In short

Whenever a current I flows through a resistance R for a time t, the resistor converts electrical energy into heat at a rate

\boxed{\;P \;=\; VI \;=\; I^{2}R \;=\; \frac{V^{2}}{R}\;}.

The total heat released in time t is

\boxed{\;H \;=\; I^{2}R\,t\;}

— Joule's law of heating. The unit is the joule (watt \times second); in household accounting, energy is billed in kilowatt-hours (1 kWh = 3.6\times10^{6} J).

Microscopically, the kinetic energy that the electric field pumps into drifting electrons between collisions is dumped, at each collision with a lattice ion, into the ion's thermal vibration. The wire warms up.

Three engineering consequences follow:

  • Transmission at high voltage. To deliver a fixed power P = VI, raising V by a factor of 10 lowers I by 10 and therefore I^{2}R losses by 100. India's 765 kV corridors exist for this reason alone.
  • Fuses and MCBs. A fuse is a deliberately weak link whose wire melts when I^{2}R raises its temperature past its melting point. A 15 A fuse lets 15 A pass without melting but fails fast above that.
  • Incandescent lamps. Drive enough current through a tungsten filament to raise its temperature past about 2700 K and roughly 5% of the electrical input emerges as visible light; the other 95% is infrared heat. LED bulbs are simply a way of beating this ratio.

Plug a 1500 W Bajaj immersion rod into a 230 V mains outlet and drop it into a bucket of water. In a quarter of an hour, one litre of water climbs from 25 °C to 70 °C — hot enough for your morning bath. Nothing visible happens inside the rod; no flame, no mechanism, no moving part. A spiral of nichrome wire sits in water and gets hot, and the water gets hot with it. That is the heating effect of current, as ordinary and as universal as it gets. The law that governs it is simple enough to fit on one line, and it was discovered, experimentally, by James Prescott Joule in 1840.

The same law runs the tandoor element in a restaurant kitchen, where a coil of Kanthal wire glows cherry-red at 900 °C to bake roomali rotis in thirty seconds. It runs the nichrome coil in a Morphy Richards pop-up toaster. It runs the tungsten filament of an old Philips 60 W bulb, which operates at about 2800 K — hotter than the melting point of iron. It also runs the filament inside a glass-tube fuse on the main switchboard of your house, which is designed to melt: when a short circuit asks for 50 A from a circuit rated for 15, the fuse vaporises before your wiring does. Every one of these devices uses the same equation in different clothes: P = I^{2}R.

And the same law, turned against the engineer, is the reason the 1200 km Raichur-Delhi transmission line at 765 kV exists. If the grid ran at 230 V (household voltage) instead, delivering the same 4 gigawatts of power would need 17 million amps of current, and the wires would melt before the power got across Maharashtra. Run it at 765 kV instead, and the current drops to 5 kA, the I^{2}R heating falls by a factor of (17{,}000{,}000/5{,}000)^{2} \approx 10{,}000{,}000, and the wires stay cool enough to be made of aluminium and steel cable. The high-voltage grid is engineering as revenge: use P = VI to keep the product constant while pushing V up and I down.

The law in one equation

Before we derive it, let us meet it cleanly.

When a potential difference V drives a current I through a conductor of resistance R, the rate at which electrical energy is converted into heat is

P \;=\; VI.

Apply Ohm's law, V = IR, to rewrite this in two other equally useful forms:

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

Why: all three forms describe the same physical quantity (power, in watts), but each uses different variables. If you know the current through a resistor, use I^{2}R. If you know the voltage across it, use V^{2}/R. If you know both, use VI. The choice is an exam shortcut, not a separate law.

The total heat released in time t is power times time:

H \;=\; P\,t \;=\; I^{2}R\,t \;=\; \frac{V^{2}}{R}\,t.

This is Joule's law of heating. The units work out: [\text{A}]^{2}\cdot[\Omega]\cdot[\text{s}] = \text{A}^{2}\cdot(\text{V/A})\cdot\text{s} = \text{A}\cdot\text{V}\cdot\text{s} = \text{watt-second} = \text{joule}.

Deriving P = VI from the definition of EMF

The shortest derivation — and the one worth seeing first — is three lines.

