The Grid: Why Transmit at High Voltage

Here is an engineering riddle with a one-line answer that reorganized civilization. A power station generates 800 kilowatts for a town 15 kilometres away. The transmission line has a resistance of 15 ohms — good thick copper, nothing shoddy. Send the power at 4,000 volts, and by the time it arrives, 600 of the 800 kilowatts have vanished — three-quarters of the station's output spent heating the countryside. Send the same power down the same wires at 40,000 volts instead, and the loss collapses to 6 kilowatts. Less than one percent. Nothing about the wire changed; only the voltage did. The riddle's answer is two lines of algebra with a square in exactly the right place. Delivering power $P$ at voltage $V$ requires current $I = P/V$ — so a higher voltage means fewer amperes carrying the same watts. And the line's toll is charged on amperes: $P_{\text{loss}} = I^2R$ . Put the two together and the loss runs as $P^2R/V^2$ — multiply the voltage by 100 and the line loss divides by 10,000. That inverse square is the entire economic case for the pylons marching across every landscape on Earth, the reason the wires overhead run at hundreds of kilovolts, and the reason the transformer — the machine that makes voltage changeable at will — won the war of the currents for AC. This module derives the lever, runs the numbers honestly, walks the grid's ladder from power station to wall socket, and answers the obvious objection ("why not just use thicker wire?") with the arithmetic that buries it.

Key takeaways

To deliver power $P$ at transmission voltage $V$ , the line must carry $I = P/V$ : raising the voltage lowers the current for the same delivered power.
The line's loss is resistive heating: $P_{\text{loss}} = I^2R = \dfrac{P^2R}{V^2}$ — inverse square in the voltage.
×10 the voltage → ÷100 the loss; ×100 the voltage → ÷10,000 the loss. This lever is why grids transmit at hundreds of kilovolts.
Transformers make the lever usable: step up at the station, step down in stages near the consumer — which is why the grid runs on AC.
Thicker wire is the wrong fix: halving $R$ costs double the copper for a mere ×2 improvement, while the $V^2$ lever gives ×100 for the price of transformers.
Key idea: the wire charges by the ampere, not by the watt — so ship the watts as volts, and spend amperes only for the last mile.

The core insight

Line loss for delivered power $P$ at voltage $V$ through resistance $R$ is $P_{\text{loss}} = P^2R/V^2$ : since the wire's toll $I^2R$ is charged on current and $I = P/V$ , stepping the voltage up slashes the current and the loss quadratically — the transformer-enabled lever that makes continent-scale electricity economic and dictated the grid's high-voltage, AC architecture.

Where this lives today

India's grid is the module's formula at subcontinental scale: five regional grids synchronized into one national machine, moving power from coal belts and hydro valleys across thousands of kilometres at 400 and 765 kV — voltages chosen by exactly the $P^2R/V^2$ bookkeeping of the worked examples, run against the price of conductor and insulation. The newest inter-regional links go further and use high-voltage DC (the ±800 kV corridors from the northeast) — power electronics stepping DC where transformers cannot, because at those distances even the AC line's incidental costs matter; the $I^2R$ law itself never stops being the enemy. The lever also explains the landscape's furniture. Pylons are tall and their insulator strings long because hundreds of kilovolts demand metres of clearance; the fat conductors are often aluminium bundles (cheap per kilogram) around steel cores (for strength) — resistance bought down to the point where the voltage does the real work. And the last rungs of the ladder are within sight of your window: the 11 kV feeder along the road, the grey transformer drum on its pole, and the 230 V service line into the house — where Household Circuits picks up the story at the meter. Within this track, the chapter now turns back to circuit craft: the LCR Numericals Ladder, the archetype gym that drills everything the AC modules built.

