switch-on; a 13 A fuse would sit silent through a 2.9 kW fault.

Step 5. The sentence that earns the reasoning mark: "The fuse rating must exceed the normal operating current (5.0 A) but be small enough to melt promptly on overload; hence 6 A."

Competitive level — the series bulbs showdown (JEE-Main/NEET pattern, ~3 minutes)

Two bulbs rated $60\ \mathrm{W},\ 220\ \mathrm{V}$ and $100\ \mathrm{W},\ 220\ \mathrm{V}$ are connected in series across a $220\ \mathrm{V}$ supply. Find the power actually dissipated in each, and state which glows brighter.

Step 1. Extract the resistances from the ratings — the rating means "this power at this voltage," so $R = V_{\text{rated}}^2/P_{\text{rated}}$ : $R_{60} = 220^2/60 = 806.7\ \Omega$ and $R_{100} = 220^2/100 = 484\ \Omega$ . Note the inversion at once: the lower-wattage bulb has the higher resistance — this is the question's entire trap.

Step 2. Series circuit → one shared current: $I = \dfrac{220}{806.7 + 484} = \dfrac{220}{1290.7} = 0.170\ \mathrm{A}$ .

Step 3. Apply the series lens $P = I^2R$ to each: $P_{60} = (0.170)^2 \times 806.7 = \boxed{23.4\ \mathrm{W}}$ and $P_{100} = (0.170)^2 \times 484 = \boxed{14.1\ \mathrm{W}}$ .

Step 4. Verdict: $\boxed{\text{the 60 W bulb glows brighter}}$ — at less than half its rating, while the "100 W" bulb limps at 14 W. Ratio shortcut worth memorising: in series, $P_{60}/P_{100} = R_{60}/R_{100} = 100/60 = 5/3$ — the powers swap in the inverse ratio of the ratings, no current computation needed.

Step 5. Sanity audit: total dissipated $= 23.4 + 14.1 = 37.5$ W, and $VI = 220 \times 0.170 = 37.5$ W ✓. Both bulbs together now convert barely a third of what the 100 W bulb alone would at full mains. Timing note: with the inverse-ratio shortcut this is a 45-second MCQ — series powers scale as the resistances, which scale as the inverse ratings. Compute actual wattages only if the options demand numbers; the "which is brighter" version needs no arithmetic at all once you can say "series → same I → bigger R wins, and the smaller rating owns the bigger R."

Exam traps

The wrong-V blunder. $P = V^2/R$ demands the voltage across that resistor, not the supply voltage of the circuit it sits in. In a series chain, no resistor sees the full supply; applying $V_{\text{supply}}^2/R$ to one element can overstate its power by orders of magnitude. When in doubt, find the current first and use $I^2R$ — current is harder to misattribute.

kW vs kWh. One is a rate, the other an amount. "A 2 kW heater used 2 kW of energy" is meaningless and mark-losing; it converts at 2 kW and consumes 2 kWh per hour of running. The meter counts kWh; the rating plate quotes kW.

Ratings treated as constants. "60 W" travels with "220 V" — detach the bulb from its rated voltage and the wattage promise is void, though the resistance $V_r^2/P_r$ survives (at operating temperature) as the useful number. Every series-bulb question is this trap in costume.

Lens-swapping mid-problem. Deciding power "increases with resistance" or "decreases with resistance" without first fixing which of $I$ or $V$ is held constant. Series (shared $I$ ): $P \propto R$ . Parallel (shared $V$ ): $P \propto 1/R$ . State the lens, then compute.

Time units in $H = I^2Rt$ . Joule's law wants seconds. Minutes and hours sneak in through the question's phrasing ("40 minutes of use") and quietly wreck the exponent; convert before substituting, and report big answers in scientific notation or kWh.

Fuse logic reversed. The fuse rating goes just above the normal current, never below (instant blowing) and never far above (no protection). Compute $I = P/V$ first; then choose from the offered set.

Constants and conventions. All boards here use $1\ \mathrm{kWh} = 3.6 \times 10^6$ J; Indian and UK papers take mains as 230 V (many classic problems still use 220 V — follow the question's value, never your memory). No board in this module needs $g$ .

Origins · 1 minJames Watt and how this was discovered

James Watt

1736–1819 · Glasgow and Birmingham

James Watt never saw an electric circuit, yet his name sits on every light bulb on Earth. The Scottish instrument-maker's improved steam engine — with its separate condenser, patented 1769 — multiplied the engine's efficiency and powered the Industrial Revolution; to sell it, Watt invented the first commercial unit of power, the horsepower, so mine owners could count how many horses each engine replaced. A century later, electrical engineers needed a unit for the new energy currency flowing through their wires, and the 1882 British Association meeting chose to honour him: one watt, one joule per second — with the electrical form given by this module's $P = VI$ . The unit outgrew its steam-age godfather a billionfold: we now rate phone chips in milliwatts and data-center campuses in gigawatts, all in the name of a man who measured power in horses.

History

1769

— Watt patents the separate-condenser steam engine; efficiency becomes an engineering quantity

1782

— Watt defines horsepower to sell engines against horses — the first commercial unit of power

1841

— Joule establishes the I²R heating law, pinning down where electrical energy goes in a resistor

1882

— The British Association adopts the watt as the unit of power, honouring Watt

1896

— Niagara Falls electricity reaches Buffalo 32 km away — high-voltage transmission proves itself at scale

Going further

Texts. NCERT Physics Class 12, Ch. 3 (electrical energy and power) · NCERT Class 10 Science, Ch. 12 (heating effect — the CBSE home of fuses and kWh) · HC Verma Vol. 2, Ch. 32–33 · CIE 9702 Topic 9.1 (P = VI, W = VIt) · IGCSE 0625 Section 4.2.5.