Let a charge dq cross a resistor from its higher-potential end to its lower-potential end. The potential difference across the resistor is V. So the electric field does work

dW \;=\; V\,dq

Why: the potential difference V between two points is, by definition, the work done per unit charge to move a test charge between them. Multiply by dq and you recover the total work done on that chunk of charge.

Since the charge emerges from the resistor at lower potential than it entered, the energy V\,dq has left the charge. But energy is conserved — and the resistor has no moving parts, no stored field, no mechanical output. The only place the energy can go is into internal random motion of the resistor's atoms, which is to say, into heat.

Divide both sides by the time dt taken for the charge to cross:

\frac{dW}{dt} \;=\; V\,\frac{dq}{dt} \;=\; VI.

Why: dq/dt is, by definition, the current I. The quantity on the left, dW/dt, is the rate at which energy is dissipated — the power.

So the power dissipated in the resistor is

\boxed{\;P \;=\; VI.\;}

This derivation uses nothing more than the definitions of potential and current, plus energy conservation. It does not use Ohm's law. That is important — it means P = VI holds for every circuit element that dissipates energy (a lamp filament, a heating coil, an electrolytic cell, even a non-ohmic conductor), not only for linear resistors. The two other forms, I^{2}R and V^{2}/R, depend on Ohm's law and therefore on R being well-defined and independent of current.

Why does the wire heat up — the microscopic picture

P = VI tells you how much heat comes out, but not why. The microscopic picture is worth telling because it makes resistance feel physical instead of algebraic.

Inside a metal, the valence electrons form a cloud — about 10^{29} per cubic metre of copper. When no voltage is applied, they wander randomly in every direction at thermal speeds of about 10^{6} m/s, with zero net drift. Apply an electric field E = V/L along the wire (where L is the wire's length), and each electron now experiences a force eE between collisions with the ions of the lattice.

Between collisions, the electron accelerates. Call the average time between collisions \tau (the relaxation time, typically \sim 10^{-14} s in copper at room temperature). In that time, the electron picks up a drift velocity

v_{d} \;=\; \frac{eE\tau}{m_{e}}

Why: this is just v = at with a = eE/m_{e} (force over mass) and t = \tau (time between collisions). The electron is accelerating from a zero-drift starting point because each collision randomises its velocity direction.

and therefore an extra kinetic energy

\Delta K \;=\; \tfrac{1}{2}m_{e}v_{d}^{2}.

Then the electron collides with a lattice ion. In the collision, the electron transfers its extra kinetic energy to the ion — which vibrates more strongly. The electron starts over, again picking up v_{d} from the field and again losing it to the ions. Every collision, every electron, dumps \tfrac{1}{2}m_{e}v_{d}^{2} into the lattice.

Stronger lattice vibration is the microscopic definition of higher temperature. So the wire heats up.

Microscopic picture of Joule heatingElectron paths drift slowly rightward through a grid of ions, picking up kinetic energy from the field between collisions, and losing it at each collision — which warms the lattice.E (electric field)electronlattice ions
An electron (red) drifts through the lattice of ions (grey). Between collisions, the field accelerates it and it picks up kinetic energy $\tfrac{1}{2}m_{e}v_{d}^{2}$. At each collision (red dots), it dumps that energy into the ions, which vibrate a little harder. This is Joule heating, atom by atom.

The macroscopic power P = I^{2}R is a sum over all these tiny energy transfers. You can even derive R itself from the electron's mean free path and relaxation time, which is what the Electric Current, Drift Velocity and Mobility chapter does. But the key insight is qualitative: resistance is the lattice's reluctance to let electrons pass without tax, and the tax is collected as heat.

Joule's historical experiment

Joule confirmed H = I^{2}Rt in the 1840s by a direct measurement: a current was passed through a coil immersed in a calorimeter of water, and the temperature rise of the water was recorded. Varying I, R, and t independently, Joule found that the heat released was proportional to the square of the current, to the resistance, and to the elapsed time. The proportionality was to within the experimental error, exact. Combined with the Newton-Joule experiment of the paddle-wheel (which established that mechanical and thermal energy are interconvertible on a 1:1 basis in SI units), this is what made "heat is energy" the standard view of thermodynamics.

For you, the take-home is: H = I^{2}Rt is an experimentally verified law, not a theoretical one. The microscopic derivation above explains the law, but Joule and his calorimeter proved it first.