The physical picture

Electricity is sold in watts but hauled in amperes. Every metre of a transmission line has resistance, and current through resistance means heat — the $I^2R$ law that Electrical Energy, Power & Heating established and every heater exploits. On a transmission line, though, that heat is pure freight cost: energy generated, paid for, and spent warming birds' feet between the station and the town. The freight cost cares only about the current. And here is the leverage: for a fixed delivery of power, the current is negotiable. Power is volts times amps — $P = VI$ — so the same 800 kW can travel as 200 amperes at 4 kV, or as 20 amperes at 40 kV, or as 2 amperes at 400 kV. The customer receives identical watts in every case. But the line, charging its toll per ampere squared, treats the three shipments completely differently: cut the current tenfold and the heating falls a hundredfold. So the shipping strategy writes itself. Convert the watts into the high-voltage, low-current form for the journey; convert back to safe, usable low voltage at the doorstep. The device that performs both conversions — cheaply, passively, at 99% efficiency — is the previous module's transformer. That is the sentence that decided the war of the currents: AC won not because AC is intrinsically superior, but because transformers only work on AC, and without transformers there is no voltage lever, and without the lever, no economic long-distance power. Edison's DC empire, locked to the low voltage of its lamps (that story is Household Circuits' ancestor, told in the current-electricity chapter), could not serve customers more than a couple of kilometres from a generator. The lever could.

Figure

The same 800 kW shipped two ways — the voltage lever in action

The flagship derivation: the P²R/V² lever

Derivation AWhy line loss falls as the square of the transmission voltage

The goal is the loss formula $P_{\text{loss}} = P^2R/V^2$ , with every symbol's meaning pinned down — because the formula is only two substitutions, and the marks (and the understanding) live in narrating them correctly. The strategy: express the line current in terms of what the customer actually needs, then feed that current to the wire's heating law. The thing to watch is which power is which: $P$ is the power being delivered; $P_{\text{loss}}$ is the extra the station must generate to pay the wire.

Step 1. Fix what the shipment must achieve: power $P$ delivered at transmission voltage $V$ . The current the line must carry follows from the definition of electrical power, $P = VI$ :

I = \frac{P}{V}

where: $P$ is the delivered power (watts), $V$ the voltage at which the line operates. This is the negotiable step — the same $P$ can ride on a large current at low voltage or a small current at high voltage. The customer cannot tell the difference; the wire can.

Step 2. Charge the wire's toll. The line, resistance $R$ , heats according to the $I^2R$ law from Electrical Energy, Power & Heating — applied here, not re-derived:

P_{\text{loss}} = I^2R = \left(\frac{P}{V}\right)^{2}R = \frac{P^2R}{V^2}

where: $R$ is the total resistance of the line (both conductors, the full round trip). Read the exponents like an engineer: the loss grows as the square of the power shipped (congestion punishes quadratically), linearly with the wire's resistance — and falls as the inverse square of the voltage. The voltage is the only lever with a square in the helpful direction.

Step 3. Price the lever. Multiply $V$ by 10: the loss divides by $10^2 = 100$ . Multiply by 100: the loss divides by 10,000. No other parameter offers anything comparable — halving $R$ (double the copper, roughly double the cost of every kilometre of conductor) merely halves the loss, once.

Step 4. Sanity-check the numbers from the overview. $P = 800$ kW, $R = 15\ \Omega$ . At $V = 4$ kV: $I = 800{,}000/4{,}000 = 200$ A, so $P_{\text{loss}} = 200^2 \times 15 = 600{,}000$ W = 600 kW — 75% of the shipment, an absurdity no utility could survive. At $V = 40$ kV: $I = 20$ A, $P_{\text{loss}} = 20^2 \times 15 = 6$ kW — 0.75%. One transformer at each end of the line buys a hundredfold saving. Verified: $800\,\mathrm{kW}/4\,\mathrm{kV} = 200$ A and $200^2 \times 15 = 600$ kW ✓; $800\,\mathrm{kW}/40\,\mathrm{kV} = 20$ A and $20^2 \times 15 = 6$ kW ✓; ratio $600/6 = 100 = (40/4)^2$ ✓ — the inverse-square lever, exactly.

One honest subtlety before using the formula at speed: the $V$ in the denominator is the transmission voltage — the voltage the line operates at — not the 230 V the customer eventually sees. And the loss percentage, $P_{\text{loss}}/P = PR/V^2$ , is the utility's real report card: it says a given line and voltage get less efficient the more power you push through them, which is why growing cities eventually force grid upgrades to the next voltage tier.