Provenance. All results standard. A verifier should recompute: derivation checks $VI = I^2R = V^2/R = 1150$ W at 230 V, 5.0 A, 46 Ω; the iron example $I = 5.0$ A, $R = 46\ \Omega$ , $H = 25 \times 46 \times 2400 = 2.76 \times 10^6$ J $= 0.77$ kWh, fuse 6 A; the bulbs example $R_{60} = 806.7\ \Omega$ , $R_{100} = 484\ \Omega$ , $I = 0.170$ A, $P_{60} = 23.4$ W, $P_{100} = 14.1$ W, total 37.5 W $= 220 \times 0.170$ ; and $1\ \mathrm{kWh} = 3.6 \times 10^6$ J.

Prerequisites. Ohm's Law & Resistance · Resistors in Series & Parallel · Electric Current & Charge Flow. Leads to. Household Circuits, kWh & Electrical Safety · Bulbs & Ratings — Brightness Logic (the archetype gym, coming up in E3) · P = V²/R Ratio Craft (the E3 technique drill) · the transmission story in the Alternating Current chapter.

Why are there three formulas for the same thing? Which one is THE formula?

$P = VI$ is the fundamental one — it comes straight from the definitions of voltage and current and holds for any component whatsoever. The other two are conveniences manufactured by substituting Ohm's law, so they carry Ohm's-law conditions: resistors only, and all three symbols belonging to the same resistor. Exams love them because each is natural in a different circuit: $I^2R$ where current is shared (series), $V^2/R$ where voltage is shared (parallel, mains). Learn one derivation and the choosing doctrine, and the "three formulas" collapse into one idea with two costumes.

Does a resistor use up current or energy?

Energy — never current. The current entering a resistor leaves it at exactly the same value; charge is conserved and has nowhere to accumulate. What each coulomb loses in transit is energy: it arrives carrying more electrical potential energy than it leaves with, and the difference — $V$ joules per coulomb — becomes lattice vibrations, i.e. heat. That is why power is $VI$ : energy lost per coulomb, times coulombs per second. "The bulb used 0.26 A" is meter-reading language; physically a 60 W bulb on 230 V mains used the 60 joules per second those 0.26 amperes delivered.

If heat is I²R, shouldn't the thick copper cable heat more than the thin heater coil? It carries the same current.

Same current, wildly different $R$ — and that is the whole design. The heater's nichrome coil is deliberately thin and resistive (tens of ohms), so $I^2R$ is large there; the copper supply cable is thick and short (hundredths of an ohm), so its share of the heating is a thousand times smaller. The same logic runs the filament bulb: the filament glows white-hot while the leads stay cool. Where engineers want the heat, they concentrate the resistance; where they don't, they fatten the conductor — Resistivity & Geometry's $R = \rho L/A$ is the design dial.

Is a kilowatt-hour a kind of kilowatt?

No — the name misleads by containing "watt." A kilowatt is a rate (1000 joules each second); a kilowatt-hour is an amount (the total delivered by that rate sustained for an hour: 3.6 million joules). Your bill charges for amounts, so it counts kWh — the "units" on an Indian meter are exactly kilowatt-hours. Quick self-test: a 100 W bulb burning for 10 hours uses 1 kWh, the same amount as a 2 kW kettle running half an hour. Different tempos, same energy, same charge on the bill.

Why are fuses rated in amperes rather than watts?

Because the fuse wire melts by its own Joule heating, $I^2R_{\text{fuse}}$ , which depends only on the current threading it — the fuse has no idea what voltage the appliance runs at or what power it converts. Current is also the quantity shared along the whole series path, so one fuse in the live wire monitors everything downstream. That is why selecting a fuse starts by converting the appliance's power rating into its operating current, $I = P/V$ , and picking the next rating above.

The 60 W bulb wins in series. Does anything change in parallel?

Everything. In parallel each bulb gets the full mains voltage — exactly the condition its rating was written for — so each delivers its rated power and the 100 W bulb glows brighter, as intuition expected. The series inversion happens because the shared quantity flips from voltage to current, and with it the deciding formula flips from $V^2/R$ to $I^2R$ . One pair of bulbs, two circuits, opposite verdicts: this is the cleanest illustration in the syllabus that power ratings describe a bulb-plus-voltage combination, not the bulb alone.

Where exactly does the electrical energy go when a wire heats up?

Into the vibrations of the metal's ion lattice, collision by collision. The field accelerates each free electron for a fraction of a picosecond; the electron slams into an ion and hands over the kinetic energy it just gained; the ion vibrates harder. Summed over some $10^{23}$ collisions per second, that microscopic stumbling is macroscopic heat — the wire's temperature rises until it sheds energy to its surroundings as fast as the current feeds it in. A filament reaches its glowing equilibrium near 2500 °C; a properly sized cable, a few degrees above room temperature.

My charger says 20 W but the phone reports charging at 15 W. Who is lying?

Nobody — you are reading two different points of the energy chain. The rating is the charger's maximum deliverable power; the phone draws what its battery-management chip requests, which varies with charge level and temperature (fast at 20%, trickling near 100%). Along the way, conversion losses in the charger and $I^2R$ heating in the cable eat a few percent — which is why both feel warm. Rated power is a ceiling; delivered power is a negotiation; the difference is mostly Joule heat, in exactly this module's formulas.