Why high voltage for transmission — the square that saves India

Take a real problem. The Vindhyachal Super Thermal Power Station in Madhya Pradesh generates 4760 MW. The load is a city 800 km away in Delhi. A transmission line runs between them with a total resistance of about 10\ \Omega. How much power is lost along the line as Joule heat?

The trick is to decide what current flows in the line. Since P_{\text{load}} = V_{\text{load}}\cdot I_{\text{line}}, the same power can be moved at low voltage + high current, or high voltage + low current. The line loss depends only on current through the line resistance:

P_{\text{lost}} \;=\; I_{\text{line}}^{2}\,R_{\text{line}}.

Why: the line's resistance is fixed (set by its copper or aluminium cross-section and length). The heat it dissipates is I^{2}R, and it depends on the current in the line, not the voltage across the load at the end of it.

Case A: transmit at 230 V. Current needed: I = P/V = (4.76\times10^{9})/230 = 2.07\times10^{7} A = 20.7 million amperes. Line loss: I^{2}R = (2.07\times10^{7})^{2} \times 10 = 4.3\times10^{15} W = 4.3 petawatts. That is a million times more than the plant generates. The wires would vaporise before they saw Maharashtra.

Case B: transmit at 765 kV (India's ultra-high-voltage grid). Current needed: I = P/V = (4.76\times10^{9})/(7.65\times10^{5}) = 6200 A. Line loss: I^{2}R = (6200)^{2}\times 10 = 3.8\times10^{8} W = 380 MW, or about 8% of the plant's output. Acceptable.

The ratio 4.3 petawatts to 380 megawatts — eleven orders of magnitude — is the square of the voltage ratio (765{,}000/230)^{2} \approx 1.1\times10^{7}. The reason transformers exist, the reason pylons are tall and insulators fat, the reason India invested in 765 kV corridors in the 2000s — it is all in the single exponent 2 in I^{2}R.

Explore the scaling yourself.

Interactive: power dissipation versus currentThe curve P = I^2 R shown for a 1 ohm resistor. Dragging a slider moves a point along the current axis and shows the instantaneous power. A small increase in current leads to a big increase in dissipated power because of the square.current I (A)power P = I² R (W), R = 1 Ω01002004505101520(for R = 1 Ω)drag to change I
Power dissipation $P = I^{2}R$ plotted against current for a fixed 1 Ω resistor. Drag the red dot: doubling the current quadruples the power. This is the square in $I^{2}R$ — the single reason long-distance transmission uses high voltage.

Fuses and MCBs — the useful side of melting

A fuse is a wire deliberately made from a low-melting-point alloy (tin-lead, for example, with melting point about 200 °C). When current I flows through it, it heats up according to I^{2}R versus the heat lost to the surroundings by conduction and convection. An equilibrium temperature sets in:

\underbrace{I^{2}R}_{\text{input}} \;=\; \underbrace{h\,A\,(T - T_{\text{ambient}})}_{\text{output to surroundings}}

Why: at steady state, the power generated inside the fuse must equal the power lost to the surroundings. The input is I^{2}R from Joule heating; the output is linear in the temperature difference (Newton's law of cooling), with h the heat-transfer coefficient and A the surface area.

Solve for T:

T \;=\; T_{\text{ambient}} + \frac{I^{2}R}{hA}.

If T stays below the melting point, the fuse survives. If I grows large enough that T crosses the melting point, the fuse wire melts, the circuit opens, and the current stops. That moment is called "blowing" the fuse. The rated current of a fuse is the current at which T equals the melting point with a safety margin — for example, a 15 A fuse nominally survives 15 A indefinitely but blows quickly at 20 A and near-instantaneously at 50 A.

Modern Indian households use miniature circuit breakers (MCBs) in place of fuses. An MCB is a switch tripped by a bimetallic strip (for overload) or an electromagnet (for short circuit), both of which respond to Joule heating indirectly. Same physics, replaceable hardware.

Worked examples

Example 1: The Bajaj 1500 W immersion heater

An Indian household runs on 230 V AC, 50 Hz. A Bajaj 1500 W immersion heater is dropped into a bucket holding 5.0 litres of water initially at 25 °C. Assume 90% of the electrical energy actually heats the water (the other 10% warms the bucket and the surrounding air). How long until the water reaches 60 °C? Take the specific heat of water as c_{w} = 4186 J/(kg·°C).