The grid's ladder — and why thick wire loses the argument

Follow the watts from turbine to toaster through the ladder every Indian city uses. Generation happens at the machine-friendly level of roughly 11–25 kV (generator insulation sets the limit). A step-up transformer at the station lifts the shipment to 220–400 kV (765 kV on the newest national corridors) for the long haul. Grid substations step down to 33 kV for regional distribution, local substations to 11 kV for the feeder network, and the neighbourhood's pole- or pad-mounted transformer makes the final drop to the 230 V of the wall socket. Count the machines: a typical joule passes through four or five transformers between generation and use, each at ~99% efficiency — the toll of the whole cascade a few percent, versus the 75% catastrophe of the low-voltage shipment. (What happens after the last transformer — the meter, the fuses, the wiring rules of the last thirty metres — is Household Circuits' territory.)

Figure

The grid's ladder — India-anchored voltage stages from turbine to wall socket

Why not skip the ladder and just build a fatter line? Run the numbers once and the argument never returns. To match what a mere ×10 voltage rise achieves (÷100 loss), wire resistance would have to fall a hundredfold — one hundred times the copper cross-section, one hundred times the conductor weight per kilometre, plus the pylons to carry it. Copper and aluminium are bought by the tonne; voltage is bought with two transformers. The economics are so lopsided that real grids do both only marginally — conductors are made as thick as mechanically sensible, and every serious efficiency gain comes from the $V^2$ lever. Thick wire is the answer to a different question ("how much current can this line carry without sagging and overheating?"), not to the loss question. Two honesty lines complete the picture. First, why not transmit at 400 kV into houses? Because insulation, clearance and safety scale brutally with voltage — 400 kV needs metres of air gap; a kitchen has centimetres. The ladder exists precisely because the optimum voltage for moving power (as high as affordable) and for using it (low enough to insulate cheaply and survive contact) are at opposite ends. Second, modern long-haul grids increasingly run their very longest links on high-voltage DC: power electronics can now step DC voltages (a transformer cannot), and DC avoids some AC-specific costs on runs of many hundreds of kilometres — a technology footnote that refines, but does not repeal, this module's law: the wins still come from the same $V^2$ lever; only the machinery performing the voltage conversion has changed.

Limiting cases

case	result	meaning
$V \times 10$ , same $P$ , $R$	loss ÷ 100	the lever: inverse square
$V \times 100$	loss ÷ 10,000	why grids reach for hundreds of kV
$R$ halved (double copper)	loss ÷ 2 only	linear — the expensive, weak fix
$P$ doubled, same line	loss × 4	congestion punishes quadratically; loss fraction doubles
$V \to$ household 230 V for the long haul	loss overwhelms delivery	the Edison problem: DC's 2-km radius
DC at high voltage (modern HVDC)	same $I^2R$ law applies	the lever survives; only the conversion machinery differs

Common trapThe higher voltage pushes the power through the wires harder, so more power gets through

Backwards in its physics, even though the conclusion sounds right. The higher voltage does not shove watts through the line more forcefully — it means the same watts travel as fewer amperes ( $I = P/V$ ), and it is amperes, squared, that the wire taxes. Get the mechanism wrong and the marks follow it: examiners at ISC and IGCSE routinely award the explanation, not the conclusion, and "high voltage forces the current through" scores zero. Three further slips cluster here. First: students compute the line loss as $V^2/R$ using the transmission voltage — that formula prices the loss of a resistor connected across $V$ , but the line is in series with the load; its loss is $I^2R$ with $I = P/V$ . (The disaster is not subtle: $V^2/R$ says loss grows with voltage — the exact opposite of the truth.) Second: the $V$ in $P^2R/V^2$ is the transmission voltage, not the 230 V at the socket; mixing the rungs of the ladder wrecks the arithmetic. Third: "raise the voltage" does not change the power delivered — $P$ is fixed by the customers; what changes is the fraction of extra generation wasted en route. Anchor the causal chain — higher $V$ → lower $I$ → far lower $I^2R$ — and every variant of the question falls in three lines.