Bucket with immersion heaterA heating coil in a bucket of water. 230 V mains supplies the coil. The rated power is 1500 watts at 230 volts, which fixes the coil resistance through P = V squared over R.230 V, 50 Hz1500 W5.0 L water
A 1500 W immersion heater in a 5 L bucket of water.

Step 1. Find the resistance of the heater.

The heater is rated 1500 W at 230 V. From P = V^{2}/R,

R \;=\; \frac{V^{2}}{P} \;=\; \frac{230^{2}}{1500} \;=\; \frac{52{,}900}{1500} \;\approx\; 35.3\ \Omega.

Why: the "rating" of a heater means the power it dissipates when operated at its rated voltage. The ratio V^{2}/P recovers the resistance of the element.

Step 2. Find the heat needed.

Mass of water: m = 5.0\ \text{kg} (since 1 L of water has mass 1 kg). Temperature rise: \Delta T = 60 - 25 = 35 °C.

Q_{\text{water}} \;=\; mc_{w}\Delta T \;=\; 5.0 \times 4186 \times 35 \;=\; 7.33\times10^{5}\ \text{J}.

Why: the standard calorimetry formula — heat absorbed equals mass times specific heat times temperature rise. It counts only the energy that ends up as heat in the water itself.

Step 3. Account for efficiency.

The heater must deliver Q_{\text{water}}/0.9 = 8.14\times10^{5} J of electrical energy to heat the water by 35 °C, because only 90% of what it dumps reaches the water.

Step 4. Find the time.

The heater dissipates P = 1500 W. Time:

t \;=\; \frac{\text{electrical energy in}}{P} \;=\; \frac{8.14\times10^{5}}{1500} \;\approx\; 543\ \text{s} \;\approx\; 9\ \text{minutes 3 seconds}.

Result. The water reaches 60 °C in roughly 9 minutes.

What this shows. A 1500 W heater dumps H = Pt = 1.5\ \text{kW}\times 0.15\ \text{h} \approx 0.23 kWh of electricity per bucket. At the Delhi-BSES tariff of roughly ₹5 per kWh, a hot bucket costs about ₹1.15 worth of electricity. Joule's law, applied to the Indian middle class, sets the price of warm water.

Example 2: The fuse that saves your wiring

A household kitchen circuit is protected by a 15 A fuse. A toaster (800 W), a microwave (1000 W), and a kettle (1500 W) are plugged into the same circuit. Mains voltage: 230 V. (a) Does the circuit's total current exceed 15 A when all three are running? (b) If a loose neutral wire creates a short that offers an effective resistance of 0.5 Ω, how much current would flow in the absence of the fuse, and how much heat would dissipate in the fuse wire (resistance 0.01 Ω) in the first 10 ms?

Fuse in kitchen circuitA 15 amp fuse in series with three appliances connected across a 230 volt source. Under overload the fuse wire heats up and melts, breaking the circuit.230V15 A fuse800 W1000 W1500 W
A 15 A fuse protects a kitchen circuit with three appliances in parallel.

Step 1. (a) Total current under normal operation.

Total power: P_{\text{total}} = 800 + 1000 + 1500 = 3300 W. Since all three devices are in parallel across 230 V,

I_{\text{total}} \;=\; \frac{P_{\text{total}}}{V} \;=\; \frac{3300}{230} \;\approx\; 14.3\ \text{A}.

Why: Kirchhoff's junction rule — currents in parallel branches add. Since each device's I = P/V at the rated voltage, the total current is the sum of the ratios, which equals the total power over voltage.

This is under 15 A, so the fuse holds — barely.

Step 2. (b) Under a short circuit. If the effective short-circuit resistance (wiring + fault) is 0.5 Ω, the current that would flow (pretending the fuse doesn't blow) is

I_{\text{short}} \;=\; \frac{V}{R_{\text{short}}} \;=\; \frac{230}{0.5} \;=\; 460\ \text{A}.

Why: Ohm's law. With the short in place, the only resistance in the path is the 0.5 Ω fault (plus the tiny 0.01 Ω fuse resistance, negligible in comparison).

Step 3. Heat dissipated in the fuse in the first 10 ms.

H_{\text{fuse}} \;=\; I^{2} R_{\text{fuse}}\,t \;=\; (460)^{2} \times 0.01 \times 0.010 \;=\; 21.2\ \text{J}.