Worked examples

Board level — the classic two-voltage audit (ISC/IGCSE pattern)

A small town draws 800 kW from a plant 15 km away. The two-wire line has a total resistance of 15 Ω. Compare the line loss when the power is transmitted at (a) 4,000 V and (b) 40,000 V, and comment.

Step 1. Part (a) current: $I = \dfrac{P}{V} = \dfrac{800{,}000}{4{,}000} = 200$ A.

Step 2. Part (a) loss: $P_{\text{loss}} = I^2R = (200)^2 \times 15 = \boxed{600\ \mathrm{kW}}$ — three-quarters of the generated power heats the line; the plant must generate 1,400 kW for the town's 800. Transmission at this voltage is economically absurd.

Step 3. Part (b) current: $I = \dfrac{800{,}000}{40{,}000} = 20$ A. Loss: $(20)^2 \times 15 = \boxed{6\ \mathrm{kW}}$ — under one percent.

Step 4. Comment for the mark: raising the voltage tenfold cut the current tenfold and the $I^2R$ loss hundredfold ( $600/6 = 100 = 10^2$ ) — which is why the grid steps voltage up with a transformer for transmission and down again for use.

Competitive level — designing to a loss budget (ISC/JEE-adjacent pattern, ~3 minutes)

A plant transmits 1.0 MW down a line of total resistance 4.0 Ω. (a) Find the line loss and its percentage at 20 kV. (b) The utility wants the loss below 0.30%. Find the minimum transmission voltage. (c) State what single piece of equipment makes the change practical.

Step 1. Part (a): $I = \dfrac{10^6}{2\times10^4} = 50$ A; $P_{\text{loss}} = 50^2 \times 4.0 = 10{,}000$ W $= \boxed{10\ \mathrm{kW}}$ , i.e. $\boxed{1.0\%}$ of the shipment.

Step 2. Part (b): loss fraction $= \dfrac{PR}{V^2}$ . Demand $\dfrac{PR}{V^2} \leq 0.0030$ :

V \geq \sqrt{\frac{PR}{0.0030}} = \sqrt{\frac{10^6 \times 4.0}{0.0030}} = \sqrt{1.333\times10^9} = 3.65\times10^4\ \mathrm{V} \approx \boxed{36.5\ \mathrm{kV}}

(Check by the lever instead: from 1.0% to 0.30% is ÷3.33, needing $V$ up by $\sqrt{3.33} = 1.83$ — and $20 \times 1.83 = 36.5$ kV ✓. Scaling beats re-derivation for speed.)

Step 3. Part (c): a step-up transformer at the plant (with matching step-downs at the receiving end) — the machine of the previous module, and the reason the scheme runs on AC. Timing note: 3 minutes. The efficient route in (b) is the loss-fraction form $PR/V^2$ — one rearrangement, one square root. The ratio check in the bracket is faster still and catches algebra slips: loss scales as $1/V^2$ , so a required loss ratio hands you the voltage ratio as its inverse square root.

Exam traps

Loss is $I^2R$ with $I = P/V$ — never $V^2/R$ . The line is in series with the load, not across the supply. The wrong formula predicts loss rising with voltage and takes the whole answer down with it.

Which $V$ ? The transmission voltage — the line's operating level — not the 230 V at the consumer. In ladder questions, use each stage's own voltage for that stage's current.

Narrate the mechanism. "High voltage → low current → low $I^2R$ " is the credited chain at ISC and IGCSE. Conclusions without the chain, or with "voltage pushes harder" folklore, lose the explanation marks.

Percentage bookkeeping. Loss fraction $= PR/V^2$ ; the station generates $P + P_{\text{loss}}$ . If a question asks what must be generated, add the loss back — delivering 800 through a 600-loss line means generating 1,400.

Doubling the load worsens the fraction. At fixed $V$ and $R$ , loss $\propto P^2$ , so the loss percentage grows linearly with $P$ — lines get less efficient as demand grows, which is why upgrades jump to the next voltage tier.

Thick-wire questions. Halving $R$ halves the loss — linear, once, at the price of double the conductor. Compare explicitly with the $V^2$ lever when asked "suggest and justify the better method."

Boards and depth. This module is ISC and IGCSE syllabus territory, tagged accordingly. ISC 861 names it explicitly: transformer applications "including transmission of power, energy losses and their minimisation" — the two-voltage audit and the minimisation argument above are that clause, worked. IGCSE physics asks for the advantages of high-voltage transmission and the grid's use of transformers — the ladder plus the lever, at formula level. CBSE and NEET's rationalized syllabi stop at "AC generator and transformer" without a transmission item (the famous NCERT village example survives as textbook colour, and the physics here remains excellent preparation); JEE's syllabus likewise leaves the grid economics untested. A-Level 9702 treats transmission advantages as assumed IGCSE background. AP-2 and IB have no AC unit at all — for those boards, this is how-the-world-works reading, not exam material.

Origins · 1 minSebastian Ziani de Ferranti and how this was discovered

Sebastian Ziani de Ferranti

1864–1930 · Deptford Power Station, London

Ferranti bet a career on this module's formula before the industry believed it. A London engineer of Italian descent who had filed his first patent at 17, he was appointed at just 22 to build the London Electric Supply Corporation's system — and he designed it backwards from the arithmetic of $P^2R/V^2$ . Instead of scattering small generators through the city as Edison's DC model demanded, Ferranti put one giant station at Deptford, on the Thames outside London where land and coal barges were cheap, and shipped the power in at an unheard-of 10,000 volts — a voltage his contemporaries considered lethal madness, carried on high-voltage cables he had to invent himself (wax-impregnated paper insulation, the ancestor of a century of cable engineering). The scheme was too far ahead of its hardware — alternator technology and London's fractured supply politics kept Deptford from full glory — but the architecture was simply the future: generate big, transmit high, step down at the edges. Every modern grid is Deptford's blueprint with better transformers. His name survives on a curiosity of the same physics — the "Ferranti effect," a voltage rise on lightly loaded long lines — and on the principle this module computes: the man who first trusted the inverse square with other people's money.

History

1882

— Edison's Pearl Street station lights Manhattan with DC — and can serve customers only within about a mile

1886

— Stanley's Great Barrington demonstration proves the step-up/step-down architecture on a working street

1889

— Ferranti begins Deptford: one huge station, 10 kV cables into London — the high-voltage bet, made first

1891

— the Lauffen–Frankfurt demonstration transmits AC 175 km at 15 kV with ~75% efficiency, settling the argument

1896

— Niagara Falls power reaches Buffalo, 32 km away, stepped up for the journey: the modern grid's template

Going further

Texts. NCERT Physics Class 12, Ch. 7 (the village-transmission example) · SL Arora, ISC Physics XII, transmission of power · Cambridge IGCSE Physics, the National Grid chapter · Hughes, Networks of Power, for Deptford and the war of the currents.

Provenance. All results standard. A verifier should recompute: $800\,\mathrm{kW}/4\,\mathrm{kV} = 200$ A, $200^2\times15 = 600$ kW (75%); $800\,\mathrm{kW}/40\,\mathrm{kV} = 20$ A, $20^2\times15 = 6$ kW (0.75%); ratio 100 = $(10)^2$ ✓; design example $50^2\times4 = 10$ kW = 1.0%; $V_{\min} = \sqrt{4\times10^6/0.003} = 36.5$ kV, cross-checked by $20\sqrt{3.33} = 36.5$ ✓; quiz numerical $I = 20$ A, $20^2\times5 = 2$ kW ✓.

Prerequisites. The Transformer (the voltage-changing machine that makes the lever usable) · Electrical Energy, Power & Heating (the $I^2R$ law, applied here). Leads to. ARCHETYPE: The LCR Numericals Ladder (back to circuit craft, next) · ARCHETYPE: Transformer & Line-Loss Numericals (this module drilled to exam speed, later in the track) · Household Circuits (the last thirty metres) · Transformers & Mutual Induction (UG deep dive).

If high voltage is so efficient, why isn't electricity delivered to houses at 400 kV?