Step 4. Sanity check — does 21 J melt the fuse wire?

A typical household fuse wire is a tin-lead alloy, 2 cm long and 0.3 mm diameter. Its volume is \pi r^{2} L = \pi (1.5\times10^{-4})^{2}(0.02) \approx 1.4\times10^{-9} m³, and its mass (density about 8000 kg/m³) is about 1.1\times10^{-5} kg. To raise it from room temperature (25 °C) to its melting point (~200 °C) requires mc\Delta T = (1.1\times10^{-5})(160)(175) \approx 0.3 J. To actually melt it requires another mL_{f} \approx (1.1\times10^{-5})(40{,}000) \approx 0.4 J for the latent heat of fusion. Total: about 0.7 J.

Why: you need both sensible heat (to warm it up) and latent heat (to melt it). 0.7 J is tiny; the short would deliver 21 J in just 10 ms. The fuse vaporises well before that.

Result. (a) 14.3 A — fuse holds. (b) Under short, 460 A would flow; the fuse absorbs 21 J in 10 ms, many times what it needs to melt. The fuse blows in well under a millisecond.

What this shows. The fuse is engineered so that the fault current heats its tiny wire to the melting point before the same current can do catastrophic damage to the household wiring (which is far more massive and harder to heat up). Joule heating, deployed deliberately, saves the house.

Example 3: Transmission-line loss

A 10 MW plant in Koyna, Maharashtra delivers power to a town 100 km away through a line of total resistance 8 Ω. Compare line losses at (a) 11 kV and (b) 220 kV transmission voltage.

Two-voltage transmission comparisonA 10 MW source supplies a town through a line of 8 ohms. The current through the line is P over V, and the line loss is I squared times the line resistance. The table under the figure shows how raising the voltage from 11 kV to 220 kV reduces the loss by a factor of 400.Plant10 MWline, R = 8 ΩTownloadtransmission voltage V ∈ {11 kV, 220 kV}
Same 10 MW, same 100 km line with 8 Ω total resistance, two different transmission voltages. The only thing that changes is the line current — and therefore, by a factor of $V^{2}$, the line loss.

Step 1. At 11 kV, the line current is

I_{A} \;=\; \frac{P}{V_{A}} \;=\; \frac{10^{7}}{1.1\times10^{4}} \;\approx\; 909\ \text{A}.

Step 2. Line loss at 11 kV:

P_{\text{loss},A} \;=\; I_{A}^{2}R \;=\; (909)^{2}\times 8 \;\approx\; 6.6\ \text{MW}.

Over 66% of the generated power is lost in the line. The scheme is unworkable.

Step 3. At 220 kV, the line current is

I_{B} \;=\; \frac{10^{7}}{2.2\times10^{5}} \;\approx\; 45.5\ \text{A}.

Step 4. Line loss at 220 kV:

P_{\text{loss},B} \;=\; I_{B}^{2}R \;=\; (45.5)^{2}\times 8 \;\approx\; 16.6\ \text{kW}.

Only 0.17% of the power is lost. Transmission at 220 kV is perfectly viable.

Why the ratio: V_{B}/V_{A} = 20, so I_{B}/I_{A} = 1/20, so losses drop by 20^{2} = 400. Exactly as the square in I^{2}R predicts.

Result. (a) At 11 kV, loss = 6.6 MW (66%). (b) At 220 kV, loss = 16.6 kW (0.17%).

What this shows. This is why every Indian city has step-up transformers on the generator side and step-down transformers at distribution. The generator makes power at 11 kV (mechanical limits of rotating machinery), it is stepped up to 220 or 400 or 765 kV for transmission, and stepped back down to 230 V for your geyser. Every conversion wastes a tiny fraction of the power, but you save far more in not-having-to-cool-the-lines.

Common confusions

If you came here to solve Joule-heating problems and understand why transmission lines are high-voltage, you have what you need. What follows covers (a) AC heating and the RMS value, (b) the Peltier effect as the "reverse" of Joule heating, (c) temperature dependence of resistance, and (d) the superconducting limit.

AC heating and the RMS convention

When a sinusoidal AC current i(t) = I_{m}\sin(\omega t) flows through a resistor, the instantaneous power dissipated is

p(t) \;=\; i(t)^{2} R \;=\; I_{m}^{2}R\sin^{2}(\omega t).

Average over one full cycle T = 2\pi/\omega:

\langle p\rangle \;=\; \frac{1}{T}\int_{0}^{T} I_{m}^{2}R\sin^{2}(\omega t)\,dt \;=\; I_{m}^{2}R \cdot \frac{1}{2} \;=\; \tfrac{1}{2}I_{m}^{2}R.

Why: the time average of \sin^{2}(\omega t) over a full period is \tfrac{1}{2}. This is a standard identity; it follows from \sin^{2}\theta = (1-\cos 2\theta)/2 and the fact that \cos 2\theta averages to zero over one period.

Define the root-mean-square (RMS) current I_{\text{rms}} = I_{m}/\sqrt{2}. Then

\langle p\rangle \;=\; I_{\text{rms}}^{2}R.

So I_{\text{rms}} is the equivalent DC current that would dissipate the same average power. The Indian grid's 230 V is V_{\text{rms}}, and the peak voltage is 230\sqrt{2} \approx 325 V. Insulation on household wiring must withstand this peak, not just the "230 V" label.

Peltier effect — Joule heating's sign-dependent cousin

Joule heating is always positive: whether current flows left or right through a resistor, the resistor gets hot. The I^{2} in the formula guarantees the sign. At the junction between two dissimilar metals, however, another effect appears: heat is absorbed or released depending on the direction of the current, and the rate scales as I, not I^{2}. This is the Peltier effect (covered in the next chapter on Thermoelectricity). Because it is linear in I, at small currents the Peltier effect dominates over Joule heating; at large currents, Joule wins. This is why Peltier coolers work — up to a point.

The real tungsten filament

A tungsten filament does not have constant resistance. At room temperature, a filament that will dissipate 60 W at 230 V has a "cold" resistance of only about 70 Ω (compared to its "hot" operating resistance of 881 Ω). The reason: metallic resistance increases with temperature, by roughly \rho(T) = \rho_{0}[1 + \alpha(T - T_{0})] with \alpha_{\text{tungsten}} \approx 4.5\times10^{-3}\ \text{K}^{-1}.

At switch-on, then, a cold bulb has low resistance and draws a large inrush current — often 10 times its steady-state current, for a few tens of milliseconds. The filament heats almost instantly, its resistance climbs, the current falls, and the bulb settles at its rated operating point. This is why incandescent bulbs usually fail at the moment of switch-on rather than during operation: the inrush current is a mechanical and thermal shock that finally snaps an already-thinning filament.

Resistance thermometers and why high-voltage wires sag

The temperature dependence of resistance is actually useful. A platinum resistance thermometer measures temperature by reading the resistance of a thin platinum wire — a linear and accurate reading from −200 °C to 600 °C. Indian Railways use platinum RTDs embedded in motor windings of electric locomotives to detect over-temperature and cut the supply before damage.

The other direction: transmission lines heated by Joule losses get longer (thermal expansion) and hotter, and their resistance grows. A busy summer afternoon in Delhi during a load peak — a 400 kV line running at 2 kA dissipates about 30 MW of heat per 100 km of line, warming the aluminium to 70 °C or more. The line sags 2 metres lower than on a cold night. Engineers must leave enough clearance over roads and rivers to account for this — the line loads itself down under its own Joule heating.

The superconducting limit — no Joule heating at all

For some materials (YBCO ceramics, niobium-titanium alloys) cooled below a critical temperature T_{c}, the resistance drops to identically zero. No resistance means no Joule heating. A persistent current, once started, flows forever without any voltage source. This is superconductivity.

Laboratories at TIFR and IIT Kanpur have run superconducting loops for months with no measurable current decay. In a superconductor, the electrons pair up (Cooper pairs) and move in a collective quantum state that cannot scatter off individual lattice vibrations — the mechanism behind ordinary resistance. The energy cost of cooling to the superconducting state is enormous (liquid nitrogen at 77 K for YBCO, liquid helium at 4 K for niobium), so superconductors are used only where I^{2}R would otherwise be catastrophic: MRI magnets, large particle accelerators, and some long-distance power cables (for example, the 138 kV HTS cable on Long Island in New York, which carries 574 MW through a cryogenic pipe).

For most terrestrial physics, though, the wire warms up. P = I^{2}R is the rule.

Where this leads next