Because using electricity and moving it optimize in opposite directions. Insulation, clearance and safety all scale punishingly with voltage: 400 kV demands metres of air gap, porcelain strings the length of a person, and switchgear the size of rooms — try fitting that behind a kitchen wall, or surviving contact with it. At 230 V, a millimetre of plastic insulates a wire and a fault is survivable with basic protections. So the grid is a deliberate compromise ladder: as high as affordable for the long haul (where the $V^2$ lever pays), stepped down in stages to as low as usable at the socket (where safety and cost of insulation dominate). The transformer's cheapness at every rung is what makes the compromise nearly free — each conversion costs about one percent.

Why does the loss formula have P² in it — doesn't the customer's power appear only once?

Because the current does double duty. The line current is set by the delivery requirement, $I = P/V$ — one factor of $P$ . The line's heating then goes as the square of that current — bringing the second factor. Physically: double the town's demand and the wires carry double the amperes, which heat four times as much. This quadratic congestion has a practical signature: the loss fraction $PR/V^2$ grows linearly with demand, so a line that wasted 1% at launch wastes 2% when the town doubles — and eventually the utility either re-conductors the line or, far more effectively, moves it up a voltage tier. The $P^2$ is why grid planning is forever chasing growth.

Does the P²R/V² argument apply to DC transmission too?

Entirely — the derivation never used anything AC-specific: $I = P/V$ and $I^2R$ hold for any current. Historically the lever belonged to AC because only transformers could climb the voltage ladder cheaply, and transformers need changing flux. Modern power electronics broke that monopoly: solid-state converters can step DC to ±800 kV and back, and at that point DC's own advantages surface on very long or undersea links (no reactance, no synchronization between the connected grids, the whole cable cross-section used steadily). That is why the longest modern corridors — including India's northeast links — are HVDC. The exam-safe summary: the loss law is universal; AC won the historical argument because the transformer was the only affordable voltage-changer for a century; HVDC is the modern refinement at the extremes, not a repeal.

What exactly is the "total resistance of the line" — the wire out, or out and back?

Both conductors, the complete circuit. Current flows out to the town and back, so the line's resistance is the sum of the go and return paths — a "15 km line at 0.5 Ω/km" means $2 \times 15 \times 0.5 = 15\ \Omega$ in careful problems (and sloppier ones quote the total directly, as this module's examples do). Watch the phrasing in board questions: "resistance per km of each conductor" doubles; "total line resistance" doesn't. The same bookkeeping explains real design tables: utilities quote line resistance per circuit-kilometre with the return path included, and three-phase systems (the real grid's arrangement) split the current across conductors in a way that belongs to engineering courses — HS problems idealize to the two-wire loop, which preserves all the physics.

The wires between pylons visibly sag and hum — is any of that related to the losses here?

The sag is mechanical (conductor weight and thermal expansion — a heavily loaded line runs hotter, from exactly this module's $I^2R$ , and sags measurably lower on a hot, high-demand day). The hum is the AC itself: corona discharge crackle and the 100 Hz magnetic forcing of hardware. Both are real engineering constraints that shape the lever's limits: sag sets how much current a conductor may carry (its thermal rating), and corona — air ionizing near very-high-voltage conductors — is an additional loss channel that grows with voltage and effectively caps how far "just raise V" can be pushed before conductor bundling and larger diameters are needed. So the grid's chosen voltages are the equilibrium of this module's inverse-square savings against insulation, clearance and corona costs — the formula wins the argument, and then engineering negotiates the terms.

My textbook says transmission losses in a real grid are 15–20% — but we just computed 0.75%. Who is wrong?

Neither — they count different things. The module's 0.75% is one well-designed high-voltage transmission line doing its job; real grids lose energy at every stage: the long-haul lines (a few percent), each transformer (~1% × four or five machines), and above all the distribution tail — the medium- and low-voltage last kilometres, where voltage is low, currents are high, and the $I^2R$ toll bites exactly as this module predicts. India's aggregate technical-and-commercial losses have historically run above 20%, but a large slice of that is "commercial" — metering gaps and theft — not physics. The physics lesson survives intact: losses concentrate where voltage is lowest, which is why utilities keep pushing the high-voltage network deeper toward the consumer, shortening the inefficient tail.